Calculus

Application of Derivatives

Medium

common

For the function $f(x) = e^{-2x}(2-x)^2$ , the point of local maxima is:

$x = 1$

$x = 2$

$x = 3$

$x = 5/2$

✅ Correct Option: 3

Given $f(x) = e^{-2x}(2-x)^2$

Using the product rule with $u = e^{-2x}$ and $v = (2-x)^2$ :

$u' = -2e^{-2x}$

$v' = 2(2-x)(-1) = -2(2-x)$

$f'(x) = u'v + uv'$

$f'(x) = -2e^{-2x}(2-x)^2 + e^{-2x}[-2(2-x)]$

Factoring out $-2e^{-2x}$ :

$f'(x) = -2e^{-2x}[(2-x)^2 + (2-x)]$

Factoring out $(2-x)$ :

$f'(x) = -2e^{-2x}(2-x)[(2-x) + 1]$

$f'(x) = -2e^{-2x}(2-x)(3-x)$

Setting $f'(x) = 0$ :

$-2e^{-2x}(2-x)(3-x) = 0$

Since $e^{-2x} \neq 0$ :

$(2-x)(3-x) = 0$

Critical points: $x = 2$ and $x = 3$

Analyzing the sign of $f'(x) = -2e^{-2x}(2-x)(3-x)$ :

Note that $-2e^{-2x}$ is always negative.

For $x < 2$ : $(2-x) > 0$ and $(3-x) > 0$ , so $f'(x) < 0$ (decreasing)

For $2 < x < 3$ : $(2-x) < 0$ and $(3-x) > 0$ , so $f'(x) > 0$ (increasing)

For $x > 3$ : $(2-x) < 0$ and $(3-x) < 0$ , so $f'(x) < 0$ (decreasing)

At $x = 2$ : $f'(x)$ changes from negative to positive, indicating a local minimum.

At $x = 3$ : $f'(x)$ changes from positive to negative, indicating a local maximum.

Therefore, the point of local maxima is $x = 3$ .

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A. Minimum value of $f(x)$ is 2

B. There is no minimum value

C. Maximum value of $f(x)$ is 4

D. Maximum value of $f(x)$ is 3

Choose the correct answer from the options given below:

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(A) is increasing on $\left[0, \frac{\pi}{6}\right]$

(B) is decreasing on $\left[\frac{\pi}{6}, \frac{\pi}{2}\right]$

(D) is decreasing on $\left[0, \frac{\pi}{2}\right]$

Choose the correct answer from the options given below:

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For the function $f(x) = \frac{1}{x^2+2x+2}$

which of the following statements are TRUE?

A. $f(x)$ is increasing on $(-\infty, -1)$

B. $f(x)$ is decreasing on $(-\infty, -1)$

C. $f(x)$ is decreasing on $(-1, \infty)$

D. maximum value of $f(x)$ is 1

Choose the correct answer from the options given below:

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The interval in which $f(x)=-2x^3-9x^2-12x+1$ is increasing, is:

For the function f(x)=e−2x(2−x)2f(x) = e^{-2x}(2-x)^2f(x)=e−2x(2−x)2, the point of local maxima is:

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

For the function $f(x) = e^{-2x}(2-x)^2$ , the point of local maxima is: