ID: 17 Type: Mcq Source: NCERT

Question 1: The rate of change of the area of a circle with respect to its radius \(r\) at \(r = 6\;{\text{cm}}\) is

• A. 10\(\pi\)
• B. 12\(\pi\)
• C. 8\(\pi\)
• D. 11\(\pi\)

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✅ Manual Solution:

The rate of change of the area of a circle with respect to its radius \(r\) at \(r = 6,\text{cm}\) is to be determined.\[\] We know that the area of a circle is: \(A = \pi r^2\) Differentiate with respect to \(r\): \[\dfrac{dA}{dr} = \dfrac{d}{dr}(\pi r^2) = 2\pi r\] Now, substitute \(r = 6\): \[\dfrac{dA}{dr} = 2\pi \cdot 6 = 12\pi,\text{cm}^2/\text{cm}\] Final Answer: The rate of change of the area at \(r = 6,\text{cm}\) is \(12\pi,\text{cm}^2/\text{cm}\).

🤖 AI Generated Solution:

Let $A$ be the area of a circle and $r$ be its radius. The formula for the area of a circle is $A = \pi r^2$. 1. **Differentiate the area formula with respect to $r$:** To find the rate of change of the area with respect to its radius, we differentiate $A$ with respect to $r$: $$\frac{dA}{dr} = \frac{d}{dr}(\pi r^2)$$ $$\frac{dA}{dr} = 2\pi r$$ 2. **Substitute the given value of $r$:** We need to find this rate of change at $r = 6\;{\text{cm}}$. Substitute $r=6$ into the derivative: $$\frac{dA}{dr}\Big|_{r=6} = 2\pi (6)$$ $$\frac{dA}{dr}\Big|_{r=6} = 12\pi$$ The rate of change of the area of a circle with respect to its radius at $r = 6\;{\text{cm}}$ is $12\pi\;{\text{cm}}^2/{\text{cm}}$. The final answer is $\boxed{12\pi}$.

ID: 18 Type: Mcq Source: NCERT

Question 2: The total revenue in Rupees received from the sale of \(x\) units of a product is given by \(R(x) = 3{x^2} + 36x + 5\). The marginal revenue, when \(x = 15\) is

• A. 116
• B. 96
• C. 90
• D. 126

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✅ Manual Solution:

The total revenue (in rupees) from selling \(x\) units of a product is given by: \[R(x) = 3x^2 + 36x + 5\] We are asked to find the marginal revenue when \(x = 15\).\[\] Marginal revenue is the rate of change of total revenue with respect to the number of units sold: \[\therefore MR = \dfrac{dR}{dx} = 3 \cdot 2x + 36 = 6x + 36\] Substituting \(x = 15\): \[MR = 6 \cdot 15 + 36 = 90 + 36 = 126\] Final Answer: The marginal revenue when \(x = 15\) is Rs. 126.

🤖 AI Generated Solution:

The marginal revenue is the derivative of the total revenue function $R(x)$ with respect to $x$. 1. Find the derivative of the revenue function $R(x)$: $$R'(x) = \frac{d}{dx}(3{x^2} + 36x + 5) = 6x + 36$$ 2. Substitute $x = 15$ into the marginal revenue function $R'(x)$: $$R'(15) = 6(15) + 36 = 90 + 36 = 126$$ The marginal revenue when $x = 15$ is 126. The final answer is $\boxed{\text{126}}$.

ID: 226 Type: Mcq Source: AISSCE(Board Exam) Year: 2025

Question 3: The absolute maximum value of function \( f(x) = x^3 - 3x + 2 \) in [0, 2] is:

• A. 0
• B. 2
• C. 4
• D. 5

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✅ Manual Solution:

\[f'(x) = \frac{d}{dx}(x^3 - 3x + 2) = 3x^2 - 3\] Set the derivative equal to zero to find the critical points:\[3x^2 - 3 = 0\]\[3(x^2 - 1) = 0\]\[x^2 = 1\]\[x = \pm 1\]The critical points are \(x = -1\) and \(x = 1\). Since the interval is \([0, 2]\), we only consider the critical point \(x = 1\), as \(x = -1\) is outside the interval.2. Evaluate Function at Critical Points and Endpoints

📍The absolute maximum value of a continuous function on a closed interval must occur at a critical point within the interval or at the endpoints of the interval.

We evaluate \(f(x)\) at \(x=1\), \(x=0\), and \(x=2\).At the critical point \(x=1\):\[f(1) = (1)^3 - 3(1) + 2 = 1 - 3 + 2 = \mathbf{0}\]At the left endpoint \(x=0\):\[f(0) = (0)^3 - 3(0) + 2 = 0 - 0 + 2 = \mathbf{2}\]At the right endpoint \(x=2\):\[f(2) = (2)^3 - 3(2) + 2 = 8 - 6 + 2 = \mathbf{4}\]

The largest value is 4.Therefore, the absolute maximum value of the function \(f(x) = x^3 - 3x + 2\) in the interval \([0, 2]\) is \(\mathbf{4}\).

🤖 AI Generated Solution:

To find the absolute maximum value, we evaluate the function at critical points within the interval and at the endpoints. 1. **Find the derivative of the function:** $f'(x) = \frac{d}{dx}(x^3 - 3x + 2) = 3x^2 - 3$. 2. **Find critical points by setting the derivative to zero:** $3x^2 - 3 = 0$ $3(x^2 - 1) = 0$ $x^2 - 1 = 0$ $(x-1)(x+1) = 0$ So, $x = 1$ or $x = -1$. 3. **Check critical points within the interval [0, 2]:** The critical point $x=1$ is within the interval $[0, 2]$. The critical point $x=-1$ is not within the interval $[0, 2]$. 4. **Evaluate the function at the critical point within the interval and at the endpoints:** * At $x = 0$ (endpoint): $f(0) = (0)^3 - 3(0) + 2 = 2$. * At $x = 1$ (critical point): $f(1) = (1)^3 - 3(1) + 2 = 1 - 3 + 2 = 0$. * At $x = 2$ (endpoint): $f(2) = (2)^3 - 3(2) + 2 = 8 - 6 + 2 = 4$. 5. **Determine the absolute maximum:** Comparing the values $2, 0, 4$, the absolute maximum value is $4$. The final answer is $\boxed{\text{4}}$.

ID: 241 Type: Mcq Source: AISSCE(Board Exam) Year: 2025

Question 4: The function \(f(x)=x^{2}-4x+6\) is increasing in the interval

• A. \((0, 2)\)
• B. \((-\infty, 2]\)
• C. \([1, 2]\)
• D. \([2, \infty)\)

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✅ Manual Solution:

\[f(x) = x^2 - 4x + 6\] \[f'(x) = 2x - 4\] For Critical Point set \(f'(x) = 0\) implies \[2x - 4 = 0 \implies 2x = 4 \implies x = 2\] The function is increasing when \(f'(x) > 0\) implies \[2x - 4 > 0\]\[2x > 4\]\[\mathbf{x > 2}\]

🤖 AI Generated Solution:

**Correct Option:** D **Reasoning:** * \(f'(x) = 2x - 4\) * For \(f(x)\) to be increasing, \(f'(x) \ge 0\). * \(2x - 4 \ge 0 \implies 2x \ge 4 \implies x \ge 2\).

ID: 250 Type: Mcq Source: AISSCE(Board Exam) Year: 2025

Question 5: A cylindrical tank of radius \(10\) cm is being filled with sugar at the rate of \(100~\pi~cm^{3}/s\). The rate, at which the height of the sugar inside the tank is increasing, is:

• A. \(0.1~cm/s\)
• B. \(0.5~cm/s\)
• C. \(1~cm/s\)
• D. \(1.1~cm/s\)

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✅ Manual Solution:

The rate at which the height of sugar inside the cylindrical tank increases can be determined using the formula for the volume of a cylinder:

\(V = \pi r^2 h\)

Given:

• Radius, \(r = 10 \text{ cm}\)
• Rate of change of volume, \(\dfrac{dV}{dt} = 100\pi \text{ cm}^3/\text{s}\)

Since the radius remains constant, differentiate both sides of the volume equation with respect to time \(t\):

\(\dfrac{dV}{dt} = \pi r^2 \dfrac{dh}{dt}\)

Substitute the known values:

\(100\pi = \pi (10)^2 \dfrac{dh}{dt}\)

\(100\pi = 100\pi \dfrac{dh}{dt}\)

Dividing both sides by \(100\pi\):

\(\dfrac{dh}{dt} = 1 \text{ cm/s}\)

Therefore, the height of the sugar in the tank is increasing at a rate of \(1 \text{ cm/s}\).

🤖 AI Generated Solution:

**Correct Option:** C **Reasoning:** * Volume of a cylinder is \(V = \pi r^2 h\). * Differentiating with respect to time, \(dV/dt = \pi r^2 (dh/dt)\). * Substitute \(dV/dt = 100\pi\) and \(r = 10\): \(100\pi = \pi (10)^2 (dh/dt) \Rightarrow 100\pi = 100\pi (dh/dt)\). * Solving for \(dh/dt\) yields \(dh/dt = 1~cm/s\).

ID: 294 Type: Mcq Source: AISSCE(Board Exam) Year: 2025

Question 6: The values of \(\lambda\) so that \(f(x)=\sin x-\cos x-\lambda x+C\) decreases for all real values of x are:

• A. \(1\lt\lambda\lt\sqrt{2}\)
• B. \(\lambda\ge1\)
• C. \(\lambda\ge\sqrt{2}\)
• D. \(\lambda\lt1\)

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✅ Manual Solution:

To determine the values of \(\lambda \) for which the function \(f(x)=\sin x-\cos x-\lambda x+C\) is always decreasing, we need to find the values of \(\lambda \) for which the derivative of \(f(x)\) is less than or equal to zero for all real \(x\). 

The derivative of \(f(x)\) with respect to \(x\) is:

\(f^{\prime }(x)=\cos x+\sin x-\lambda \)

For \(f(x)\) to be a decreasing function for all real values of \(x\), we must have:\(f^{\prime }(x)\le 0\)\(\cos x+\sin x-\lambda \le 0\) that is

\(\cos x+\sin x\le \lambda \)

Note that for an expression of the form \(a\cos x+b\sin x\), the maximum value is \(\sqrt{a^{2}+b^{2}}\).

Here, \(a=1\) and \(b=1\). So the maximum value of \(\cos x+\sin x\) is:\(\sqrt{1^{2}+1^{2}}=\sqrt{2}\)

Now the condition \(\cos x+\sin x\le \lambda \) for all real \(x\).implies that \(\lambda \) must be greater than or equal to the maximum possible value of \(\cos x+\sin x\).

Therefore, we must have:\(\lambda \ge \sqrt{2}\)

🤖 AI Generated Solution:

**Correct Option:** C **Reasoning:** * For \(f(x)\) to decrease for all x, \(f'(x) \le 0\). * \(f'(x) = \cos x + \sin x - \lambda\). Thus, \(\cos x + \sin x \le \lambda\). * The maximum value of \(\cos x + \sin x\) is \(\sqrt{1^2+1^2} = \sqrt{2}\). Hence, \(\lambda \ge \sqrt{2}\).

ID: 302 Type: Mcq Source: AISSCE(Board Exam) Year: 2025

Question 7: If \(f(x)=2x+\cos x\), then f(x):

• A. has a maxima at \(x=\pi\)
• B. has a minima at \(x=\pi\)
• C. is an increasing function
• D. is a decreasing function

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✅ Manual Solution:

We differentiate $f(x)$ with respect to $x$:$$f'(x) = 2 - \sin x$$ We know that the range of the sine function is $-1 \le \sin x \le 1$. This means that $f'(x)$ is always positive ($f'(x) > 0$) for all real values of $x$.

Since the first derivative $f'(x)$ is strictly positive for all $x$, the function $f(x) = 2x + \cos x$ is an increasing function throughout its domain. This also rules out (A) and (B) because an increasing function has no local maxima or minima.

🤖 AI Generated Solution:

**Correct Option:** C **Reasoning:** * Calculate the first derivative: \(f'(x) = 2 - \sin x\). * Since \(-1 \le \sin x \le 1\), the range of \(f'(x)\) is \([2-1, 2-(-1)] = [1, 3]\). * As \(f'(x) \ge 1\) for all x, \(f'(x)\) is always positive, indicating \(f(x)\) is an increasing function.

ID: 317 Type: Mcq Source: AISSCE(Board Exam) Year: 2025

Question 8: The slope of the curve \(y=-x^{3}+3x^{2}+8x-20\) is maximum at:

• A. (1,-10)
• B. (1,10)
• C. (10, 1)
• D. (-10, 1)

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🤖 AI Generated Solution:

**Correct Option:** A **Reasoning:** * First derivative: \( \frac{dy}{dx} = -3x^2 + 6x + 8 \) * Set second derivative to zero: \( \frac{d^2y}{dx^2} = -6x + 6 = 0 \implies x = 1 \) * Substitute \( x=1 \) into original equation: \( y = -(1)^3 + 3(1)^2 + 8(1) - 20 = -10 \). The point is (1, -10).

ID: 334 Type: Mcq Source: AISSCE(Board Exam) Year: 2025

Question 9: Let \(f(x)=|x|\), \(x\in R\). Then, which of the following statements is **incorrect**?

• A. f has a minimum value at \(x=0\).
• B. f has no maximum value in R.
• C. f is continuous at \(x=0\).
• D. f is differentiable at \(x=0\).

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🤖 AI Generated Solution:

**Correct Option:** D **Reasoning:** * \(f(x)=|x|\) has \(f(x) \ge 0\) and \(f(0)=0\), so minimum at \(x=0\). * \( \lim_{x \to \pm\infty} |x| = \infty \), so no maximum value. * \( \lim_{x \to 0} |x| = 0 = f(0) \), so continuous at \(x=0\). * LHD at \(x=0\) is -1, RHD at \(x=0\) is 1. Not differentiable at \(x=0\).

ID: 353 Type: Mcq Source: AISSCE(Board Exam) Year: 2025

Question 10: A spherical ball has a variable diameter \(\frac{5}{2}(3x+1).\) The rate of change of its volume w.r.t. x, when \(x=1\), is :

• A. \(225\pi\)
• B. \(300\pi\)
• C. \(375\pi\)
• D. \(125\pi\)

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🤖 AI Generated Solution:

The diameter of the spherical ball is given by $D = \frac{5}{2}(3x+1)$. The radius of the ball is $r = \frac{D}{2} = \frac{1}{2} \cdot \frac{5}{2}(3x+1) = \frac{5}{4}(3x+1)$. The volume of a sphere is $V = \frac{4}{3}\pi r^3$. Substitute the expression for $r$: $V = \frac{4}{3}\pi \left(\frac{5}{4}(3x+1)\right)^3$ $V = \frac{4}{3}\pi \left(\frac{125}{64}(3x+1)^3\right)$ $V = \frac{125\pi}{48}(3x+1)^3$ To find the rate of change of its volume with respect to $x$, we differentiate $V$ with respect to $x$: $\frac{dV}{dx} = \frac{d}{dx}\left(\frac{125\pi}{48}(3x+1)^3\right)$ $\frac{dV}{dx} = \frac{125\pi}{48} \cdot 3(3x+1)^{3-1} \cdot \frac{d}{dx}(3x+1)$ $\frac{dV}{dx} = \frac{125\pi}{48} \cdot 3(3x+1)^2 \cdot 3$ $\frac{dV}{dx} = \frac{125\pi \cdot 9}{48}(3x+1)^2$ $\frac{dV}{dx} = \frac{125\pi \cdot 3}{16}(3x+1)^2$ $\frac{dV}{dx} = \frac{375\pi}{16}(3x+1)^2$ Now, we evaluate this rate of change when $x=1$: $\left.\frac{dV}{dx}\right|_{x=1} = \frac{375\pi}{16}(3(1)+1)^2$ $\left.\frac{dV}{dx}\right|_{x=1} = \frac{375\pi}{16}(4)^2$ $\left.\frac{dV}{dx}\right|_{x=1} = \frac{375\pi}{16} \cdot 16$ $\left.\frac{dV}{dx}\right|_{x=1} = 375\pi$ The final answer is $\boxed{\text{375\pi}}$.

ID: 473 Type: Mcq Source: AISSCE(Board Exam) Year: 2024

Question 11: Let \(f(x)\) be a continuous function on [a, b] and differentiable on (a, b). Then, this function \(f(x)\) is strictly increasing in (a, b) if

• A. \(f^{\prime}(x)\lt;0\), \(\forall x\in(a,b)\)
• B. \(f^{\prime}(x)\gt;0\), \(\forall x\in(a,b)\)
• C. \(f^{\prime}(x)=0\), \(\forall x\in(a,b)\)
• D. \(f(x)\gt;0\), \(\forall x\in(a,b)\)

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🤖 AI Generated Solution:

The solution is as follows: **Step 1:** Recall the relationship between the derivative of a function and its increasing/decreasing behavior. A function is strictly increasing on an interval if its derivative is strictly positive on that interval. **Step 2:** Analyze the given options in light of this relationship. Option A states \(f^{\prime}(x) \lt 0\), which implies the function is strictly decreasing. Option B states \(f^{\prime}(x) \gt 0\), which implies the function is strictly increasing. Option C states \(f^{\prime}(x) = 0\), which implies the function is constant. Option D is about the function's value, not its rate of change. **Step 3:** Conclude based on the definition of a strictly increasing function. For \(f(x)\) to be strictly increasing in \((a, b)\), its derivative \(f^{\prime}(x)\) must be greater than 0 for all \(x\) in \((a, b)\). The final answer is $\boxed{B}$.

ID: 492 Type: Mcq Source: AISSCE(Board Exam) Year: 2024

Question 12: The function \(f(x)=x^{3}-3x^{2}+12x-18\) is:

• A. strictly decreasing on R
• B. strictly increasing on R
• C. neither strictly increasing nor strictly decreasing on R
• D. strictly decreasing on \((-\infty, 0)\)

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🤖 AI Generated Solution:

Here's the step-by-step solution: 1. **Find the first derivative of the function:** $f'(x) = \frac{d}{dx}(x^3 - 3x^2 + 12x - 18) = 3x^2 - 6x + 12$. 2. **Analyze the sign of the first derivative:** The derivative is a quadratic function. To determine its sign, we can find its discriminant. The discriminant of $ax^2 + bx + c$ is $b^2 - 4ac$. For $3x^2 - 6x + 12$, the discriminant is $(-6)^2 - 4(3)(12) = 36 - 144 = -108$. 3. **Interpret the discriminant:** Since the discriminant is negative and the leading coefficient (3) is positive, the quadratic $f'(x) = 3x^2 - 6x + 12$ is always positive for all real values of $x$. 4. **Conclude the function's behavior:** Because $f'(x) > 0$ for all $x \in \mathbb{R}$, the function $f(x)$ is strictly increasing on $\mathbb{R}$. The final answer is $\boxed{B}$.

ID: 510 Type: Mcq Source: AISSCE(Board Exam) Year: 2024

Question 13: If the sides of a square are decreasing at the rate of \(1.5~cm/s\) the rate of decrease of its perimeter is:

• A. \(1.5~cm/s\)
• B. \(6~cm/s\)
• C. \(3~cm/s\)
• D. \(2.25~cm/s\)

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🤖 AI Generated Solution:

Let the side of the square be $s$. The perimeter of the square is $P = 4s$. We are given that the rate of decrease of the side is $\frac{ds}{dt} = -1.5~cm/s$. We need to find the rate of decrease of its perimeter, which is $\frac{dP}{dt}$. Differentiate the perimeter equation with respect to time $t$: $\frac{dP}{dt} = \frac{d}{dt}(4s)$ $\frac{dP}{dt} = 4 \frac{ds}{dt}$ Substitute the given value of $\frac{ds}{dt}$: $\frac{dP}{dt} = 4 (-1.5~cm/s)$ $\frac{dP}{dt} = -6~cm/s$ The rate of decrease of its perimeter is $6~cm/s$. The final answer is $\boxed{6~cm/s}$.

ID: 541 Type: Mcq Source: AISSCE(Board Exam) Year: 2024

Question 14: The function \(f(x)=kx-\sin~x\) is strictly increasing for

• A. \(k\gt1\)
• B. \(k\lt1\)
• C. \(k\gt-1\)
• D. \(k\lt-1\)

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🤖 AI Generated Solution:

The solution proceeds as follows: **Step 1: Find the derivative of the function.** The derivative of \(f(x) = kx - \sin~x\) is \(f'(x) = k - \cos~x\). **Step 2: Determine the condition for a strictly increasing function.** For a function to be strictly increasing, its derivative must be strictly positive, i.e., \(f'(x) > 0\). **Step 3: Apply the condition to the derivative.** We need \(k - \cos~x > 0\) for all \(x\). This means \(k > \cos~x\) for all \(x\). **Step 4: Find the maximum value of $\cos~x$.** The maximum value of $\cos~x$ is 1. **Step 5: Determine the condition for $k$.** For \(k > \cos~x\) to hold for all \(x\), \(k\) must be greater than the maximum value of $\cos~x$. Therefore, \(k > 1\). **Final Answer:** The function \(f(x)=kx-\sin~x\) is strictly increasing for \(k\gt1\). The final answer is $\boxed{k\gt1}$.

ID: 550 Type: Mcq Source: AISSCE(Board Exam) Year: 2024

Question 15: The function \(f(x)=\frac{x}{2}+\frac{2}{x}\) has a local minima at x equal to:

• A. 2
• B. 1
• C. 0
• D. -2

ID: 551 Type: Mcq Source: AISSCE(Board Exam) Year: 2024

Question 16: Given a curve \(y=7x-x^{3}\) and x increases at the rate of 2 units per second. The rate at which the slope of the curve is changing, when \(x=5\) is:

• A. \(-60~units/sec\)
• B. \(60~units/sec\)
• C. \(-70~units/sec\)
• D. \(-140~units/sec\)

ID: 5 Type: Very short (VSA)

Question 17: A stone is dropped into a quiet lake and waves move in circles at the speed of \(5\;{\text{cm}}/{\text{s}}.\) At the instant when the radius of the circular wave is \(8\;{\text{cm}}\), how fast is the enclosed area increasing?

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✅ Manual Solution:

We are asked to find how fast the area enclosed by the wave is increasing when the radius is \(8\,\text{cm}\). We know that the area of a circle is given by \(A = \pi r^2\). \[\]Differentiating both sides with respect to time \(t\):\[\] \[\frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = \frac{d}{dr}(\pi r^2) \cdot \frac{dr}{dt} = 2\pi r \cdot \frac{dr}{dt}\] Given \(\frac{dr}{dt} = 5\,\text{cm/s}\), we substitute the values: \(\frac{dA}{dt} = 2\pi (8)(5) = 80\pi\,\text{cm}^2/\text{s}\)

🤖 AI Generated Solution:

Here's a step-by-step solution: 1. Let $r$ be the radius of the circular wave and $A$ be the enclosed area. The area of a circle is given by $A = \pi r^2$. 2. We are given the rate at which the radius is increasing, $\frac{dr}{dt} = 5\;{\text{cm}}/{\text{s}}$. We need to find how fast the enclosed area is increasing, $\frac{dA}{dt}$, when $r = 8\;{\text{cm}}$. 3. Differentiate the area formula with respect to time $t$: $$ \frac{dA}{dt} = \frac{d}{dt}(\pi r^2) $$ Using the chain rule, this becomes: $$ \frac{dA}{dt} = 2\pi r \frac{dr}{dt} $$ 4. Substitute the given values $r = 8\;{\text{cm}}$ and $\frac{dr}{dt} = 5\;{\text{cm}}/{\text{s}}$ into the equation: $$ \frac{dA}{dt} = 2\pi (8\;{\text{cm}}) (5\;{\text{cm}}/{\text{s}}) $$ 5. Calculate the final value: $$ \frac{dA}{dt} = 80\pi\;{\text{cm}}^2/{\text{s}} $$ The enclosed area is increasing at a rate of $80\pi\;{\text{cm}}^2/{\text{s}}$.

ID: 197 Type: Very short (VSA) Source: AISSCE(Board Exam) Year: 2025

Question 18: Find the intervals in which function \(f(x) = 5x^{\frac{3}{2}} - 3x^{\frac{5}{3}}\) is:(i) increasing (ii) decreasing.

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🤖 AI Generated Solution:

To find the intervals where the function $f(x)$ is increasing or decreasing, we first find its derivative, $f'(x)$, and then analyze its sign. 1. **Find the derivative** $f'(x)$: Given $f(x) = 5x^{\frac{3}{2}} - 3x^{\frac{5}{3}}$. Using the power rule, $(x^n)' = nx^{n-1}$: $f'(x) = 5 \cdot \frac{3}{2} x^{\frac{3}{2}-1} - 3 \cdot \frac{5}{3} x^{\frac{5}{3}-1}$ $f'(x) = \frac{15}{2} x^{\frac{1}{2}} - 5 x^{\frac{2}{3}}$ 2. **Determine the domain of $f(x)$**: For $x^{\frac{3}{2}}$ to be defined, $x \ge 0$. The term $x^{\frac{5}{3}}$ is defined for all real $x$. Thus, the domain of $f(x)$ is $x \in [0, \infty)$. 3. **Find critical points**: Set $f'(x) = 0$: $\frac{15}{2} x^{\frac{1}{2}} - 5 x^{\frac{2}{3}} = 0$ Factor out $5x^{\frac{1}{2}}$: $5x^{\frac{1}{2}} \left(\frac{3}{2} - x^{\frac{2}{3}-\frac{1}{2}}\right) = 0$ $5x^{\frac{1}{2}} \left(\frac{3}{2} - x^{\frac{1}{6}}\right) = 0$ This gives two possibilities: (a) $5x^{\frac{1}{2}} = 0 \implies x = 0$. (b) $\frac{3}{2} - x^{\frac{1}{6}} = 0 \implies x^{\frac{1}{6}} = \frac{3}{2}$ Raise both sides to the power of 6: $x = \left(\frac{3}{2}\right)^6 = \frac{3^6}{2^6} = \frac{729}{64}$. The critical points are $x=0$ and $x=\frac{729}{64}$. 4. **Analyze the sign of $f'(x)$** in intervals using test points: The domain is $[0, \infty)$, and the critical points divide it into intervals $(0, \frac{729}{64})$ and $(\frac{729}{64}, \infty)$. * **Interval $(0, \frac{729}{64})$**: Choose a test point, e.g., $x=1$. $f'(1) = 5(1)^{\frac{1}{2}} \left(\frac{3}{2} - (1)^{\frac{1}{6}}\right) = 5(1)\left(\frac{3}{2} - 1\right) = 5\left(\frac{1}{2}\right) = \frac{5}{2}$. Since $f'(1) > 0$, $f(x)$ is increasing in this interval. * **Interval $(\frac{729}{64}, \infty)$**: Choose a test point, e.g., $x=64$ (since $x^{\frac{1}{6}} = (64)^{\frac{1}{6}} = 2$). $f'(64) = 5(64)^{\frac{1}{2}} \left(\frac{3}{2} - (64)^{\frac{1}{6}}\right) = 5(8)\left(\frac{3}{2} - 2\right) = 40\left(-\frac{1}{2}\right) = -20$. Since $f'(64) < 0$, $f(x)$ is decreasing in this interval. 5. **Conclusion**: (i) $f(x)$ is increasing on $\left[0, \frac{729}{64}\right]$. (ii) $f(x)$ is decreasing on $\left[\frac{729}{64}, \infty\right)$.

ID: 262 Type: Very short (VSA) Source: AISSCE(Board Exam) Year: 2025

Question 19: Find the values of 'a' for which \(f(x)=\sin x-ax+b\) is increasing on R.

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🤖 AI Generated Solution:

To find the values of 'a' for which $f(x)=\sin x-ax+b$ is increasing on R, we need its first derivative to be non-negative for all $x \in R$. 1. Find the derivative $f'(x)$: $f'(x) = \frac{d}{dx}(\sin x - ax + b) = \cos x - a$. 2. For $f(x)$ to be increasing on R, $f'(x) \ge 0$ for all $x \in R$: $\cos x - a \ge 0$ $\cos x \ge a$ 3. We know that the range of $\cos x$ is $[-1, 1]$, meaning $-1 \le \cos x \le 1$ for all $x \in R$. 4. For $\cos x \ge a$ to hold true for *all* $x \in R$, $a$ must be less than or equal to the minimum value of $\cos x$. The minimum value of $\cos x$ is $-1$. Therefore, $a \le -1$. The values of 'a' for which $f(x)$ is increasing on R are $a \le -1$.

ID: 364 Type: Very short (VSA) Source: AISSCE(Board Exam) Year: 2025

Question 20: Find the least value of 'a' so that \(f(x)=2x^{2}-ax+3\) is an increasing function on [2, 4].

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🤖 AI Generated Solution:

To find the least value of 'a' such that $f(x)$ is an increasing function on $[2, 4]$, we first find the derivative of $f(x)$. 1. Given function: $f(x) = 2x^2 - ax + 3$. 2. Find the first derivative: $f'(x) = \frac{d}{dx}(2x^2 - ax + 3) = 4x - a$. 3. For $f(x)$ to be an increasing function on the interval $[2, 4]$, its derivative $f'(x)$ must be non-negative for all $x$ in that interval. So, $f'(x) \geq 0 \implies 4x - a \geq 0$ for all $x \in [2, 4]$. 4. Rearranging the inequality, we get $a \leq 4x$. 5. This condition $a \leq 4x$ must hold for *all* $x \in [2, 4]$. Therefore, 'a' must be less than or equal to the minimum value of $4x$ on the interval $[2, 4]$. 6. The function $g(x) = 4x$ is an increasing function. Its minimum value on $[2, 4]$ occurs at $x=2$. $\min_{x \in [2, 4]} (4x) = 4(2) = 8$. 7. Thus, we must have $a \leq 8$. 8. The set of all values of 'a' for which $f(x)$ is increasing on $[2, 4]$ is $(-\infty, 8]$. The question asks for the "least value of 'a' so that $f(x)$ is an increasing function". In this context, it refers to the maximum possible value 'a' can take while satisfying the condition, which is the upper bound of the solution set. The least value of 'a' for which $f(x)$ is increasing on $[2, 4]$ is $8$. The final answer is $\boxed{8}$.

ID: 365 Type: Very short (VSA) Source: AISSCE(Board Exam) Year: 2025

Question 21: If \(f(x)=x+\frac{1}{x}, x\ge1\), show that f is an increasing function.

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🤖 AI Generated Solution:

**Correct Option:** N/A **Reasoning:** * \(f'(x) = 1 - \frac{1}{x^2}\). * For \(x \ge 1\), \(x^2 \ge 1 \implies \frac{1}{x^2} \le 1\). * Therefore, \(f'(x) = 1 - \frac{1}{x^2} \ge 0\), proving f is an increasing function.

ID: 369 Type: Very short (VSA) Source: AISSCE(Board Exam) Year: 2025

Question 22: For the curve \(y=5x-2x^{3}\) if x increases at the rate of \(2~\text{units/s}\), then how fast is the slope of the curve changing when \(x=2\) ?

ID: 387 Type: Very short (VSA) Source: AISSCE(Board Exam) Year: 2025

Question 23: Surface area of a balloon (spherical), when air is blown into it, increases at a rate of \(5\text{ mm}^{2}/\text{s}\). When the radius of the balloon is 8 mm, find the rate at which the volume of the balloon is increasing.

ID: 428 Type: Very short (VSA) Source: AISSCE(Board Exam) Year: 2025

Question 24: Determine the values of x for which \(f(x)=\frac{x-4}{x+1}\), \(x\ne-1\) is an increasing or a decreasing function.

ID: 581 Type: Very short (VSA) Source: AISSCE(Board Exam) Year: 2024

Question 25: Find the interval in which the function \(f(x)=x^{4}-4x^{3}+10\) is strictly decreasing.

ID: 588 Type: Very short (VSA) Source: AISSCE(Board Exam) Year: 2024

Question 26: If M and m denote the local maximum and local minimum values of the function \(f(x)=x+\frac{1}{x}(x\ne0)\) respectively, find the value of \((M-m)\)

ID: 591 Type: Very short (VSA) Source: AISSCE(Board Exam) Year: 2024

Question 27: Show that \(f(x)=e^{x}-e^{-x}+x-\tan^{-1}x\) is strictly increasing in its domain.

ID: 592 Type: Very short (VSA) Source: AISSCE(Board Exam) Year: 2024

Question 28: Show that the function \(f(x)=4x^{3}-18x^{2}+27x-7\) has neither maxima nor minima.

ID: 593 Type: Very short (VSA) Source: AISSCE(Board Exam) Year: 2024

Question 29: The area of the circle is increasing at a uniform rate of \(2~cm^{2}/sec\). How fast is the circumference of the circle increasing when the radius \(r=5\) cm?

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🤖 AI Generated Solution:

The area of a circle is given by $A = \pi r^2$. Differentiating $A$ with respect to time $t$: $$ \frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = 2\pi r \frac{dr}{dt} $$ Given $\frac{dA}{dt} = 2~cm^2/sec$ and $r = 5~cm$: $$ 2 = 2\pi (5) \frac{dr}{dt} \implies 2 = 10\pi \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{2}{10\pi} = \frac{1}{5\pi}~cm/sec $$ The circumference of a circle is given by $C = 2\pi r$. Differentiating $C$ with respect to time $t$: $$ \frac{dC}{dt} = \frac{d}{dt}(2\pi r) = 2\pi \frac{dr}{dt} $$ Substitute the value of $\frac{dr}{dt}$: $$ \frac{dC}{dt} = 2\pi \left(\frac{1}{5\pi}\right) = \frac{2}{5}~cm/sec $$ The circumference of the circle is increasing at a rate of $\frac{2}{5}~cm/sec$.

ID: 594 Type: Very short (VSA) Source: AISSCE(Board Exam) Year: 2024

Question 30: The volume of a cube is increasing at the rate of \(6~cm^{3}/s.\) How fast is the surface area of cube increasing, when the length of an edge is 8 cm?

ID: 1 Type: Short Source: NCERT

Question 31: Find the rate of change of the area of a circle with respect to its radius \( r \) when (a) \( r = 3 \) cm (b) \( r = 4 \) cm

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✅ Manual Solution:

The area of a circle, denoted by \(A\), with a radius \(r\), is given by the formula:\[A = \pi r^2\]To find the rate of change of the area with respect to its radius, we differentiate the area formula with respect to \(r\):\[\frac{dA}{dr} = \frac{d}{dr}(\pi r^2) = 2\pi r\]Now, let's calculate this rate of change for specific values of the radius: \[\] 1. When \(r = 3\) cm: Substitute \(r=3\) into the derivative: \[\frac{dA}{dr} = 2\pi (3) = 6\pi\] Therefore, when the radius is 3 cm, the area of the circle is changing at a rate of \(6\pi\) cm\(^2\)/s. \[\] 2. When \(r = 4\) cm: Substitute \(r=4\) into the derivative: \[\frac{dA}{dr} = 2\pi (4) = 8\pi\] Therefore, when the radius is 4 cm, the area of the circle is changing at a rate of \(8\pi\) cm\(^2\)/s.

ID: 3 Type: Short Source: NCERT

Question 32: The radius of a circle is increasing uniformly at the rate of \(3\;{\text{cm}}/{\text{s}}\). Find the rate at which the area of the circle is increasing when the radius is \(10\;{\text{cm}}/{\text{s}}\)

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✅ Manual Solution:

We know that \(A = \pi {r^2}\) \[\therefore \dfrac{{dA}}{{dt}} = \dfrac{d}{{dr}}\left( {\pi {r^2}} \right)\dfrac{{dr}}{{dt}} = 2\pi r\dfrac{{dr}}{{dt}}\] Given that \(\dfrac{{dr}}{{dt}} = 3\;{\text{cm}}/{\text{s}}\) \[\therefore \dfrac{{dA}}{{dt}} = 2\pi r(3) = 6\pi r\] So, when \(r = 10\;{\text{cm}}\), \(\dfrac{{dA}}{{dt}} = 6\pi (10) = 60\pi {\text{c}}{{\text{m}}^2}/{\text{s}}\)

ID: 4 Type: Short Source: NCERT

Question 33: An edge of a variable cube is increasing at the rate of \(3\;{\text{cm}}/{\text{s}}\). How fast is the volume of the cube increasing when the edge is \(10\;{\text{cm}}\) long?

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✅ Manual Solution:

Let the side length of the cube be \(x\), and its volume be \(V\). \[\] Since \(V = x^3\), we differentiate with respect to time \(t\):\[\] \[\frac{dV}{dt} = \frac{d}{dt}(x^3) = \frac{d}{dx}(x^3) \cdot \frac{dx}{dt} = 3x^2 \cdot \frac{dx}{dt}\] Given \(\frac{dx}{dt} = 3\,\text{cm/s}\), we substitute: \[\frac{dV}{dt} = 3x^2 \cdot 3 = 9x^2\] When \(x = 10\,\text{cm}\), \[\]we find: \(\frac{dV}{dt} = 9(10)^2 = 900\,\text{cm}^3/\text{s}\) \[\] Hence, the volume of the cube is increasing at the rate of \(900\,\text{cm}^3/\text{s}\) when the edge is 10 cm long.

ID: 6 Type: Short Source: NCERT

Question 34: The radius of a circle is increasing at the rate of \(0.7,\text{cm/s}\). What is the rate of increase of its circumference?

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✅ Manual Solution:

We know that \(C = 2\pi r\). Differentiating with respect to time \(t\): \[\frac{dC}{dt} = \frac{d}{dr}(2\pi r) \cdot \frac{dr}{dt} = 2\pi \cdot \frac{dr}{dt}\] Given \(\frac{dr}{dt} = 0.7,\text{cm/s}\): \[\] \(\frac{dC}{dt} = 2\pi (0.7) = 1.4\pi,\text{cm/s}\)

ID: 7 Type: Short Source: NCERT

Question 35: The length \(x\) of a rectangle is decreasing at the rate of \(5,\text{cm/min}\), and the width \(y\) is increasing at the rate of \(4,\text{cm/min}\). When \(x = 8,\text{cm}\) and \(y = 6,\text{cm}\), find the rate of change of:(a) Perimeter (b) Area

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✅ Manual Solution:

(a) Perimeter: The perimeter is \(P = 2(x + y)\) \[\frac{dP}{dt} = 2\left(\frac{dx}{dt} + \frac{dy}{dt}\right) = 2(-5 + 4) = -2,\text{cm/min}\] (b) Area: The area is \(A = xy\) \[\frac{dA}{dt} = \frac{dx}{dt} \cdot y + x \cdot \frac{dy}{dt} = (-5)(6) + (8)(4) = -30 + 32 = 2,\text{cm}^2/\text{min}\]

ID: 8 Type: Short Source: NCERT

Question 36: A spherical balloon is being inflated at \(900,\text{cm}^3/\text{s}\). Find the rate of change of radius when \(r = 15,\text{cm}\).

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✅ Manual Solution:

Volume of a sphere: \(V = \frac{4}{3}\pi r^3\) \[\] Differentiating: \[\frac{dV}{dt} = \frac{d}{dr}\left(\frac{4}{3}\pi r^3\right) \cdot \frac{dr}{dt} = 4\pi r^2 \cdot \frac{dr}{dt}\] Given \(\frac{dV}{dt} = 900\): This implies \[900 = 4\pi (15)^2 \cdot \frac{dr}{dt} \Rightarrow \frac{dr}{dt} = \frac{900}{4\pi (225)} = \frac{1}{\pi} \text{cm/s}\]

ID: 9 Type: Short Source: NCERT

Question 37: A spherical balloon has a variable radius. Find the rate at which its volume is increasing with respect to radius when \(r = 10,\text{cm}\).

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✅ Manual Solution:

Volume of a sphere: \(V = \frac{4}{3}\pi r^3\)\[\] Differentiating with respect to \(r\): \[\frac{dV}{dr} = \frac{d}{dr}\left(\frac{4}{3}\pi r^3\right) = 4\pi r^2\] When \(r = 10,\text{cm}\): \[\frac{dV}{dr} = 4\pi (10)^2 = 400\pi \quad\text{cm}^3/\text{cm}\]

ID: 10 Type: Short Source: NCERT

Question 38: A ladder \(5,\text{m}\) long leans against a wall. The bottom slides away at \(2,\text{cm/s}\). How fast is the top of the ladder sliding down when the bottom is \(4,\text{m}\) from the wall?

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✅ Manual Solution:

Let \(x\) be the horizontal distance from the wall, and \(y\) be the height on the wall. \[\] Using Pythagoras: \(x^2 + y^2 = 25\)\[\] Differentiate: \[\frac{dy}{dt} = \frac{d}{dt}\left(\sqrt{25 - x^2}\right) = \frac{-x}{\sqrt{25 - x^2}} \cdot \frac{dx}{dt}\] Given \(\frac{dx}{dt} = 2,\text{cm/s} = 0.02,\text{m/s}\) and \(x = 4\): \[\frac{dy}{dt} = \frac{-4}{\sqrt{25 - 16}} \cdot 0.02 = \frac{-4}{3} \cdot 0.02 = -\frac{0.08}{3} \approx -0.0267,\text{m/s}\] So, the top of the ladder is sliding down at approximately \(0.0267,\text{m/s}\).

ID: 11 Type: Short Source: NCERT

Question 39: A particle is moving along the curve \(6y = x^3 + 2\). Find the points on the curve where the \(y\)-coordinate is changing 8 times as fast as the \(x\)-coordinate.

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✅ Manual Solution:

We start with the equation of the curve: \[6y = x^3 + 2\] Differentiating both sides with respect to time \(t\) gives: \[6\frac{dy}{dt} = 3x^2 \frac{dx}{dt}\] Dividing both sides by 3: \[2\frac{dy}{dt} = x^2 \frac{dx}{dt}\] According to the problem, the \(y\)-coordinate is changing 8 times as fast as the \(x\)-coordinate, i.e., \[\frac{dy}{dt} = 8\frac{dx}{dt}\] Substituting this into the equation: \[2(8\frac{dx}{dt}) = x^2 \frac{dx}{dt}\] \[\Rightarrow 16\frac{dx}{dt} = x^2 \frac{dx}{dt}\] Assuming \(\frac{dx}{dt} \ne 0\), we cancel it out. So : \[x^2 = 16 \Rightarrow x = \pm 4\] Now, we substitute these \(x\) values back into the original equation to find \(y\): \[\] When \(x = 4\): \[y = \frac{4^3 + 2}{6} = \frac{64 + 2}{6} = \frac{66}{6} = 11\] When \(x = -4\): \[y = \frac{(-4)^3 + 2}{6} = \frac{-64 + 2}{6} = \frac{-62}{6} = -\frac{31}{3}\] Final Answer: The required points on the curve are \((4,\ 11)\) and \(\left(-4,\ -\frac{31}{3}\right)\)

ID: 12 Type: Short Source: NCERT

Question 40: The radius of an air bubble is increasing at the rate of \(\dfrac{1}{2}\;{\text{cm}}/{\text{s}}\). At what rate is the volume of the bubble increasing when the radius is \(1\;{\text{cm}}\) ?

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✅ Manual Solution:

We are given that the radius of an air bubble is increasing at the rate of \(\frac{1}{2}\) cm/s. We want to find the rate at which the volume of the bubble is increasing when the radius is 1 cm. Assuming the bubble is spherical, the volume is given by: \[V = \frac{4}{3}\pi r^3\] Differentiating both sides with respect to time \(t\): \[\frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right) = \frac{d}{dr}\left(\frac{4}{3}\pi r^3\right) \cdot \frac{dr}{dt} = 4\pi r^2 \cdot \frac{dr}{dt}\] Given: \(\frac{dr}{dt} = \frac{1}{2}\) cm/s At \(r = 1\) cm: \[\frac{dV}{dt} = 4\pi (1)^2 \cdot \frac{1}{2} = 2\pi \text{ cm}^3/\text{s}\] Final Answer: The volume is increasing at a rate of \(2\pi\) cm\(^3\)/s when the radius is 1 cm.

ID: 13 Type: Short Source: NCERT

Question 41: A balloon, which always remains spherical, has a variable diameter \(\dfrac{3}{2}(2x + 1)\). Find the rate of change of its volume with respect to \(x\).

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✅ Manual Solution:

A balloon, which always remains spherical, has a variable diameter given by \[d = \dfrac{3}{2}(2x + 1).\] We are to find the rate of change of its volume with respect to \(x\). The volume of a sphere is given by: \[V = \dfrac{4}{3}\pi r^3\] Since the diameter is \(d = \dfrac{3}{2}(2x + 1)\), the radius is: \[r = \dfrac{d}{2} = \dfrac{3}{4}(2x + 1)\] Substitute into the volume formula: \[V = \dfrac{4}{3}\pi \left( \dfrac{3}{4}(2x + 1) \right)^3 = \dfrac{4}{3}\pi \cdot \dfrac{27}{64} (2x + 1)^3 = \dfrac{9}{16}\pi (2x + 1)^3\] Now differentiate with respect to \(x\): \[\dfrac{dV}{dx} = \dfrac{9}{16}\pi \cdot \dfrac{d}{dx}(2x + 1)^3 = \dfrac{9}{16}\pi \cdot 3(2x + 1)^2 \cdot 2 = \dfrac{27}{8}\pi (2x + 1)^2\] Final Answer: \[\dfrac{dV}{dx} = \dfrac{27}{8}\pi (2x + 1)^2\]

ID: 15 Type: Short Source: NCERT

Question 42: The total cost \(C(x)\) in Rupees associated with the production of \(x\) units of an item is given by \(C(x) = 0.007{x^3} - 0.003{x^2} + 15x + 4000\). Find the marginal cost when 17 units are produced.

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✅ Manual Solution:

The total cost \(C(x)\) (in rupees) for producing \(x\) units of an item is given by: \[C(x) = 0.007x^3 - 0.003x^2 + 15x + 4000\] We are asked to find the marginal cost when \(x = 17\).\[\] Marginal cost is the rate of change of total cost with respect to the number of units produced, i.e., \(MC = \dfrac{dC}{dx}\) \[\] Differentiate \(C(x)\): \[\dfrac{dC}{dx} = 0.007 \cdot 3x^2 - 0.003 \cdot 2x + 15 = 0.021x^2 - 0.006x + 15\] Now, substitute \(x = 17\): \[MC = 0.021 \cdot 17^2 - 0.006 \cdot 17 + 15 = 0.021 \cdot 289 - 0.006 \cdot 17 + 15\] \[= 6.069 - 0.102 + 15 = 20.967\] Final Answer: When 17 units are produced, the marginal cost is Rs. 20.967

ID: 16 Type: Short Source: NCERT

Question 43: The total revenue in Rupees received from the sale of \(x\) units of a product is given by \(R(x) = 13{x^2} + 26x + 15\). Find the marginal revenue when \(x = 7\).

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✅ Manual Solution:

The total revenue (in rupees) from selling \(x\) units of a product is given by: \[R(x) = 13x^2 + 26x + 15\] We are asked to find the marginal revenue when \(x = 7\).\[\] Marginal revenue is the rate of change of total revenue with respect to the number of units sold, i.e., \(MR = \dfrac{dR}{dx}\) \[\] Differentiate \(R(x)\): \[\dfrac{dR}{dx} = 13 \cdot 2x + 26 = 26x + 26\] Now substitute \(x = 7\): \[MR = 26 \cdot 7 + 26 = 182 + 26 = 208\] Final Answer: The marginal revenue when \(x = 7\) is Rs. 208

ID: 19 Type: Short Source: NCERT

Question 44: Show that the function given by \(f(x) = 3x + 17\) is strictly increasing on \(\mathbb{R}\).

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✅ Manual Solution:

Let \(x_1\) and \(x_2\) be any two real numbers such that \(x_1 < x_2\). Then, \[f(x_1) = 3x_1 + 17 < 3x_2 + 17 = f(x_2)\] \[\Rightarrow f(x_1) < f(x_2)\] Hence, \(f\) is strictly increasing on \(\mathbb{R}\). \[\] \( \bf Alternate Method:\) Differentiate the function: \[f'(x) = \dfrac{d}{dx}(3x + 17) = 3\] Since \(f'(x) = 3 > 0\) for all \(x \in \mathbb{R}\), the function is strictly increasing on \(\mathbb{R}\).

ID: 20 Type: Short Source: NCERT

Question 45: Show that the function \(f(x) = e^{2x}\) is strictly increasing on \(\mathbb{R}\).

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✅ Manual Solution:

Let \(x_1\) and \(x_2\) be any two real numbers such that \(x_1 < x_2\). Then: \[x_1 < x_2 \Rightarrow 2x_1 < 2x_2\] \[\Rightarrow e^{2x_1} < e^{2x_2}\] \[\Rightarrow f(x_1) < f(x_2)\] Hence, \(f(x)\) is strictly increasing on \(\mathbb{R}\). \[\] \(\bf Alternate Method:\) Differentiating w.r.x. we get \[f'(x) =2 e^{2x}\] It is always positive for any value of x. So \(f(x)\) is strictly increasing on \(\mathbb{R}\).

ID: 21 Type: Short Source: NCERT

Question 46: Show that the function given by \(f(x) = \sin x\) is \[\](a) Strictly increasing in \(\left( {0,\dfrac{\pi }{2}} \right)\) \[\](b) Strictly decreasing \(\left( {\dfrac{\pi }{2},\pi } \right)\) \[\] (c) Neither increasing nor decreasing in \((0,\pi )\).

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✅ Manual Solution:

We are given: \(f(x) = \sin x \Rightarrow f'(x) = \cos x\) \[\] In the interval \(x \in \left(0,\dfrac{\pi}{2}\right)\), we have \(\cos x > 0 \Rightarrow f'(x) > 0\) \(\Rightarrow f(x)\) is strictly increasing in \(\left(0,\dfrac{\pi}{2}\right)\) \[\] In the interval \(x \in \left(\dfrac{\pi}{2},\pi\right)\), we have \(\cos x < 0 \Rightarrow f'(x) < 0\) \(\Rightarrow f(x)\) is strictly decreasing in \(\left(\dfrac{\pi}{2},\pi\right)\) \[\] Since \(f\) increases in part of the interval and decreases in another, \(f\) is neither increasing nor decreasing in the entire interval \((0,\pi)\).

ID: 22 Type: Short Source: NCERT

Question 47: Find the intervals in which the function \(f(x) = 2x^2 - 3x\) is: (a) strictly increasing (b) strictly decreasing

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✅ Manual Solution:

We are given: \(f(x) = 2x^2 - 3x\). \[\] Differentiate: \[f'(x) = \dfrac{d}{dx}(2x^2 - 3x) = 4x - 3\] Set \(f'(x) = 0\) to find critical point: \[4x - 3 = 0 \Rightarrow x = \dfrac{3}{4}\] Now analyze the sign of \(f'(x)\): In the interval \(\left(-\infty,\dfrac{3}{4}\right)\), we have \(f'(x) = 4x - 3 < 0\) \(\Rightarrow f(x)\) is strictly decreasing on \(\left(-\infty,\dfrac{3}{4}\right)\) \[\] In the interval \(\left(\dfrac{3}{4},\infty\right)\), we have \(f'(x) = 4x - 3 > 0\) \(\Rightarrow f(x)\) is strictly increasing on \(\left(\dfrac{3}{4},\infty\right)\) \[\]Final Answer: (a) \(f(x)\) is strictly increasing in \(\left(\dfrac{3}{4}, \infty\right)\) (b) \(f(x)\) is strictly decreasing in \(\left(-\infty, \dfrac{3}{4}\right)\)

ID: 23 Type: Short Source: NCERT

Question 48: Find the intervals in which the function \(f(x) = 2x^3 - 3x^2 - 36x + 7\) is: (a) Strictly increasing (b) Strictly decreasing

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✅ Manual Solution:

Differentiate the function: \[f'(x) = \dfrac{d}{dx}(2x^3 - 3x^2 - 36x + 7) = 6x^2 - 6x - 36\] Factor the derivative: \[f'(x) = 6(x^2 - x - 6) = 6(x + 2)(x - 3)\] Set \(f'(x) = 0\) to find critical points: \[6(x + 2)(x - 3) = 0 \Rightarrow x = -2,,3\] Now analyze the sign of \(f'(x)\) in the intervals: \[\] In \((-\infty, -2)\): Both \((x + 2) < 0\) and \((x - 3) < 0 \Rightarrow f'(x) > 0\) \(\Rightarrow f(x)\) is strictly increasing on \((-\infty, -2)\) \[\] In \((-2, 3)\): \((x + 2) > 0\) and \((x - 3) < 0 \Rightarrow f'(x) < 0\) \(\Rightarrow f(x)\) is strictly decreasing on \((-2, 3)\) \[\] In \((3, \infty)\): Both \((x + 2) > 0\) and \((x - 3) > 0 \Rightarrow f'(x) > 0\) \(\Rightarrow f(x)\) is strictly increasing on \((3, \infty)\) \[\] Final Answer: \(f(x)\) is strictly increasing on \((-\infty, -2)\) and \((3, \infty)\) and\[\] \(f(x)\) is strictly decreasing on \((-2, 3)\)

ID: 24 Type: Short Source: NCERT

Question 49: Find the intervals in which the function \(f(x) = x^2 + 2x - 5\) is strictly increasing or strictly decreasing.

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✅ Manual Solution:

We have: \[f'(x) = 2x + 2\] Setting \(f'(x) = 0\): \[2x + 2 = 0 \Rightarrow x = -1\] In \((-\infty, -1)\), \[f'(x) < 0 ⟹ \text {strictly decreasing}\] In \((-1, \infty)\), \[f'(x) > 0 ⟹ \text{strictly increasing}\]

ID: 25 Type: Short Source: NCERT

Question 50: Find the intervals in which the function \(f(x) = 10 - 6x - 2x^2\) is strictly increasing or strictly decreasing.

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✅ Manual Solution:

We have: \[f'(x) = -6 - 4x\] Setting \(f'(x) = 0\): \[-6 - 4x = 0 \Rightarrow x = -\dfrac{3}{2}\] In \(\left(-\infty, -\dfrac{3}{2}\right)\), \(f'(x) > 0\) ⟹ strictly increasing \[\] In \(\left(-\dfrac{3}{2}, \infty\right)\), \(f'(x) < 0\) ⟹ strictly decreasing

ID: 26 Type: Short Source: NCERT

Question 51: Find the intervals in which the function \(f(x) = -2x^3 - 9x^2 - 12x + 1\) is strictly increasing or strictly decreasing.

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✅ Manual Solution:

We have: \[f'(x) = -6x^2 - 18x - 12 = -6(x + 1)(x + 2)\] Setting \(f'(x) = 0\): \[x = -1,\ -2\] Intervals: \((-\infty, -2),\ (-2, -1),\ (-1, \infty)\) \[\] In \((-\infty, -2)\) and \((-1, \infty)\), \(f'(x) < 0\) ⟹ strictly decreasing \[\] In \((-2, -1)\), \(f'(x) > 0\) ⟹ strictly increasing

ID: 27 Type: Short Source: NCERT

Question 52: Find the intervals in which the function \(f(x) = 6 - 9x - x^2\) is strictly increasing or strictly decreasing.

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✅ Manual Solution:

We have: \[f'(x) = -9 - 2x\] Setting \(f'(x) = 0\): \[x = -\dfrac{9}{2}\] In \(\left(-\infty, -\dfrac{9}{2}\right)\), \(f'(x) > 0\) ⟹ strictly increasing \[\] In \(\left(-\dfrac{9}{2}, \infty\right)\), \(f'(x) < 0\) ⟹ strictly decreasing

ID: 28 Type: Short Source: NCERT

Question 53: Find the intervals in which the function \(f(x) = (x + 1)^3(x - 3)^3\) is strictly increasing or strictly decreasing.

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✅ Manual Solution:

We have: \[ f(x) = (x + 1)^3(x - 3)^3 \] Differentiate using product rule: \[ f'(x) = 3(x + 1)^2(x - 3)^3 + 3(x - 3)^2(x + 1)^3 \] Factor common terms: \[ f'(x) = 3(x + 1)^2(x - 3)^2[(x - 3) + (x + 1)] \] \[ = 3(x + 1)^2(x - 3)^2(2x - 2) \] \[ = 6(x + 1)^2(x - 3)^2(x - 1) \] Now, set derivative equal to zero: \[ f'(x) = 0 \Rightarrow (x + 1)^2 = 0 \text{ or } (x - 3)^2 = 0 \text{ or } (x - 1) = 0 \] \[ \Rightarrow x = -1,\ 1,\ 3 \] These values divide the number line into intervals: \[ (-\infty, -1),\ (-1, 1),\ (1, 3),\ (3, \infty) \] Now we analyze the sign of \(f'(x)\) in each interval: \[\] In \((- \infty, -1)\), choose \(x = -2\): \(f'(x) = 6(-1)^2(-5)^2(-3) < 0\) So, \(f(x)\) is \(\textbf{strictly decreasing}\). \[\] In \((-1, 1)\), choose \(x = 0\): \(f'(x) = 6(1)^2(-3)^2(-1) < 0\) So, \(f(x)\) is \(\textbf{strictly decreasing}\). \[\] In \((1, 3)\), choose \(x = 2\): \(f'(x) = 6(3)^2(-1)^2(1) > 0\) So, \(f(x)\) is \(\textbf{strictly increasing}\). \[\] In \((3, \infty)\), choose \(x = 4\): \(f'(x) = 6(5)^2(1)^2(3) > 0\) So, \(f(x)\) is \(\textbf{strictly increasing}\). \[\] Conclusion: \[ f(x) \text{ is strictly decreasing in } (-\infty, -1) \cup (-1, 1) \] \[ f(x) \text{ is strictly increasing in } (1, 3) \cup (3, \infty) \]

ID: 32 Type: Short Source: NCERT

Question 54: Prove that the logarithmic function is strictly increasing on \( (0, \infty) \).

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✅ Manual Solution:

Let \( f(x) = \log x \). Then, \[ f'(x) = \frac{d}{dx}(\log x) = \frac{1}{x} \] For \( x > 0 \), we have: \[ f'(x) = \frac{1}{x} > 0 \] Since the derivative is positive for all \( x > 0 \), the function \( f(x) = \log x \) is strictly increasing on \( (0, \infty) \).

ID: 33 Type: Short Source: NCERT

Question 55: Prove that the function \( f \) given by \( f(x) = x^2 - x + 1 \) is neither strictly increasing nor strictly decreasing on \( (-1, 1) \).

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✅ Manual Solution:

We have: \[ f(x) = x^2 - x + 1 \] Differentiating with respect to \( x \), we get: \[ f'(x) = 2x - 1 \] Now, \[ f'(x) = 0 \Rightarrow 2x - 1 = 0 \Rightarrow x = \frac{1}{2} \] So, \( x = \frac{1}{2} \) divides the interval \( (-1, 1) \) into two parts: \[ (-1, \tfrac{1}{2}) \quad \text{and} \quad (\tfrac{1}{2}, 1) \] In \( (-1, \tfrac{1}{2}) \), we have: \[ f'(x) = 2x - 1 < 0 \Rightarrow f \text{ is strictly decreasing} \] In \( (\tfrac{1}{2}, 1) \), we have: \[ f'(x) = 2x - 1 > 0 \Rightarrow f \text{ is strictly increasing} \] \(\textbf{Conclusion:} \) Since \( f \) is decreasing in one part and increasing in another, \[\text{the function } f(x) = x^2 - x + 1 \text{ is neither strictly increasing nor strictly decreasing on } (-1, 1). \]

ID: 34 Type: Short Source: NCERT

Question 56: Find the least value of \( a \) such that the function \( f(x) = x^2 + ax + 1 \) is strictly increasing on the interval \( (1, 2) \).

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✅ Manual Solution:

We know a function is strictly increasing on an interval if its derivative is positive on that interval. \[ f'(x) = \frac{d}{dx}(x^2 + ax + 1) = 2x + a \] We require: \[ f'(x) > 0 \quad \text{for all } x \in (1, 2) \] This means: \[ 2x + a > 0 \quad \text{for all } x \in (1, 2) \] The minimum value of \( f'(x) = 2x + a \) in the interval \( (1, 2) \) occurs at the left endpoint \( x = 1 \). \[ \text{So, we need: } 2(1) + a > 0 \Rightarrow 2 + a > 0 \Rightarrow a > -2 \] Thus, the least value of \( a \) such that \( f(x) \) is strictly increasing on \( (1, 2) \) is: \[ \boxed{-2} \]

ID: 38 Type: Short Source: NCERT

Question 57: Prove that the function given by \( f(x) = x^3 - 3x^2 + 3x = 100 \) is increasing on \( \mathbb{R} \).

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✅ Manual Solution:

Let \[ f(x) = x^3 - 3x^2 + 3x \] Differentiate \( f(x) \) with respect to \( x \): \[ f'(x) = \frac{d}{dx}(x^3 - 3x^2 + 3x) = 3x^2 - 6x + 3 \] Factor the derivative: \[ f'(x) = 3x^2 - 6x + 3 = 3(x^2 - 2x + 1) = 3(x - 1)^2 \] Now, for all \( x \in \mathbb{R} \), we know: \[ (x - 1)^2 \geq 0 \Rightarrow f'(x) = 3(x - 1)^2 \geq 0 \] Also, \( f'(x) = 0 \) only at \( x = 1 \), and \( f'(x) > 0 \) elsewhere. Thus, \( f'(x) \geq 0 \) for all \( x \in \mathbb{R} \), and \( f'(x) > 0 \) for all \( x \ne 1 \). \[\] \(\textbf{Conclusion:}\) \( f(x) = x^3 - 3x^2 + 3x \text{ is an increasing function on } \mathbb{R}. \)

ID: 39 Type: Short Source: NCERT

Question 58: Find the maximum and minimum values, if any, of the function \( f(x) = (2x - 1)^2 + 3 \)

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✅ Manual Solution:

We have: \[ (2x - 1)^2 \geq 0 \text{ for all } x \in \mathbb{R} \] \[ f(x) = (2x - 1)^2 + 3 \geq 3 \text{ for all } x \in \mathbb{R} \] Minimum value occurs when \(2x - 1 = 0 \Rightarrow x = \frac{1}{2}\). \[ \text{Minimum value of } f\left(\frac{1}{2}\right) = (2 \cdot \frac{1}{2} - 1)^2 + 3 = 3 \] No maximum value exists since the function is unbounded above.

ID: 40 Type: Short Source: NCERT

Question 59: Find the maximum and minimum values, if any, of the function \( f(x) = 9x^2 + 12x + 2 \)

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✅ Manual Solution:

Complete the square: \[ f(x) = 9x^2 + 12x + 2 = 9\left(x^2 + \frac{4}{3}x\right) + 2 = 9\left(x + \frac{2}{3}\right)^2 - 4 + 2 = 9\left(x + \frac{2}{3}\right)^2 - 2 \] \[ \text{Minimum value occurs at } x = -\frac{2}{3} \] \[ f\left(-\frac{2}{3}\right) = -2 \] No maximum value exists since the parabola opens upwards.

ID: 41 Type: Short Source: NCERT

Question 60: Find the maximum and minimum values, if any, of the function \( f(x) = - (x - 1)^2 + 10 \)

View Solution

✅ Manual Solution:

We know: \[ (x - 1)^2 \geq 0 \Rightarrow - (x - 1)^2 \leq 0 \Rightarrow f(x) = - (x - 1)^2 + 10 \leq 10 \] Maximum value occurs when \(x = 1\): \[ f(1) = - (1 - 1)^2 + 10 = 10 \] No minimum value exists since the function decreases indefinitely.

ID: 42 Type: Short Source: NCERT

Question 61: Find the maximum and minimum values, if any, of the function \( g(x) = x^3 + 1 \)

View Solution

✅ Manual Solution:

Since \(x^3\) is strictly increasing on \(\mathbb{R}\), the function \(g(x) = x^3 + 1\) is also strictly increasing. \[ \text{Hence, } g(x) \text{ has neither a maximum nor a minimum value on } \mathbb{R}. \]

ID: 43 Type: Short Source: NCERT

Question 62: Find the maximum and minimum values of the function \( f(x) = |x + 2| - 1 \)

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✅ Manual Solution:

Since \(|x + 2| \geq 0\) for all \(x \in \mathbb{R}\), \[ f(x) = |x + 2| - 1 \geq -1 \] Equality occurs when \(|x + 2| = 0 \Rightarrow x = -2\). \[ \text{Minimum value of } f = f(-2) = | -2 + 2 | - 1 = -1 \] There is no maximum value since \(|x + 2|\) grows unbounded as \(x \to \infty\).

ID: 44 Type: Short Source: NCERT

Question 63: Find the maximum and minimum values of the function \( g(x) = -|x + 1| + 3 \)

View Solution

✅ Manual Solution:

Since \(-|x + 1| \leq 0\), \[ g(x) = -|x + 1| + 3 \leq 3 \] Equality occurs when \(|x + 1| = 0 \Rightarrow x = -1\). \[ \text{Maximum value of } g = g(-1) = -| -1 + 1 | + 3 = 3 \] There is no minimum value since \(-|x + 1|\) decreases without bound.

ID: 45 Type: Short Source: NCERT

Question 64: Find the maximum and minimum values of the function \( h(x) = \sin (2x) + 5 \)

View Solution

✅ Manual Solution:

Since \(-1 \leq \sin(2x) \leq 1\), \[ 4 \leq \sin(2x) + 5 \leq 6 \] \[ \text{Minimum value of } h(x) = 4, \quad \text{Maximum value of } h(x) = 6 \]

ID: 46 Type: Short Source: NCERT

Question 65: Find the maximum and minimum values of the function \( f(x) = |\sin(4x) + 3| \)

View Solution

✅ Manual Solution:

Since \(-1 \leq \sin(4x) \leq 1\), \[ 2 \leq \sin(4x) + 3 \leq 4 \] \[ \Rightarrow 2 \leq |\sin(4x) + 3| \leq 4 \] \[ \text{Minimum value of } f(x) = 2, \quad \text{Maximum value of } f(x) = 4 \]

ID: 47 Type: Short Source: NCERT

Question 66: Find the maximum and minimum values of the function \( h(x) = x + 4, \quad x \in (-1, 1) \)

View Solution

✅ Manual Solution:

This is a linear function defined on an open interval. \[\]As \(x \to -1\), \(h(x) \to 3\), and as \(x \to 1\), \(h(x) \to 5\). \[\] However, neither 3 nor 5 is attained in the open interval \((-1, 1)\). \[ \text{Therefore, } h(x) \text{ has neither a maximum nor a minimum value in } (-1, 1). \]

ID: 48 Type: Short Source: NCERT

Question 67: Find the local maxima and local minima, if any, of the function \( f(x) = x^2 \). Find also the local maximum and the local minimum values.

View Solution

✅ Manual Solution:

We have: \[ f(x) = x^2 \Rightarrow f'(x) = 2x \] Setting \( f'(x) = 0 \Rightarrow x = 0 \) Also, \[ f''(x) = 2 > 0 \] So, \( x = 0 \) is a point of local minima. Local minimum value is: \[ f(0) = 0 \] There is no local maximum.

ID: 49 Type: Short Source: NCERT

Question 68: Find the local maxima and local minima, if any, of the function \( g(x) = x^3 - 3x \).

View Solution

✅ Manual Solution:

\[ g(x) = x^3 - 3x \Rightarrow g'(x) = 3x^2 - 3 = 3(x^2 - 1) = 3(x - 1)(x + 1) \] Setting \( g'(x) = 0 \Rightarrow x = \pm 1 \) \[ g''(x) = 6x \Rightarrow g''(1) = 6 > 0 \Rightarrow \text{Local minimum at } x = 1 \] \[ g(1) = 1 - 3 = -2 \] \[ g''(-1) = -6 < 0 \Rightarrow \text{Local maximum at } x = -1 \] \[ g(-1) = -1 + 3 = 2 \]

ID: 50 Type: Short Source: NCERT

Question 69: Find the local maxima and minima of \( h(x) = \sin x + \cos x \) in \( \left( 0, \frac{\pi}{2} \right) \)

View Solution

✅ Manual Solution:

\[ h(x) = \sin x + \cos x \Rightarrow h'(x) = \cos x - \sin x \] \[ h'(x) = 0 \Rightarrow \cos x = \sin x \Rightarrow \tan x = 1 \Rightarrow x = \frac{\pi}{4} \] \[ h''(x) = -\sin x - \cos x < 0 \Rightarrow \text{Local maximum at } x = \frac{\pi}{4} \] \[ h\left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2} \]

ID: 51 Type: Short Source: NCERT

Question 70: Find the local maxima and minima of \( f(x) = \sin x - \cos x \) in \( \left(0, 2\pi \right) \).

View Solution

✅ Manual Solution:

\[ f(x) = \sin x - \cos x \Rightarrow f'(x) = \cos x + \sin x \] \[ f'(x) = 0 \Rightarrow \cos x = -\sin x \Rightarrow \tan x = -1 \Rightarrow x = \frac{3\pi}{4}, \frac{7\pi}{4} \] Second derivative: \[ f''(x) = -\sin x + \cos x \] At \( x = \frac{3\pi}{4} \), \[ f''\left( \frac{3\pi}{4} \right) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\sqrt{2} < 0 \Rightarrow \text{Local maximum} \] \[ f\left( \frac{3\pi}{4} \right) = \frac{1}{\sqrt{2}} - \left( -\frac{1}{\sqrt{2}} \right) = \sqrt{2} \] At \( x = \frac{7\pi}{4} \), \[ f''\left( \frac{7\pi}{4} \right) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2} > 0 \Rightarrow \text{Local minimum} \] \[ f\left( \frac{7\pi}{4} \right) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\sqrt{2} \]

ID: 52 Type: Short Source: NCERT

Question 71: Find the local maxima and minima of \( f(x) = x^3 - 6x^2 + 9x + 15 \).

View Solution

✅ Manual Solution:

\[ f'(x) = 3x^2 - 12x + 9 = 3(x - 1)(x - 3) \Rightarrow f'(x) = 0 \Rightarrow x = 1, 3 \] \[ f''(x) = 6x - 12 \Rightarrow f''(1) = -6 < 0 \Rightarrow \text{Local maximum} \Rightarrow f(1) = 1 - 6 + 9 + 15 = 19 \] \[ f''(3) = 6 > 0 \Rightarrow \text{Local minimum} \Rightarrow f(3) = 27 - 54 + 27 + 15 = 15 \]

ID: 53 Type: Short Source: NCERT

Question 72: Find the local maxima and minima of \(g(x) = \frac{x}{2} + \frac{2}{x}, \, x > 0 \).

View Solution

✅ Manual Solution:

\[ g'(x) = \frac{1}{2} - \frac{2}{x^2} \Rightarrow g'(x) = 0 \Rightarrow \frac{1}{2} = \frac{2}{x^2} \Rightarrow x^2 = 4 \Rightarrow x = 2 \, (\text{since } x > 0) \] \[ g''(x) = \frac{4}{x^3} \Rightarrow g''(2) = \frac{4}{8} = \frac{1}{2} > 0 \Rightarrow \text{Local minimum} \Rightarrow g(2) = 1 + 1 = 2 \]

ID: 54 Type: Short Source: NCERT

Question 73: Find the local maxima and minima of \( g(x) = \frac{1}{x^2 + 2} \).

View Solution

✅ Manual Solution:

\[ g'(x) = \frac{-2x}{(x^2 + 2)^2} \Rightarrow g'(x) = 0 \Rightarrow x = 0 \] Check sign change: - For \( x < 0 \), \( g'(x) > 0 \) - For \( x > 0 \), \( g'(x) < 0 \) So by first derivative test: \[ x = 0 \text{ is a local maximum} \Rightarrow g(0) = \frac{1}{2} \]

ID: 55 Type: Short Source: NCERT

Question 74: Find the local maxima and minima of \( f(x) = x \sqrt{1 - x}, \, x > 0 \).

View Solution

✅ Manual Solution:

Let us differentiate: \[ f(x) = x \sqrt{1 - x} \Rightarrow f'(x) = \sqrt{1 - x} + x \cdot \left( \frac{-1}{2\sqrt{1 - x}} \right) = \frac{2(1 - x) - x}{2\sqrt{1 - x}} = \frac{2 - 3x}{2\sqrt{1 - x}} \] Set \( f'(x) = 0 \Rightarrow 2 - 3x = 0 \Rightarrow x = \frac{2}{3} \) Now second derivative: \[ f''(x) = \frac{d}{dx} \left( \frac{2 - 3x}{2\sqrt{1 - x}} \right) = \frac{3x - 4}{4(1 - x)^{3/2}} \Rightarrow f''\left( \frac{2}{3} \right) = \frac{2 - 4}{2(1/3)^{3/2}} < 0 \Rightarrow \text{Local maximum} \] \[ f\left( \frac{2}{3} \right) = \frac{2}{3} \cdot \sqrt{\frac{1}{3}} = \frac{2\sqrt{3}}{9} \]

ID: 56 Type: Short Source: NCERT

Question 75: Prove that the function \( f(x) = e^x \) does not have any maximum or minimum.

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✅ Manual Solution:

We have, \[ f(x) = e^x \Rightarrow f'(x) = e^x \] Since \( e^x > 0 \) for all \( x \in \mathbb{R} \), \( f'(x) \neq 0 \) for any real \( x \). Therefore, there is no critical point \( c \in \mathbb{R} \) such that \( f'(c) = 0 \). Hence, the function \( f(x) = e^x \) does not have any maximum or minimum.

ID: 57 Type: Short Source: NCERT

Question 76: Prove that the function \( g(x) = \log x \) does not have any maximum or minimum.

View Solution

✅ Manual Solution:

We have, \[ g(x) = \log x \Rightarrow g'(x) = \frac{1}{x} \] The function \( \log x \) is defined for \( x > 0 \), and for all such \( x \), \[ g'(x) = \frac{1}{x} > 0 \] So, \( g(x) \) is strictly increasing and \( g'(x) \neq 0 \) for any \( x > 0 \). Therefore, there does not exist any \( c \in (0, \infty) \) such that \( g'(c) = 0 \). Hence, the function \( g(x) = \log x \) does not have any maximum or minimum.

ID: 58 Type: Short Source: NCERT

Question 77: Prove that the function \( h(x) = x^3 + x^2 + x + 1 \) does not have any maximum or minimum.

View Solution

✅ Manual Solution:

We have, \[ h(x) = x^3 + x^2 + x + 1 \Rightarrow h'(x) = 3x^2 + 2x + 1 \] Now, the discriminant of \( h'(x) = 3x^2 + 2x + 1 \) is: \[ \Delta = (2)^2 - 4(3)(1) = 4 - 12 = -8 < 0 \] So, the derivative has no real roots. This implies \( h'(x) \neq 0 \) for all \( x \in \mathbb{R} \). Also, since the leading coefficient is positive, \( h'(x) > 0 \) for all \( x \in \mathbb{R} \). Therefore, the function is strictly increasing and has no maximum or minimum. Hence, \( h(x) = x^3 + x^2 + x + 1 \) does not have any local maxima or minima.

ID: 59 Type: Short Source: NCERT

Question 78: Find the absolute maximum and minimum values of the function \( f(x) = x^3 \) in the interval \([ -2, 2 ]\).

View Solution

✅ Manual Solution:

We have \( f(x) = x^3 \) \[ f'(x) = 3x^2 \] Setting \( f'(x) = 0 \Rightarrow x = 0 \) Now evaluate \( f \) at endpoints and critical points: \[ f(-2) = (-2)^3 = -8 \\ f(0) = 0 \\ f(2) = 2^3 = 8 \] \(\textbf{Absolute maximum}\) is \( f(2) = 8 \) at \( x = 2 \) and \(\textbf{Absolute minimum}\) is \( f(-2) = -8 \) at \( x = -2 \)

ID: 60 Type: Short Source: NCERT

Question 79: Find the absolute maximum and minimum values of the function \( f(x) = \sin x + \cos x \) in the interval \( [0, \pi] \).

View Solution

✅ Manual Solution:

We have \( f(x) = \sin x + \cos x \) \[ f'(x) = \cos x - \sin x \\ f'(x) = 0 \Rightarrow \sin x = \cos x \Rightarrow \tan x = 1 \Rightarrow x = \frac{\pi}{4} \] Now evaluate \( f \) at endpoints and critical points: \[ f\left( \frac{\pi}{4} \right) = \sin \frac{\pi}{4} + \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2} \\ f(0) = \sin 0 + \cos 0 = 1 \\ f(\pi) = \sin \pi + \cos \pi = -1 \] \(\textbf{Absolute maximum}\) is \( \sqrt{2} \) at \( x = \frac{\pi}{4} \) and \(\textbf{Absolute minimum}\) is \( -1 \) at \( x = \pi \)

ID: 61 Type: Short Source: NCERT

Question 80: Find the absolute maximum and minimum values of the function \( f(x) = 4x - \frac{1}{2}x^2 \) in the interval \( \left[ -2, \frac{9}{2} \right] \).

View Solution

✅ Manual Solution:

We have \( f(x) = 4x - \frac{1}{2}x^2 \) \[ f'(x) = 4 - x \\ f'(x) = 0 \Rightarrow x = 4 \] Now evaluate \( f \) at endpoints and critical point: \[ f(-2) = 4(-2) - \frac{1}{2}(-2)^2 = -8 - 2 = -10 \] \[f(4) = 4(4) - \frac{1}{2}(16) = 16 - 8 = 8 \] \[f\left( \frac{9}{2} \right) = 4 \cdot \frac{9}{2} - \frac{1}{2} \cdot \left( \frac{9}{2} \right)^2 = 18 - \frac{81}{8} = 18 - 10.125 = 7.875 \] \(\textbf{Absolute maximum}\) is \( 8 \) at \( x = 4 \) and \(\textbf{Absolute minimum}\) is \( -10 \) at \( x = -2 \)

ID: 62 Type: Short Source: NCERT

Question 81: Find the absolute maximum and minimum values of the function \( f(x) = (x - 1)^2 + 3 \) in the interval \( [-3, 1] \).

View Solution

✅ Manual Solution:

We have \( f(x) = (x - 1)^2 + 3 \) \[ f'(x) = 2(x - 1) \Rightarrow f'(x) = 0 \Rightarrow x = 1 \] Now evaluate \( f \) at endpoints and critical point: \[ f(-3) = (-3 - 1)^2 + 3 = 16 + 3 = 19 \\ f(1) = (1 - 1)^2 + 3 = 0 + 3 = 3 \] \(\textbf{Absolute maximum}\) is \( 19 \) at \( x = -3 \) and \(\textbf{Absolute minimum}\) is \( 3 \) at \( x = 1 \)

ID: 63 Type: Short Source: NCERT

Question 82: Find the maximum profit that a company can make, if the profit function is given by \( p(x) = 41 - 24x - 18x^2 \)

View Solution

✅ Manual Solution:

We have: \[ p(x) = 41 - 24x - 18x^2 \Rightarrow p'(x) = -24 - 36x \Rightarrow p''(x) = -36 < 0 \] Set derivative to zero: \[ p'(x) = 0 \Rightarrow -24 - 36x = 0 \Rightarrow x = -\frac{2}{3} \] By second derivative test, local maximum occurs at \( x = -\frac{2}{3} \) Now, compute: \[ p\left( -\frac{2}{3} \right) = 41 - 24\left( -\frac{2}{3} \right) - 18\left( -\frac{2}{3} \right)^2 = 41 + 16 - 8 = 49 \] \(\textbf{Maximum profit}\) = 49

ID: 64 Type: Short Source: NCERT

Question 83: Find the intervals in which the function \( f(x) = x^3 + \frac{1}{x^3}, \quad x \ne 0 \) is (i) Increasing (ii) Decreasing.

View Solution

✅ Manual Solution:

\[ f'(x) = 3x^2 - \frac{3}{x^4} = \frac{3x^6 - 3}{x^4} \] Set derivative to zero: \[ f'(x) = 0 \Rightarrow x^6 = 1 \Rightarrow x = \pm 1 \] Sign of derivative: \[\] \( f'(x) > 0 \) in \( (-\infty, -1) \cup (1, \infty) \) ⟹ Increasing \[\] \( f'(x) < 0 \) in \( (-1, 1) \) ⟹ Decreasing

ID: 65 Type: Short Source: NCERT

Question 84: At what points in the interval \([0, 2\pi]\), does the function \( f(x) = \sin 2x \) attain its maximum value?

View Solution

✅ Manual Solution:

We are given: \[ f(x) = \sin 2x \Rightarrow f'(x) = 2\cos 2x \] Set the derivative to zero: \[ f'(x) = 0 \Rightarrow \cos 2x = 0 \Rightarrow 2x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2} \Rightarrow x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \] Now evaluate \( f(x) \) at these critical points: \[ f\left( \frac{\pi}{4} \right) = \sin\left( \frac{\pi}{2} \right) = 1 \] \[f\left( \frac{3\pi}{4} \right) = \sin\left( \frac{3\pi}{2} \right) = -1 \] \[f\left( \frac{5\pi}{4} \right) = \sin\left( \frac{5\pi}{2} \right) = 1 \] \[f\left( \frac{7\pi}{4} \right) = \sin\left( \frac{7\pi}{2} \right) = -1 \] Also check endpoints: \[ f(0) = \sin 0 = 0, \quad f(2\pi) = \sin(4\pi) = 0 \] Thus, the absolute maximum value of \( f(x) \) on \([0, 2\pi]\) is \( \boxed{1} \quad \text{at} \quad x = \frac{\pi}{4} \text{ and } x = \frac{5\pi}{4} \)

ID: 68 Type: Short Source: NCERT

Question 85: It is given that at \(x = 1\), the function \(f(x) = x^4 - 62x^2 + ax + 9\) attains its maximum value on the interval \([0, 2]\). Find the value of \(a\).

View Solution

✅ Manual Solution:

We are given: \[ f(x) = x^4 - 62x^2 + ax + 9 \] Differentiate: \[ f'(x) = 4x^3 - 124x + a \] Since \(f(x)\) has a maximum at \(x = 1\), we have: \[ f'(1) = 0 \] Substitute \(x = 1\): \[ f'(1) = 4(1)^3 - 124(1) + a = 4 - 124 + a = 0 \] Solve for \(a\): \[ a = 120 \] \[ \boxed{a = 120} \]

ID: 200 Type: Short (SA) Source: AISSCE(Board Exam) Year: 2025

Question 86: The side of an equilateral triangle is increasing at the rate of \(3\) cm/s. At what rate is its area increasing when the side of the triangle is \(15\) cm?

View Solution

✅ Manual Solution:

The area of an equilateral triangle is \(A = \frac{\sqrt{3}}{4} s^2\).

Differentiate w.r.t. Time (\(t\)):\[\frac{dA}{dt} = \frac{d}{dt} \left( \frac{\sqrt{3}}{4} s^2 \right) = \frac{\sqrt{3}}{4} \cdot 2s \cdot \frac{ds}{dt}\]\[\frac{dA}{dt} = \frac{\sqrt{3}}{2} s \frac{ds}{dt}\] Substitute the given values: \(s = 15 \text{ cm}\) and \(\frac{ds}{dt} = 3 \text{ cm/s}\).\[\frac{dA}{dt} = \frac{\sqrt{3}}{2} (15) (3)\]\[\frac{dA}{dt} = \frac{45\sqrt{3}}{2}\]The rate of increase of the area is \(\mathbf{\frac{45\sqrt{3}}{2} \text{ cm}^2\text{/s}}\).

ID: 393 Type: Short (SA) Source: AISSCE(Board Exam) Year: 2025

Question 87: Find the value of 'a' for which \(f(x)=\sqrt{3}\sin x-\cos x-2ax+6\) is decreasing in R.

ID: 459 Type: Short (SA) Source: AISSCE(Board Exam) Year: 2025

Question 88: Amongst all pairs of positive integers with product as 289, find which of the two numbers add up to the least.

ID: 601 Type: Short (SA) Source: AISSCE(Board Exam) Year: 2024

Question 89: Find the intervals in which the function \(f(x)=\frac{log~x}{x}\) is strictly increasing or strictly decreasing.

ID: 614 Type: Short (SA) Source: AISSCE(Board Exam) Year: 2024

Question 90: Find the absolute maximum and absolute minimum values of the function f given by \(f(x)=\frac{x}{2}+\frac{2}{x}\) on the interval [1, 2].

ID: 2 Type: Long Source: NCERT

Question 91: The volume of a cube is increasing at ( 8 ) cm³/s. How fast is the surface area increasing when the edge length is ( 12 ) cm?

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✅ Manual Solution:

Let the length of a side of the cube be \(x\), its volume be \(V\), and its surface area be \(S\). We know that \(x\) is a function of time \(t\). The formulas for the volume and surface area of a cube are: \[\] Volume: \(V = x^3\) Surface Area: \(S = 6x^2\)\[\] It is given that the rate of change of the volume with respect to time is 8 cm\(^3\)/s, i.e.:\[ \frac{dV}{dt} = 8 \text{ cm}^3/\text{s} \]To relate the rate of change of volume to the rate of change of the side length, we use the chain rule on the volume formula:\[ \frac{dV}{dt} = \frac{d}{dt}(x^3) = \frac{d}{dx}(x^3) \cdot \frac{dx}{dt} \]\[ 8 = 3x^2 \cdot \frac{dx}{dt} \]From this, we can express the rate of change of the side length, \(\frac{dx}{dt}\):\[ \frac{dx}{dt} = \frac{8}{3x^2} \quad \text{(1)} \]Now, we need to find the rate of change of the surface area with respect to time, \(\frac{dS}{dt}\). We apply the chain rule to the surface area formula:\[ \frac{dS}{dt} = \frac{d}{dt}(6x^2) = \frac{d}{dx}(6x^2) \cdot \frac{dx}{dt} \]\[ \frac{dS}{dt} = 12x \cdot \frac{dx}{dt} \]Substitute the expression for \(\frac{dx}{dt}\) from equation (1) into this equation:\[ \frac{dS}{dt} = 12x \cdot \left(\frac{8}{3x^2}\right) \]\[ \frac{dS}{dt} = \frac{96x}{3x^2} = \frac{32}{x} \text{ cm}^2/\text{s} \]Finally, we calculate the rate of change of the surface area when the length of the edge of the cube is 12 cm.When \(x = 12\) cm:\[ \frac{dS}{dt} = \frac{32}{12} = \frac{8}{3} \text{ cm}^2/\text{s} \]Therefore, if the length of the edge of the cube is 12 cm, its surface area is increasing at the rate of \(\frac{8}{3}\) cm\(^2\)/s.

ID: 14 Type: Long Source: NCERT

Question 92: Sand is pouring from a pipe at the rate of \(12\;{\text{c}}{{\text{m}}^3}/{\text{s}}\). The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is \(4\;{\text{cm}}\) ?

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✅ Manual Solution:

Sand is pouring from a pipe at the rate of \(12,\text{cm}^3/\text{s}\). The falling sand forms a cone on the ground such that the height is always one-sixth of the radius of the base. We are to find how fast the height of the cone is increasing when the height is \(4,\text{cm}\).\[\] The volume of a cone is given by: \[V = \dfrac{1}{3} \pi r^2 h\] Since the height is one-sixth of the radius, \[h = \dfrac{1}{6}r \Rightarrow r = 6h\] Substitute into the volume formula: \[V = \dfrac{1}{3} \pi (6h)^2 h = \dfrac{1}{3} \pi \cdot 36h^2 \cdot h = 12\pi h^3\] Differentiate both sides with respect to time \(t\): \[\dfrac{dV}{dt} = \dfrac{d}{dt}(12\pi h^3) = 12\pi \cdot 3h^2 \cdot \dfrac{dh}{dt} = 36\pi h^2 \dfrac{dh}{dt}\] We are given: At \(h = 4,\text{cm}\): \(\dfrac{dV}{dt} = 12,\text{cm}^3/\text{s}\), So \[12 = 36\pi (4)^2 \dfrac{dh}{dt} = 36\pi \cdot 16 \cdot \dfrac{dh}{dt}\] \[\Rightarrow \dfrac{dh}{dt} = \dfrac{12}{576\pi} = \dfrac{1}{48\pi},\text{cm/s}\] Final Answer: The height of the cone is increasing at a rate of \(\dfrac{1}{48\pi},\text{cm/s}\) when the height is \(4,\text{cm}\).

ID: 29 Type: Long Source: NCERT

Question 93: Show that \( y = \log (1 + x) - \dfrac{2x}{2 + x},\ x > -1 \), is an increasing function throughout its domain.

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✅ Manual Solution:

We are given: \[ y = \log (1 + x) - \dfrac{2x}{2 + x} \] Differentiate with respect to \(x\): \[ \dfrac{dy}{dx} = \dfrac{1}{1 + x} - \dfrac{(2 + x)(2) - 2x(1)}{(2 + x)^2} = \dfrac{1}{1 + x} - \dfrac{4}{(2 + x)^2} \] Take LCM and simplify: \[ \dfrac{dy}{dx} = \dfrac{x^2}{(1 + x)(2 + x)^2} \] Now, consider the domain \(x > -1\). Let us examine the sign of the derivative: \[\] The numerator \(x^2 \geq 0\), and is 0 only when \(x = 0\) \[\] The denominator \((1 + x)(2 + x)^2 > 0\) for all \(x > -1\), since all terms are positive. Therefore, \[ \dfrac{dy}{dx} = \dfrac{x^2}{(1 + x)(2 + x)^2} \geq 0 \text{ for all } x > -1, \] \(\textbf{Conclusion:} \) Since \(\dfrac{dy}{dx} > 0\) for all \(x > -1\), the function is \(\textbf{increasing throughout its domain}\).

ID: 30 Type: Long Source: NCERT

Question 94: Find the values of \(x\) for which \( y = [x(x - 2)]^2 \) is an increasing function.

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✅ Manual Solution:

We start with: \[ y = [x(x - 2)]^2 = \left(x^2 - 2x\right)^2 \] Differentiate using chain rule: \[ \frac{dy}{dx} = 2(x^2 - 2x) \cdot \frac{d}{dx}(x^2 - 2x) = 2(x^2 - 2x)(2x - 2) \] \[ \Rightarrow \frac{dy}{dx} = 4x(x - 2)(x - 1) \] Now set the derivative equal to zero: \[ \frac{dy}{dx} = 0 \Rightarrow x = 0,\ 1,\ 2 \] These critical points divide the real number line into the intervals: \[ (-\infty, 0),\ (0, 1),\ (1, 2),\ (2, \infty) \] Now test the sign of \(\frac{dy}{dx}\) in each interval: \[\] In \((- \infty, 0)\), choose \(x = -1\): \[ \frac{dy}{dx} = 4(-1)(-3)(-2) = -24 < 0 \] So, \(y\) is decreasing. \[\] In \((0, 1)\), choose \(x = 0.5\): \[ \frac{dy}{dx} = 4(0.5)(-1.5)(-0.5) = 1.5 > 0 \] So, \(y\) is increasing. \[\] In \((1, 2)\), choose \(x = 1.5\): \[ \frac{dy}{dx} = 4(1.5)(-0.5)(0.5) = -1.5 < 0 \] So, \(y\) is decreasing. \[\] In \((2, \infty)\), choose \(x = 3\): \[ \frac{dy}{dx} = 4(3)(1)(2) = 24 > 0 \] So, \(y\) is increasing. \[\] \(\textbf{Conclusion:}\) The function \( y = [x(x - 2)]^2 \) is increasing in the intervals: \[ (0, 1) \quad \text{and} \quad (2, \infty) \]

ID: 31 Type: Long Source: NCERT

Question 95: Prove that \( y = \dfrac{4\sin \theta}{2 + \cos \theta} - \theta \) is an increasing function of \( \theta \) in \( \left[0, \dfrac{\pi}{2} \right] \).

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✅ Manual Solution:

We are given: \[ y = \frac{4\sin \theta}{2 + \cos \theta} - \theta \] Differentiate with respect to \( \theta \): \[ \frac{dy}{d\theta} = \frac{(2 + \cos \theta)(4\cos \theta) - 4\sin \theta(-\sin \theta)}{(2 + \cos \theta)^2} - 1 \] \[ = \frac{8\cos \theta + 4\cos^2 \theta + 4\sin^2 \theta}{(2 + \cos \theta)^2} - 1 \] Use the identity \( \sin^2 \theta + \cos^2 \theta = 1 \): \[ = \frac{8\cos \theta + 4(\cos^2 \theta + \sin^2 \theta)}{(2 + \cos \theta)^2} - 1 = \frac{8\cos \theta + 4}{(2 + \cos \theta)^2} - 1 \] Now, set the derivative equal to zero: \[ \frac{dy}{d\theta} = 0 \Rightarrow \frac{8\cos \theta + 4}{(2 + \cos \theta)^2} = 1 \] Multiply both sides by \( (2 + \cos \theta)^2 \): \[ 8\cos \theta + 4 = (2 + \cos \theta)^2 = 4 + 4\cos \theta + \cos^2 \theta \] Simplify: \[ 8\cos \theta + 4 = 4 + 4\cos \theta + \cos^2 \theta \Rightarrow 4\cos \theta = \cos^2 \theta \] \[ \Rightarrow \cos \theta (4 - \cos \theta) = 0 \] So, \( \cos \theta = 0 \) or \( \cos \theta = 4 \). Since \( \cos \theta \in [0,1] \), the only valid solution is \( \cos \theta = 0 \Rightarrow \theta = \frac{\pi}{2} \). Now express the derivative in a factored form: \[ \frac{dy}{d\theta} = \frac{8\cos \theta + 4 - (4 + 4\cos \theta + \cos^2 \theta)}{(2 + \cos \theta)^2} = \frac{4\cos \theta - \cos^2 \theta}{(2 + \cos \theta)^2} \] \[ = \frac{\cos \theta(4 - \cos \theta)}{(2 + \cos \theta)^2} \] Now analyze the sign of \( \frac{dy}{d\theta} \) in \( \left[0, \frac{\pi}{2} \right] \): \[\] In this interval, \( \cos \theta > 0 \), and \( 4 - \cos \theta > 0 \) since \( \cos \theta \le 1 \) \[\] Also, \( (2 + \cos \theta)^2 > 0 \) So, \[ \frac{dy}{d\theta} = \frac{\cos \theta (4 - \cos \theta)}{(2 + \cos \theta)^2} > 0 \quad \text{for all } \theta \in \left(0, \frac{\pi}{2} \right) \] The derivative is positive throughout the open interval, and since the function is continuous on the closed interval \( \left[0, \frac{\pi}{2} \right] \), it is strictly increasing on the entire interval.

ID: 35 Type: Long Source: NCERT

Question 96: Let \( \mathbf{I} \) be any interval disjoint from \( (-1,1) \). Prove that the function \( f \) given by \( f(x) = x + \frac{1}{x} \) is strictly increasing on \( \mathbf{I} \).

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✅ Manual Solution:

We are given: \[ f(x) = x + \frac{1}{x} \] Differentiate with respect to \( x \): \[ f'(x) = \frac{d}{dx} \left( x + \frac{1}{x} \right) = 1 - \frac{1}{x^2} \] Set \( f'(x) = 0 \): \[ 1 - \frac{1}{x^2} = 0 \Rightarrow \frac{1}{x^2} = 1 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1 \] So, the critical points are \( x = -1 \) and \( x = 1 \), which divide the real line into intervals: \[ (-\infty, -1),\quad (-1, 1),\quad (1, \infty) \] Now analyze the sign of \( f'(x) \) in each interval: \[\] In \( (-\infty, -1) \): \( x < -1 \Rightarrow x^2 > 1 \Rightarrow \frac{1}{x^2} < 1 \Rightarrow f'(x) = 1 - \frac{1}{x^2} > 0 \) So, \( f \) is strictly increasing on \( (-\infty, -1) \). \[\] In \( (-1, 1) \): \( -1 < x < 1,\ x \neq 0 \Rightarrow x^2 < 1 \Rightarrow \frac{1}{x^2} > 1 \Rightarrow f'(x) = 1 - \frac{1}{x^2} < 0 \) So, \( f \) is strictly decreasing on \( (-1, 1) \). \[\] In \( (1, \infty) \): \( x > 1 \Rightarrow x^2 > 1 \Rightarrow \frac{1}{x^2} < 1 \Rightarrow f'(x) = 1 - \frac{1}{x^2} > 0 \) So, \( f \) is strictly increasing on \( (1, \infty) \). \[\] Hence, \( f(x) = x + \frac{1}{x} \) is strictly increasing on any interval \( \mathbf{I} \) such that \( \mathbf{I} \cap (-1, 1) = \emptyset \). That is, \( f \) is strictly increasing on any interval disjoint from \( (-1, 1) \).

ID: 36 Type: Long Source: NCERT

Question 97: Prove that the function \( f \) given by \( f(x) = \log \sin x \) is strictly increasing on \( \left(0, \dfrac{\pi}{2} \right) \) and strictly decreasing on \( \left( \dfrac{\pi}{2}, \pi \right) \).

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✅ Manual Solution:

We are given: \[ f(x) = \log \sin x \] Differentiate with respect to \( x \): \[ f'(x) = \frac{d}{dx} \left( \log \sin x \right) = \frac{1}{\sin x} \cdot \cos x = \cot x \] Now analyze the sign of \( f'(x) = \cot x \) in the given intervals: \[\] In \( \left(0, \dfrac{\pi}{2} \right) \), \( \cot x > 0 \), since both \( \sin x > 0 \) and \( \cos x > 0 \), \( \Rightarrow f'(x) > 0 \) \( \Rightarrow f \) is strictly increasing on \( \left(0, \dfrac{\pi}{2} \right) \). \[\] In \( \left( \dfrac{\pi}{2}, \pi \right) \), \( \cot x < 0 \), since \( \sin x > 0 \) and \( \cos x < 0 \), \( \Rightarrow f'(x) < 0 \) \( \Rightarrow f \) is strictly decreasing on \( \left( \dfrac{\pi}{2}, \pi \right) \). \[\] \(\textbf{Conclusion:}\) \[ f(x) = \log \sin x \text{ is strictly increasing on } \left(0, \dfrac{\pi}{2} \right) \text{ and strictly decreasing on } \left( \dfrac{\pi}{2}, \pi \right). \]

ID: 37 Type: Long Source: NCERT

Question 98: Prove that the function \( f \) given by \( f(x) = \log \cos x \) is strictly decreasing on \( \left(0, \dfrac{\pi}{2} \right) \) and strictly increasing on \( \left( \dfrac{\pi}{2}, \pi \right) \).

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✅ Manual Solution:

We are given: \[ f(x) = \log \cos x \] Differentiate with respect to \( x \): \[ f'(x) = \frac{d}{dx} \left( \log \cos x \right) = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x \] Now analyze the sign of \( f'(x) = -\tan x \) in the given intervals: \[\] In \( \left(0, \dfrac{\pi}{2} \right) \), \( \tan x > 0 \Rightarrow -\tan x < 0 \Rightarrow f'(x) < 0 \) \( \Rightarrow f \) is strictly decreasing on \( \left(0, \dfrac{\pi}{2} \right) \). \[\] In \( \left( \dfrac{\pi}{2}, \pi \right) \), \( \tan x < 0 \Rightarrow -\tan x > 0 \Rightarrow f'(x) > 0 \) \( \Rightarrow f \) is strictly increasing on \( \left( \dfrac{\pi}{2}, \pi \right) \). \[\] \(\textbf{Conclusion:}\) \[ f(x) = \log \cos x \text{ is strictly decreasing on } \left(0, \dfrac{\pi}{2} \right) \text{ and strictly increasing on } \left( \dfrac{\pi}{2}, \pi \right). \]

ID: 66 Type: Long Source: NCERT

Question 99: What is the maximum value of the function \(\sin x + \cos x\) ?

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✅ Manual Solution:

Let \[ f(x) = \sin x + \cos x \] Differentiate: \[ f'(x) = \cos x - \sin x \] Set \( f'(x) = 0 \): \[ \cos x - \sin x = 0 \Rightarrow \sin x = \cos x \Rightarrow \tan x = 1 \Rightarrow x = \frac{\pi}{4}, \frac{5\pi}{4}, \ldots \] Now compute the second derivative: \[ f''(x) = -\sin x - \cos x = -(\sin x + \cos x) \] To determine if it's a maximum, evaluate at \( x = \frac{\pi}{4} \): \[ f''\left( \frac{\pi}{4} \right) = -\left( \sin \frac{\pi}{4} + \cos \frac{\pi}{4} \right) = -\left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right) = -\sqrt{2} < 0 \] Since \( f''(x) < 0 \), this is a local maximum. Now compute the value: \[ f\left( \frac{\pi}{4} \right) = \sin \frac{\pi}{4} + \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2} \] \[ \boxed{\text{Maximum value of } \sin x + \cos x \text{ is } \sqrt{2}, \text{ at } x = \frac{\pi}{4}} \]

ID: 67 Type: Long Source: NCERT

Question 100: Find the maximum value of \(2x^3 - 24x + 107\) in the interval \([1,3]\). Find the maximum value of the same function in \([-3, -1]\).

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✅ Manual Solution:

Let \[ f(x) = 2x^3 - 24x + 107 \] Differentiate: \[ f'(x) = 6x^2 - 24 = 6(x^2 - 4) \] Set \( f'(x) = 0 \): \[ x^2 - 4 = 0 \Rightarrow x = \pm 2 \] For the Interval \([1, 3]\): Evaluate the function at: \(x = 1\): \[ f(1) = 2(1)^3 - 24(1) + 107 = 2 - 24 + 107 = 85 \] and at \(x = 2\): \[ f(2) = 2(8) - 24(2) + 107 = 16 - 48 + 107 = 75 \] At \(x = 3\): \[ f(3) = 2(27) - 24(3) + 107 = 54 - 72 + 107 = 89 \] \[ \boxed{\text{Maximum value in } [1,3] \text{ is } 89 \text{ at } x = 3} \] For the Interval \([-3, -1]\): Evaluate the function at: \(x = -3\): \[ f(-3) = 2(-27) - 24(-3) + 107 = -54 + 72 + 107 = 125 \] and at \(x = -2\): \[ f(-2) = 2(-8) - 24(-2) + 107 = -16 + 48 + 107 = 139 \] and at \(x = -1\): \[ f(-1) = 2(-1) - 24(-1) + 107 = -2 + 24 + 107 = 129 \] \[ \boxed{\text{Maximum value in } [-3, -1] \text{ is } 139 \text{ at } x = -2} \]

ID: 69 Type: Long Source: NCERT

Question 101: Find the maximum and minimum values of \( f(x) = x + \sin 2x \) on the interval \([0, 2\pi]\).

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✅ Manual Solution:

Let \[ f(x) = x + \sin 2x \] Differentiate: \[ f'(x) = 1 + 2\cos 2x \] Set derivative to zero for critical points: \[ f'(x) = 0 \Rightarrow 1 + 2\cos 2x = 0 \Rightarrow \cos 2x = -\frac{1}{2} \] Now solve \( \cos 2x = -\frac{1}{2} \): \[ 2x = \frac{2\pi}{3}, \frac{4\pi}{3} \Rightarrow x = \frac{\pi}{3}, \frac{2\pi}{3} \] Also in the interval \([0, 2\pi]\), the values of \(x\) satisfying \( \cos 2x = -\frac{1}{2} \) are: \[ x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3} \] Evaluate \(f(x)\) at critical points and endpoints: \[ f\left(0\right) = 0 + \sin 0 = 0 \] \[ f\left(2\pi\right) = 2\pi + \sin 4\pi = 2\pi + 0 = 2\pi \] \[ f\left(\frac{\pi}{3}\right) = \frac{\pi}{3} + \sin\left(\frac{2\pi}{3}\right) = \frac{\pi}{3} + \frac{\sqrt{3}}{2} \] \[ f\left(\frac{2\pi}{3}\right) = \frac{2\pi}{3} + \sin\left(\frac{4\pi}{3}\right) = \frac{2\pi}{3} - \frac{\sqrt{3}}{2} \] \[ f\left(\frac{4\pi}{3}\right) = \frac{4\pi}{3} + \sin\left(\frac{8\pi}{3}\right) = \frac{4\pi}{3} + \frac{\sqrt{3}}{2} \] \[ f\left(\frac{5\pi}{3}\right) = \frac{5\pi}{3} + \sin\left(\frac{10\pi}{3}\right) = \frac{5\pi}{3} - \frac{\sqrt{3}}{2} \] Compare all values. The smallest is at \(x = 0\), and the largest is at \(x = 2\pi\). \[ \boxed{ \text{Maximum value of } f(x) \text{ on } [0, 2\pi] \text{ is } 2\pi \text{ at } x = 2\pi } \] \[ \boxed{ \text{Minimum value of } f(x) \text{ on } [0, 2\pi] \text{ is } 0 \text{ at } x = 0 } \]

ID: 70 Type: Long Source: NCERT

Question 102: Find two numbers whose sum is 24 and whose product is as large as possible.

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✅ Manual Solution:

Let one of the numbers be \( x \). Then, the other number is \( 24 - x \). Let the product of the two numbers be: \[ P(x) = x(24 - x) = 24x - x^2 \] Differentiate \( P(x) \): \[ P'(x) = 24 - 2x, \quad P''(x) = -2 \] Set the first derivative to zero for critical points: \[ P'(x) = 0 \Rightarrow 24 - 2x = 0 \Rightarrow x = 12 \] Check the second derivative: \[ P''(12) = -2 < 0 \] Since \( P''(x) < 0 \), the function attains a **maximum** at \( x = 12 \). Therefore, the two numbers are: \[ x = 12, \quad 24 - x = 12 \] The two numbers are 12 and 12. The product is maximum when both are equal.

ID: 71 Type: Long Source: NCERT

Question 103: Find two positive numbers \( x \) and \( y \) such that \( x + y = 60 \) and \( x y^3 \) is maximum.

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✅ Manual Solution:

We are given: \[ x + y = 60 \Rightarrow y = 60 - x \] We want to maximize: \[ f(x) = x y^3 = x(60 - x)^3 \] Differentiate using the product rule: \[ f'(x) = \frac{d}{dx} \left[x(60 - x)^3\right] = (60 - x)^3 + x \cdot 3(60 - x)^2 \cdot (-1) \] \[ f'(x) = (60 - x)^2 \left[(60 - x) - 3x\right] = (60 - x)^2(60 - 4x) \] Set derivative to zero: \[ f'(x) = 0 \Rightarrow (60 - x)^2(60 - 4x) = 0 \] \[ \Rightarrow x = 60 \text{ or } x = 15 \] Check second derivative at \( x = 15 \): \[ f''(x) = \frac{d}{dx}\left[ (60 - x)^2(60 - 4x) \right] \] Using the product rule: \[ f''(x) = 2(60 - x)(-1)(60 - 4x) + (60 - x)^2(-4) \] \[ f''(x) = -2(60 - x)(60 - 4x) - 4(60 - x)^2 \] \[ f''(x) = -2(60 - x)[(60 - 4x) + 2(60 - x)] = -2(60 - x)(60 - 4x + 2(60 - x)) \] \[ = -2(60 - x)(180 - 6x) = -12(60 - x)(30 - x) \] Now evaluate at \( x = 15 \): \[ f''(15) = -12(60 - 15)(30 - 15) = -12(45)(15) < 0 \] Since \( f''(15) < 0 \), \( x = 15 \) is a point of **local maximum**. Hence, \[ x = 15, \quad y = 60 - 15 = 45 \] The two positive numbers are 15 and 45.

ID: 72 Type: Long Source: NCERT

Question 104: Find two positive numbers \( x \) and \( y \) such that their sum is 35 and the product \( x^2 y^5 \) is maximum.

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✅ Manual Solution:

We are given: \[ x + y = 35 \Rightarrow y = 35 - x \] We want to maximize: \[ P(x) = x^2 (35 - x)^5 \] Differentiate using the product rule: \[ P'(x) = \frac{d}{dx}\left[ x^2 (35 - x)^5 \right] = 2x(35 - x)^5 + x^2 \cdot 5(35 - x)^4 \cdot (-1) \] \[ P'(x) = 2x(35 - x)^5 - 5x^2(35 - x)^4 \] Factor common terms: \[ P'(x) = x(35 - x)^4 [2(35 - x) - 5x] \] \[ = x(35 - x)^4 (70 - 2x - 5x) = x(35 - x)^4 (70 - 7x) \] \[ = 7x(35 - x)^4 (10 - x) \] Set derivative to zero: \[ P'(x) = 0 \Rightarrow x = 0,\ 35,\ \text{or}\ 10 \] Check valid values: \[ x = 0 \Rightarrow y = 35 \Rightarrow P = 0 \] \[ x = 35 \Rightarrow y = 0 \Rightarrow P = 0 \] \[ x = 10 \Rightarrow y = 25 \] Hence valid critical point.\[\] Now compute second derivative to verify maximum: \[ P''(x) = \frac{d}{dx} \left[ 7x(35 - x)^4(10 - x) \right] \] Use product rule: Let \( u = x,\ v = (35 - x)^4(10 - x) \), then \[ P''(x) = 7[u'v + uv'] \] Compute: \[ u' = 1,\quad v = (35 - x)^4(10 - x) \] \[ v' = \frac{d}{dx}[(35 - x)^4(10 - x)] = (35 - x)^3[-4(10 - x) + (35 - x)(-1)] \] \[ = (35 - x)^3 [-4(10 - x) - (35 - x)] = (35 - x)^3 [-40 + 4x - 35 + x] = (35 - x)^3(-75 + 5x) \] So: \[ P''(x) = 7[(35 - x)^4(10 - x) + x(35 - x)^3(-75 + 5x)] \] Evaluate at \( x = 10 \): \[ (35 - 10) = 25,\quad (10 - 10) = 0\] \[\Rightarrow P''(10) = 7[25^4 \cdot 0 + 10 \cdot 25^3(-75 + 50)] = 7 \cdot 10 \cdot 25^3(-25) \] \[ = -7 \cdot 10 \cdot 25^3 \cdot 25 < 0 \] Since \( P''(10) < 0 \), the function attains a \({\bf maximum}\) at \( x = 10 \). Thus, the required numbers are: \[ \boxed{x = 10,\quad y = 35-x=25} \]

ID: 73 Type: Long Source: NCERT

Question 105: Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.

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✅ Manual Solution:

Let one number be \( x \). Then the other number is \( 16 - x \). We want to minimize the sum of their cubes: \[ S(x) = x^3 + (16 - x)^3 \] Differentiate: \[ S'(x) = 3x^2 - 3(16 - x)^2 \] Set \( S'(x) = 0 \) for critical points: \[ 3x^2 = 3(16 - x)^2 \Rightarrow x^2 = (16 - x)^2 \] Expand: \[ x^2 = 256 - 32x + x^2 \Rightarrow -32x + 256 = 0 \Rightarrow x = \frac{256}{32} = 8 \] Now check second derivative: \[ S''(x) = 6x + 6(16 - x) = 6x + 96 - 6x = 96 > 0 \] Since \( S''(8) > 0 \), this confirms a local minimum. \[\] Final Answer: \[ \boxed{\text{The two numbers are } 8 \text{ and } 16-8=8.} \]

ID: 74 Type: Long Source: NCERT

Question 106: A square piece of tin of side \(18\,\text{cm}\) is to be made into a box without a top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible?

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Let the side of the square cut off from each corner be \( x \,\text{cm} \). \[\] Then, the length and breadth of the resulting box will each be \( (18 - 2x)\,\text{cm} \), \[\] and the height will be \( x \,\text{cm} \). \[\] So, the volume \( V(x) \) of the box is: \[ V(x) = x(18 - 2x)^2 \] Differentiate: \[ V'(x) = (18 - 2x)^2 + x \cdot 2(18 - 2x)(-2) = (18 - 2x)^2 - 4x(18 - 2x) \] Factor: \[ V'(x) = (18 - 2x)[(18 - 2x) - 4x] = (18 - 2x)(18 - 6x) \] Set derivative to zero: \[ V'(x) = 0 \Rightarrow (18 - 2x)(18 - 6x) = 0 \Rightarrow x = 9 \text{ or } x = 3 \] If \( x = 9 \), then \( 18 - 2x = 0 \), so length and breadth become zero. This gives zero volume which is not acceptable. \[\] So, \( x = 3 \) is a critical point. \[\] Check second derivative: \[ V''(x) = \frac{d}{dx}\left[(18 - 2x)(18 - 6x)\right] = -2(18 - 6x) + (18 - 2x)(-6) = -2(18 - 6x) - 6(18 - 2x) \] Evaluate at \( x = 3 \): \[ V''(3) = -2(18 - 18) - 6(18 - 6) = -6 \cdot 12 = -72 < 0 \] Since \( V''(3) < 0 \), the volume is \(\bf maximum\) at \( x = 3 \). \[\] Final Answer: \[ \boxed{\text{The side of the square to be cut off is } 3\,\text{cm.}} \]

ID: 77 Type: Long Source: NCERT

Question 107: A rectangular sheet of tin \(45\,\text{cm} \times 24\,\text{cm}\) is to be made into a box without a top, by cutting off a square from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is the maximum possible?

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Let the side of each square cut off be \( x\,\text{cm} \). - After folding, the height of the box = \( x \) - Length of the box = \( 45 - 2x \) - Breadth of the box = \( 24 - 2x \) So, the volume \( V(x) \) of the box is: \[ V(x) = x(45 - 2x)(24 - 2x) \] Expand: \[ V(x) = x[(45)(24) - 2x(24) - 2x(45) + 4x^2] = x[1080 - 48x - 90x + 4x^2] = x[1080 - 138x + 4x^2] = 4x^3 - 138x^2 + 1080x \] Differentiate: \[ V'(x) = 12x^2 - 276x + 1080 \] Factor: \[ V'(x) = 12(x^2 - 23x + 90) = 12(x - 18)(x - 5) \] Set derivative to zero: \[ V'(x) = 0 \Rightarrow x = 18 \text{ or } x = 5 \] Now, \( x = 18 \) is not valid since: - \( 2x = 36 > 24 \) → length or breadth becomes negative or zero. So, only possible value is \( x = 5 \). Check second derivative: \[ V''(x) = \frac{d}{dx}(12x^2 - 276x + 1080) = 24x - 276 \] Evaluate at \( x = 5 \): \[ V''(5) = 24(5) - 276 = 120 - 276 = -156 < 0 \] So, \( x = 5 \) is a point of **local maximum**. --- ### **Final Answer**: \[ \boxed{\text{The side of the square to be cut off is } 5\,\text{cm.}} \]

ID: 211 Type: Long (LA) Source: AISSCE(Board Exam) Year: 2025

Question 108: Find the absolute maximum and absolute minimum of function \(f(x) = 2x^3 - 15x^2 + 36x + 1\) on \([1, 5]\).

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\(f(x) = 2x^3 - 15x^2 + 36x + 1\) implies

\[f'(x) = \frac{d}{dx}(2x^3 - 15x^2 + 36x + 1)\]\[f'(x) = 6x^2 - 30x + 36\]Set \(f'(x) = 0\) to find critical points:\[6x^2 - 30x + 36 = 0\] \[\implies x^2 - 5x + 6 = 0\]\[\implies(x - 2)(x - 3) = 0\implies x=2, 3\]The critical points are \(x=2\) and \(x=3\).

Note that both the critical points are les in the interval \([1, 5]\)

Now evaluate \(f(x)\) at the critical points that lie in the interval and at the endpoints.

For (\(x=1\)) we have \(f(1) = 2(1)^3 - 15(1)^2 + 36(1) + 1\)=\(24\)

For (\(x=2\)) we have \(f(2) = 2(8) - 15(4) + 36(2) + 1 = 16 - 60 + 72 + 1\)=\(29\)

For (\(x=3\)) we have \(f(3) = 2(27) - 15(9) + 36(3) + 1 = 54 - 135 + 108 + 1\)=\(28\)

For (\(x=5\)) we have \(f(5) = 2(125) - 15(25) + 36(5) + 1 = 250 - 375 + 180 + 1\)=\(56\)

Hence Absolute Maximum Value: \(\mathbf{56}\) (Occurs at \(x=5\)) and Absolute Minimum Value: \(\mathbf{24}\) (Occurs at \(x=1\))

🤖 AI Generated Solution:

To find the absolute maximum and absolute minimum of the function $f(x) = 2x^3 - 15x^2 + 36x + 1$ on the closed interval $[1, 5]$, we follow these steps: **Step 1: Find the critical points of the function.** First, we find the derivative of the function $f(x)$ with respect to $x$: $$f'(x) = \frac{d}{dx}(2x^3 - 15x^2 + 36x + 1)$$ $$f'(x) = 6x^2 - 30x + 36$$ To find the critical points, we set $f'(x) = 0$: $$6x^2 - 30x + 36 = 0$$ Divide the entire equation by 6: $$x^2 - 5x + 6 = 0$$ Factor the quadratic equation: $$(x - 2)(x - 3) = 0$$ This gives us two critical points: $$x = 2 \quad \text{or} \quad x = 3$$ Both critical points, $x=2$ and $x=3$, lie within the given interval $[1, 5]$. **Step 2: Evaluate the function at the critical points that lie within the interval.** For $x = 2$: $$f(2) = 2(2)^3 - 15(2)^2 + 36(2) + 1$$ $$f(2) = 2(8) - 15(4) + 72 + 1$$ $$f(2) = 16 - 60 + 72 + 1$$ $$f(2) = 29$$ For $x = 3$: $$f(3) = 2(3)^3 - 15(3)^2 + 36(3) + 1$$ $$f(3) = 2(27) - 15(9) + 108 + 1$$ $$f(3) = 54 - 135 + 108 + 1$$ $$f(3) = 28$$ **Step 3: Evaluate the function at the endpoints of the interval.** For the left endpoint $x = 1$: $$f(1) = 2(1)^3 - 15(1)^2 + 36(1) + 1$$ $$f(1) = 2 - 15 + 36 + 1$$ $$f(1) = 24$$ For the right endpoint $x = 5$: $$f(5) = 2(5)^3 - 15(5)^2 + 36(5) + 1$$ $$f(5) = 2(125) - 15(25) + 180 + 1$$ $$f(5) = 250 - 375 + 180 + 1$$ $$f(5) = 56$$ **Step 4: Compare all the values obtained in Step 2 and Step 3.** The values of the function at the critical points and endpoints are: * $f(2) = 29$ * $f(3) = 28$ * $f(1) = 24$ * $f(5) = 56$ Comparing these values, we can identify the absolute maximum and minimum: The smallest value among these is 24. The largest value among these is 56. **Conclusion:** The absolute maximum value of the function $f(x)$ on the interval $[1, 5]$ is $56$, which occurs at $x = 5$. The absolute minimum value of the function $f(x)$ on the interval $[1, 5]$ is $24$, which occurs at $x = 1$.

ID: 380 Type: Long (LA) Source: AISSCE(Board Exam) Year: 2025

Question 109: The relation between the height of the plant (y cm) with respect to exposure to sunlight is governed by the equation \(y=4x-\frac{1}{2}x^{2},\) where x is the number of days exposed to sunlight.

Find the rate of growth of the plant with respect to sunlight.

In how many days will the plant attain its maximum height ?

What is the maximum height ?

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**Solution:** Let the height of the plant be $y$ cm and the number of days exposed to sunlight be $x$. The relation between the height of the plant and the exposure to sunlight is given by: $$y = 4x - \frac{1}{2}x^2$$ **1. Rate of growth of the plant with respect to sunlight:** The rate of growth of the plant is given by the derivative of $y$ with respect to $x$, i.e., $\frac{dy}{dx}$. Differentiating the given equation with respect to $x$: $$\frac{dy}{dx} = \frac{d}{dx}\left(4x - \frac{1}{2}x^2\right)$$ $$\frac{dy}{dx} = \frac{d}{dx}(4x) - \frac{d}{dx}\left(\frac{1}{2}x^2\right)$$ $$\frac{dy}{dx} = 4(1) - \frac{1}{2}(2x)$$ $$\frac{dy}{dx} = 4 - x$$ Thus, the rate of growth of the plant with respect to sunlight is $4-x$ cm/day. **2. In how many days will the plant attain its maximum height?** To find the maximum height, we set the rate of growth equal to zero: $$\frac{dy}{dx} = 0$$ $$4 - x = 0$$ $$x = 4$$ To confirm that this value of $x$ corresponds to a maximum, we find the second derivative of $y$ with respect to $x$: $$\frac{d^2y}{dx^2} = \frac{d}{dx}(4-x)$$ $$\frac{d^2y}{dx^2} = -1$$ Since $\frac{d^2y}{dx^2} = -1 < 0$, the height of the plant is maximum when $x=4$. Therefore, the plant will attain its maximum height in **4 days**. **3. What is the maximum height?** To find the maximum height, substitute $x=4$ into the original equation for $y$: $$y = 4x - \frac{1}{2}x^2$$ $$y_{max} = 4(4) - \frac{1}{2}(4)^2$$ $$y_{max} = 16 - \frac{1}{2}(16)$$ $$y_{max} = 16 - 8$$ $$y_{max} = 8$$ Therefore, the maximum height of the plant is **8 cm**. **Summary of Answers:** * The rate of growth of the plant with respect to sunlight is $\mathbf{4-x}$ **cm/day**. * The plant will attain its maximum height in **4 days**. * The maximum height is **8 cm**.

ID: 646 Type: Long (LA) Source: AISSCE(Board Exam) Year: 2024

Question 110: It is given that function \(f(x)=x^{4}-62x^{2}+ax+9\) attains local maximum value at \(x=1\) Find the value of 'a', hence obtain all other points where the given function f(x) attains local maximum or local minimum values.

ID: 659 Type: Long (LA) Source: AISSCE(Board Exam) Year: 2024

Question 111: The perimeter of a rectangular metallic sheet is 300 cm. It is rolled along one of its sides to form a cylinder. Find the dimensions of the rectangular sheet so that volume of cylinder so formed is maximum.

ID: 215 Type: Case Study Source: AISSCE(Board Exam) Year: 2025

Question 112: A technical company is designing a rectangular solar panel installation on a roof using 300 metres of boundary material. The design includes a partition running parallel to one of the sides dividing the area (roof) into two sections.Let the length of the side perpendicular to the partition be \(x\) metres and with parallel to the partition be \(y\) metres. \[\]Based on this information, answer the following questions: \[\] (i) Write the equation for the total boundary material used in the boundary and parallel to the partition in terms of \(x\) and \(y\). \[\] (ii) Write the area of the solar panel as a function of \(x\).\[\] (iii) (a) Find the critical points of the area function. Use second derivative test to determine critical points at the maximum area. Also, find the maximum area.\[\] OR\[\] (iii) (b) Using first derivative test, calculate the maximum area the company can enclose with the 300 metres of boundary material, considering the parallel partition.

ID: 281 Type: Case Study Source: AISSCE(Board Exam) Year: 2025

Question 113: A small town is analyzing the pattern of a new street light installation. The lights are set up in such a way that the intensity of light at any point \(x\) metres from the start of the street can be modelled by \(f(x)=e^{x} \sin x,\) where \(x\) is in metres.

Based on the above, answer the folowing

(i) Find the intervals on which the \(f(x)\) is increasing or decreasing, \(x\in[0,\pi]\). (2 marks)

(ii) Verify, whether each critical point when \(x\in[0,\pi]\) is a point of local maximum or local minimum or a point of inflexion. (2 marks)

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