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MathHardAdv 2017 • Paper 1

Identify Correct Column Match for Function Behavior Analysis I

Comprehension Passage

Answer Q.52, Q.53, and Q.54 by appropriately matching the details provided in the three columns of the given table.

Consider f(x)=x+log⁡ex−xlog⁡exf(x) = x + \log_e x - x \log_e x for x∈(0,∞)x \in (0, \infty).

  • Column 1 contains information regarding zeros of f(x),f′(x),f(x), f'(x), and f′′(x)f''(x).
  • Column 2 provides information about the limiting behavior of f(x),f′(x),f(x), f'(x), and f′′(x)f''(x) as x→∞x \to \infty.
  • Column 3 contains information about the increasing or decreasing nature of f(x)f(x) and f′(x)f'(x).
Column 1Column 2Column 3(I) f(x)=0 for some x∈(1,e2)(i) lim⁡x→∞f(x)=0(P) f is increasing in (0,1)(II) f′(x)=0 for some x∈(1,e)(ii) lim⁡x→∞f(x)=−∞(Q) f is decreasing in (e,e2)(III) f′(x)=0 for some x∈(0,1)(iii) lim⁡x→∞f′(x)=−∞(R) f′ is increasing in (0,1)(IV) f′′(x)=0 for some x∈(1,e)(iv) lim⁡x→∞f′′(x)=0(S) f′ is decreasing in (e,e2)\begin{array}{|l|l|l|} \hline \text{Column 1} & \text{Column 2} & \text{Column 3} \\ \hline (\text{I})\ f(x) = 0 \text{ for some } x \in (1, e^2) & (\text{i})\ \lim_{x \to \infty} f(x) = 0 & (\text{P})\ f \text{ is increasing in } (0, 1) \\ \hline (\text{II})\ f'(x) = 0 \text{ for some } x \in (1, e) & (\text{ii})\ \lim_{x \to \infty} f(x) = -\infty & (\text{Q})\ f \text{ is decreasing in } (e, e^2) \\ \hline (\text{III})\ f'(x) = 0 \text{ for some } x \in (0, 1) & (\text{iii})\ \lim_{x \to \infty} f'(x) = -\infty & (\text{R})\ f' \text{ is increasing in } (0, 1) \\ \hline (\text{IV})\ f''(x) = 0 \text{ for some } x \in (1, e) & (\text{iv})\ \lim_{x \to \infty} f''(x) = 0 & (\text{S})\ f' \text{ is decreasing in } (e, e^2) \\ \hline \end{array}

Which of the options listed below represents the only CORRECT combination?

Options

A

(I) (i) (P)

B

(II) (ii) (Q)

Correct
C

(III) (iii) (R)

D

(IV) (iv) (S)

Topics & Concepts

Application of DerivativesMonotonicityLimits

Step-by-Step Solution

To determine the correct combination, we analyze the function f(x)f(x) and its derivatives:

f(x)=x+log⁡ex−xlog⁡ex,x∈(0,∞)f(x) = x + \log_e x - x \log_e x, \quad x \in (0, \infty)

1. Derivatives of f(x)f(x)

Differentiating f(x)f(x) with respect to xx: f′(x)=ddx(x+log⁡ex−xlog⁡ex)f'(x) = \frac{d}{dx} \left( x + \log_e x - x \log_e x \right) f′(x)=1+1x−(log⁡ex+x⋅1x)=1x−log⁡exf'(x) = 1 + \frac{1}{x} - \left( \log_e x + x \cdot \frac{1}{x} \right) = \frac{1}{x} - \log_e x

Differentiating f′(x)f'(x) with respect to xx: f′′(x)=−1x2−1x=−1+xx2f''(x) = -\frac{1}{x^2} - \frac{1}{x} = -\frac{1 + x}{x^2}

2. Analysis of Column 1

  • Statement (I): f(1)=1+log⁡e1−1⋅log⁡e1=1>0f(1) = 1 + \log_e 1 - 1 \cdot \log_e 1 = 1 > 0 f(e2)=e2+log⁡e(e2)−e2log⁡e(e2)=e2+2−2e2=2−e2<0(since e2≈7.389)f(e^2) = e^2 + \log_e(e^2) - e^2 \log_e(e^2) = e^2 + 2 - 2e^2 = 2 - e^2 < 0 \quad (\text{since } e^2 \approx 7.389) By the Intermediate Value Theorem, since f(x)f(x) is continuous on [1,e2][1, e^2], f(x)=0f(x) = 0 for some x∈(1,e2)x \in (1, e^2). Thus, (I) is TRUE.

  • Statement (II): f′(1)=11−log⁡e1=1>0f'(1) = \frac{1}{1} - \log_e 1 = 1 > 0 f′(e)=1e−log⁡ee=1e−1<0f'(e) = \frac{1}{e} - \log_e e = \frac{1}{e} - 1 < 0 By the Intermediate Value Theorem, since f′(x)f'(x) is continuous on [1,e][1, e], f′(x)=0f'(x) = 0 for some x∈(1,e)x \in (1, e). Thus, (II) is TRUE.

  • Statement (III): For x∈(0,1)x \in (0, 1), 1x>1\frac{1}{x} > 1 and log⁡ex<0\log_e x < 0, which gives f′(x)=1x−log⁡ex>1f'(x) = \frac{1}{x} - \log_e x > 1. Thus, f′(x)≠0f'(x) \neq 0 for any x∈(0,1)x \in (0, 1), making (III) FALSE.

  • Statement (IV): For any x∈(0,∞)x \in (0, \infty), f′′(x)=−1+xx2<0f''(x) = -\frac{1+x}{x^2} < 0. Thus, f′′(x)≠0f''(x) \neq 0 for any x∈(1,e)x \in (1, e), making (IV) FALSE.

3. Analysis of Column 2

  • Statement (i) & (ii): lim⁡x→∞f(x)=lim⁡x→∞(x(1−log⁡ex)+log⁡ex)=−∞\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \left( x(1 - \log_e x) + \log_e x \right) = -\infty Therefore, (i) is FALSE and (ii) is TRUE.

  • Statement (iii): lim⁡x→∞f′(x)=lim⁡x→∞(1x−log⁡ex)=0−∞=−∞\lim_{x \to \infty} f'(x) = \lim_{x \to \infty} \left( \frac{1}{x} - \log_e x \right) = 0 - \infty = -\infty Thus, (iii) is TRUE.

  • Statement (iv): lim⁡x→∞f′′(x)=lim⁡x→∞(−1+xx2)=0\lim_{x \to \infty} f''(x) = \lim_{x \to \infty} \left( -\frac{1+x}{x^2} \right) = 0 Thus, (iv) is TRUE.

4. Analysis of Column 3

  • Statement (P): For x∈(0,1)x \in (0, 1), f′(x)>0f'(x) > 0, so f(x)f(x) is increasing in (0,1)(0, 1). Thus, (P) is TRUE.

  • Statement (Q): For x∈(e,e2)x \in (e, e^2), 1x<1e<1\frac{1}{x} < \frac{1}{e} < 1 and log⁡ex>1\log_e x > 1, which implies f′(x)=1x−log⁡ex<0f'(x) = \frac{1}{x} - \log_e x < 0. Hence, f(x)f(x) is decreasing in (e,e2)(e, e^2). Thus, (Q) is TRUE.

  • Statement (R): Since f′′(x)<0f''(x) < 0 for all x>0x > 0, f′(x)f'(x) is decreasing everywhere in (0,∞)(0, \infty). Thus, (R) is FALSE.

  • Statement (S): Since f′′(x)<0f''(x) < 0 for all x>0x > 0, f′(x)f'(x) is decreasing in (e,e2)(e, e^2). Thus, (S) is TRUE.

5. Evaluating the Options

  • (A) (I) (i) (P): Incorrect, because (i) is false.
  • (B) (II) (ii) (Q): Correct, because (II) is true, (ii) is true, and (Q) is true.
  • (C) (III) (iii) (R): Incorrect, because (III) and (R) are false.
  • (D) (IV) (iv) (S): Incorrect, because (IV) is false.

Thus, the only correct combination is (B).

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