This section comprises of **18 multiple choice questions (MCQs)** (Q1 to Q18) and **2 Assertion-Reason based questions** (Q19, Q20) of 1 mark each.
- If \(A=\begin{bmatrix}5&0&0\\ 0&5&0\\ 0&0&5\end{bmatrix},\) then \(A^{3}\) is:
- \(3\begin{bmatrix}5&0&0\\ 0&5&0\\ 0&0&5\end{bmatrix}\)
- \(\begin{bmatrix}125&0&0\\ 0&125&0\\ 0&0&125\end{bmatrix}\)
- \(\begin{bmatrix}15&0&0\\ 0&15&0\\ 0&0&15\end{bmatrix}\)
- \(\begin{bmatrix}5^{3}&0&0\\ 0&5&0\\ 0&0&5\end{bmatrix}\)
- If \(P(A\cup B)=0.9\) and \(P(A\cap B)=0\cdot4,\) then \(P(\overline{A})+P(\overline{B})\) is:
- 0.3
- 1
- 1.3
- 0.7
- If \(A=\begin{bmatrix}1&2&3\\ -4&3&7\end{bmatrix}\) and \(B=\begin{bmatrix}4&3\\ -1&2\\ 0&5\end{bmatrix},\) then the correct statement is:
- Only AB is defined.
- Only BA is defined.
- AB and BA, both are defined.
- AB and BA, both are not defined.
- If \(\begin{vmatrix}2x&5\\ 12&x\end{vmatrix}=\begin{vmatrix}6&-5\\ 4&3\end{vmatrix}\) then the value of x is:
- 3
- 7
- \(\pm7\)
- \(\pm3\)
- If \(f(x)=\begin{cases}\frac{\sin^{2}ax}{x^{2}},&x\ne0\\ 1,&x=0\end{cases}\) is continuous at \(x=0\), then the value of a is:
- 1
- -1
- \(\pm1\)
- 0
- If \(A=[a_{ij}]\) is a \(3\times3\) diagonal matrix such that \(a_{11}=1\), \(a_{22}=5\) and \(a_{33}=-2\), then \(|A|\) is:
- 0
- -10
- 10
- 1
- The principal value of \(\cot^{-1}(-\frac{1}{\sqrt{3}})\) is:
- \(-\frac{\pi}{3}\)
- \(-\frac{2\pi}{3}\)
- \(\frac{\pi}{3}\)
- \(\frac{2\pi}{3}\)
- If \(\begin{bmatrix}4+x&x-1\\ -2&3\end{bmatrix}\) is a singular matrix, then the value of x is:
- 0
- 1
- -2
- 4
- If \(f(x)=\{[x],x\in R\}\) is the greatest integer function, then the correct statement is:
- f is continuous but not differentiable at \(x=2\).
- f is neither continuous nor differentiable at \(x=2\).
- f is continuous as well as differentiable at \(x=2\).
- f is not continuous but differentiable at \(x=2\).
- The slope of the curve \(y=-x^{3}+3x^{2}+8x-20\) is maximum at:
- (1,-10)
- (1,10)
- (10, 1)
- (-10, 1)
- \(\int\sqrt{1+\sin x}dx\) is equal to :
- \(2(-\sin\frac{x}{2}+\cos\frac{x}{2})+C\)
- \(2(\sin\frac{x}{2}-\cos\frac{x}{2})+C\)
- \(-2(\sin\frac{x}{2}+\cos\frac{x}{2})+C\)
- \(2(\sin\frac{x}{2}+\cos\frac{x}{2})+C\)
- \(\int_{0}^{\pi/2}\cos x\cdot e^{\sin x}dx\) is equal to:
- 0
- \(1-e\)
- \(e-1\)
- e
- The area of the region enclosed between the curve \(y=x|x|\), x-axis, \(x=-2\) and \(x=2\) is:
- \(\frac{8}{3}\)
- \(\frac{16}{3}\)
- 0
- 8
- The integrating factor of the differential equation \((\frac{e^{-2\sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}})\frac{dx}{dy}=1\) is:
- \(e^{-1/\sqrt{x}}\)
- \(e^{2/\sqrt{x}}\)
- \(e^{2\sqrt{x}}\)
- \(e^{-2\sqrt{x}}\)
- The sum of the order and degree of the differential equation \([1+(\frac{dy}{dx})^{2}]^{3}=\frac{d^{2}y}{dx^{2}}\) is:
- 2
- \(\frac{5}{2}\)
- 3
- 4
- For a Linear Programming Problem (LPP), the given objective function \(Z=3x+2y\) is subject to constraints: \(x+2y\le10\), \(3x+y\le15\), \(x, y\ge0\). The correct feasible region is:
- ABC
- AOEC
- CED
- Open unbounded region BCD
- Let \(\vec{a}\) be a position vector whose tip is the point \((2,-3)\). If \(\vec{AB}=\vec{a}\), where coordinates of A are \((-4, 5)\), then the coordinates of B are:
- \((-2,-2)\)
- \((2,-2)\)
- \((-2,2)\)
- \((2, 2)\)
- The respective values of \(|\vec{a}|\) and \(|\vec{b}|\), if given \((\vec{a}-\vec{b})\cdot(\vec{a}+\vec{b})=512\) and \(|\vec{a}|=3|\vec{b}|\), are:
- 48 and 16
- 3 and 1
- 24 and 8
- 6 and 2
-
Assertion (A): The shaded portion of the graph represents the feasible region for the given Linear Programming Problem (LPP). Min \(Z=50x+70y\) subject to constraints \(2x+y\ge8\), \(x+2y\ge10\), \(x, y\ge0\)
Reason (R): \(Z=50x+70y\) has a minimum value \(=380\) at \(B(2,4)\). The region representing \(50x+70y<380\) does not have any point common with the feasible region.
- Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
- Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
- Assertion (A) is true, but Reason (R) is false.
- Assertion (A) is false, but Reason (R) is true.
-
Assertion (A): Let \(A=\{x\in R:-1\le x\le1\}\). If \(f:A\rightarrow A\) be defined as \(f(x)=x^{2}\) then f is not an onto function.
Reason (R): If \(y=-1\in A\) then \(x=\pm\sqrt{-1}\notin A.\)
- Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
- Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
- Assertion (A) is true, but Reason (R) is false.
- Assertion (A) is false, but Reason (R) is true.
This section comprises of **5 Very Short Answer (VSA)** type questions of 2 marks each.
- Find the domain of the function \(f(x)=\cos^{-1}(x^{2}-4).\)
- 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.
-
(a) Differentiate \(\frac{\sin x}{\sqrt{\cos x}}\) with respect to x.
OR
(b) If \(y=5\cos x-3\sin x\), prove that \(\frac{d^{2}y}{dx^{2}}+y=0.\)
-
(a) Find a vector of magnitude 5 which is perpendicular to both the vectors \(3\hat{i}-2\hat{j}+\hat{k}\) and \(4\hat{i}+3\hat{j}-2\hat{k}\).
OR
(b) Let \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) be three vectors such that \(\vec{a}\cdot\vec{b}=\vec{a}\cdot\vec{c}\) and \(\vec{a}\times\vec{b}=\vec{a}\times\vec{c}\), \(\vec{a}\ne0\). Show that \(\vec{b}=\vec{c}\).
- A man needs to hang two lanterns on a straight wire whose end points have coordinates \(A(4,1,-2)\) and \(B(6,2,-3)\). Find the coordinates of the points where he hangs the lanterns such that these points trisect the wire AB.
This section comprises of **6 Short Answer (SA)** type questions of 3 marks each.
- Find the value of 'a' for which \(f(x)=\sqrt{3}\sin x-\cos x-2ax+6\) is decreasing in R.
-
(a) Find: \(\int\frac{2x}{(x^{2}+3)(x^{2}-5)}dx\)
OR
(b) Evaluate: \(\int_{1}^{4}(|x-2|+|x-4|)dx\)
- Find the particular solution of the differential equation \([x\sin^{2}(\frac{y}{x})-y]dx+x dy=0\) given that \(y=\frac{\pi}{4}\) when \(x=1\).
- In the Linear Programming Problem (LPP), find the point/points giving maximum value for \(Z=5x+10y\) subject to constraints \(x+2y\le120\), \(x+y\ge60\), \(x-2y\ge0\), \(x, y\ge0\).
-
(a) If \(\vec{a}+\vec{b}+\vec{c}=\vec{0}\) such that \(|\vec{a}|=3\), \(|\vec{b}|=5\), \(|\vec{c}|=7\), then find the angle between \(\vec{a}\) and \(\vec{b}\).
OR
(b) If \(\vec{a}\) and \(\vec{b}\) are unit vectors inclined with each other at an angle \(\theta\), then prove that \(\frac{1}{2}|\vec{a}-\vec{b}|=\sin\frac{\theta}{2}\).
-
(a) The probability that a student buys a colouring book is 0.7 and that she buys a box of colours is 0.2. The probability that she buys a colouring book, given that she buys a box of colours, is 0.3. Find the probability that the student:
- Buys both the colouring book and the box of colours.
- Buys a box of colours given that she buys the colouring book.
OR
(b) A person has a fruit box that contains 6 apples and 4 oranges. He picks out a fruit three times, one after the other, after replacing the previous one in the box. Find:
- The probability distribution of the number of oranges he draws.
- The expectation of the random variable (number of oranges).
- Sketch a graph of \(y=x^{2}.\) Using integration, find the area of the region bounded by \(y=9\), \(x=0\) and \(y=x^{2}\).
This section comprises of **4 Long Answer (LA)** type questions of 5 marks each.
- Sketch a graph of \(y=x^{2}.\) Using integration, find the area of the region bounded by \(y=9\), \(x=0\) and \(y=x^{2}\).
- A furniture workshop produces three types of furniture chairs, tables and beds each day. On a particular day the total number of furniture pieces produced is 45. It was also found that production of beds exceeds that of chairs by 8, while the total production of beds and chairs together is twice the production of tables. Determine the units produced of each type of furniture, using **matrix method**.
-
(a) For a positive constant 'a', differentiate \(a^{(t+\frac{1}{t})}\) with respect to \(t+\frac{1}{t}\) where t is a non-zero real number.
OR
(b) Find \(\frac{dy}{dx}\) if \(y^{x}+x^{y}+x^{x}=a^{b}\) where a and b are constants.
-
(a) Find the foot of the perpendicular drawn from the point \((1, 1, 4)\) on the line \(\frac{x+2}{5}=\frac{y+1}{2}=\frac{-z+4}{-3}\).
OR
(b) Find the point on the line \(\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-4}{3}\) at a distance of \(2\sqrt{2}\) units from the point \((-1, -1, 2)\).
This section comprises of **3 case study based questions** of 4 marks each.
-
Case Study 1: Wooden Cuboidal Box 📦
A carpenter needs to make a wooden cuboidal box, closed from all sides, which has a square base and fixed volume. Since he is short of the paint required to paint the box on completion, he wants the surface area to be minimum.
- (i) Taking length = breadth = x m and height = y m, express the surface area (S) of the box in terms of x and its volume (V), which is constant. (1 mark)
- (ii) Find \(\frac{dS}{dx}\). (1 mark)
-
(iii) (a) Find a relation between x and y such that the surface area (S) is minimum. (2 marks)
OR
(iii) (b) If surface area (S) is constant, the volume \((V)=\frac{1}{4}(Sx-2x^{3})\) x being the edge of base. Show that volume (V) is maximum for \(x=\sqrt{\frac{S}{6}}\). (2 marks)
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Case Study 2: Function and Roll Numbers 🧑🎓
Let A be the set of 30 students of class XII in a school. Let \(f:A\rightarrow N\), N is a set of natural numbers such that function \(f(x)=\) Roll Number of student x.
- (i) Is f a bijective function ? (2 marks)
- (ii) Give reasons to support your answer to (i). (2 marks)
-
Case Study 3: Traffic Flow (Missing Content) 🚧
This question is not fully provided in the source. Based on typical paper structure, it is the third case study question (Q38, 4 marks).
- (i) [Part 1 of Q38, 1 mark]
- (ii) [Part 2 of Q38, 1 mark]
-
(iii) (a) [Part 3 (a) of Q38, 2 marks]
OR
(iii) (b) [Part 3 (b) of Q38, 2 marks]