First, find the first derivative \(y_1 = \frac{dy}{dx}\):
\(y_1 = e^{m\sin^{-1}x} \cdot \frac{m}{\sqrt{1-x^2}} = \frac{my}{\sqrt{1-x^2}}\)
Rearranging and squaring both sides:
\(y_1\sqrt{1-x^2} = my \implies y_1^2(1-x^2) = m^2y^2\)
Differentiating again with respect to \(x\):
\(2y_1 y_2 (1-x^2) - 2x y_1^2 = m^2 (2y y_1)\)
Dividing throughout by \(2y_1\):
\(y_2(1-x^2) - x y_1 = m^2 y\)
At \(x = 0\): \(y(0) = e^0 = 1\).
Substitute \(x=0\) into the differential equation:
\(y_2(0)(1 - 0) - 0 = m^2(1) \implies y_2(0) = m^2\).
Answer: (C)
Given \(\frac{dx}{dy} = u\), we have \(\frac{dy}{dx} = \frac{1}{u} = u^{-1}\).
Now, differentiate with respect to \(x\):
\(\frac{d^2y}{dx^2} = \frac{d}{dx}(u^{-1})\)
Using the chain rule, this becomes \(\frac{d}{dy}(u^{-1}) \cdot \frac{dy}{dx}\).
\(= -u^{-2} \frac{du}{dy} \cdot (u^{-1})\)
Since \(u = \frac{dx}{dy}\), its derivative \(\frac{du}{dy} = \frac{d^2x}{dy^2} = v\).
Substituting back:
\(= -u^{-2} \cdot v \cdot u^{-1} = -\frac{v}{u^3}\).
Answer: (C)
Surface area of a sphere \(S = 4\pi r^2\).
Differentiating with respect to time \(t\): \(\frac{dS}{dt} = 8\pi r \frac{dr}{dt}\).
Given \(\frac{dS}{dt} = 2\) and \(r = 6\):
\(2 = 8\pi(6) \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{2}{48\pi} = \frac{1}{24\pi}\).
Volume of a sphere \(V = \frac{4}{3}\pi r^3\).
Differentiating with respect to time \(t\): \(\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}\).
Substitute \(r = 6\) and \(\frac{dr}{dt} = \frac{1}{24\pi}\):
\(\frac{dV}{dt} = 4\pi(36) \left( \frac{1}{24\pi} \right) = \frac{144\pi}{24\pi} = 6\).
Answer: (D)
First, find the derivative \(f'(x)\):
\(f'(x) = 4x - \frac{1}{x} = \frac{4x^2 - 1}{x}\).
Factorizing the numerator: \(\frac{(2x - 1)(2x + 1)}{x}\).
Critical points are \(x = -\frac{1}{2}, 0, \frac{1}{2}\).
Check the sign of \(f'(x)\) in the intervals:
1. \((-\infty, -1/2)\): \(f'(x) < 0\) (Decreasing)
2. \((-1/2, 0)\): \(f'(x) > 0\) (Increasing)
3. \((0, 1/2)\): \(f'(x) < 0\) (Decreasing)
4. \((1/2, \infty)\): \(f'(x) > 0\) (Increasing)
The function is monotonically increasing in \((-\frac{1}{2}, 0) \cup (\frac{1}{2}, \infty)\).
Answer: (C)
Find the derivative \(f'(x)\):
\(f'(x) = 6x^2 - 6x - 36 = 6(x^2 - x - 6)\).
Factorizing: \(f'(x) = 6(x - 3)(x + 2)\).
Critical points are \(x = -2\) and \(x = 3\).
For strict decrease, \(f'(x) < 0\). This happens when \((x - 3)(x + 2) < 0\), which gives \(-2 < x < 3\).
Therefore, the function is strictly decreasing in the interval \((-2, 3)\).
Answer: (D)
Let the distance from origin be \(D\). We want to minimize \(D^2 = x^2 + y^2\).
From the curve equation, \(x = y^{-2}\). Substitute this into the distance function:
\(f(y) = (y^{-2})^2 + y^2 = y^{-4} + y^2\).
Differentiate to find the minimum:
\(f'(y) = -4y^{-5} + 2y = 0 \implies 2y = \frac{4}{y^5} \implies y^6 = 2 \implies y = \pm 2^{1/6}\).
Now find \(x\):
\(x = \frac{1}{y^2} = \frac{1}{(2^{1/6})^2} = \frac{1}{2^{1/3}} = 2^{-1/3}\).
The coordinates are \((2^{-1/3}, 2^{1/6})\).
Answer: (C)
Differentiate the curve \(y^2 = x^3\):
\(2y \frac{dy}{dx} = 3x^2 \implies \frac{dy}{dx} = \frac{3x^2}{2y}\).
Slope of tangent at \((m^2, m^3)\): \(m_1 = \frac{3(m^2)^2}{2m^3} = \frac{3}{2}m\).
Slope of tangent at \((M^2, M^3)\): \(\frac{3}{2}M\).
Slope of normal at \((M^2, M^3)\): \(m_2 = -\frac{2}{3M}\).
Since the tangent at the first point is the same line as the normal at the second point, their slopes must be equal:
\(\frac{3}{2}m = -\frac{2}{3M} \implies mM = -\frac{4}{9}\).
Answer: (D)
The equation of the normal to the parabola \(y^2 = 4ax\) at point \(t\) is \(y = -tx + 2at + at^3\). Here \(a = b\).
For the normal at \(t_1\) to intersect the parabola again at \(t_2\), substitute \((bt_2^2, 2bt_2)\) into the normal equation:
\(2bt_2 = -t_1(bt_2^2) + 2bt_1 + bt_1^3\)
\(2b(t_2 - t_1) = -bt_1(t_2^2 - t_1^2)\).
Dividing by \(b(t_2 - t_1)\) (since \(t_2 \neq t_1\)):
\(2 = -t_1(t_2 + t_1) \implies t_2 + t_1 = -\frac{2}{t_1}\).
Therefore, \(t_2 = -t_1 - \frac{2}{t_1}\).
Answer: (D)
Critical points occur where \(f'(x) = 0\).
\(f'(x) = \frac{d}{dx}\left(\frac{2}{3}x^3 - \frac{3}{2}x^2 - 2x + 5\right) = 2x^2 - 3x - 2\).
Set the derivative to zero and solve the quadratic equation:
\(2x^2 - 4x + x - 2 = 0 \implies 2x(x - 2) + 1(x - 2) = 0\)
\((2x + 1)(x - 2) = 0\).
The roots are \(x = -\frac{1}{2}\) and \(x = 2\).
Answer: (B)
Let \(P(A) = \frac{1}{2}, P(B) = \frac{1}{3}, P(C) = \frac{1}{4}\).
The probability that none of them solves the problem is the product of their failure probabilities:
\(P(\text{none}) = P(A') \cdot P(B') \cdot P(C') = \left(1 - \frac{1}{2}\right)\left(1 - \frac{1}{3}\right)\left(1 - \frac{1}{4}\right)\)
\(= \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} = \frac{1}{4}\).
The probability that the problem is solved (at least one solves it) is:
\(1 - P(\text{none}) = 1 - \frac{1}{4} = \frac{3}{4}\).
Answer: (D)
Use the addition rule to find \(P(A \cap B)\):
\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
\(\frac{5}{6} = \frac{2}{3} + \frac{1}{2} - P(A \cap B)\)
\(P(A \cap B) = \frac{4}{6} + \frac{3}{6} - \frac{5}{6} = \frac{2}{6} = \frac{1}{3}\).
Now, calculate conditional probability \(P(B/A)\):
\(P(B/A) = \frac{P(A \cap B)}{P(A)} = \frac{1/3}{2/3} = \frac{1}{2}\).
Answer: (A)
| X | -2 | -1 | 0 | 1 | 2 | 3 |
|---|---|---|---|---|---|---|
| P(x) | 0.1 | k | 0.2 | 2k | 0.3 | 3k |
The sum of all probabilities must equal 1:
\(\sum P(X) = 0.1 + k + 0.2 + 2k + 0.3 + 3k = 1\)
\(0.6 + 6k = 1 \implies 6k = 0.4 \implies k = \frac{0.4}{6} = \frac{2}{30}\).
We need to find \(P(X \ge 2) = P(X = 2) + P(X = 3)\):
\(P(X \ge 2) = 0.3 + 3k = 0.3 + 3\left(\frac{2}{30}\right) = 0.3 + \frac{6}{30}\).
Since \(\frac{6}{30} = 0.2\), we have \(0.3 + 0.2 = 0.5\).
Answer: (B)
Let S = Event of failing in Statistics, M = Event of failing in Mathematics.
Given: \(P(S) = 0.25\), \(P(M) = 0.30\), \(P(S \cup M) = 0.45\).
First, find the probability of failing both:
\(P(S \cap M) = P(S) + P(M) - P(S \cup M) = 0.25 + 0.30 - 0.45 = 0.10\).
We need the probability of passing Statistics given failure in Mathematics. Passing Statistics is the complement event \(S'\).
\(P(S'/M) = \frac{P(S' \cap M)}{P(M)} = \frac{P(M) - P(S \cap M)}{P(M)}\)
\(= \frac{0.30 - 0.10}{0.30} = \frac{0.20}{0.30} = \frac{2}{3}\).
Answer: (B)
Statement (A): R is not an equivalence relation.
Reason (R): R is not symmetric.
Reflexivity: Every number is a factor of itself (\(n | n\)). True.
Transitivity: If \(n | m\) and \(m | p\), then \(n | p\). True.
Symmetry: If \(n\) is a factor of \(m\), \(m\) is NOT necessarily a factor of \(n\) (e.g., 2 is a factor of 4, but 4 is not a factor of 2). False.
Since the relation is not symmetric, it fails to be an equivalence relation. Both the statement and the reason are true, and the reason correctly explains the statement.
Answer: (A)
Reason (R): \(|adj A| = |A|^{n-1}\)
Using the property provided in the Reason (R), substitute matrix \(X = adj A\):
\(|adj X| = |X|^{n-1}\)
\(|adj(adj A)| = |adj A|^{n-1}\).
Substituting \(|adj A| = |A|^{n-1}\) again:
\(= (|A|^{n-1})^{n-1} = |A|^{(n-1)(n-1)} = |A|^{(n-1)^2}\).
Both statements are mathematically correct standard properties of matrices, and the derivation shows R is indeed used to prove A.
Answer: (A)
In an office, three employees A, B and C process incoming copies of a certain form. A processes 50% of the forms, B processes 20% and C processes the remaining 30%. The error rate of A is 0.06, the error rate of B is 0.04 and the error rate of C is 0.03.
Let \(E_2\) be the event that B processes the form, and \(E\) be the event of an error.
We are given: \(P(E_2) = 0.20\) and \(P(E/E_2) = 0.04\).
The probability of both events occurring is their intersection:
\(P(E_2 \cap E) = P(E_2) \cdot P(E/E_2) = 0.20 \times 0.04 = 0.008\).
Answer: (C)
Let \(E_1, E_2, E_3\) denote processing by A, B, and C respectively. Event \(E\) is an error.
\(P(E_1) = 0.50, P(E/E_1) = 0.06\)
\(P(E_2) = 0.20, P(E/E_2) = 0.04\)
\(P(E_3) = 0.30, P(E/E_3) = 0.03\)
Using the Law of Total Probability:
\(P(E) = P(E_1)P(E/E_1) + P(E_2)P(E/E_2) + P(E_3)P(E/E_3)\)
\(= (0.50)(0.06) + (0.20)(0.04) + (0.30)(0.03)\)
\(= 0.030 + 0.008 + 0.009 = 0.047\).
Answer: (B)
First, find the probability that it WAS processed by A given that there is an error, using Bayes' Theorem:
\(P(E_1/E) = \frac{P(E_1)P(E/E_1)}{P(E)} = \frac{0.030}{0.047} = \frac{30}{47}\).
The probability that it was NOT processed by A is the complement:
\(P(E_1'/E) = 1 - P(E_1/E) = 1 - \frac{30}{47} = \frac{17}{47}\).
Answer: (A)
The events \(E_1, E_2,\) and \(E_3\) are mutually exclusive and exhaustive (they form a partition of the sample space because every form is processed by exactly one of the three employees).
For any valid partition, the sum of posterior probabilities given an event \(E\) is always equal to 1.
Mathematically: \(\sum P(E_i/E) = \frac{\sum P(E_i \cap E)}{P(E)} = \frac{P(E)}{P(E)} = 1\).
Answer: (D)
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