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Mathematics - 1 — MH-CET Full Length Paper MCQs with Solutions
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Sample questions with solutions
Q1 — easy
The contrapositive of the statement: "If two triangles are congruent, then they are similar" is:
A. If two triangles are not similar, then they are not congruent. ✓ Correct
B. If two triangles are similar, then they are congruent.
C. If two triangles are not congruent, then they are not similar.
D. Two triangles are congruent if and only if they are similar.
Solution: For conditional $p \to q$, the contrapositive is $\sim q \to \sim p$. So: "If not similar, then not congruent."
Q2 — easy
Which of the following compound propositions is a tautology?
A. $(p \vee q) \wedge \sim q$
B. $p \to (p \vee q)$ ✓ Correct
C. $(p \wedge q) \to \sim p$
D. $p \wedge (\sim p \vee q)$
Solution: $p \to (p \vee q) \equiv \sim p \vee (p \vee q) \equiv (\sim p \vee p) \vee q \equiv T \vee q \equiv T$. Hence tautology.
Q3 — easy
If $A = \begin{bmatrix} 2 & -1 \\ 4 & 3 \end{bmatrix}$ and $A^{-1} = \frac{1}{k}\begin{bmatrix} 3 & 1 \\ -4 & 2 \end{bmatrix}$, then the value of the scalar $k$ is:
A. $2$
B. $5$
C. $10$ ✓ Correct
D. $8$
Solution: $\det(A) = (2)(3) - (-1)(4) = 6 + 4 = 10$. Since $A^{-1} = \frac{1}{\det(A)}\text{adj}(A)$, $k = 10$.
Q4 — easy
The direction cosines of the vector $\vec{r} = 2\hat{i} - 3\hat{j} + 6\hat{k}$ are:
A. $\frac{2}{7}, -\frac{3}{7}, \frac{6}{7}$ ✓ Correct
B. $\frac{2}{49}, -\frac{3}{49}, \frac{6}{49}$
C. $\frac{2}{\sqrt{11}}, -\frac{3}{\sqrt{11}}, \frac{6}{\sqrt{11}}$
D. $\frac{1}{7}, -\frac{2}{7}, \frac{3}{7}$
Solution: $|\vec{r}| = \sqrt{4 + 9 + 36} = 7$. Direction cosines: $\frac{2}{7}, -\frac{3}{7}, \frac{6}{7}$.
Q5 — easy
The perpendicular distance of the point $P(1, 2, 3)$ from the plane $2x - y + 2z + 3 = 0$ is:
A. $1$ unit
B. $2$ units
C. $3$ units ✓ Correct
D. $4$ units
Solution: $d = \frac{|2(1) - 1(2) + 2(3) + 3|}{\sqrt{4 + 1 + 4}} = \frac{9}{3} = 3$ units.
Q6 — easy
If $y = \tan^{-1}\left(\frac{\cos x - \sin x}{\cos x + \sin x}\right)$, then $\frac{dy}{dx}$ is:
A. $1$
B. $-1$ ✓ Correct
C. $\frac{1}{1+x^2}$
D. $0$
Solution: $y = \tan^{-1}(\tan(\frac{\pi}{4} - x)) = \frac{\pi}{4} - x$. Thus $\frac{dy}{dx} = -1$.
Q7 — easy
The equation of the tangent to the curve $y = 2x^2 - 3x - 2$ at the point where the curve crosses the positive $x$-axis is:
A. $5x - y - 10 = 0$ ✓ Correct
B. $5x + y - 10 = 0$
C. $x - 5y - 2 = 0$
D. $3x - y - 6 = 0$
Solution: At $y = 0$: $2x^2 - 3x - 2 = 0 \implies x = 2$. Slope $= 4(2) - 3 = 5$. Tangent: $y = 5(x-2) \implies 5x - y - 10 = 0$.
Q8 — easy
The function $f(x) = x^3 - 3x^2 + 3x - 100$ is:
A. Strictly increasing on $\mathbb{R}$ ✓ Correct
B. Strictly decreasing on $\mathbb{R}$
C. Increasing for $x > 1$ and decreasing for $x < 1$
D. Neither increasing nor decreasing
Solution: $f'(x) = 3x^2 - 6x + 3 = 3(x-1)^2 \ge 0$ for all $x$, vanishing only at $x = 1$. Hence strictly increasing on $\mathbb{R}$.
Q9 — easy
The value of the definite integral $\int_0^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$ is:
A. $\pi$
B. $\frac{\pi}{2}$
C. $\frac{\pi}{4}$ ✓ Correct
D. $0$
Solution: Using the property $I = \int_0^a f(x)dx = \int_0^a f(a-x)dx$: $2I = \int_0^{\pi/2} 1\,dx = \frac{\pi}{2} \implies I = \frac{\pi}{4}$.
Q10 — easy
The value of the definite integral $\int_{-1}^1 \ln\left(\frac{2-x}{2+x}\right) dx$ is:
A. $0$ ✓ Correct
B. $2\ln 2$
C. $\ln 3$
D. $1$
Solution: $f(x) = \ln\frac{2-x}{2+x}$ is odd since $f(-x) = -f(x)$. So $\int_{-1}^1 f(x)\,dx = 0$.
Q11 — easy
The area of the region bounded by the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ is:
A. $12\pi$ sq. units ✓ Correct
B. $24\pi$ sq. units
C. $7\pi$ sq. units
D. $144\pi$ sq. units
Solution: Area of ellipse $= \pi ab = \pi(4)(3) = 12\pi$ sq. units.
Q12 — easy
The order and degree of the differential equation $\left[1 + \left(\frac{dy}{dx}\right)^2\right]^{3/2} = 5\frac{d^2y}{dx^2}$ are respectively:
A. $2$ and $3$
B. $2$ and $2$ ✓ Correct
C. $3$ and $2$
D. $1$ and $3$
Solution: Squaring: $[1 + (y')^2]^3 = 25(y'')^2$. Highest derivative is $y''$ (order 2), its power is 2 (degree 2).
Q13 — easy
The integrating factor of the linear differential equation $\frac{dy}{dx} + y\sec x = \tan x$ is:
A. $\sec x + \tan x$ ✓ Correct
B. $\sec x \tan x$
C. $\ln|\sec x + \tan x|$
D. $\cos x$
Solution: I.F. $= e^{\int \sec x\,dx} = e^{\ln|\sec x + \tan x|} = \sec x + \tan x$.
Q14 — easy
The rate of decay of a radioactive substance is proportional to the amount $N$ present at that time. If the half-life of the substance is $1600$ years, the percentage of substance remaining after $4800$ years is:
A. $25\%$
B. $12.5\%$ ✓ Correct
C. $6.25\%$
D. $50\%$
Solution: Number of half-lives $= 4800/1600 = 3$. Fraction remaining $= (1/2)^3 = 1/8 = 12.5\%$.
Q15 — easy
A random variable $X$ has the probability distribution given by $P(X = x) = c(x+1)$ for $x = 0, 1, 2, 3$. The value of constant $c$ is:
A. $\frac{1}{10}$ ✓ Correct
B. $\frac{1}{6}$
C. $\frac{1}{4}$
D. $\frac{1}{8}$
Solution: $\sum P = c(1 + 2 + 3 + 4) = 10c = 1 \implies c = 1/10$.
Q16 — easy
For a discrete random variable $X$, if $E(X) = 3$ and $E(X^2) = 11$, then the standard deviation $\sigma_X$ is:
A. $2$
B. $\sqrt{2}$ ✓ Correct
C. $8$
D. $4$
Solution: $\text{Var}(X) = E(X^2) - [E(X)]^2 = 11 - 9 = 2$. $\sigma = \sqrt{2}$.
Q17 — easy
A fair die is thrown 6 times independently. The probability of getting an odd number exactly 4 times is:
A. $\frac{15}{64}$ ✓ Correct
B. $\frac{5}{32}$
C. $\frac{15}{32}$
D. $\frac{3}{16}$
Solution: $P = \binom{6}{4}(1/2)^4(1/2)^2 = 15/64$.
Q18 — easy
If the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and $\frac{x-2}{3} = \frac{y-4}{4} = \frac{z-5}{5}$ are coplanar, the shortest distance between them is:
A. $1$ unit
B. $0$ units ✓ Correct
C. $2$ units
D. $\sqrt{6}$ units
Solution: For coplanar lines, the shortest distance is zero since they either intersect or are parallel in the same plane.
Q19 — easy
If $\int \frac{2x+3}{(x-1)(x+2)} dx = A\ln|x-1| + B\ln|x+2| + C$, then the ordered pair $(A, B)$ is:
A. $\left(\frac{5}{3}, \frac{1}{3}\right)$ ✓ Correct
B. $\left(\frac{5}{3}, -\frac{1}{3}\right)$
C. $(2, 3)$
D. $(1, 1)$
Solution: Partial fractions: $A = \frac{2(1)+3}{1+2} = \frac{5}{3}$, $B = \frac{2(-2)+3}{-2-1} = \frac{-1}{-3} = \frac{1}{3}$.
Q20 — easy
The perpendicular distance between the two parallel straight lines $3x + 4y - 9 = 0$ and $6x + 8y + 12 = 0$ is:
A. $3$ units ✓ Correct
B. $1.5$ units
C. $2.1$ units
D. $4.2$ units
Solution: Rewrite line 2: $3x + 4y + 6 = 0$. Distance $= \frac{|-9-6|}{\sqrt{9+16}} = \frac{15}{5} = 3$ units.
Q21 — easy
The length of the intercept intercepted on the $y$-axis by the circle $x^2 + y^2 - 4x - 6y - 12 = 0$ is:
A. $\sqrt{21}$
B. $2\sqrt{21}$ ✓ Correct
C. $6$
D. $8$
Solution: $y$-intercept $= 2\sqrt{f^2 - c}$ where $f = -3$, $c = -12$. $= 2\sqrt{9 + 12} = 2\sqrt{21}$.
Q22 — easy
The eccentricity of the ellipse $9x^2 + 25y^2 = 225$ is:
A. $\frac{4}{5}$ ✓ Correct
B. $\frac{3}{5}$
C. $\frac{16}{25}$
D. $\frac{2}{3}$
Solution: $\frac{x^2}{25} + \frac{y^2}{9} = 1$. $e = \sqrt{1 - 9/25} = \sqrt{16/25} = 4/5$.
Q23 — easy
The equation of the directrix of the parabola $y^2 - 4y - 8x - 4 = 0$ is:
A. $x = -3$ ✓ Correct
B. $x = 1$
C. $x = -1$
D. $x = 3$
Solution: $(y-2)^2 = 8(x+1)$. Form $Y^2 = 4aX$ with $a = 2$. Directrix: $X = -a \implies x + 1 = -2 \implies x = -3$.
Q24 — easy
If the coefficient of variation (C.V.) of a distribution is $40\%$ and its mean is $25$, then the variance of the distribution is:
A. $10$
B. $100$ ✓ Correct
C. $16$
D. $25$
Solution: C.V. $= (\sigma/\bar{x}) \times 100 \implies 40 = (\sigma/25) \times 100 \implies \sigma = 10$. Variance $= 100$.
Q25 — easy
Two cards are drawn successively without replacement from a well-shuffled pack of 52 cards. The probability that both drawn cards are Aces is:
A. $\frac{1}{221}$ ✓ Correct
B. $\frac{1}{169}$
C. $\frac{4}{663}$
D. $\frac{2}{221}$
Solution: $P = \frac{4}{52} \times \frac{3}{51} = \frac{12}{2652} = \frac{1}{221}$.
Q26 — easy
The total number of 4-digit numbers that can be formed using the digits $1, 2, 3, 4, 5, 6$ without repetition such that each number is divisible by 5 is:
A. $120$
B. $60$ ✓ Correct
C. $24$
D. $36$
Solution: Units digit must be 5 (only choice). Remaining 3 positions from 5 digits: $^5P_3 = 60$.
Q27 — easy
The value of the limit $\lim_{x \to 0} \frac{e^{3x} - \cos 2x}{x}$ is:
A. $1$
B. $3$ ✓ Correct
C. $5$
D. $0$
Solution: $\lim \frac{e^{3x}-1}{x} + \lim \frac{1-\cos 2x}{x} = 3 + 0 = 3$.
Q28 — hard
If the acute angle between the pair of straight lines given by $ax^2 + 2hxy + by^2 = 0$ is $\frac{\pi}{4}$, then $4h^2$ is equal to:
A. $(a + b)^2 + 4ab$
B. $(a + 3b)(3a + b)$ ✓ Correct
C. $a^2 + 6ab + b^2$
D. $(a - b)^2$
Solution: $\tan\frac{\pi}{4} = 1 = \frac{2\sqrt{h^2 - ab}}{|a+b|}$. Squaring: $(a+b)^2 = 4(h^2 - ab)$, so $4h^2 = (a+b)^2 + 4ab = (a+3b)(3a+b)$.
Q29 — hard
If $x = a(\theta - \sin\theta)$ and $y = a(1 - \cos\theta)$, then the value of $\frac{d^2y}{dx^2}$ at $\theta = \frac{\pi}{2}$ is:
A. $\frac{1}{a}$
B. $-\frac{1}{a}$ ✓ Correct
C. $\frac{1}{2a}$
D. $-\frac{1}{2a}$
Solution: $\frac{dy}{dx} = \cot(\theta/2)$. $\frac{d^2y}{dx^2} = -\frac{\csc^2(\theta/2)}{2a(1-\cos\theta)}$. At $\theta = \pi/2$: $= -\frac{2}{2a \cdot 1} = -\frac{1}{a}$.
Q30 — hard
The integral $\int e^x \left(\frac{1 + \sin x}{1 + \cos x}\right) dx$ is equal to:
A. $e^x \tan x + C$
B. $e^x \tan\left(\frac{x}{2}\right) + C$ ✓ Correct
C. $e^x \sec^2\left(\frac{x}{2}\right) + C$
D. $\frac{1}{2}e^x \tan\left(\frac{x}{2}\right) + C$
Solution: The expression simplifies to $\int e^x[\tan(x/2) + \frac{1}{2}\sec^2(x/2)]dx$, which is of the form $\int e^x[f + f']dx = e^x f + C = e^x\tan(x/2) + C$.