MathematicsQ1–60 of 60 questions
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1
MathematicsDifficultMCQKCET · 2026
The probability of at least one of the events $A$ and $B$ occurring is $0.6$. If $A$ and $B$ occur simultaneously with probability $0.2$, then $P(\overline{A}) + P(\overline{B})$ is
A
$1$
B
$0.8$
C
$0.6$
D
$1.2$
Solution
(D) Given: $P(A \cup B) = 0.6$ and $P(A \cap B) = 0.2$.
Using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$:
$0.6 = P(A) + P(B) - 0.2$
$P(A) + P(B) = 0.8$
We know that $P(\overline{A}) = 1 - P(A)$ and $P(\overline{B}) = 1 - P(B)$.
Therefore, $P(\overline{A}) + P(\overline{B}) = (1 - P(A)) + (1 - P(B)) = 2 - (P(A) + P(B))$.
Substituting the value: $2 - 0.8 = 1.2$.
Using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$:
$0.6 = P(A) + P(B) - 0.2$
$P(A) + P(B) = 0.8$
We know that $P(\overline{A}) = 1 - P(A)$ and $P(\overline{B}) = 1 - P(B)$.
Therefore, $P(\overline{A}) + P(\overline{B}) = (1 - P(A)) + (1 - P(B)) = 2 - (P(A) + P(B))$.
Substituting the value: $2 - 0.8 = 1.2$.
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2
MathematicsDifficultMCQKCET · 2026
The maximum value of $\sin(x + \pi/6) + \cos(x + \pi/6)$ is attained at $x =$
A
$\pi/2$
B
$\pi/4$
C
$\pi/6$
D
$\pi/12$
Solution
(D) Let $f(x) = \sin(x + \pi/6) + \cos(x + \pi/6)$.
Using the identity $\sin \theta + \cos \theta = \sqrt{2} \sin(\theta + \pi/4)$, we have $f(x) = \sqrt{2} \sin(x + \pi/6 + \pi/4)$.
The maximum value of the sine function is $1$, which occurs when the argument is $\pi/2$.
Set $x + \pi/6 + \pi/4 = \pi/2$.
$x + 5\pi/12 = \pi/2$.
$x = \pi/2 - 5\pi/12 = 6\pi/12 - 5\pi/12 = \pi/12$.
Using the identity $\sin \theta + \cos \theta = \sqrt{2} \sin(\theta + \pi/4)$, we have $f(x) = \sqrt{2} \sin(x + \pi/6 + \pi/4)$.
The maximum value of the sine function is $1$, which occurs when the argument is $\pi/2$.
Set $x + \pi/6 + \pi/4 = \pi/2$.
$x + 5\pi/12 = \pi/2$.
$x = \pi/2 - 5\pi/12 = 6\pi/12 - 5\pi/12 = \pi/12$.
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3
MathematicsDifficultMCQKCET · 2026
The angles of a triangle are in $A.P.$ and the greatest angle is double the least angle. Find the sine of the third angle.
A
$\frac{\sqrt{3}}{2}$
B
$\frac{1}{\sqrt{2}}$
C
$\frac{1}{2}$
D
$0$
Solution
(A) Let the angles of the triangle be $(A-d)$, $A$, and $(A+d)$.
Since the sum of angles in a triangle is $180^\circ$, we have $(A-d) + A + (A+d) = 180^\circ$, which gives $3A = 180^\circ$, so $A = 60^\circ$.
The angles are $(60^\circ-d)$, $60^\circ$, and $(60^\circ+d)$.
The greatest angle is $(60^\circ+d)$ and the least angle is $(60^\circ-d)$.
Given that the greatest angle is double the least angle: $60^\circ+d = 2(60^\circ-d)$.
$60^\circ+d = 120^\circ - 2d \implies 3d = 60^\circ \implies d = 20^\circ$.
The angles are $60^\circ-20^\circ = 40^\circ$, $60^\circ$, and $60^\circ+20^\circ = 80^\circ$.
The third angle is $60^\circ$.
Therefore, $\sin(60^\circ) = \frac{\sqrt{3}}{2}$.
Since the sum of angles in a triangle is $180^\circ$, we have $(A-d) + A + (A+d) = 180^\circ$, which gives $3A = 180^\circ$, so $A = 60^\circ$.
The angles are $(60^\circ-d)$, $60^\circ$, and $(60^\circ+d)$.
The greatest angle is $(60^\circ+d)$ and the least angle is $(60^\circ-d)$.
Given that the greatest angle is double the least angle: $60^\circ+d = 2(60^\circ-d)$.
$60^\circ+d = 120^\circ - 2d \implies 3d = 60^\circ \implies d = 20^\circ$.
The angles are $60^\circ-20^\circ = 40^\circ$, $60^\circ$, and $60^\circ+20^\circ = 80^\circ$.
The third angle is $60^\circ$.
Therefore, $\sin(60^\circ) = \frac{\sqrt{3}}{2}$.
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4
MathematicsDifficultMCQKCET · 2026
The mean and standard deviation of $100$ items are $50$ and $4$, respectively. Then the sum of the squares of all the items is:
A
$250000$
B
$251600$
C
$256100$
D
$265100$
Solution
(B) Given: $n = 100$, mean $\bar{x} = 50$, and standard deviation $\sigma = 4$.
The formula for variance is $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.
Substitute the given values: $4^2 = \frac{\sum x_i^2}{100} - (50)^2$.
$16 = \frac{\sum x_i^2}{100} - 2500$.
$16 + 2500 = \frac{\sum x_i^2}{100}$.
$2516 = \frac{\sum x_i^2}{100}$.
$\sum x_i^2 = 2516 \times 100 = 251600$.
The formula for variance is $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.
Substitute the given values: $4^2 = \frac{\sum x_i^2}{100} - (50)^2$.
$16 = \frac{\sum x_i^2}{100} - 2500$.
$16 + 2500 = \frac{\sum x_i^2}{100}$.
$2516 = \frac{\sum x_i^2}{100}$.
$\sum x_i^2 = 2516 \times 100 = 251600$.
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5
MathematicsMediumMCQKCET · 2026
The probability of occurrence of an event $A$ is $1/2$ and that of $B$ is $3/10$. If $A$ and $B$ are mutually exclusive, then the probability of occurrence of neither $A$ nor $B$ is (in $/5$)
A
$4$
B
$3$
C
$2$
D
$1$
Solution
(D) Given: $P(A) = 1/2$ and $P(B) = 3/10$.
Since $A$ and $B$ are mutually exclusive, $P(A \cap B) = 0$.
The probability of occurrence of either $A$ or $B$ is $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 1/2 + 3/10 - 0$.
$P(A \cup B) = 5/10 + 3/10 = 8/10 = 4/5$.
The probability of occurrence of neither $A$ nor $B$ is $P(A' \cap B') = 1 - P(A \cup B)$.
$P(A' \cap B') = 1 - 4/5 = 1/5$.
Since $A$ and $B$ are mutually exclusive, $P(A \cap B) = 0$.
The probability of occurrence of either $A$ or $B$ is $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 1/2 + 3/10 - 0$.
$P(A \cup B) = 5/10 + 3/10 = 8/10 = 4/5$.
The probability of occurrence of neither $A$ nor $B$ is $P(A' \cap B') = 1 - P(A \cup B)$.
$P(A' \cap B') = 1 - 4/5 = 1/5$.
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6
MathematicsMediumMCQKCET · 2026
The probability of obtaining an even prime number on each die when a pair of dice is rolled is
A
$0$
B
$\frac{1}{6}$
C
$\frac{1}{12}$
D
$\frac{1}{36}$
Solution
(D) The only even prime number on a standard die is $2$.
The probability of getting $2$ on one die is $P(2) = \frac{1}{6}$.
Since the two dice are rolled independently, the probability of getting $2$ on both dice is $P(2 \text{ and } 2) = P(2) \times P(2) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$.
The probability of getting $2$ on one die is $P(2) = \frac{1}{6}$.
Since the two dice are rolled independently, the probability of getting $2$ on both dice is $P(2 \text{ and } 2) = P(2) \times P(2) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$.
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7
MathematicsDifficultMCQKCET · 2026
The probability that a man and his wife will be alive after $20 \text{ years}$ are $\frac{1}{4}$ and $\frac{1}{3}$ respectively. The probability that neither the man nor his wife will be alive after $20 \text{ years}$ is
A
$\frac{3}{4}$
B
$\frac{1}{12}$
C
$\frac{7}{12}$
D
$\frac{1}{2}$
Solution
(D) Let $P(M)$ be the probability that the man is alive and $P(W)$ be the probability that the wife is alive.
Given $P(M) = \frac{1}{4}$ and $P(W) = \frac{1}{3}$.
The probability that the man is not alive is $P(M') = 1 - P(M) = 1 - \frac{1}{4} = \frac{3}{4}$.
The probability that the wife is not alive is $P(W') = 1 - P(W) = 1 - \frac{1}{3} = \frac{2}{3}$.
Since the events are independent, the probability that neither is alive is $P(M' \cap W') = P(M') \times P(W')$.
$P(M' \cap W') = \frac{3}{4} \times \frac{2}{3} = \frac{6}{12} = \frac{1}{2}$.
Given $P(M) = \frac{1}{4}$ and $P(W) = \frac{1}{3}$.
The probability that the man is not alive is $P(M') = 1 - P(M) = 1 - \frac{1}{4} = \frac{3}{4}$.
The probability that the wife is not alive is $P(W') = 1 - P(W) = 1 - \frac{1}{3} = \frac{2}{3}$.
Since the events are independent, the probability that neither is alive is $P(M' \cap W') = P(M') \times P(W')$.
$P(M' \cap W') = \frac{3}{4} \times \frac{2}{3} = \frac{6}{12} = \frac{1}{2}$.
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8
MathematicsDifficultMCQKCET · 2026
If $\alpha$ and $\beta$ are acute angles such that $\alpha+\beta$ and $\alpha-\beta$ satisfy the equation $\tan^2\theta - 4\tan\theta + 1 = 0$, then $\alpha$ and $\beta$ are respectively:
A
$45^\circ, 30^\circ$
B
$30^\circ, 45^\circ$
C
$30^\circ, 60^\circ$
D
$60^\circ, 45^\circ$
Solution
(A) Given the quadratic equation $\tan^2\theta - 4\tan\theta + 1 = 0$.
Using the quadratic formula $\tan\theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we get $\tan\theta = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}$.
Since $\alpha$ and $\beta$ are acute angles, $\alpha+\beta$ and $\alpha-\beta$ are also acute. We know $\tan 75^\circ = 2+\sqrt{3}$ and $\tan 15^\circ = 2-\sqrt{3}$.
Thus, $\alpha+\beta = 75^\circ$ and $\alpha-\beta = 15^\circ$.
Adding the two equations: $2\alpha = 90^\circ \implies \alpha = 45^\circ$.
Subtracting the two equations: $2\beta = 60^\circ \implies \beta = 30^\circ$.
Using the quadratic formula $\tan\theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we get $\tan\theta = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}$.
Since $\alpha$ and $\beta$ are acute angles, $\alpha+\beta$ and $\alpha-\beta$ are also acute. We know $\tan 75^\circ = 2+\sqrt{3}$ and $\tan 15^\circ = 2-\sqrt{3}$.
Thus, $\alpha+\beta = 75^\circ$ and $\alpha-\beta = 15^\circ$.
Adding the two equations: $2\alpha = 90^\circ \implies \alpha = 45^\circ$.
Subtracting the two equations: $2\beta = 60^\circ \implies \beta = 30^\circ$.
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9
MathematicsMediumMCQKCET · 2026
$\sum_{n=1}^{4} (\sqrt{-1})^{2n} = $
A
$2$
B
$-i$
C
$0$
D
$i$
Solution
(C) We know that $\sqrt{-1} = i$.
Therefore, the expression becomes $\sum_{n=1}^{4} (i)^{2n} = i^2 + i^4 + i^6 + i^8$.
Since $i^2 = -1$, $i^4 = 1$, $i^6 = -1$, and $i^8 = 1$.
Substituting these values: $-1 + 1 - 1 + 1 = 0$.
Therefore, the expression becomes $\sum_{n=1}^{4} (i)^{2n} = i^2 + i^4 + i^6 + i^8$.
Since $i^2 = -1$, $i^4 = 1$, $i^6 = -1$, and $i^8 = 1$.
Substituting these values: $-1 + 1 - 1 + 1 = 0$.
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10
MathematicsDifficultMCQKCET · 2026
The solution of the inequality $3(x-1) \leq 2(x-3)$ is
A
$x \leq -3$
B
$x \geq -3$
C
$x \leq 3$
D
$x \geq 3$
Solution
(A) Step $1$: Expand both sides of the inequality: $3x - 3 \leq 2x - 6$.
Step $2$: Subtract $2x$ from both sides: $3x - 2x - 3 \leq -6$.
Step $3$: Add $3$ to both sides: $x \leq -6 + 3$.
Step $4$: Simplify to get the final result: $x \leq -3$.
Step $2$: Subtract $2x$ from both sides: $3x - 2x - 3 \leq -6$.
Step $3$: Add $3$ to both sides: $x \leq -6 + 3$.
Step $4$: Simplify to get the final result: $x \leq -3$.
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11
MathematicsDifficultMCQKCET · 2026
$10$ distinct points are taken on a circle. Then using these points:
Statement $I$: The number of triangles that can be formed is $100$
Statement $II$: The number of chords that can be formed is $45$
Which of the following is correct?
Statement $I$: The number of triangles that can be formed is $100$
Statement $II$: The number of chords that can be formed is $45$
Which of the following is correct?
A
Both statement $I$ and statement $II$ are true
B
Both statement $I$ and statement $II$ are false
C
Statement $I$ is true and statement $II$ is false
D
Statement $I$ is false and statement $II$ is true
Solution
(D) Step $1$: The number of triangles formed by $n$ points on a circle is given by $^{n}C_3$. For $n=10$, number of triangles = $^{10}C_3 = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$. Thus, Statement $I$ is false.
Step $2$: The number of chords formed by $n$ points on a circle is given by $^{n}C_2$. For $n=10$, number of chords = $^{10}C_2 = \frac{10 \times 9}{2} = 45$. Thus, Statement $II$ is true.
Step $3$: Comparing the results, Statement $I$ is false and Statement $II$ is true.
Step $2$: The number of chords formed by $n$ points on a circle is given by $^{n}C_2$. For $n=10$, number of chords = $^{10}C_2 = \frac{10 \times 9}{2} = 45$. Thus, Statement $II$ is true.
Step $3$: Comparing the results, Statement $I$ is false and Statement $II$ is true.
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12
MathematicsDifficultMCQKCET · 2026
How many ways can you arrange all the characters in "$KCET$ $2025$" such that the arrangement starts with $K$ and ends with $5$?
A
$720$
B
$360$
C
$120$
D
$180$
Solution
(B) The string "$KCET$ $2025$" contains $8$ characters: $K, C, E, T, \text{space}, 2, 0, 2, 5$. However, assuming the space is ignored and we consider the $8$ characters: $K, C, E, T, 2, 0, 2, 5$.
Fix $K$ at the first position and $5$ at the last position.
The remaining $6$ characters are $C, E, T, 2, 0, 2$.
Among these $6$ characters, the digit $2$ repeats twice.
The number of ways to arrange these $6$ characters is $\frac{6!}{2!} = \frac{720}{2} = 360$ ways.
Fix $K$ at the first position and $5$ at the last position.
The remaining $6$ characters are $C, E, T, 2, 0, 2$.
Among these $6$ characters, the digit $2$ repeats twice.
The number of ways to arrange these $6$ characters is $\frac{6!}{2!} = \frac{720}{2} = 360$ ways.
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13
MathematicsMediumMCQKCET · 2026
The value of the expression $\frac{x^3 + 3x^2 + 3x + 1}{x^4 + 4x^3 + 6x^2 + 4x + 1}$ at $x = 2$ is:
A
$3$
B
$\frac{25}{61}$
C
$\frac{1}{3}$
D
$\frac{19}{73}$
Solution
(C) The given expression is $\frac{x^3 + 3x^2 + 3x + 1}{x^4 + 4x^3 + 6x^2 + 4x + 1}$.
Using the binomial expansion formula $(a+b)^n$, we can write the numerator as $(x+1)^3$ and the denominator as $(x+1)^4$.
Thus, the expression simplifies to $\frac{(x+1)^3}{(x+1)^4} = \frac{1}{x+1}$.
Substituting $x = 2$ into the simplified expression, we get $\frac{1}{2+1} = \frac{1}{3}$.
Using the binomial expansion formula $(a+b)^n$, we can write the numerator as $(x+1)^3$ and the denominator as $(x+1)^4$.
Thus, the expression simplifies to $\frac{(x+1)^3}{(x+1)^4} = \frac{1}{x+1}$.
Substituting $x = 2$ into the simplified expression, we get $\frac{1}{2+1} = \frac{1}{3}$.
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14
MathematicsDifficultMCQKCET · 2026
If we insert two numbers between $\sqrt{2}$ and $4$ so that the resulting sequence is in $G.P.$, then the inserted numbers in the order are
A
$8, \sqrt{2}$
B
$2, \sqrt{8}$
C
$\sqrt{8}, 2$
D
$\sqrt{2}, 8$
Solution
(B) Let the sequence be $\sqrt{2}, g_1, g_2, 4$.
Here, the first term $a = \sqrt{2}$ and the fourth term $ar^3 = 4$.
$r^3 = \frac{4}{\sqrt{2}} = \frac{2^2}{2^{1/2}} = 2^{2 - 1/2} = 2^{3/2}$.
Taking the cube root on both sides, $r = (2^{3/2})^{1/3} = 2^{1/2} = \sqrt{2}$.
Thus, the inserted numbers are:
$g_1 = a \cdot r = \sqrt{2} \cdot \sqrt{2} = 2$.
$g_2 = a \cdot r^2 = \sqrt{2} \cdot (\sqrt{2})^2 = \sqrt{2} \cdot 2 = 2\sqrt{2} = \sqrt{8}$.
Here, the first term $a = \sqrt{2}$ and the fourth term $ar^3 = 4$.
$r^3 = \frac{4}{\sqrt{2}} = \frac{2^2}{2^{1/2}} = 2^{2 - 1/2} = 2^{3/2}$.
Taking the cube root on both sides, $r = (2^{3/2})^{1/3} = 2^{1/2} = \sqrt{2}$.
Thus, the inserted numbers are:
$g_1 = a \cdot r = \sqrt{2} \cdot \sqrt{2} = 2$.
$g_2 = a \cdot r^2 = \sqrt{2} \cdot (\sqrt{2})^2 = \sqrt{2} \cdot 2 = 2\sqrt{2} = \sqrt{8}$.
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15
MathematicsDifficultMCQKCET · 2026
The line $L_1$ passes through the points $(-1, 2)$ and $(3, 6)$. If $L_1$ divides the line $L_2$ (which passes through $(3, -1)$) in the ratio $1:3$ internally, and the point of intersection lies on $L_1$, find the equation of $L_2$ given that it is perpendicular to $L_1$.
A
$4x - 3y - 15 = 0$
B
$x + y - 2 = 0$
C
$x + y + 2 = 0$
D
$x - y - 4 = 0$
Solution
(B) Step $1$: Find the slope of $L_1$. The slope $m_1 = \frac{6 - 2}{3 - (-1)} = \frac{4}{4} = 1$.
Step $2$: Since $L_2$ is perpendicular to $L_1$, the slope of $L_2$ is $m_2 = -\frac{1}{m_1} = -1$.
Step $3$: Use the point-slope form for $L_2$ passing through $(3, -1)$ with slope $-1$: $y - (-1) = -1(x - 3)$.
Step $4$: Simplify the equation: $y + 1 = -x + 3$, which gives $x + y - 2 = 0$.
Step $2$: Since $L_2$ is perpendicular to $L_1$, the slope of $L_2$ is $m_2 = -\frac{1}{m_1} = -1$.
Step $3$: Use the point-slope form for $L_2$ passing through $(3, -1)$ with slope $-1$: $y - (-1) = -1(x - 3)$.
Step $4$: Simplify the equation: $y + 1 = -x + 3$, which gives $x + y - 2 = 0$.
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16
MathematicsMediumMCQKCET · 2026
In the figure, Statement $I$: when $\alpha > \beta \ge 0$, the section is a hyperbola. Statement $II$: when $\beta > 90^\circ$, the section is an ellipse.
A
Statement $I$ is true, statement $II$ is false
B
Statement $I$ is false, statement $II$ is true
C
Both the statements are true
D
Both the statements are false
Solution
(A) For a hyperbola, the condition is $\alpha > \beta \ge 0$.
For an ellipse, the condition is $0 \le \alpha < \beta < 90^\circ$.
Thus, Statement $I$ is true and Statement $II$ is false.
For an ellipse, the condition is $0 \le \alpha < \beta < 90^\circ$.
Thus, Statement $I$ is true and Statement $II$ is false.
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17
MathematicsDifficultMCQKCET · 2026
If $\lim_{x \to 3} \frac{x^2 - ax - 3b}{x - 3} = 5$, then $a + b =$
A
$1$
B
$2$
C
$3$
D
$4$
Solution
(C) Given $\lim_{x \to 3} \frac{x^2 - ax - 3b}{x - 3} = 5$.
Since the denominator approaches $0$ as $x \to 3$, the limit exists only if the numerator also approaches $0$ at $x = 3$.
Substituting $x = 3$ in the numerator: $3^2 - a(3) - 3b = 0$.
$9 - 3a - 3b = 0$.
$3(3 - a - b) = 0$.
$3 - a - b = 0$.
Therefore, $a + b = 3$.
Since the denominator approaches $0$ as $x \to 3$, the limit exists only if the numerator also approaches $0$ at $x = 3$.
Substituting $x = 3$ in the numerator: $3^2 - a(3) - 3b = 0$.
$9 - 3a - 3b = 0$.
$3(3 - a - b) = 0$.
$3 - a - b = 0$.
Therefore, $a + b = 3$.
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18
MathematicsDifficultMCQKCET · 2026
The area enclosed by the curve $x = \sqrt{3} \cos \theta, y = \sqrt{3} \sin \theta$ is
A
$\sqrt{3} \pi \text{ sq. units}$
B
$9\pi \text{ sq. units}$
C
$6\pi \text{ sq. units}$
D
$3\pi \text{ sq. units}$
Solution
(D) Given the parametric equations $x = \sqrt{3} \cos \theta$ and $y = \sqrt{3} \sin \theta$.
Squaring and adding both equations: $x^2 + y^2 = (\sqrt{3} \cos \theta)^2 + (\sqrt{3} \sin \theta)^2$.
$x^2 + y^2 = 3 \cos^2 \theta + 3 \sin^2 \theta = 3(\cos^2 \theta + \sin^2 \theta) = 3$.
This represents a circle with radius $r = \sqrt{3}$.
The area of a circle is given by $A = \pi r^2$.
$A = \pi (\sqrt{3})^2 = 3\pi \text{ sq. units}$.
Squaring and adding both equations: $x^2 + y^2 = (\sqrt{3} \cos \theta)^2 + (\sqrt{3} \sin \theta)^2$.
$x^2 + y^2 = 3 \cos^2 \theta + 3 \sin^2 \theta = 3(\cos^2 \theta + \sin^2 \theta) = 3$.
This represents a circle with radius $r = \sqrt{3}$.
The area of a circle is given by $A = \pi r^2$.
$A = \pi (\sqrt{3})^2 = 3\pi \text{ sq. units}$.
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19
MathematicsMediumMCQKCET · 2026
If $A = \{a, b, c, d, e, f\}$, then the number of subsets of $A$ which contains at least $2$ elements is
A
$64$
B
$65$
C
$57$
D
$59$
Solution
(C) The total number of subsets of a set with $n$ elements is $2^n$.
Here, $n = 6$, so the total number of subsets is $2^6 = 64$.
Subsets with $0$ elements (the empty set) $= \binom{6}{0} = 1$.
Subsets with $1$ element $= \binom{6}{1} = 6$.
Subsets with at least $2$ elements $=$ Total subsets $-$ (Subsets with $0$ elements $+$ Subsets with $1$ element).
Subsets with at least $2$ elements $= 64 - (1 + 6) = 64 - 7 = 57$.
Here, $n = 6$, so the total number of subsets is $2^6 = 64$.
Subsets with $0$ elements (the empty set) $= \binom{6}{0} = 1$.
Subsets with $1$ element $= \binom{6}{1} = 6$.
Subsets with at least $2$ elements $=$ Total subsets $-$ (Subsets with $0$ elements $+$ Subsets with $1$ element).
Subsets with at least $2$ elements $= 64 - (1 + 6) = 64 - 7 = 57$.
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20
MathematicsMediumMCQKCET · 2026
If $A = \{1, 2, 3, 4, \dots, 10\}$, then the number of non-empty subsets of $A$ containing only even numbers is
A
$31$
B
$32$
C
$30$
D
$29$
Solution
(A) The even numbers in set $A$ are $\{2, 4, 6, 8, 10\}$.
There are $5$ such even numbers.
The total number of subsets that can be formed using these $5$ elements is $2^5 = 32$.
Since we need non-empty subsets, we exclude the empty set $\emptyset$.
Therefore, the number of non-empty subsets is $32 - 1 = 31$.
There are $5$ such even numbers.
The total number of subsets that can be formed using these $5$ elements is $2^5 = 32$.
Since we need non-empty subsets, we exclude the empty set $\emptyset$.
Therefore, the number of non-empty subsets is $32 - 1 = 31$.
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21
MathematicsMediumMCQKCET · 2026
If $n(A) = 2$ and the number of relations from set $A$ to set $B$ is $1024$, then $n(B)$ is
A
$2$
B
$5$
C
$2^5$
D
$5^2$
Solution
(B) The number of relations from set $A$ to set $B$ is given by $2^{n(A) \times n(B)}$.
Given that $2^{n(A) \times n(B)} = 1024$.
Since $1024 = 2^{10}$, we have $n(A) \times n(B) = 10$.
Given $n(A) = 2$, we substitute the value: $2 \times n(B) = 10$.
Therefore, $n(B) = \frac{10}{2} = 5$.
Given that $2^{n(A) \times n(B)} = 1024$.
Since $1024 = 2^{10}$, we have $n(A) \times n(B) = 10$.
Given $n(A) = 2$, we substitute the value: $2 \times n(B) = 10$.
Therefore, $n(B) = \frac{10}{2} = 5$.
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22
MathematicsMediumMCQKCET · 2026
Let $R$ be a relation in the set of natural numbers $N$ defined by $R = \{(a, b) : a = b - 2, b > 6\}$. Which of the following is correct?
A
$(2, 4) \in R$
B
$(3, 8) \in R$
C
$(6, 8) \in R$
D
$(8, 7) \in R$
Solution
(C) The relation is defined as $R = \{(a, b) : a = b - 2, b > 6\}$.
For $(2, 4)$: $b = 4$, which is not greater than $6$. Thus, $(2, 4) \notin R$.
For $(3, 8)$: $a = 3$ and $b = 8$. Checking the condition $a = b - 2$: $3 = 8 - 2 = 6$, which is false.
For $(6, 8)$: $a = 6$ and $b = 8$. Checking the condition $a = b - 2$: $6 = 8 - 2 = 6$, which is true. Also, $b = 8 > 6$ is true. Thus, $(6, 8) \in R$.
For $(8, 7)$: $a = 8$ and $b = 7$. Checking the condition $a = b - 2$: $8 = 7 - 2 = 5$, which is false.
Therefore, the correct option is $(6, 8) \in R$.
For $(2, 4)$: $b = 4$, which is not greater than $6$. Thus, $(2, 4) \notin R$.
For $(3, 8)$: $a = 3$ and $b = 8$. Checking the condition $a = b - 2$: $3 = 8 - 2 = 6$, which is false.
For $(6, 8)$: $a = 6$ and $b = 8$. Checking the condition $a = b - 2$: $6 = 8 - 2 = 6$, which is true. Also, $b = 8 > 6$ is true. Thus, $(6, 8) \in R$.
For $(8, 7)$: $a = 8$ and $b = 7$. Checking the condition $a = b - 2$: $8 = 7 - 2 = 5$, which is false.
Therefore, the correct option is $(6, 8) \in R$.
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23
MathematicsDifficultMCQKCET · 2026
Find the sum of the series: $\tan^{-1} \left( \frac{1}{1 + 1 \times 2} \right) + \tan^{-1} \left( \frac{1}{1 + 2 \times 3} \right) + \dots + \tan^{-1} \left( \frac{1}{1 + n(n + 1)} \right) =$
A
$\tan^{-1} \left( \frac{n}{n + 2} \right)$
B
$\tan^{-1} \left( \frac{n + 1}{n} \right)$
C
$\tan^{-1} \left( \frac{n}{n + 1} \right)$
D
$\tan^{-1} \left( \frac{n + 2}{n} \right)$
Solution
(A) The general term of the series is $T_k = \tan^{-1} \left( \frac{1}{1 + k(k + 1)} \right)$.
We can rewrite the argument as $\frac{(k + 1) - k}{1 + k(k + 1)}$.
Using the identity $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \left( \frac{x - y}{1 + xy} \right)$, we get $T_k = \tan^{-1}(k + 1) - \tan^{-1}(k)$.
The sum $S_n = \sum_{k=1}^{n} (\tan^{-1}(k + 1) - \tan^{-1}(k))$.
This is a telescoping series: $S_n = (\tan^{-1} 2 - \tan^{-1} 1) + (\tan^{-1} 3 - \tan^{-1} 2) + \dots + (\tan^{-1}(n + 1) - \tan^{-1} n)$.
All intermediate terms cancel out, leaving $S_n = \tan^{-1}(n + 1) - \tan^{-1}(1)$.
Using the identity again, $S_n = \tan^{-1} \left( \frac{(n + 1) - 1}{1 + (n + 1)(1)} \right) = \tan^{-1} \left( \frac{n}{n + 2} \right)$.
We can rewrite the argument as $\frac{(k + 1) - k}{1 + k(k + 1)}$.
Using the identity $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \left( \frac{x - y}{1 + xy} \right)$, we get $T_k = \tan^{-1}(k + 1) - \tan^{-1}(k)$.
The sum $S_n = \sum_{k=1}^{n} (\tan^{-1}(k + 1) - \tan^{-1}(k))$.
This is a telescoping series: $S_n = (\tan^{-1} 2 - \tan^{-1} 1) + (\tan^{-1} 3 - \tan^{-1} 2) + \dots + (\tan^{-1}(n + 1) - \tan^{-1} n)$.
All intermediate terms cancel out, leaving $S_n = \tan^{-1}(n + 1) - \tan^{-1}(1)$.
Using the identity again, $S_n = \tan^{-1} \left( \frac{(n + 1) - 1}{1 + (n + 1)(1)} \right) = \tan^{-1} \left( \frac{n}{n + 2} \right)$.
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24
MathematicsDifficultMCQKCET · 2026
The corner points of the feasible region determined by the system of linear constraints are $(0, 10)$, $(5, 5)$, $(15, 15)$, and $(0, 20)$. Let $z = px + qy$ where $p, q > 0$. The condition on $p$ and $q$ such that the maximum value of $z$ occurs at both points $(15, 15)$ and $(0, 20)$ is
A
$p = q$
B
$p = 2q$
C
$q = 2p$
D
$q = 3p$
Solution
(D) For the maximum value of $z$ to occur at both points $(15, 15)$ and $(0, 20)$, the value of $z$ must be equal at these two points.
Substitute the coordinates into the objective function $z = px + qy$:
At $(15, 15)$, $z_1 = p(15) + q(15) = 15p + 15q$.
At $(0, 20)$, $z_2 = p(0) + q(20) = 20q$.
Equating $z_1$ and $z_2$:
$15p + 15q = 20q$.
Subtract $15q$ from both sides:
$15p = 5q$.
Divide by $5$:
$q = 3p$.
Substitute the coordinates into the objective function $z = px + qy$:
At $(15, 15)$, $z_1 = p(15) + q(15) = 15p + 15q$.
At $(0, 20)$, $z_2 = p(0) + q(20) = 20q$.
Equating $z_1$ and $z_2$:
$15p + 15q = 20q$.
Subtract $15q$ from both sides:
$15p = 5q$.
Divide by $5$:
$q = 3p$.
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25
MathematicsMediumMCQKCET · 2026
In a linear programming problem $(LPP)$, if the objective function $Z = ax + by$ has the same maximum value at two distinct corner points, then the number of points at which $Z_{max}$ occurs is
A
$1$
B
$2$
C
$0$
D
Infinitely many
Solution
(D) $1$. In an $LPP$, the objective function $Z = ax + by$ is a linear function.
$2$. If $Z$ attains the same maximum value at two distinct corner points, then by the property of linear functions, $Z$ will attain the same maximum value at every point on the line segment joining these two corner points.
$3$. Since a line segment contains an infinite number of points, the objective function $Z$ attains its maximum value at infinitely many points.
$2$. If $Z$ attains the same maximum value at two distinct corner points, then by the property of linear functions, $Z$ will attain the same maximum value at every point on the line segment joining these two corner points.
$3$. Since a line segment contains an infinite number of points, the objective function $Z$ attains its maximum value at infinitely many points.
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26
MathematicsDifficultMCQKCET · 2026
The integrating factor of the differential equation $(1 - x^2) \frac{dy}{dx} - xy = 1$ is
A
$1 - x^2$
B
$\frac{1}{2} \log |1 - x^2|$
C
$\frac{x}{1 + x^2}$
D
$\sqrt{1 - x^2}$
Solution
(D) The given differential equation is $(1 - x^2) \frac{dy}{dx} - xy = 1$.
Divide by $(1 - x^2)$ to get the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$:
$\frac{dy}{dx} - \frac{x}{1 - x^2}y = \frac{1}{1 - x^2}$.
Here, $P(x) = -\frac{x}{1 - x^2}$.
The integrating factor $(I.F.)$ is given by $e^{\int P(x) dx} = e^{\int -\frac{x}{1 - x^2} dx}$.
Let $t = 1 - x^2$, then $dt = -2x dx$, which implies $-x dx = \frac{dt}{2}$.
$I.F. = e^{\int \frac{dt}{2t}} = e^{\frac{1}{2} \log |t|} = e^{\log |t|^{1/2}} = \sqrt{t} = \sqrt{1 - x^2}$.
Divide by $(1 - x^2)$ to get the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$:
$\frac{dy}{dx} - \frac{x}{1 - x^2}y = \frac{1}{1 - x^2}$.
Here, $P(x) = -\frac{x}{1 - x^2}$.
The integrating factor $(I.F.)$ is given by $e^{\int P(x) dx} = e^{\int -\frac{x}{1 - x^2} dx}$.
Let $t = 1 - x^2$, then $dt = -2x dx$, which implies $-x dx = \frac{dt}{2}$.
$I.F. = e^{\int \frac{dt}{2t}} = e^{\frac{1}{2} \log |t|} = e^{\log |t|^{1/2}} = \sqrt{t} = \sqrt{1 - x^2}$.
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27
MathematicsDifficultMCQKCET · 2026
Recent studies suggest that $12\%$ of the world population is left-handed. Depending on parents' hand usage, the chances of having left-handed children are as follows: $A$. Both parents are left-handed, chances of having left-handed children = $24\%$. $B$. Both parents are right-handed, chance of having left-handed children = $9\%$. $C$. Father left-handed and mother right-handed, chances of having left-handed children = $17\%$. $D$. Father right-handed and mother left-handed, chances of having left-handed children = $22\%$. Given $P(A) = P(B) = P(C) = P(D) = 1/4$ and $L$ denotes the event that the child is left-handed. What is the probability $P(A|L)$?
A
$\frac{17}{100}$
B
$\frac{19}{25}$
C
$\frac{1}{3}$
D
$\frac{2}{3}$
Solution
(C) Using Bayes' theorem: $P(A|L) = \frac{P(L|A)P(A)}{P(L|A)P(A) + P(L|B)P(B) + P(L|C)P(C) + P(L|D)P(D)}$.
Given $P(A) = P(B) = P(C) = P(D) = 1/4$, the terms $1/4$ cancel out:
$P(A|L) = \frac{P(L|A)}{P(L|A) + P(L|B) + P(L|C) + P(L|D)}$.
Substituting the given probabilities: $P(A|L) = \frac{0.24}{0.24 + 0.09 + 0.17 + 0.22}$.
$P(A|L) = \frac{0.24}{0.72} = \frac{1}{3}$.
Given $P(A) = P(B) = P(C) = P(D) = 1/4$, the terms $1/4$ cancel out:
$P(A|L) = \frac{P(L|A)}{P(L|A) + P(L|B) + P(L|C) + P(L|D)}$.
Substituting the given probabilities: $P(A|L) = \frac{0.24}{0.24 + 0.09 + 0.17 + 0.22}$.
$P(A|L) = \frac{0.24}{0.72} = \frac{1}{3}$.
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28
MathematicsMediumMCQKCET · 2026
Match List-$I$ with List-$II$:
| List-$I$ | List-$II$ |
| a) $A$ matrix which is not a square matrix | i) Symmetric matrix |
| b) $A$ square matrix $A' = A$ | ii) Null matrix |
| c) The diagonal elements of a diagonal matrix are same | iii) Rectangular matrix |
| d) $A$ matrix which is both symmetric and skew symmetric | iv) Scalar matrix |
A
$a-iii, b-i, c-ii, d-iv$
B
$a-iii, b-ii, c-iv, d-i$
C
$a-iii, b-i, c-iv, d-ii$
D
$a-iii, b-iv, c-i, d-ii$
Solution
(C) Step $1$: $A$ matrix where the number of rows is not equal to the number of columns is a $Rectangular \ matrix$ (a-iii).
Step $2$: $A$ square matrix $A$ is symmetric if $A' = A$ (b-i).
Step $3$: $A$ diagonal matrix where all diagonal elements are equal is a $Scalar \ matrix$ (c-iv).
Step $4$: $A$ matrix that is both symmetric $(A' = A)$ and skew-symmetric $(A' = -A)$ must satisfy $A = -A$, which implies $2A = 0$, so $A$ is a $Null \ matrix$ (d-ii).
Therefore, the correct matching is $a-iii, b-i, c-iv, d-ii$.
Step $2$: $A$ square matrix $A$ is symmetric if $A' = A$ (b-i).
Step $3$: $A$ diagonal matrix where all diagonal elements are equal is a $Scalar \ matrix$ (c-iv).
Step $4$: $A$ matrix that is both symmetric $(A' = A)$ and skew-symmetric $(A' = -A)$ must satisfy $A = -A$, which implies $2A = 0$, so $A$ is a $Null \ matrix$ (d-ii).
Therefore, the correct matching is $a-iii, b-i, c-iv, d-ii$.
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29
MathematicsMediumMCQKCET · 2026
Consider the following statements:
Statement $I$: If $A$ is a non-singular matrix, then $A^{-1}$ exists.
Statement $II$: If $A$ and $B$ are symmetric matrices of the same order, then $(AB - BA)$ is a skew-symmetric matrix.
Choose the correct option.
Statement $I$: If $A$ is a non-singular matrix, then $A^{-1}$ exists.
Statement $II$: If $A$ and $B$ are symmetric matrices of the same order, then $(AB - BA)$ is a skew-symmetric matrix.
Choose the correct option.
A
Statement $I$ is true and statement $II$ is false
B
Statement $I$ is false and statement $II$ is false
C
Statement $I$ is true and statement $II$ is true
D
Statement $I$ is false and statement $II$ is true
Solution
(C) Step $1$: Statement $I$ is true because for a non-singular matrix, $|A| \neq 0$, which is the necessary and sufficient condition for the existence of the inverse matrix $A^{-1}$.
Step $2$: Statement $II$ is true. Given $A$ and $B$ are symmetric, $A' = A$ and $B' = B$. Consider the transpose of $(AB - BA)$:
$(AB - BA)' = (AB)' - (BA)' = B'A' - A'B' = BA - AB = -(AB - BA)$.
Since the transpose of $(AB - BA)$ is its negative, $(AB - BA)$ is a skew-symmetric matrix.
Step $2$: Statement $II$ is true. Given $A$ and $B$ are symmetric, $A' = A$ and $B' = B$. Consider the transpose of $(AB - BA)$:
$(AB - BA)' = (AB)' - (BA)' = B'A' - A'B' = BA - AB = -(AB - BA)$.
Since the transpose of $(AB - BA)$ is its negative, $(AB - BA)$ is a skew-symmetric matrix.
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30
MathematicsEasyMCQKCET · 2026
$A$ row matrix has only
A
One element
B
One row with one or more columns
C
One column with one or more rows
D
One row and one column
Solution
(B) matrix is called a row matrix if it has only one row. The number of columns can be any positive integer $n \ge 1$. Thus, a row matrix is of the order $1 \times n$.
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31
MathematicsMediumMCQKCET · 2026
Let $X$ be a matrix of order $2 \times n$ and $Z$ be a matrix of order $2 \times p$. If $n = p$, then the order of the matrix $8X - 9Z$ is
A
$2 \times n$
B
$p \times 2$
C
$n \times 3$
D
$pn \times 1$
Solution
(A) For the addition or subtraction of two matrices, they must have the same order.
Given that matrix $X$ has order $2 \times n$ and matrix $Z$ has order $2 \times p$.
Since $n = p$, both matrices have the same order $2 \times n$.
The resulting matrix obtained by the operation $8X - 9Z$ will also have the same order, which is $2 \times n$.
Given that matrix $X$ has order $2 \times n$ and matrix $Z$ has order $2 \times p$.
Since $n = p$, both matrices have the same order $2 \times n$.
The resulting matrix obtained by the operation $8X - 9Z$ will also have the same order, which is $2 \times n$.
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32
MathematicsEasyMCQKCET · 2026
Which of the following statements is correct regarding a determinant?
A
Determinant is a square matrix.
B
Determinant is a number associated with a matrix.
C
Determinant is a unique number associated with a square matrix.
D
Determinant is not defined for a square matrix.
Solution
(C) $1$. $A$ determinant is defined only for square matrices.
$2$. For every square matrix $A$, there exists a unique scalar value associated with it, which is called its determinant, denoted by $|A|$ or $\det(A)$.
$3$. Therefore, the correct statement is that a determinant is a unique number associated with a square matrix.
$2$. For every square matrix $A$, there exists a unique scalar value associated with it, which is called its determinant, denoted by $|A|$ or $\det(A)$.
$3$. Therefore, the correct statement is that a determinant is a unique number associated with a square matrix.
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33
MathematicsMediumMCQKCET · 2026
If $A$ and $B$ are invertible matrices of the same order, then which of the following is not correct?
A
$A(\text{adj } A) = (\text{adj } A)A = AI$
B
$A(\text{adj } A) = (\text{adj } A)A = |A|I$
C
$(AB)^{-1} = B^{-1}A^{-1}$
D
$|A| \neq 0, |B| \neq 0$
Solution
(A) Step $1$: The fundamental property of the adjoint of a matrix $A$ is $A(\text{adj } A) = (\text{adj } A)A = |A|I$, where $|A|$ is the determinant of $A$ and $I$ is the identity matrix.
Step $2$: Option $(A)$ states $A(\text{adj } A) = (\text{adj } A)A = AI$. This is incorrect because it omits the determinant factor $|A|$.
Step $3$: Option $(B)$ is the correct identity. Option $(C)$ is the standard reversal law for the inverse of a product of matrices. Option $(D)$ is the condition for a matrix to be invertible.
Step $4$: Therefore, the incorrect statement is $(A)$.
Step $2$: Option $(A)$ states $A(\text{adj } A) = (\text{adj } A)A = AI$. This is incorrect because it omits the determinant factor $|A|$.
Step $3$: Option $(B)$ is the correct identity. Option $(C)$ is the standard reversal law for the inverse of a product of matrices. Option $(D)$ is the condition for a matrix to be invertible.
Step $4$: Therefore, the incorrect statement is $(A)$.
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34
MathematicsMediumMCQKCET · 2026
If $A$ and $B$ are invertible square matrices of order $n$, then which of the following is not correct?
A
$det(AB) = det(A) \cdot det(B)$
B
$det(kA) = k^n det(A)$
C
$det(A + B) = det(A) + det(B)$
D
$det(A') = 1 / det(A^{-1})$
Solution
(C) $1$. The property $det(AB) = det(A) \cdot det(B)$ is true for all square matrices.
$2$. The property $det(kA) = k^n det(A)$ is true for a matrix of order $n$.
$3$. The property $det(A') = det(A) = 1 / det(A^{-1})$ is true for invertible matrices.
$4$. The determinant of a sum is not generally the sum of the determinants, i.e., $det(A + B) \neq det(A) + det(B)$. Thus, option $C$ is incorrect.
$2$. The property $det(kA) = k^n det(A)$ is true for a matrix of order $n$.
$3$. The property $det(A') = det(A) = 1 / det(A^{-1})$ is true for invertible matrices.
$4$. The determinant of a sum is not generally the sum of the determinants, i.e., $det(A + B) \neq det(A) + det(B)$. Thus, option $C$ is incorrect.
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35
MathematicsDifficultMCQKCET · 2026
The area of the triangle with vertices $(3, 8)$, $(-4, 2)$ and $(5, 1)$ is $\frac{P}{4}$. The value of $P$ is:
A
$\frac{61}{2}$
B
$\frac{2}{61}$
C
$122$
D
$\frac{1}{122}$
Solution
(C) The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$ and $(x_3, y_3)$ is given by $\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Substituting the given vertices $(3, 8)$, $(-4, 2)$ and $(5, 1)$:
$\text{Area} = \frac{1}{2} |3(2 - 1) + (-4)(1 - 8) + 5(8 - 2)|$
$\text{Area} = \frac{1}{2} |3(1) - 4(-7) + 5(6)|$
$\text{Area} = \frac{1}{2} |3 + 28 + 30| = \frac{61}{2}$.
Given that $\text{Area} = \frac{P}{4}$, we have $\frac{P}{4} = \frac{61}{2}$.
Multiplying both sides by $4$, we get $P = \frac{61 \times 4}{2} = 122$.
Substituting the given vertices $(3, 8)$, $(-4, 2)$ and $(5, 1)$:
$\text{Area} = \frac{1}{2} |3(2 - 1) + (-4)(1 - 8) + 5(8 - 2)|$
$\text{Area} = \frac{1}{2} |3(1) - 4(-7) + 5(6)|$
$\text{Area} = \frac{1}{2} |3 + 28 + 30| = \frac{61}{2}$.
Given that $\text{Area} = \frac{P}{4}$, we have $\frac{P}{4} = \frac{61}{2}$.
Multiplying both sides by $4$, we get $P = \frac{61 \times 4}{2} = 122$.
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36
MathematicsMediumMCQKCET · 2026
The system of equations $x + 2y = 3$ and $2x + 3y = 3$ has
A
No solution
B
Unique solution
C
Infinite solutions
D
Only two solutions
Solution
(B) For a system of linear equations $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$, the condition for a unique solution is $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$.
Here, $a_1 = 1, b_1 = 2, c_1 = 3$ and $a_2 = 2, b_2 = 3, c_2 = 3$.
Calculating the ratios: $\frac{a_1}{a_2} = \frac{1}{2}$ and $\frac{b_1}{b_2} = \frac{2}{3}$.
Since $\frac{1}{2} \neq \frac{2}{3}$, the system has a unique solution.
Here, $a_1 = 1, b_1 = 2, c_1 = 3$ and $a_2 = 2, b_2 = 3, c_2 = 3$.
Calculating the ratios: $\frac{a_1}{a_2} = \frac{1}{2}$ and $\frac{b_1}{b_2} = \frac{2}{3}$.
Since $\frac{1}{2} \neq \frac{2}{3}$, the system has a unique solution.
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37
MathematicsDifficultMCQKCET · 2026
If $\vec{a} = 2\hat{i} + 2\hat{j} - \hat{k}$, $\vec{b} = \alpha\hat{i} + \beta\hat{j} + 2\hat{k}$ and $|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}|$, then $\alpha + \beta$ is equal to
A
$2$
B
$-1$
C
$0$
D
$1$
Solution
(D) Given $|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}|$.
Squaring both sides, we get $|\vec{a} + \vec{b}|^2 = |\vec{a} - \vec{b}|^2$.
Using the property $|\vec{u} \pm \vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 \pm 2(\vec{u} \cdot \vec{v})$, we have $|\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b})$.
This simplifies to $4(\vec{a} \cdot \vec{b}) = 0$, which implies $\vec{a} \cdot \vec{b} = 0$.
Calculating the dot product: $(2\hat{i} + 2\hat{j} - \hat{k}) \cdot (\alpha\hat{i} + \beta\hat{j} + 2\hat{k}) = 0$.
$2\alpha + 2\beta - 2 = 0$.
Dividing by $2$, we get $\alpha + \beta - 1 = 0$, so $\alpha + \beta = 1$.
Squaring both sides, we get $|\vec{a} + \vec{b}|^2 = |\vec{a} - \vec{b}|^2$.
Using the property $|\vec{u} \pm \vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 \pm 2(\vec{u} \cdot \vec{v})$, we have $|\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b})$.
This simplifies to $4(\vec{a} \cdot \vec{b}) = 0$, which implies $\vec{a} \cdot \vec{b} = 0$.
Calculating the dot product: $(2\hat{i} + 2\hat{j} - \hat{k}) \cdot (\alpha\hat{i} + \beta\hat{j} + 2\hat{k}) = 0$.
$2\alpha + 2\beta - 2 = 0$.
Dividing by $2$, we get $\alpha + \beta - 1 = 0$, so $\alpha + \beta = 1$.
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38
MathematicsDifficultMCQKCET · 2026
If $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = \hat{j} - \hat{k}$ and $\vec{a} \times \vec{c} = \vec{b}$, $\vec{a} \cdot \vec{c} = 3$, then $\vec{c}$ is
A
$\frac{5}{3}\hat{i} + \frac{2}{3}\hat{j} - \frac{2}{3}\hat{k}$
B
$-\frac{2}{3}\hat{i} + \frac{2}{3}\hat{j} - \hat{k}$
C
$\frac{5}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}$
D
$\frac{5}{3}\hat{i} - \frac{2}{3}\hat{j} - \frac{2}{3}\hat{k}$
Solution
(C) Let $\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}$.
Given $\vec{a} \times \vec{c} = \vec{b}$, we have:
$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ x & y & z \end{vmatrix} = \hat{j} - \hat{k}$
$(z-y)\hat{i} + (x-z)\hat{j} + (y-x)\hat{k} = 0\hat{i} + 1\hat{j} - 1\hat{k}$.
Comparing components: $z-y=0 \Rightarrow z=y$, $x-z=1$, and $y-x=-1$.
Given $\vec{a} \cdot \vec{c} = 3$, we have $x+y+z=3$.
Substituting $z=y$ into $x-z=1$, we get $x-y=1 \Rightarrow x=y+1$.
Substituting $x=y+1$ and $z=y$ into $x+y+z=3$:
$(y+1) + y + y = 3 \Rightarrow 3y = 2 \Rightarrow y = \frac{2}{3}$.
Then $z = \frac{2}{3}$ and $x = \frac{2}{3} + 1 = \frac{5}{3}$.
Thus, $\vec{c} = \frac{5}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}$.
Given $\vec{a} \times \vec{c} = \vec{b}$, we have:
$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ x & y & z \end{vmatrix} = \hat{j} - \hat{k}$
$(z-y)\hat{i} + (x-z)\hat{j} + (y-x)\hat{k} = 0\hat{i} + 1\hat{j} - 1\hat{k}$.
Comparing components: $z-y=0 \Rightarrow z=y$, $x-z=1$, and $y-x=-1$.
Given $\vec{a} \cdot \vec{c} = 3$, we have $x+y+z=3$.
Substituting $z=y$ into $x-z=1$, we get $x-y=1 \Rightarrow x=y+1$.
Substituting $x=y+1$ and $z=y$ into $x+y+z=3$:
$(y+1) + y + y = 3 \Rightarrow 3y = 2 \Rightarrow y = \frac{2}{3}$.
Then $z = \frac{2}{3}$ and $x = \frac{2}{3} + 1 = \frac{5}{3}$.
Thus, $\vec{c} = \frac{5}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}$.
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39
MathematicsDifficultMCQKCET · 2026
The value of $\lambda$ for which the vectors $\vec{a} = 2\hat{i} + \lambda\hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$ are orthogonal is
A
$\frac{5}{2}$
B
$-\frac{5}{2}$
C
$\frac{2}{5}$
D
$-\frac{2}{5}$
Solution
(B) Two vectors are orthogonal if their dot product is zero, i.e., $\vec{a} \cdot \vec{b} = 0$.
Given $\vec{a} = 2\hat{i} + \lambda\hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$.
Calculating the dot product: $(2)(1) + (\lambda)(2) + (1)(3) = 0$.
$2 + 2\lambda + 3 = 0$.
$2\lambda + 5 = 0$.
$2\lambda = -5$.
$\lambda = -\frac{5}{2}$.
Given $\vec{a} = 2\hat{i} + \lambda\hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$.
Calculating the dot product: $(2)(1) + (\lambda)(2) + (1)(3) = 0$.
$2 + 2\lambda + 3 = 0$.
$2\lambda + 5 = 0$.
$2\lambda = -5$.
$\lambda = -\frac{5}{2}$.
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40
MathematicsDifficultMCQKCET · 2026
The angle between the lines whose direction ratios are $a, b, c$ and $b - c, c - a, a - b$ is (in $^\circ$)
A
$90$
B
$60$
C
$30$
D
$0$
Solution
(A) Let the two direction vectors be $\vec{v_1} = a\hat{i} + b\hat{j} + c\hat{k}$ and $\vec{v_2} = (b-c)\hat{i} + (c-a)\hat{j} + (a-b)\hat{k}$.
Calculate the dot product $\vec{v_1} \cdot \vec{v_2} = a(b-c) + b(c-a) + c(a-b)$.
Expand the terms: $ab - ac + bc - ba + ca - cb$.
Simplify the expression: $(ab - ba) + (bc - cb) + (ca - ac) = 0 + 0 + 0 = 0$.
Since the dot product is $0$, the vectors are perpendicular, and the angle between the lines is $90^\circ$.
Calculate the dot product $\vec{v_1} \cdot \vec{v_2} = a(b-c) + b(c-a) + c(a-b)$.
Expand the terms: $ab - ac + bc - ba + ca - cb$.
Simplify the expression: $(ab - ba) + (bc - cb) + (ca - ac) = 0 + 0 + 0 = 0$.
Since the dot product is $0$, the vectors are perpendicular, and the angle between the lines is $90^\circ$.
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41
MathematicsDifficultMCQKCET · 2026
The measure of the angle between the lines $x = k + 1, y = 2k - 1, z = 2k + 3, k \in R$ and $\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-3}{1}$ is
A
$\cos^{-1}(\frac{2}{3})$
B
$\cos^{-1}(\sqrt{\frac{2}{3}})$
C
$\cos^{-1}(\sqrt{\frac{3}{2}})$
D
$\cos^{-1}(\frac{3}{2})$
Solution
(B) The first line is given by $x = 1 + 1k, y = -1 + 2k, z = 3 + 2k$. Its direction ratios are $\vec{a} = (1, 2, 2)$.
The second line is $\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-3}{1}$. Its direction ratios are $\vec{b} = (2, 1, 1)$.
The angle $\theta$ between the lines is given by $\cos \theta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|}$.
$\vec{a} \cdot \vec{b} = (1)(2) + (2)(1) + (2)(1) = 2 + 2 + 2 = 6$.
$|\vec{a}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.
$|\vec{b}| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$.
$\cos \theta = \frac{6}{3 \cdot \sqrt{6}} = \frac{2}{\sqrt{6}} = \sqrt{\frac{4}{6}} = \sqrt{\frac{2}{3}}$.
Therefore, $\theta = \cos^{-1}(\sqrt{\frac{2}{3}})$.
The second line is $\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-3}{1}$. Its direction ratios are $\vec{b} = (2, 1, 1)$.
The angle $\theta$ between the lines is given by $\cos \theta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|}$.
$\vec{a} \cdot \vec{b} = (1)(2) + (2)(1) + (2)(1) = 2 + 2 + 2 = 6$.
$|\vec{a}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.
$|\vec{b}| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$.
$\cos \theta = \frac{6}{3 \cdot \sqrt{6}} = \frac{2}{\sqrt{6}} = \sqrt{\frac{4}{6}} = \sqrt{\frac{2}{3}}$.
Therefore, $\theta = \cos^{-1}(\sqrt{\frac{2}{3}})$.
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42
MathematicsDifficultMCQKCET · 2026
The three points $A(2, 4, 3)$, $B(4, a, 9)$ and $C(10, -1, 7)$ form a right-angled triangle with $\angle B = 90^\circ$. The value of $a$ is
A
$1$ or $4$
B
$-1$ or $4$
C
$1$ or $-4$
D
$-1$ or $-4$
Solution
(B) Step $1$: Find the vectors $\vec{AB}$ and $\vec{BC}$.
$\vec{AB} = (4-2)\hat{i} + (a-4)\hat{j} + (9-3)\hat{k} = 2\hat{i} + (a-4)\hat{j} + 6\hat{k}$.
$\vec{BC} = (10-4)\hat{i} + (-1-a)\hat{j} + (7-9)\hat{k} = 6\hat{i} - (1+a)\hat{j} - 2\hat{k}$.
Step $2$: Since $\angle B = 90^\circ$, the dot product $\vec{AB} \cdot \vec{BC} = 0$.
Step $3$: Calculate the dot product: $2(6) + (a-4)(-(1+a)) + 6(-2) = 0$.
Step $4$: Simplify the equation: $12 - (a-4)(a+1) - 12 = 0$.
Step $5$: Solve for $a$: $-(a-4)(a+1) = 0$, which gives $(a-4)(a+1) = 0$.
Therefore, $a = 4$ or $a = -1$.
$\vec{AB} = (4-2)\hat{i} + (a-4)\hat{j} + (9-3)\hat{k} = 2\hat{i} + (a-4)\hat{j} + 6\hat{k}$.
$\vec{BC} = (10-4)\hat{i} + (-1-a)\hat{j} + (7-9)\hat{k} = 6\hat{i} - (1+a)\hat{j} - 2\hat{k}$.
Step $2$: Since $\angle B = 90^\circ$, the dot product $\vec{AB} \cdot \vec{BC} = 0$.
Step $3$: Calculate the dot product: $2(6) + (a-4)(-(1+a)) + 6(-2) = 0$.
Step $4$: Simplify the equation: $12 - (a-4)(a+1) - 12 = 0$.
Step $5$: Solve for $a$: $-(a-4)(a+1) = 0$, which gives $(a-4)(a+1) = 0$.
Therefore, $a = 4$ or $a = -1$.
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43
MathematicsDifficultMCQKCET · 2026
If $f(x) = \begin{cases} x^2 - 1, & \text{if } x \ge 2 \\ x + 1, & \text{if } x < 2 \end{cases}$, then $\lim_{x \to 1} f(x) + \lim_{x \to 2} f(x) =$
A
$3$
B
$5$
C
$7$
D
$9$
Solution
(B) Step $1$: Calculate $\lim_{x \to 1} f(x)$. Since $1 < 2$, we use $f(x) = x + 1$. Thus, $\lim_{x \to 1} (x + 1) = 1 + 1 = 2$.
Step $2$: Calculate $\lim_{x \to 2} f(x)$. We check the left-hand and right-hand limits.
Left-hand limit: $\lim_{x \to 2^-} f(x) = \lim_{x \to 2} (x + 1) = 2 + 1 = 3$.
Right-hand limit: $\lim_{x \to 2^+} f(x) = \lim_{x \to 2} (x^2 - 1) = 2^2 - 1 = 4 - 1 = 3$.
Since both limits are equal, $\lim_{x \to 2} f(x) = 3$.
Step $3$: The sum is $\lim_{x \to 1} f(x) + \lim_{x \to 2} f(x) = 2 + 3 = 5$.
Step $2$: Calculate $\lim_{x \to 2} f(x)$. We check the left-hand and right-hand limits.
Left-hand limit: $\lim_{x \to 2^-} f(x) = \lim_{x \to 2} (x + 1) = 2 + 1 = 3$.
Right-hand limit: $\lim_{x \to 2^+} f(x) = \lim_{x \to 2} (x^2 - 1) = 2^2 - 1 = 4 - 1 = 3$.
Since both limits are equal, $\lim_{x \to 2} f(x) = 3$.
Step $3$: The sum is $\lim_{x \to 1} f(x) + \lim_{x \to 2} f(x) = 2 + 3 = 5$.
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44
MathematicsDifficultMCQKCET · 2026
If $y = \sqrt[3]{\tan x + y}$, then $\frac{dy}{dx} =$
A
$\frac{\sec^2 x}{3y^2 + 1}$
B
$\frac{\sec^2 x}{3y^2 - 1}$
C
$\frac{\tan x}{3y^2 - 1}$
D
$\frac{\sec^2 x}{3y - 1}$
Solution
(B) Given $y = (\tan x + y)^{1/3}$.
Cube both sides: $y^3 = \tan x + y$.
Differentiating both sides with respect to $x$: $\frac{d}{dx}(y^3) = \frac{d}{dx}(\tan x + y)$.
$3y^2 \frac{dy}{dx} = \sec^2 x + \frac{dy}{dx}$.
Rearranging the terms: $3y^2 \frac{dy}{dx} - \frac{dy}{dx} = \sec^2 x$.
$\frac{dy}{dx}(3y^2 - 1) = \sec^2 x$.
Therefore, $\frac{dy}{dx} = \frac{\sec^2 x}{3y^2 - 1}$.
Cube both sides: $y^3 = \tan x + y$.
Differentiating both sides with respect to $x$: $\frac{d}{dx}(y^3) = \frac{d}{dx}(\tan x + y)$.
$3y^2 \frac{dy}{dx} = \sec^2 x + \frac{dy}{dx}$.
Rearranging the terms: $3y^2 \frac{dy}{dx} - \frac{dy}{dx} = \sec^2 x$.
$\frac{dy}{dx}(3y^2 - 1) = \sec^2 x$.
Therefore, $\frac{dy}{dx} = \frac{\sec^2 x}{3y^2 - 1}$.
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45
MathematicsDifficultMCQKCET · 2026
If $f(x) = \begin{cases} ax + 7, & \text{if } x < 1 \\ 3x - 1, & \text{if } x = 1 \\ \frac{x + 3}{b}, & \text{if } x > 1 \end{cases}$ is continuous at $x = 1$, then
A
$a = 5, b = 2$
B
$a = -5, b = -2$
C
$a = 5, b = -2$
D
$a = -5, b = 2$
Solution
(D) For $f(x)$ to be continuous at $x = 1$, the condition $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)$ must hold.
Step $1$: Calculate $f(1) = 3(1) - 1 = 2$.
Step $2$: Calculate the left-hand limit: $\lim_{x \to 1^-} (ax + 7) = a(1) + 7 = a + 7$. Equating to $f(1)$, we get $a + 7 = 2$, which implies $a = -5$.
Step $3$: Calculate the right-hand limit: $\lim_{x \to 1^+} \frac{x + 3}{b} = \frac{1 + 3}{b} = \frac{4}{b}$. Equating to $f(1)$, we get $\frac{4}{b} = 2$, which implies $b = 2$.
Thus, $a = -5$ and $b = 2$.
Step $1$: Calculate $f(1) = 3(1) - 1 = 2$.
Step $2$: Calculate the left-hand limit: $\lim_{x \to 1^-} (ax + 7) = a(1) + 7 = a + 7$. Equating to $f(1)$, we get $a + 7 = 2$, which implies $a = -5$.
Step $3$: Calculate the right-hand limit: $\lim_{x \to 1^+} \frac{x + 3}{b} = \frac{1 + 3}{b} = \frac{4}{b}$. Equating to $f(1)$, we get $\frac{4}{b} = 2$, which implies $b = 2$.
Thus, $a = -5$ and $b = 2$.
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46
MathematicsDifficultMCQKCET · 2026
The second order derivative of $\cos^{-1}(4x^3 - 3x)$ with respect to $\cos^{-1}(2x^2 - 1)$, where $\frac{1}{2} < x < 1$ is
A
$0$
B
$\frac{-1}{\sqrt{1 - x^2}}$
C
$\frac{3}{2}$
D
$\frac{-3}{2}$
Solution
(A) Let $u = \cos^{-1}(4x^3 - 3x)$ and $v = \cos^{-1}(2x^2 - 1)$.
Given $\frac{1}{2} < x < 1$, we can substitute $x = \cos \theta$, where $0 < \theta < \frac{\pi}{3}$.
Then $u = \cos^{-1}(\cos 3\theta) = 3\theta$ and $v = \cos^{-1}(\cos 2\theta) = 2\theta$.
We need to find $\frac{d^2u}{dv^2}$.
Since $u = 3\theta$ and $v = 2\theta$, we have $\theta = \frac{v}{2}$.
Substituting $\theta$ in $u$, we get $u = 3(\frac{v}{2}) = \frac{3}{2}v$.
Now, $\frac{du}{dv} = \frac{3}{2}$.
Therefore, the second order derivative $\frac{d^2u}{dv^2} = \frac{d}{dv}(\frac{3}{2}) = 0$.
Given $\frac{1}{2} < x < 1$, we can substitute $x = \cos \theta$, where $0 < \theta < \frac{\pi}{3}$.
Then $u = \cos^{-1}(\cos 3\theta) = 3\theta$ and $v = \cos^{-1}(\cos 2\theta) = 2\theta$.
We need to find $\frac{d^2u}{dv^2}$.
Since $u = 3\theta$ and $v = 2\theta$, we have $\theta = \frac{v}{2}$.
Substituting $\theta$ in $u$, we get $u = 3(\frac{v}{2}) = \frac{3}{2}v$.
Now, $\frac{du}{dv} = \frac{3}{2}$.
Therefore, the second order derivative $\frac{d^2u}{dv^2} = \frac{d}{dv}(\frac{3}{2}) = 0$.
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47
MathematicsDifficultMCQKCET · 2026
If $f(x) = \sin^{-1} \left( \frac{2x}{1 + x^2} \right)$, then $f' \left( \frac{1}{2} \right) =$
A
$\frac{8}{5}$
B
$\frac{5}{8}$
C
$\frac{4}{5}$
D
$0$
Solution
(A) Let $x = \tan \theta$. For $|x| \le 1$, $f(x) = \sin^{-1}(\sin 2\theta) = 2\theta = 2\tan^{-1} x$.
Then, $f'(x) = \frac{d}{dx}(2\tan^{-1} x) = \frac{2}{1 + x^2}$.
Substituting $x = \frac{1}{2}$:
$f' \left( \frac{1}{2} \right) = \frac{2}{1 + (1/2)^2} = \frac{2}{1 + 1/4} = \frac{2}{5/4} = \frac{8}{5}$.
Then, $f'(x) = \frac{d}{dx}(2\tan^{-1} x) = \frac{2}{1 + x^2}$.
Substituting $x = \frac{1}{2}$:
$f' \left( \frac{1}{2} \right) = \frac{2}{1 + (1/2)^2} = \frac{2}{1 + 1/4} = \frac{2}{5/4} = \frac{8}{5}$.
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48
MathematicsDifficultMCQKCET · 2026
If $x^{1/2} y^{1/3} = (x + y)^n$ and $x \frac{dy}{dx} - y = 0$, then $n =$
A
$1$
B
$\frac{6}{5}$
C
$\frac{5}{6}$
D
$\frac{4}{9}$
Solution
(C) Given $x^{1/2} y^{1/3} = (x + y)^n$.
Taking natural logarithm on both sides: $\frac{1}{2} \ln x + \frac{1}{3} \ln y = n \ln(x + y)$.
Differentiating with respect to $x$: $\frac{1}{2x} + \frac{1}{3y} \frac{dy}{dx} = \frac{n}{x + y} (1 + \frac{dy}{dx})$.
Given $x \frac{dy}{dx} - y = 0 \implies \frac{dy}{dx} = \frac{y}{x}$.
Substituting $\frac{dy}{dx} = \frac{y}{x}$ into the differentiated equation: $\frac{1}{2x} + \frac{1}{3y} (\frac{y}{x}) = \frac{n}{x + y} (1 + \frac{y}{x})$.
$\frac{1}{2x} + \frac{1}{3x} = \frac{n}{x + y} (\frac{x + y}{x})$.
$\frac{3 + 2}{6x} = \frac{n}{x}$.
$\frac{5}{6x} = \frac{n}{x} \implies n = \frac{5}{6}$.
Taking natural logarithm on both sides: $\frac{1}{2} \ln x + \frac{1}{3} \ln y = n \ln(x + y)$.
Differentiating with respect to $x$: $\frac{1}{2x} + \frac{1}{3y} \frac{dy}{dx} = \frac{n}{x + y} (1 + \frac{dy}{dx})$.
Given $x \frac{dy}{dx} - y = 0 \implies \frac{dy}{dx} = \frac{y}{x}$.
Substituting $\frac{dy}{dx} = \frac{y}{x}$ into the differentiated equation: $\frac{1}{2x} + \frac{1}{3y} (\frac{y}{x}) = \frac{n}{x + y} (1 + \frac{y}{x})$.
$\frac{1}{2x} + \frac{1}{3x} = \frac{n}{x + y} (\frac{x + y}{x})$.
$\frac{3 + 2}{6x} = \frac{n}{x}$.
$\frac{5}{6x} = \frac{n}{x} \implies n = \frac{5}{6}$.
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49
MathematicsDifficultMCQKCET · 2026
In a Maha Kumbh, a drone camera is moving along the curve $3y = x^3 - 3$. The camera captures high-quality pictures when the $y$-coordinate changes $9$ times as fast as the $x$-coordinate. Which of the following is a precise position of the drone at that instant?
A
$(-3, -8)$
B
$(3, -8)$
C
$(3, 8)$
D
$(-3, 8)$
Solution
(C) Given the equation of the path: $3y = x^3 - 3$.
Differentiating both sides with respect to time $t$: $3 \frac{dy}{dt} = 3x^2 \frac{dx}{dt}$.
Given the condition: $\frac{dy}{dt} = 9 \frac{dx}{dt}$.
Substituting the condition into the differentiated equation: $3(9 \frac{dx}{dt}) = 3x^2 \frac{dx}{dt}$.
This simplifies to $27 \frac{dx}{dt} = 3x^2 \frac{dx}{dt}$, which implies $x^2 = 9$, so $x = \pm 3$.
If $x = 3$, then $3y = (3)^3 - 3 = 27 - 3 = 24$, so $y = 8$. The point is $(3, 8)$.
If $x = -3$, then $3y = (-3)^3 - 3 = -27 - 3 = -30$, so $y = -10$. The point is $(-3, -10)$.
Comparing with the given options, $(3, 8)$ is the correct position.
Differentiating both sides with respect to time $t$: $3 \frac{dy}{dt} = 3x^2 \frac{dx}{dt}$.
Given the condition: $\frac{dy}{dt} = 9 \frac{dx}{dt}$.
Substituting the condition into the differentiated equation: $3(9 \frac{dx}{dt}) = 3x^2 \frac{dx}{dt}$.
This simplifies to $27 \frac{dx}{dt} = 3x^2 \frac{dx}{dt}$, which implies $x^2 = 9$, so $x = \pm 3$.
If $x = 3$, then $3y = (3)^3 - 3 = 27 - 3 = 24$, so $y = 8$. The point is $(3, 8)$.
If $x = -3$, then $3y = (-3)^3 - 3 = -27 - 3 = -30$, so $y = -10$. The point is $(-3, -10)$.
Comparing with the given options, $(3, 8)$ is the correct position.
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50
MathematicsDifficultMCQKCET · 2026
$A$ YouTube short video is going viral according to the function $f(t) = -2t^3 + 3t^2 + 5$. At what time $t$ (in hours) does the video get the maximum number of shares?
A
$1 \text{ hour}$
B
$2 \text{ hours}$
C
$0 \text{ hours}$
D
$0.5 \text{ hours}$
Solution
(A) Step $1$: Find the first derivative of the function $f(t) = -2t^3 + 3t^2 + 5$ with respect to $t$.
$f'(t) = \frac{d}{dt}(-2t^3 + 3t^2 + 5) = -6t^2 + 6t$.
Step $2$: Set the first derivative to zero to find the critical points.
$-6t^2 + 6t = 0 \Rightarrow -6t(t - 1) = 0$.
This gives $t = 0$ and $t = 1$.
Step $3$: Find the second derivative to determine the nature of the critical points.
$f''(t) = \frac{d}{dt}(-6t^2 + 6t) = -12t + 6$.
Step $4$: Evaluate the second derivative at the critical points.
For $t = 1$, $f''(1) = -12(1) + 6 = -6$.
Since $f''(1) < 0$, the function has a local maximum at $t = 1$ hour.
$f'(t) = \frac{d}{dt}(-2t^3 + 3t^2 + 5) = -6t^2 + 6t$.
Step $2$: Set the first derivative to zero to find the critical points.
$-6t^2 + 6t = 0 \Rightarrow -6t(t - 1) = 0$.
This gives $t = 0$ and $t = 1$.
Step $3$: Find the second derivative to determine the nature of the critical points.
$f''(t) = \frac{d}{dt}(-6t^2 + 6t) = -12t + 6$.
Step $4$: Evaluate the second derivative at the critical points.
For $t = 1$, $f''(1) = -12(1) + 6 = -6$.
Since $f''(1) < 0$, the function has a local maximum at $t = 1$ hour.
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51
MathematicsDifficultMCQKCET · 2026
If $\int x f(x) dx + \frac{f(x)}{2} = 0$, then $f(x)$ is equal to
A
$C e^{-x^2}$
B
$C e^{x^2}$
C
$C e^{-2x^2}$
D
$C e^{2x^2}$
Solution
(A) Given the equation: $\int x f(x) dx + \frac{f(x)}{2} = 0$.
Differentiating both sides with respect to $x$ using the Fundamental Theorem of Calculus:
$x f(x) + \frac{1}{2} f'(x) = 0$.
Rearranging the terms to separate variables:
$\frac{f'(x)}{f(x)} = -2x$.
Integrating both sides with respect to $x$:
$\int \frac{f'(x)}{f(x)} dx = \int -2x dx$.
$\ln|f(x)| = -x^2 + C_1$.
Exponentiating both sides:
$f(x) = e^{-x^2 + C_1} = e^{C_1} \cdot e^{-x^2}$.
Letting $e^{C_1} = C$, we get $f(x) = C e^{-x^2}$.
Differentiating both sides with respect to $x$ using the Fundamental Theorem of Calculus:
$x f(x) + \frac{1}{2} f'(x) = 0$.
Rearranging the terms to separate variables:
$\frac{f'(x)}{f(x)} = -2x$.
Integrating both sides with respect to $x$:
$\int \frac{f'(x)}{f(x)} dx = \int -2x dx$.
$\ln|f(x)| = -x^2 + C_1$.
Exponentiating both sides:
$f(x) = e^{-x^2 + C_1} = e^{C_1} \cdot e^{-x^2}$.
Letting $e^{C_1} = C$, we get $f(x) = C e^{-x^2}$.
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52
MathematicsDifficultMCQKCET · 2026
One of the possible functions $f(x)$ which satisfies $\int_{-2}^{2} f(x) dx = 0$ is
A
$\log \left( \frac{2+x}{2-x} \right)$
B
$\sin(2+x)$
C
$2x^3 + 2x + 1$
D
$2x \tan x$
Solution
(A) The integral of an odd function over a symmetric interval $[-a, a]$ is zero.
Let $f(x) = \log \left( \frac{2+x}{2-x} \right)$.
Then $f(-x) = \log \left( \frac{2-x}{2-(-x)} \right) = \log \left( \frac{2-x}{2+x} \right) = \log \left( \left( \frac{2+x}{2-x} \right)^{-1} \right) = -\log \left( \frac{2+x}{2-x} \right) = -f(x)$.
Since $f(-x) = -f(x)$, the function is odd, and therefore $\int_{-2}^{2} f(x) dx = 0$.
Let $f(x) = \log \left( \frac{2+x}{2-x} \right)$.
Then $f(-x) = \log \left( \frac{2-x}{2-(-x)} \right) = \log \left( \frac{2-x}{2+x} \right) = \log \left( \left( \frac{2+x}{2-x} \right)^{-1} \right) = -\log \left( \frac{2+x}{2-x} \right) = -f(x)$.
Since $f(-x) = -f(x)$, the function is odd, and therefore $\int_{-2}^{2} f(x) dx = 0$.
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53
MathematicsDifficultMCQKCET · 2026
$\int_{a-6}^{b-6} f(x+6) dx$ is equal to
A
$\int_{a}^{b} f(x-6) dx$
B
$\int_{a}^{b} f(x+6) dx$
C
$\int_{a}^{b} f(x) dx$
D
$\int_{a}^{b} f(-x) dx$
Solution
(C) Let $t = x+6$, then $dt = dx$.
When $x = a-6$, $t = (a-6)+6 = a$.
When $x = b-6$, $t = (b-6)+6 = b$.
Substituting these into the integral, we get $\int_{a}^{b} f(t) dt$.
Since the variable of integration is a dummy variable, $\int_{a}^{b} f(t) dt = \int_{a}^{b} f(x) dx$.
When $x = a-6$, $t = (a-6)+6 = a$.
When $x = b-6$, $t = (b-6)+6 = b$.
Substituting these into the integral, we get $\int_{a}^{b} f(t) dt$.
Since the variable of integration is a dummy variable, $\int_{a}^{b} f(t) dt = \int_{a}^{b} f(x) dx$.
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54
MathematicsDifficultMCQKCET · 2026
If $n$ is a natural number, then $\int \frac{\sin^n x}{\cos^{n+2} x} dx =$
A
$\frac{\tan^{n-1} x}{n-1} + C$
B
$\frac{\tan^n x}{n} + C$
C
$\frac{\tan^{n+2} x}{n+2} + C$
D
$\frac{\tan^{n+1} x}{n+1} + C$
Solution
(D) We have the integral $I = \int \frac{\sin^n x}{\cos^{n+2} x} dx$.
Rewrite the integrand as $I = \int \left( \frac{\sin x}{\cos x} \right)^n \cdot \frac{1}{\cos^2 x} dx$.
Since $\frac{\sin x}{\cos x} = \tan x$ and $\frac{1}{\cos^2 x} = \sec^2 x$, we get $I = \int \tan^n x \sec^2 x dx$.
Let $t = \tan x$, then $dt = \sec^2 x dx$.
Substituting these into the integral, we get $I = \int t^n dt$.
Using the power rule for integration, $I = \frac{t^{n+1}}{n+1} + C$.
Substituting $t = \tan x$ back, we get $I = \frac{\tan^{n+1} x}{n+1} + C$.
Rewrite the integrand as $I = \int \left( \frac{\sin x}{\cos x} \right)^n \cdot \frac{1}{\cos^2 x} dx$.
Since $\frac{\sin x}{\cos x} = \tan x$ and $\frac{1}{\cos^2 x} = \sec^2 x$, we get $I = \int \tan^n x \sec^2 x dx$.
Let $t = \tan x$, then $dt = \sec^2 x dx$.
Substituting these into the integral, we get $I = \int t^n dt$.
Using the power rule for integration, $I = \frac{t^{n+1}}{n+1} + C$.
Substituting $t = \tan x$ back, we get $I = \frac{\tan^{n+1} x}{n+1} + C$.
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55
MathematicsDifficultMCQKCET · 2026
$\int e^{-x \log 2} 2^x dx =$
A
$\log x + C$
B
$x + C$
C
$\frac{1}{x} + C$
D
$\frac{x^2}{2} + C$
Solution
(B) Step $1$: Simplify the integrand using the property $e^{\log a} = a$. We have $e^{-x \log 2} = (e^{\log 2})^{-x} = 2^{-x}$.
Step $2$: Substitute this into the integral: $\int 2^{-x} \cdot 2^x dx$.
Step $3$: Simplify the product: $\int 2^{-x+x} dx = \int 2^0 dx = \int 1 dx$.
Step $4$: Integrate with respect to $x$: $\int 1 dx = x + C$.
Step $2$: Substitute this into the integral: $\int 2^{-x} \cdot 2^x dx$.
Step $3$: Simplify the product: $\int 2^{-x+x} dx = \int 2^0 dx = \int 1 dx$.
Step $4$: Integrate with respect to $x$: $\int 1 dx = x + C$.
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56
MathematicsDifficultMCQKCET · 2026
The area of the region bounded by the curve $y^2 = x^3$, the $y$-axis and the lines $y = 1$ and $y = 8$ is
A
$\frac{155}{3} \text{ sq. units}$
B
$\frac{93}{5} \text{ sq. units}$
C
$93 \text{ sq. units}$
D
$155 \text{ sq. units}$
Solution
(B) Given the curve $y^2 = x^3$, we express $x$ in terms of $y$ as $x = y^{2/3}$.
The area $A$ bounded by the curve, the $y$-axis, and the lines $y = 1$ and $y = 8$ is given by the integral:
$A = \int_{1}^{8} x \, dy = \int_{1}^{8} y^{2/3} \, dy$
Integrating $y^{2/3}$ with respect to $y$:
$A = \left[ \frac{y^{(2/3) + 1}}{(2/3) + 1} \right]_{1}^{8} = \left[ \frac{y^{5/3}}{5/3} \right]_{1}^{8} = \left[ \frac{3}{5} y^{5/3} \right]_{1}^{8}$
Evaluating the definite integral:
$A = \frac{3}{5} [8^{5/3} - 1^{5/3}]$
Since $8^{5/3} = (2^3)^{5/3} = 2^5 = 32$ and $1^{5/3} = 1$:
$A = \frac{3}{5} [32 - 1] = \frac{3}{5} \times 31 = \frac{93}{5} \text{ sq. units}$.
The area $A$ bounded by the curve, the $y$-axis, and the lines $y = 1$ and $y = 8$ is given by the integral:
$A = \int_{1}^{8} x \, dy = \int_{1}^{8} y^{2/3} \, dy$
Integrating $y^{2/3}$ with respect to $y$:
$A = \left[ \frac{y^{(2/3) + 1}}{(2/3) + 1} \right]_{1}^{8} = \left[ \frac{y^{5/3}}{5/3} \right]_{1}^{8} = \left[ \frac{3}{5} y^{5/3} \right]_{1}^{8}$
Evaluating the definite integral:
$A = \frac{3}{5} [8^{5/3} - 1^{5/3}]$
Since $8^{5/3} = (2^3)^{5/3} = 2^5 = 32$ and $1^{5/3} = 1$:
$A = \frac{3}{5} [32 - 1] = \frac{3}{5} \times 31 = \frac{93}{5} \text{ sq. units}$.
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57
MathematicsDifficultMCQKCET · 2026
The sum of the squares of the order and degree (if defined) of the differential equation $2y' + (y'')^2 = \sqrt{y'' - 3}$ is
A
$3$
B
$20$
C
$8$
D
$16$
Solution
(B) Given differential equation: $2y' + (y'')^2 = \sqrt{y'' - 3}$.
To define the degree, we must eliminate the radical by squaring both sides: $(2y' + (y'')^2)^2 = y'' - 3$.
The highest order derivative present is $y''$, so the order is $2$.
The highest power of the highest order derivative after making the equation a polynomial in derivatives is $4$ (since $(y'')^2$ squared becomes $(y'')^4$). Thus, the degree is $4$.
The sum of the squares of the order and degree is $2^2 + 4^2 = 4 + 16 = 20$.
To define the degree, we must eliminate the radical by squaring both sides: $(2y' + (y'')^2)^2 = y'' - 3$.
The highest order derivative present is $y''$, so the order is $2$.
The highest power of the highest order derivative after making the equation a polynomial in derivatives is $4$ (since $(y'')^2$ squared becomes $(y'')^4$). Thus, the degree is $4$.
The sum of the squares of the order and degree is $2^2 + 4^2 = 4 + 16 = 20$.
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58
MathematicsMediumMCQKCET · 2026
The domain of the function $f(x) = \sqrt{\frac{x-7}{9-x}}$ is
A
$(7, 9)$
B
$[7, 9)$
C
$[7, 9]$
D
$(7, 9]$
Solution
(B) For the function to be defined, the expression under the square root must be non-negative: $\frac{x-7}{9-x} \ge 0$.
Multiply both sides by $-1$ and reverse the inequality sign: $\frac{x-7}{x-9} \le 0$.
The critical points are $x = 7$ and $x = 9$.
Since the denominator cannot be zero, $x \neq 9$.
Testing the intervals $(-\infty, 7]$, $[7, 9)$, and $[9, \infty)$, the inequality $\frac{x-7}{x-9} \le 0$ holds for $x \in [7, 9)$.
Thus, the domain is $[7, 9)$.
Multiply both sides by $-1$ and reverse the inequality sign: $\frac{x-7}{x-9} \le 0$.
The critical points are $x = 7$ and $x = 9$.
Since the denominator cannot be zero, $x \neq 9$.
Testing the intervals $(-\infty, 7]$, $[7, 9)$, and $[9, \infty)$, the inequality $\frac{x-7}{x-9} \le 0$ holds for $x \in [7, 9)$.
Thus, the domain is $[7, 9)$.
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59
MathematicsMediumMCQKCET · 2026
If $f(x) = (x + 1)^2$ for $x \ge 1$, and $g(x)$ is a function whose graph is the reflection of the graph of $f(x)$ in the line $y = x$, then $g(x)$ is:
A
$-\sqrt{x} - 1$
B
$\sqrt{x} + 1$
C
$\sqrt{x} - 1$
D
$\sqrt{x - 1}$
Solution
(C) The reflection of the graph of a function $f(x)$ in the line $y = x$ is its inverse function $f^{-1}(x)$.
Let $y = f(x) = (x + 1)^2$.
To find the inverse, solve for $x$ in terms of $y$:
$y = (x + 1)^2$
$\sqrt{y} = x + 1$ (since $x \ge 1$, $x+1 \ge 2$, so we take the positive root)
$x = \sqrt{y} - 1$.
Thus, $g(x) = f^{-1}(x) = \sqrt{x} - 1$.
Let $y = f(x) = (x + 1)^2$.
To find the inverse, solve for $x$ in terms of $y$:
$y = (x + 1)^2$
$\sqrt{y} = x + 1$ (since $x \ge 1$, $x+1 \ge 2$, so we take the positive root)
$x = \sqrt{y} - 1$.
Thus, $g(x) = f^{-1}(x) = \sqrt{x} - 1$.
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60
MathematicsMediumMCQKCET · 2026
If $\sin^{-1} x + \sin^{-1} y = \pi/2$, then $x^2$ is equal to
A
$1 - y^2$
B
$1 + y^2$
C
$\sqrt{1 - y^2}$
D
$\sqrt{1 + y^2}$
Solution
(A) Given $\sin^{-1} x + \sin^{-1} y = \pi/2$.
We know that $\sin^{-1} y + \cos^{-1} y = \pi/2$, so $\sin^{-1} y = \pi/2 - \cos^{-1} y$.
Substituting this into the given equation: $\sin^{-1} x = \pi/2 - \sin^{-1} y = \cos^{-1} y$.
Taking $\sin$ on both sides: $x = \sin(\cos^{-1} y)$.
Using the identity $\sin(\cos^{-1} y) = \sqrt{1 - y^2}$, we get $x = \sqrt{1 - y^2}$.
Squaring both sides, we get $x^2 = 1 - y^2$.
We know that $\sin^{-1} y + \cos^{-1} y = \pi/2$, so $\sin^{-1} y = \pi/2 - \cos^{-1} y$.
Substituting this into the given equation: $\sin^{-1} x = \pi/2 - \sin^{-1} y = \cos^{-1} y$.
Taking $\sin$ on both sides: $x = \sin(\cos^{-1} y)$.
Using the identity $\sin(\cos^{-1} y) = \sqrt{1 - y^2}$, we get $x = \sqrt{1 - y^2}$.
Squaring both sides, we get $x^2 = 1 - y^2$.
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