Let $f : R \to R$ be defined by $f (x) = ∣ x + 1 ∣ e^{- x^{2}}$ . Then which of the following statements is true?

Answer (Detailed Solution Below)View answer ▾

Answer: Option B — $f$ has a point of maxima in $(- 2, - 1)$

Detailed Solution

Step 1 ( $x < - 1$ ): $f = - (x + 1) e^{- x^{2}}$ , $f' (x) = e^{- x^{2}} (2 x^{2} + 2 x - 1)$ . Zero at $x = \frac{- 1 - 3}{2} \approx - 1.37 \in (- 2, - 1)$ .

$f'$ changes from $+$ to $-$ there (since $2 x^{2} + 2 x - 1 > 0$ to the left and $< 0$ to the right), so it is a local maximum. (B) is true.

Step 3 ( $x > - 1$ ): $f = (x + 1) e^{- x^{2}}$ , $f' = e^{- x^{2}} (1 - 2 x - 2 x^{2})$ , zero at $x = \frac{- 1 + 3}{2} \approx 0.37$ : the global maximum ( $f \approx 1.19$ ). It is in $(0, 1)$ , not $(1, 2)$ , so (A) is false.

The global minimum is $f (- 1) = 0$ , not in $(0, 1)$ ; and on $(0, 1)$ there is only a maximum. (C), (D) are false.

Quick tip

Split $∣ x + 1∣$ at $x = - 1$ ; check sign change of $f^{'}$

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Parabola — Preparation

This question tests your understanding of Parabola, a key topic in the Quantitative Aptitude section. The detailed solution above uses a step-by-step approach to show the reasoning, and where available, a shortcut that helps you solve similar questions in under 90 seconds during the actual exam.

A parabola is the set of points equidistant from a fixed point (the focus) and a fixed line (the directrix), so its eccentricity is exactly 1. The standard form y^2 = 4ax has its vertex at the origin, focus at (a, 0), directrix x = -a, and latus rectum of length 4a. The other standard forms open left, up or down. Exam questions test identifying the focus, directrix, axis and latus rectum, the parametric point (at^2, 2at), focal distances, and equations of tangents and normals. A key result is that the line y = mx + c touches y^2 = 4ax when c = a/m. Drawing a quick sketch of which way the parabola opens prevents most sign errors. Practise the parabola questions below, each with a step-by-step solution.

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Let $f : R \to R$ be defined by $f (x) = ∣ x + 1 ∣ e^{- x^{2}}$ . Then which of the following statements is true?

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Let f:R→R be defined by f(x)=∣x+1∣e−x2. Then which of the following statements is true?

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Let $f : R \to R$ be defined by $f (x) = ∣ x + 1 ∣ e^{- x^{2}}$ . Then which of the following statements is true?