Differentiation (Part-38) | S N De

Q30. Match the functions in Column-I with their intervals of increase in Column-II and choose the correct option:

Column-I	Column-II
(i) \(3x^4 + 4x^3 - 12x^2 + 12\)	(a) \([0, 2]\)
(ii) \(-x^2 + 6x - 3\)	(b) \([-3, 0] \cup [3, \infty)\)
(iii) \((9 - x^2)^2\)	(c) \([-2, 0] \cup [1, \infty)\)
(iv) \(x^2e^{-x}\)	(d) \((-\infty, 3]\)

• (A) (i)-(d), (ii)-(c), (iii)-(b), (iv)-(a)
• (B) (i)-(c), (ii)-(d), (iii)-(a), (iv)-(b)
• (C) (i)-(d), (ii)-(c), (iii)-(a), (iv)-(b)
• (D) (i)-(c), (ii)-(d), (iii)-(b), (iv)-(a)

Solution

A function is monotonically increasing where its first derivative is \(f'(x) \ge 0\).

Evaluating (i): \(f(x) = 3x^4 + 4x^3 - 12x^2 + 12\)
\(f'(x) = 12x^3 + 12x^2 - 24x = 12x(x^2 + x - 2) = 12x(x+2)(x-1)\).
Setting \(f'(x) \ge 0\) yields the intervals \([-2, 0] \cup [1, \infty)\). So, (i) \(\to\) (c).

Evaluating (ii): \(f(x) = -x^2 + 6x - 3\)
\(f'(x) = -2x + 6 \ge 0 \implies x \le 3\).
Interval is \((-\infty, 3]\). So, (ii) \(\to\) (d).

Evaluating (iii): \(f(x) = (9 - x^2)^2\)
\(f'(x) = 2(9 - x^2)(-2x) = 4x(x^2 - 9) = 4x(x - 3)(x + 3)\).
Setting \(f'(x) \ge 0\) yields \([-3, 0] \cup [3, \infty)\). So, (iii) \(\to\) (b).

Evaluating (iv): \(f(x) = x^2e^{-x}\)
\(f'(x) = 2xe^{-x} - x^2e^{-x} = xe^{-x}(2 - x)\).
Since \(e^{-x} > 0\), \(x(2-x) \ge 0 \implies x \in [0, 2]\). So, (iv) \(\to\) (a).

Answer: (D)

Q31. The minimum value of \(f(x) = (x^2 - 3)^3 + 27\) is:

• (A) 0
• (B) 3
• (C) 2
• (D) 1

Solution

To find the minimum value, we calculate the first derivative and find critical points:
\(f'(x) = 3(x^2 - 3)^2 \cdot (2x) = 6x(x^2 - 3)^2\).

Setting \(f'(x) = 0\), we get \(x = 0\) and \(x = \pm\sqrt{3}\).

Using the second derivative test:
\(f''(x) = 6(x^2 - 3)^2 + 6x \cdot 2(x^2 - 3) \cdot 2x = 6(x^2 - 3)[(x^2 - 3) + 4x^2] = 6(x^2 - 3)(5x^2 - 3)\).

At \(x = 0\), \(f''(0) = 6(-3)(-3) = 54 > 0\). This confirms a local minimum exists at \(x = 0\).

Substitute \(x = 0\) into the original function:
\(f(0) = (0^2 - 3)^3 + 27 = (-3)^3 + 27 = -27 + 27 = 0\).

Answer: (A)

Q32. In the adjacent figure, AFDE is a parallelogram where \(AF \parallel ED\), \(AE \parallel FD\), \(AE = y\) units, and \(DE = x\) units. If the area of the parallelogram AFDE is maximum, then its value will be:

• (A) \(\frac{1}{2} \times \text{Area of } \Delta ABC\)
• (B) \(\frac{1}{4} \times \text{Area of } \Delta ABC\)
• (C) \(\frac{1}{6} \times \text{Area of } \Delta ABC\)
• (D) \(\frac{1}{8} \times \text{Area of } \Delta ABC\)

Solution

A well-known geometric property states that when a parallelogram is inscribed in a triangle such that they share a common angle, the maximum area of the parallelogram is achieved when the vertex on the opposite side of the shared angle lies exactly at the midpoint of that side.

In this case, when D is the midpoint of BC, F and E become the midpoints of AB and AC respectively.

The area of the inscribed parallelogram is then exactly half the area of the original triangle.

Answer: (A)

Q33. A and B are two events such that \(P(A \cup B) = \frac{5}{6}\), \(P(A \cap B) = \frac{1}{3}\), and \(P(\bar{B}) = \frac{1}{2}\). Then the events A and B are mutually:

• (A) Independent
• (B) Not independent
• (C) Mutually exclusive
• (D) None of these

Solution

First, find \(P(B)\):
\(P(B) = 1 - P(\bar{B}) = 1 - \frac{1}{2} = \frac{1}{2}\).

Using the union probability formula to find \(P(A)\):
\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
\(\frac{5}{6} = P(A) + \frac{1}{2} - \frac{1}{3}\)
\(\frac{5}{6} = P(A) + \frac{1}{6} \implies P(A) = \frac{4}{6} = \frac{2}{3}\).

Now, test for independence. Two events are independent if \(P(A \cap B) = P(A) \times P(B)\):
\(P(A) \times P(B) = \frac{2}{3} \times \frac{1}{2} = \frac{1}{3}\).

Since this equals the given \(P(A \cap B) = \frac{1}{3}\), the events are independent.

Answer: (A)

Q34. A and B are two events such that \(P(A) = \frac{3}{8}\), \(P(B) = \frac{1}{2}\) and \(P(A \cap B) = \frac{1}{4}\). Then the value of \(P(\bar{A} / \bar{B})\) is:

• (A) \(\frac{3}{8}\)
• (B) \(\frac{1}{2}\)
• (C) \(\frac{3}{5}\)
• (D) \(\frac{3}{4}\)

Solution

By definition of conditional probability:
\(P(\bar{A} / \bar{B}) = \frac{P(\bar{A} \cap \bar{B})}{P(\bar{B})}\).

Using De Morgan's Law, \(P(\bar{A} \cap \bar{B}) = P(\overline{A \cup B}) = 1 - P(A \cup B)\).

First, calculate \(P(A \cup B)\):
\(P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{3}{8} + \frac{1}{2} - \frac{1}{4} = \frac{3}{8} + \frac{4}{8} - \frac{2}{8} = \frac{5}{8}\).

Now, calculate the numerator and denominator:
Numerator: \(1 - P(A \cup B) = 1 - \frac{5}{8} = \frac{3}{8}\).
Denominator: \(P(\bar{B}) = 1 - P(B) = 1 - \frac{1}{2} = \frac{1}{2}\).

\(P(\bar{A} / \bar{B}) = \frac{3/8}{1/2} = \frac{3}{8} \times 2 = \frac{3}{4}\).

Answer: (D)

Q35. If k is a natural number such that \(k \le 5\), then the probability that the quadratic equation \(2x^2 + 2kx - x + 1 = 0\) has real roots is:

• (A) \(\frac{3}{5}\)
• (B) \(\frac{2}{3}\)
• (C) \(\frac{4}{5}\)
• (D) \(\frac{2}{5}\)

Solution

Rewrite the quadratic equation in standard form:
\(2x^2 + (2k - 1)x + 1 = 0\).

For the equation to have real roots, the discriminant \(D\) must be greater than or equal to 0:
\(D = (2k - 1)^2 - 4(2)(1) \ge 0 \implies (2k - 1)^2 \ge 8\).

Since k is a natural number (\(k \in \mathbb{N}\)) and \(k \le 5\), the possible outcomes for k are \(\{1, 2, 3, 4, 5\}\).
Total possible outcomes = 5.

Let's check the condition for each k:
If \(k = 1\): \((2(1)-1)^2 = 1 \not\ge 8\) (False)
If \(k = 2\): \((2(2)-1)^2 = 9 \ge 8\) (True)
If \(k = 3, 4, 5\), the square will be even larger and obviously \(\ge 8\).

The favorable outcomes are \(k \in \{2, 3, 4, 5\}\), which gives 4 favorable outcomes.

Probability = \(\frac{\text{Favorable}}{\text{Total}} = \frac{4}{5}\).

Answer: (C)

Q36. If \(P(E_1) = \frac{4}{5}\), \(P(E_2) = \frac{1}{5}\), \(P(A/E_1) = \frac{1}{8}\), and \(P(A/E_2) = \frac{1}{5}\), then the value of \(P(E_2/A)\) is:

• (A) \(\frac{2}{7}\)
• (B) \(\frac{3}{7}\)
• (C) \(\frac{4}{7}\)
• (D) \(\frac{5}{7}\)

Solution

Use Bayes' Theorem to find \(P(E_2/A)\):
\(P(E_2/A) = \frac{P(E_2) \cdot P(A/E_2)}{P(E_1) \cdot P(A/E_1) + P(E_2) \cdot P(A/E_2)}\)

First, calculate the total probability of event A (the denominator):
\(P(A) = \left(\frac{4}{5} \times \frac{1}{8}\right) + \left(\frac{1}{5} \times \frac{1}{5}\right) = \frac{4}{40} + \frac{1}{25} = \frac{1}{10} + \frac{1}{25}\).
Finding a common denominator (50):
\(P(A) = \frac{5}{50} + \frac{2}{50} = \frac{7}{50}\).

Now, calculate the numerator:
\(P(E_2) \cdot P(A/E_2) = \frac{1}{5} \times \frac{1}{5} = \frac{1}{25} = \frac{2}{50}\).

\(P(E_2/A) = \frac{2/50}{7/50} = \frac{2}{7}\).

Answer: (A)

Q37. The value of \(var(5x + 3)\) is:

• (A) \(25var(x)\)
• (B) \(5var(x)\)
• (C) \(5var(x) + 3\)
• (D) 0

Solution

The property of variance for a linear transformation \(aX + b\) is given by:
\(Var(aX + b) = a^2Var(X)\).

Applying this property to \(5x + 3\):
\(Var(5x + 3) = 5^2Var(x) = 25var(x)\).

Answer: (A)

Q38. A discrete random variable 'X' takes values \(x_1, x_2, \dots, x_n\) with probabilities \(p_1, p_2, \dots, p_n\) respectively; then the mean of X is \(\bar{X}\).
Statement (A): For a frequency distribution, \(\bar{X} = \frac{1}{N}(f_1x_1 + f_2x_2 + \dots + f_nx_n)\).
Reason (R): \(\bar{X} = \sum_{i=1}^{n} p_ix_i\).
Which of the following options correctly explains Statement (A) and Reason (R)?

• (A) Both A and R are true and R is the correct explanation of A
• (B) Both A and R are true but R is NOT the correct explanation of A
• (C) A is true but R is false
• (D) A is false but R is true

Solution

Statement (A) represents the formula for calculating the arithmetic mean for a grouped frequency distribution (sample mean), which is mathematically correct.

Reason (R) gives the formula for calculating the expected value (mean) of a discrete probability distribution (population mean), which is also mathematically correct and directly addresses the scenario presented in the premise.

However, Statement A refers strictly to a frequency distribution context, whereas Reason R defines the expected value in a theoretical probability context. While conceptually analogous (\(p_i\) is the limit of relative frequency \(f_i/N\)), one does not serve as the direct algebraic explanation of the other in basic statistical contexts; they belong to slightly different scopes (sample vs population).

Therefore, both statements are true, but R is not the definitive explanation of A.

Answer: (B)

Q39. The probability distribution of a random variable 'X' is as follows:

\(x_i\)	-2	-1	0	1	2	3
\(P_i\)	0.1	k	0.2	2k	0.3	k

The value of k will be:

• (A) 0.2
• (B) 0.1
• (C) 0.3
• (D) 0.4

Solution

For a valid probability distribution, the sum of all probabilities must equal 1.
\(\sum P_i = 1\)

\(0.1 + k + 0.2 + 2k + 0.3 + k = 1\)
Combine the constants and the \(k\) terms:
\(4k + 0.6 = 1\)
\(4k = 1 - 0.6 = 0.4\)
\(k = 0.1\)

Answer: (B)

Q40. \(var(x) = \)

• (A) \(E(x^2) - E(x)\)
• (B) \(E(x^2) - \{E(x)\}^2\)
• (C) \(E(x^2) + E(x)\)
• (D) \(\{E(x)\}^2 - E(x^2)\)

Solution

The formula for the variance of a random variable X is defined as the expected value of the squared deviation from the mean. Mathematically, it expands to:

\(Var(X) = E[(X - \mu)^2] = E(X^2) - [E(X)]^2\).

Answer: (B)