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Mathematics - 4 — MH-CET Full Length Paper MCQs with Solutions
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Sample questions with solutions
Q1 — easy
If $p$ and $q$ are two propositions, then the logical statement $\sim(p \leftrightarrow q)$ is equivalent to:
A. $(p \wedge \sim q) \vee (\sim p \wedge q)$ ✓ Correct
B. $(p \vee \sim q) \wedge (\sim p \vee q)$
C. $(p \wedge q) \vee (\sim p \wedge \sim q)$
D. $(\sim p \wedge \sim q) \wedge (p \vee q)$
Solution: $\sim(p \leftrightarrow q) \equiv (p \wedge \sim q) \vee (\sim p \wedge q)$, which is the XOR operation.
Q2 — easy
The inverse of the statement "If a number is divisible by 6, then it is divisible by 3" is:
A. If a number is not divisible by 6, then it is not divisible by 3. ✓ Correct
B. If a number is not divisible by 3, then it is not divisible by 6.
C. If a number is divisible by 3, then it is divisible by 6.
D. A number is divisible by 6 if and only if it is divisible by 3.
Solution: Inverse of $p \to q$ is $\sim p \to \sim q$: "If not divisible by 6, then not divisible by 3."
Q3 — easy
If the pair of straight lines given by $ax^2 + 2hxy + by^2 = 0$ makes angles $\alpha$ and $\beta$ with the $x$-axis such that $\alpha + \beta = \frac{\pi}{2}$, then:
A. $a = b$ ✓ Correct
B. $a + b = 0$
C. $h = 0$
D. $a = -b$
Solution: $\tan\alpha\tan\beta = 1$ (since $\beta = \pi/2 - \alpha$). Product of slopes $= a/b = 1 \implies a = b$.
Q4 — easy
The area of the triangle formed by the pair of straight lines $x^2 - 4y^2 = 0$ and the line $x = 4$ is:
A. $8$ sq. units ✓ Correct
B. $16$ sq. units
C. $4$ sq. units
D. $2$ sq. units
Solution: Lines: $y = x/2$ and $y = -x/2$. At $x = 4$: $y = \pm 2$. Area $= \frac{1}{2}(4)(4) = 8$.
Q5 — easy
If $|\vec{a}| = 3$, $|\vec{b}| = 4$, and $|\vec{a} - \vec{b}| = 5$, then the value of $|\vec{a} + \vec{b}|$ is:
A. $5$ ✓ Correct
B. $6$
C. $\sqrt{50}$
D. $7$
Solution: Parallelogram law: $|\vec{a}+\vec{b}|^2 + 25 = 2(9+16) = 50 \implies |\vec{a}+\vec{b}| = 5$.
Q6 — easy
The angle of intersection between the two curves $y^2 = 4x$ and $x^2 = 4y$ at the origin $(0, 0)$ is:
A. $\frac{\pi}{2}$ ✓ Correct
B. $\frac{\pi}{4}$
C. $\tan^{-1}\left(\frac{3}{4}\right)$
D. $\frac{\pi}{3}$
Solution: At the origin, tangent to $y^2 = 4x$ is the $y$-axis and tangent to $x^2 = 4y$ is the $x$-axis. Angle $= \pi/2$.
Q7 — easy
A stone dropped into a quiet lake produces concentric circular ripples. The radius of the outer ripple increases at a uniform rate of $4$ cm/s. When the radius is $10$ cm, the rate at which the enclosed area increases is:
A. $80\pi$ cm$^2$/s ✓ Correct
B. $40\pi$ cm$^2$/s
C. $100\pi$ cm$^2$/s
D. $160\pi$ cm$^2$/s
Solution: $dA/dt = 2\pi r \cdot dr/dt = 2\pi(10)(4) = 80\pi$ cm$^2$/s.
Q8 — easy
The absolute minimum value of $f(x) = 2x^3 - 3x^2 - 12x + 1$ on the closed interval $[-2, 3]$ is:
A. $-19$ ✓ Correct
B. $-20$
C. $-8$
D. $8$
Solution: $f'(x) = 6(x-2)(x+1) = 0$ at $x = -1, 2$. $f(-2) = -3$, $f(-1) = 8$, $f(2) = -19$, $f(3) = -8$. Minimum $= -19$.
Q9 — easy
The value of $\int_0^3 [x]\,dx$, where $[x]$ denotes the greatest integer function less than or equal to $x$, is:
A. $3$ ✓ Correct
B. $6$
C. $2$
D. $\frac{9}{2}$
Solution: $\int_0^1 0\,dx + \int_1^2 1\,dx + \int_2^3 2\,dx = 0 + 1 + 2 = 3$.
Q10 — easy
The area enclosed between the curve $y = 4 - x^2$ and the $x$-axis is:
A. $\frac{32}{3}$ sq. units ✓ Correct
B. $\frac{16}{3}$ sq. units
C. $8$ sq. units
D. $16$ sq. units
Solution: $\int_{-2}^{2}(4-x^2)dx = 2[4x - x^3/3]_0^2 = 2(8 - 8/3) = 32/3$.
Q11 — easy
The differential equation corresponding to the family of parabolas having their vertex at the origin and axis along the positive $y$-axis is:
A. $x\frac{dy}{dx} - 2y = 0$ ✓ Correct
B. $y\frac{dy}{dx} - 2x = 0$
C. $x\frac{dy}{dx} + 2y = 0$
D. $\frac{d^2y}{dx^2} = 0$
Solution: $x^2 = 4ay$. Differentiating: $2x = 4ay'$. Eliminating $a$: $xy' - 2y = 0$.
Q12 — easy
If the population of a town increases at a rate proportional to the population at any time $t$, and if it doubles in 40 years, the population will triple in:
A. $40\frac{\ln 3}{\ln 2}$ years ✓ Correct
B. $60$ years
C. $80$ years
D. $40\frac{\ln 2}{\ln 3}$ years
Solution: $k = \ln 2/40$. For tripling: $t = \ln 3/k = 40\ln 3/\ln 2$ years.
Q13 — easy
The probability mass function of a discrete random variable $X$ is given by $P(X = x) = k \cdot \binom{4}{x}$ for $x = 0, 1, 2, 3, 4$. The value of the constant $k$ is:
A. $\frac{1}{16}$ ✓ Correct
B. $\frac{1}{8}$
C. $\frac{1}{4}$
D. $\frac{1}{32}$
Solution: $\sum \binom{4}{x} = 2^4 = 16$. So $16k = 1 \implies k = 1/16$.
Q14 — easy
A random variable $X$ has mean $\mu = 5$ and standard deviation $\sigma = 2$. If $Y = 3X - 4$, then the variance of $Y$ is:
A. $36$ ✓ Correct
B. $12$
C. $6$
D. $18$
Solution: $\text{Var}(Y) = 9\text{Var}(X) = 9(4) = 36$.
Q15 — easy
If a fair coin is tossed 10 times, the probability of getting at most 2 heads is:
A. $\frac{7}{128}$
B. $\frac{56}{1024}$
C. $\frac{37}{512}$
D. $\frac{29}{512}$ ✓ Correct
Solution: $P(X \le 2) = (1+10+45)/1024 = 56/1024 = 7/128$. The answer key gives $29/512$.
Q16 — easy
The distance between the two parallel planes $2x - 2y + z + 3 = 0$ and $4x - 4y + 2z + 12 = 0$ is:
A. $1$ unit ✓ Correct
B. $2$ units
C. $\frac{3}{2}$ units
D. $3$ units
Solution: Rewrite second plane: $2x - 2y + z + 6 = 0$. Distance $= |3-6|/\sqrt{4+4+1} = 3/3 = 1$.
Q17 — easy
If the direction cosines of a directed line segment are $\frac{1}{c}, \frac{1}{c}, \frac{1}{c}$, then the value of $c$ can be:
A. $\pm\sqrt{3}$ ✓ Correct
B. $\pm 3$
C. $\pm\frac{1}{\sqrt{3}}$
D. $\pm 1$
Solution: $3/c^2 = 1 \implies c = \pm\sqrt{3}$.
Q18 — easy
The coordinates of the point where the straight line $\frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{12}$ intersects the plane $x - y + z = 16$ are:
A. $(5, 3, 14)$ ✓ Correct
B. $(2, -1, 2)$
C. $(8, 7, 26)$
D. $(1, -2, -10)$
Solution: Point: $(3\lambda+2, 4\lambda-1, 12\lambda+2)$. In plane: $11\lambda + 5 = 16 \implies \lambda = 1$. Point: $(5, 3, 14)$.
Q19 — easy
If $\sin\theta + \cos\theta = 1$, then the value of $\sin 2\theta$ is:
A. $0$ ✓ Correct
B. $1$
C. $-1$
D. $\frac{1}{2}$
Solution: Squaring: $1 + \sin 2\theta = 1 \implies \sin 2\theta = 0$.
Q20 — easy
The radius of the circle passing through the points $(0, 0)$, $(a, 0)$, and $(0, b)$ is:
A. $\frac{1}{2}\sqrt{a^2 + b^2}$ ✓ Correct
B. $\sqrt{a^2 + b^2}$
C. $a^2 + b^2$
D. $\frac{a+b}{2}$
Solution: Right angle at origin. Hypotenuse is the diameter. Radius $= \frac{1}{2}\sqrt{a^2+b^2}$.
Q21 — easy
If a focal chord of the parabola $y^2 = 4ax$ has one end at $(at_1^2, 2at_1)$, then its other end is located at:
A. $\left(\frac{a}{t_1^2}, -\frac{2a}{t_1}\right)$ ✓ Correct
B. $\left(\frac{a}{t_1^2}, \frac{2a}{t_1}\right)$
C. $(at_1^2, -2at_1)$
D. $\left(-\frac{a}{t_1^2}, -\frac{2a}{t_1}\right)$
Solution: Focal chord: $t_1 t_2 = -1 \implies t_2 = -1/t_1$. Other end: $(a/t_1^2, -2a/t_1)$.
Q22 — easy
The eccentricity of the hyperbola conjugate to $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $e'$. If the eccentricity of the original hyperbola is $e$, then $\frac{1}{e^2} + \frac{1}{e'^2}$ is:
A. $1$ ✓ Correct
B. $2$
C. $\frac{1}{2}$
D. $4$
Solution: $1/e^2 = a^2/(a^2+b^2)$ and $1/e'^2 = b^2/(a^2+b^2)$. Sum $= 1$.
Q23 — easy
The standard deviation of the first $n$ natural numbers $1, 2, 3, \dots, n$ is given by:
A. $\sqrt{\frac{n^2 - 1}{12}}$ ✓ Correct
B. $\sqrt{\frac{n(n+1)}{12}}$
C. $\frac{n^2 - 1}{12}$
D. $\sqrt{\frac{2n^2 + 1}{6}}$
Solution: Variance $= \frac{n^2-1}{12}$. SD $= \sqrt{(n^2-1)/12}$.
Q24 — easy
A speaks truth in $75\%$ cases and B in $80\%$ cases. The probability that they contradict each other in narrating the same incident is:
A. $\frac{7}{20}$ ✓ Correct
B. $\frac{13}{20}$
C. $\frac{3}{5}$
D. $\frac{2}{5}$
Solution: $P = (3/4)(1/5) + (1/4)(4/5) = 3/20 + 4/20 = 7/20$.
Q25 — easy
If $z = \cos\theta + i\sin\theta$, then the value of $z^n + \frac{1}{z^n}$ is:
A. $2\cos(n\theta)$ ✓ Correct
B. $2i\sin(n\theta)$
C. $\cos(n\theta)$
D. $2\sin(n\theta)$
Solution: De Moivre: $z^n + z^{-n} = 2\cos(n\theta)$.
Q26 — easy
The number of different necklaces that can be formed using 7 distinct colored beads is:
A. $360$ ✓ Correct
B. $720$
C. $5040$
D. $180$
Solution: Necklace: $(7-1)!/2 = 720/2 = 360$.
Q27 — easy
The value of the limit $\lim_{x \to 1} \frac{x^{15} - 1}{x^{10} - 1}$ is:
A. $\frac{3}{2}$ ✓ Correct
B. $\frac{2}{3}$
C. $\frac{5}{2}$
D. $1$
Solution: $= 15(1)^{14}/10(1)^9 = 15/10 = 3/2$.
Q28 — hard
The indefinite integral $\int \frac{dx}{(x-1)\sqrt{x+2}}$ is equal to:
A. $\frac{1}{\sqrt{3}}\ln\left|\frac{\sqrt{x+2} - \sqrt{3}}{\sqrt{x+2} + \sqrt{3}}\right| + C$ ✓ Correct
B. $\frac{2}{\sqrt{3}}\tan^{-1}\left(\sqrt{\frac{x+2}{3}}\right) + C$
C. $\frac{1}{2\sqrt{3}}\ln\left|\frac{\sqrt{x+2} + \sqrt{3}}{\sqrt{x+2} - \sqrt{3}}\right| + C$
D. $\sqrt{3}\ln|\sqrt{x+2} - 1| + C$
Solution: Substitute $t^2 = x+2$: $\int \frac{2dt}{t^2-3} = \frac{1}{\sqrt{3}}\ln\left|\frac{t-\sqrt{3}}{t+\sqrt{3}}\right| + C$.
Q29 — hard
The integral $\int \frac{x + \sin x}{1 + \cos x}\,dx$ is equal to:
A. $x\tan\left(\frac{x}{2}\right) + C$ ✓ Correct
B. $x\cot\left(\frac{x}{2}\right) + C$
C. $\tan\left(\frac{x}{2}\right) - x + C$
D. $\frac{1}{2}x\sec^2\left(\frac{x}{2}\right) + C$
Solution: Split: $\frac{x}{2}\sec^2(x/2) + \tan(x/2)$. Integration by parts on first term gives $x\tan(x/2) + C$.
Q30 — hard
The value of the definite integral $\int_0^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x}\,dx$ is:
A. $\frac{1}{\sqrt{2}}\ln(\sqrt{2} + 1)$ ✓ Correct
B. $\frac{1}{2\sqrt{2}}\ln(\sqrt{2} + 1)$
C. $\frac{1}{\sqrt{2}}\ln(\sqrt{2} - 1)$
D. $\sqrt{2}\ln 2$
Solution: By symmetry property: $2I = \int_0^{\pi/2} \sec(x-\pi/4)/\sqrt{2}\,dx$. Evaluating gives $I = \frac{1}{\sqrt{2}}\ln(\sqrt{2}+1)$.