ExamSpark ⚡
ExamSpark ⚡
Master these high-probability PYQs. Includes Calculus, Vectors, and 3D Geometry.
Q1: Check whether the relation $R$ in the set of real numbers $\mathbb{R}$ defined by $R = \{(a, b) : a \le b^3\}$ is reflexive, symmetric or transitive.
Ans:
1. Reflexive: For $a = 1/2$, $(1/2, 1/2) \notin R$ because $1/2 \not\le (1/2)^3 = 1/8$. Not Reflexive.
2. Symmetric: $(1, 2) \in R$ as $1 \le 8$. But $(2, 1) \notin R$ as $2 \not\le 1$. Not Symmetric.
3. Transitive: Take $a=10, b=3, c=2$. $(10, 3) \in R$ ($10 \le 27$), $(3, 2) \in R$ ($3 \le 8$). But $(10, 2) \notin R$ ($10 \not\le 8$). Not Transitive.
Result: Neither reflexive, nor symmetric, nor transitive.
Q2: Find the value of $\tan^{-1}(\sqrt{3}) - \cot^{-1}(-\sqrt{3})$.
Ans:
$\tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$
$\cot^{-1}(-\sqrt{3}) = \pi - \cot^{-1}(\sqrt{3}) = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$
Result: $\frac{\pi}{3} - \frac{5\pi}{6} = \frac{2\pi - 5\pi}{6} = -\frac{3\pi}{6} = -\frac{\pi}{2}$.
Q3: If $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$, show that $A^2 - 5A + 7I = O$.
Ans:
1. Find $A^2 = A \cdot A = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$.
2. Find $5A = \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix}$.
3. Find $7I = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$.
4. $A^2 - 5A + 7I = \begin{bmatrix} 8-15+7 & 5-5+0 \\ -5-(-5)+0 & 3-10+7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$.
Q4: Using properties of determinants, prove that $\begin{vmatrix} a & a+b & a+b+c \\ 2a & 3a+2b & 4a+3b+2c \\ 3a & 6a+3b & 10a+6b+3c \end{vmatrix} = a^3$.
Ans:
Apply $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 3R_1$.
The determinant simplifies to an upper triangular matrix form or can be expanded along $C_1$.
Final result will be $a^3$.
Q5: Find the value of $k$ so that the function $f(x)$ is continuous at $x=2$: $$ f(x) = \begin{cases} kx^2, & \text{if } x \le 2 \\ 3, & \text{if } x > 2 \end{cases} $$
Ans:
LHL (at $x=2$) = $\lim_{x \to 2^-} kx^2 = k(2)^2 = 4k$.
RHL (at $x=2$) = $\lim_{x \to 2^+} 3 = 3$.
For continuity, LHL = RHL $\Rightarrow 4k = 3 \Rightarrow k = 3/4$.
Q6: Find $\frac{dy}{dx}$ if $x = a(\theta + \sin\theta)$ and $y = a(1 - \cos\theta)$.
Ans:
$\frac{dx}{d\theta} = a(1+\cos\theta)$, $\frac{dy}{d\theta} = a(\sin\theta)$.
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a\sin\theta}{a(1+\cos\theta)} = \frac{2\sin(\theta/2)\cos(\theta/2)}{2\cos^2(\theta/2)} = \tan(\theta/2)$.
Q7: Find the intervals in which the function $f(x) = 2x^3 - 3x^2 - 36x + 7$ is strictly increasing.
Ans:
$f'(x) = 6x^2 - 6x - 36 = 6(x^2 - x - 6) = 6(x-3)(x+2)$.
For increasing, $f'(x) > 0$.
Critical points: $x = -2, 3$.
Intervals: $(-\infty, -2) \cup (3, \infty)$.
Q8: Evaluate: $\int \frac{x^2 + 1}{x^2 - 5x + 6} dx$.
Ans:
Since degree of numerator = degree of denominator, divide first.
Integrand becomes $1 + \frac{5x - 5}{x^2 - 5x + 6}$.
Use Partial Fractions on $\frac{5x-5}{(x-2)(x-3)}$.
Answer: $x - 5\log|x-2| + 10\log|x-3| + C$.
Q9: Evaluate: $\int_0^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$.
Ans:
Using property $\int_0^a f(x)dx = \int_0^a f(a-x)dx$.
Let $I = \int \dots$. Then $I = \int \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx$.
Adding both equations: $2I = \int_0^{\pi/2} 1 dx = [x]_0^{\pi/2} = \pi/2$.
Therefore, $I = \frac{\pi}{4}$.
Q10: Find the area of the region bounded by the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$.
Ans:
Area of ellipse = $\pi a b$.
Here $a=4, b=3$.
Area = $12\pi$ sq units. (For Board exam, you must show integration method: $4 \int_0^4 \frac{3}{4}\sqrt{16-x^2} dx$).
Q11: Find the general solution of the differential equation: $\frac{dy}{dx} + y \cot x = 2x + x^2 \cot x$.
Ans:
It is a Linear D.E. of form $\frac{dy}{dx} + Py = Q$.
I.F. = $e^{\int \cot x dx} = e^{\log \sin x} = \sin x$.
Solution: $y(\sin x) = \int (2x + x^2 \cot x)\sin x dx$.
$y\sin x = x^2 \sin x + C \Rightarrow y = x^2 + C \csc x$.
Q12: Find the projection of vector $\vec{a} = 2\hat{i} + 3\hat{j} + 2\hat{k}$ on vector $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$.
Ans:
Projection = $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$.
$\vec{a} \cdot \vec{b} = (2)(1) + (3)(2) + (2)(1) = 10$.
$|\vec{b}| = \sqrt{1+4+1} = \sqrt{6}$.
Projection = $10/\sqrt{6}$.
Q13: Find the area of a parallelogram whose adjacent sides are given by $\vec{a} = 3\hat{i} + \hat{j} + 4\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$.
Ans:
Area = $|\vec{a} \times \vec{b}|$.
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 4 \\ 1 & -1 & 1 \end{vmatrix} = 5\hat{i} + \hat{j} - 4\hat{k}$.
Magnitude = $\sqrt{25 + 1 + 16} = \sqrt{42}$ sq units.
Q14: Find the shortest distance between the lines $\vec{r} = (\hat{i} + \hat{j}) + \lambda(2\hat{i} - \hat{j} + \hat{k})$ and $\vec{r} = (2\hat{i} + \hat{j} - \hat{k}) + \mu(3\hat{i} - 5\hat{j} + 2\hat{k})$.
Ans:
Formula: $d = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right|$.
Calculate cross product of direction vectors, then dot product with difference of points.
Ans: $10/\sqrt{59}$.
Q15: A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.
Ans: (Bayes' Theorem)
Let $E_1$: Bag I chosen, $E_2$: Bag II chosen, $A$: Red ball drawn.
$P(E_1) = 1/2, P(E_2) = 1/2$.
$P(A|E_1) = 4/8 = 1/2$.
$P(A|E_2) = 2/8 = 1/4$.
$P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)} = \frac{(1/2)(1/2)}{(1/2)(1/2) + (1/2)(1/4)} = \frac{1/4}{3/8} = 2/3$.
Q16: A fair die is rolled. If the outcome is an odd number, what is the probability that it is prime?
Ans:
Sample Space (Odd) = $\{1, 3, 5\}$. Total = 3.
Favorable (Prime in Odd) = $\{3, 5\}$. Total = 2.
Probability = $2/3$.
Q17: Express the matrix $A = \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}$ as the sum of a symmetric and a skew-symmetric matrix.
Ans:
Let $P = \frac{1}{2}(A + A')$ (Symmetric) and $Q = \frac{1}{2}(A - A')$ (Skew-Symmetric).
$A' = \begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix}$.
$P = \frac{1}{2} \begin{bmatrix} 6 & 6 \\ 6 & -2 \end{bmatrix} = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix}$.
$Q = \frac{1}{2} \begin{bmatrix} 0 & 4 \\ -4 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$.
Check: $P+Q = A$.
Q18: If area of triangle is 35 sq units with vertices $(2, -6), (5, 4)$ and $(k, 4)$. Find $k$.
Ans:
Area $\Delta = \frac{1}{2} | \det | = 35 \Rightarrow \det = \pm 70$.
$\begin{vmatrix} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{vmatrix} = \pm 70$.
Expanding along $C_3$: $1(20-4k) - 1(8+6k) + 1(\dots)$... simpler to expand $C_2$ or $R_1$.
Solving gives $k = 12$ or $k = -2$.
Q19: Solve for $x$: $\tan^{-1} 2x + \tan^{-1} 3x = \frac{\pi}{4}$.
Ans:
Formula: $\tan^{-1} \frac{2x+3x}{1-6x^2} = \frac{\pi}{4}$.
$\frac{5x}{1-6x^2} = \tan(\frac{\pi}{4}) = 1$.
$6x^2 + 5x - 1 = 0 \Rightarrow (6x-1)(x+1) = 0$.
$x = 1/6$ or $x = -1$.
Since $x=-1$ makes LHS negative, reject it. Ans: $x = 1/6$.
Q20: Discuss the continuity of the function $f(x) = |x-1| + |x-2|$ at $x=1$ and $x=2$.
Ans:
The modulus function is continuous everywhere.
$|x-1|$ is continuous at $x=1$. $|x-2|$ is continuous at $x=2$.
Sum of continuous functions is continuous.
Hence, $f(x)$ is continuous at both points (though not differentiable there).
Q21: If $y = (\log x)^x + x^{\log x}$, find $\frac{dy}{dx}$.
Ans:
Let $u = (\log x)^x$ and $v = x^{\log x}$.
Take log on both sides for $u$ and $v$ separately.
$\frac{du}{dx} = (\log x)^x [\frac{1}{\log x} + \log(\log x)]$.
$\frac{dv}{dx} = x^{\log x} [\frac{2\log x}{x}]$.
Add them for final answer.
Q22: Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.
Ans:
Let radius $R$, sides $x, y$. Relation: $x^2 + y^2 = (2R)^2$.
Maximize Area $A = xy$. Better to max $Z = A^2 = x^2y^2 = x^2(4R^2 - x^2)$.
Differentiate $Z$ w.r.t $x$, set to 0. You get $x^2 = 2R^2$.
Then $y^2 = 2R^2$. Since $x=y$, it is a square.
Q23: Evaluate: $\int \frac{x e^x}{(1+x)^2} dx$.
Ans:
Rewrite num: $x = (1+x) - 1$.
$I = \int e^x [ \frac{1+x}{(1+x)^2} - \frac{1}{(1+x)^2} ] dx$.
$I = \int e^x [ \frac{1}{1+x} + (\frac{-1}{(1+x)^2}) ] dx$.
Form: $\int e^x [f(x) + f'(x)] dx = e^x f(x)$.
Ans: $\frac{e^x}{1+x} + C$.
Q24: Evaluate: $\int_0^{\pi} \frac{x \sin x}{1 + \cos^2 x} dx$.
Ans:
Let $I = \dots$ (1). Use property $\int_0^a f(x) = \int_0^a f(a-x)$.
$I = \int_0^{\pi} \frac{(\pi-x)\sin x}{1+\cos^2 x} dx$ (2).
Add (1) & (2): $2I = \pi \int_0^{\pi} \frac{\sin x}{1+\cos^2 x} dx$.
Put $\cos x = t, -\sin x dx = dt$. Limits $1$ to $-1$.
Ans: $I = \frac{\pi^2}{4}$.
Q25: Find the particular solution of: $\frac{dy}{dx} - 3y \cot x = \sin 2x$, given $y=2$ when $x = \pi/2$.
Ans:
Linear DE. $P = -3\cot x$. $I.F. = e^{\int -3\cot x} = \frac{1}{\sin^3 x}$.
Sol: $y(\frac{1}{\sin^3 x}) = \int \frac{\sin 2x}{\sin^3 x} dx = \int 2\cot x \csc x dx$.
$y \csc^3 x = -2 \csc x + C$.
Apply condition $y(\pi/2)=2$ to find $C=4$.
Q26: Find the area of the region bounded by the parabola $y = x^2$ and the line $y = |x|$.
Ans:
Due to symmetry, Area = $2 \times$ (Area in 1st quadrant).
Intersect at $(0,0)$ and $(1,1)$.
Area = $2 \int_0^1 (y_{line} - y_{parabola}) dx = 2 \int_0^1 (x - x^2) dx$.
$= 2 [\frac{x^2}{2} - \frac{x^3}{3}]_0^1 = 2(\frac{1}{2} - \frac{1}{3}) = 1/3$ sq units.
Q27: Find a unit vector perpendicular to both $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + \hat{j}$.
Ans:
Required vector $\vec{n} = \vec{a} \times \vec{b}$.
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & 1 & 0 \end{vmatrix} = -\hat{i} + \hat{j}$.
Unit vector = $\frac{-\hat{i} + \hat{j}}{\sqrt{2}}$.
Q28: Find the vector equation of the line passing through $(1, 2, -4)$ and perpendicular to the two lines: $\frac{x-8}{3} = \frac{y+19}{-16} = \frac{z-10}{7}$ and $\frac{x-15}{3} = \frac{y-29}{8} = \frac{z-5}{-5}$.
Ans:
Direction of required line $\vec{b} = \vec{b_1} \times \vec{b_2}$.
$\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix}$.
Calculate determinants to get direction ratios.
Eqn: $\vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(\vec{b})$.
Q29: Find the coordinates of the foot of the perpendicular drawn from the origin to the plane $2x - 3y + 4z - 6 = 0$.
Ans:
Normal vector $\vec{n} = (2, -3, 4)$. Equation of line through origin along normal: $\frac{x}{2} = \frac{y}{-3} = \frac{z}{4} = k$.
General point $P(2k, -3k, 4k)$.
Put $P$ in plane eqn: $2(2k) - 3(-3k) + 4(4k) = 6 \Rightarrow 29k = 6 \Rightarrow k = 6/29$.
Foot: $(12/29, -18/29, 24/29)$.
Q30: If $|\vec{a}|=3, |\vec{b}|=4, |\vec{c}|=5$ and each is perpendicular to the sum of the other two, find $|\vec{a}+\vec{b}+\vec{c}|$.
Ans:
Given: $\vec{a} \cdot (\vec{b}+\vec{c}) = 0 \Rightarrow \vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c} = 0$.
Similarly $\vec{b}\cdot\vec{c} + \vec{b}\cdot\vec{a} = 0$ etc. Summing gives $\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a} = 0$.
$|\vec{a}+\vec{b}+\vec{c}|^2 = a^2 + b^2 + c^2 + 2(0) = 9 + 16 + 25 = 50$.
Result: $\sqrt{50} = 5\sqrt{2}$.
Q31: Probability of solving a problem by A is 1/2 and by B is 1/3. If both try independently, find the probability that the problem is solved.
Ans:
$P(A \cup B) = 1 - P(A' \cap B')$.
$P(A') = 1/2, P(B') = 2/3$.
$P(\text{Not Solved}) = (1/2)(2/3) = 1/3$.
$P(\text{Solved}) = 1 - 1/3 = 2/3$.
Q32: Solve LPP: Maximize $Z = 3x + 4y$ subject to $x+y \le 4, x \ge 0, y \ge 0$.
Ans:
Corner points of feasible region: $O(0,0), A(4,0), B(0,4)$.
$Z(0,0) = 0$.
$Z(4,0) = 12$.
$Z(0,4) = 16$.
Max value is 16 at $(0,4)$.
Q33: Find the equation of the tangent to the curve $y = x^3 - 11x + 5$ at the point where the tangent is $y = x - 11$.
Ans:
Slope of tangent $m = 1$.
$\frac{dy}{dx} = 3x^2 - 11$.
Set $3x^2 - 11 = 1 \Rightarrow 3x^2 = 12 \Rightarrow x = \pm 2$.
Find corresponding $y$ values and verify which point lies on $y=x-11$.
Point is $(2, -9)$. Tangent eqn: $y - (-9) = 1(x-2) \Rightarrow y = x - 11$.
Q34: If $A$ is a square matrix of order 3 and $|A| = 5$, find the value of $|2A'|$.
Ans:
Formula: $|kA| = k^n |A|$. Here $n=3, k=2$.
$|2A'| = 2^3 |A'|$.
Since $|A'| = |A| = 5$.
Value $= 8 \times 5 = 40$.
Q35: Evaluate: $\int \frac{1}{\cos(x-a)\cos(x-b)} dx$.
Ans:
Multiply and divide by $\sin(a-b)$.
Write $a-b = (x-b) - (x-a)$ in numerator.
Expand using $\sin(A-B)$ formula.
Ans: $\frac{1}{\sin(a-b)} \log |\frac{\cos(x-a)}{\cos(x-b)}| + C$.
Q36: Differentiate $\sin^2 x$ with respect to $e^{\cos x}$.
Ans:
Let $u = \sin^2 x$ and $v = e^{\cos x}$.
$\frac{du}{dx} = 2\sin x \cos x = \sin 2x$.
$\frac{dv}{dx} = e^{\cos x}(-\sin x)$.
$\frac{du}{dv} = \frac{2\sin x \cos x}{-e^{\cos x}\sin x} = -2\cos x e^{-\cos x}$.
Q37: A card is drawn from a pack of 52 cards. Find the probability that it is a King or a Heart.
Ans:
$P(K) = 4/52$. $P(H) = 13/52$.
$P(K \cap H) = 1/52$ (The King of Hearts).
$P(K \cup H) = 4/52 + 13/52 - 1/52 = 16/52 = 4/13$.
Q38: If $A$ is a skew-symmetric matrix of odd order $n$, find $|A|$.
Ans:
For skew-symmetric, $A' = -A$.
$|A'| = |-A| = (-1)^n |A|$.
Since $|A'|=|A|$ and $n$ is odd, $|A| = -|A| \Rightarrow 2|A|=0 \Rightarrow |A|=0$.
Q39: Find the area bounded by $y^2 = 4x$ and $y=2x$.
Ans:
Intersection points: $(2x)^2 = 4x \Rightarrow 4x^2 - 4x = 0 \Rightarrow x=0, 1$.
Limits $x=0$ to $1$.
Area = $\int_0^1 (\sqrt{4x} - 2x) dx = \int_0^1 (2\sqrt{x} - 2x) dx$.
$= [2 \frac{x^{3/2}}{3/2} - x^2]_0^1 = \frac{4}{3} - 1 = 1/3$ sq unit.
Q40: Find the angle between the line $\frac{x+1}{2} = \frac{y}{3} = \frac{z-3}{6}$ and the plane $10x + 2y - 11z = 3$.
Ans:
Line direction $\vec{b} = (2, 3, 6)$. Plane normal $\vec{n} = (10, 2, -11)$.
Angle is $\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}||\vec{n}|}$.
$\vec{b}\cdot\vec{n} = 20 + 6 - 66 = -40$.
$|\vec{b}|=7, |\vec{n}|=15$.
$\sin \theta = |-40| / (7 \times 15) = 8/21 \Rightarrow \theta = \sin^{-1}(8/21)$.
Q41: The radius of a circle is increasing at 0.7 cm/s. What is the rate of increase of its circumference?
Ans:
$dr/dt = 0.7$.
$C = 2\pi r$.
$dC/dt = 2\pi (dr/dt) = 2\pi (0.7) = 1.4\pi$ cm/s.
Q42: Solve for $X$ if $2X + \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 3 & 8 \\ 7 & 2 \end{bmatrix}$.
Ans:
$2X = \begin{bmatrix} 3 & 8 \\ 7 & 2 \end{bmatrix} - \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 2 & 6 \\ 4 & -2 \end{bmatrix}$.
$X = \frac{1}{2} \begin{bmatrix} 2 & 6 \\ 4 & -2 \end{bmatrix} = \begin{bmatrix} 1 & 3 \\ 2 & -1 \end{bmatrix}$.
Q43: Find $\int \frac{1}{x^2 + 16} dx$.
Ans:
Standard form: $\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \tan^{-1} \frac{x}{a}$.
Here $a=4$.
Ans: $\frac{1}{4} \tan^{-1} \frac{x}{4} + C$.
Q44: Evaluate $\int \log x dx$.
Ans:
Use Integration by Parts. Take $u=\log x, v=1$.
$= \log x \int 1 dx - \int (\frac{d}{dx}\log x \int 1 dx) dx$.
$= x \log x - \int \frac{1}{x} \cdot x dx = x \log x - x + C$.
Q45: If $\vec{a}$ is a unit vector and $(\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = 8$, find $|\vec{x}|$.
Ans:
$(\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = |\vec{x}|^2 - |\vec{a}|^2$.
$|\vec{x}|^2 - 1 = 8$ (since $|\vec{a}|=1$).
$|\vec{x}|^2 = 9 \Rightarrow |\vec{x}| = 3$.
Q46: Is $f: \mathbb{R} \to \mathbb{R}$ given by $f(x) = x^3$ a bijective function?
Ans:
1. One-one: Let $x_1^3 = x_2^3 \Rightarrow x_1 = x_2$. Yes.
2. Onto: For any real $y$, there exists $x = y^{1/3}$ in $\mathbb{R}$ such that $f(x)=y$. Yes.
Result: Yes, it is bijective.