Calculus

Application of Derivatives

Medium

applied

The maximum and the minimum value of the function $f(x) = -||x+3|-5| + 2$ are :

$Max\,[f(x)] = 7$ ; $Min\,[f(x)] = 2$

$Max\,[f(x)]=$ doesn't exist; $Min\,[f(x)] = -3$

$Max\,[f(x)]=$ doesn't exist; $Min\,[f(x)] = 2$

$Max\,[f(x)] = 2$ ; $Min\,[f(x)]$ doesn't exist

✅ Correct Option: 4

Since $\left||x+3|-5\right|\ge0$ , we have $f(x)\le2$ , with equality when $|x+3|=5$ , i.e. $x=2$ or $x=-8$ . So the maximum is 2. As $|x|$ grows the inner modulus grows without bound, so $f(x)\to-\infty$ and no minimum exists.

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Match the LIST-I with LIST-II

	LIST-I		LIST-II
A.	$x = 0$	I.	Local minimum value of $f(x)$
B.	$x = -2$	II.	Local maximum of value of $f(x)$
C.	$f(0)$	III.	Point of local minima
D.	$f(1)$	IV.	Point of local maxima

Choose the correct answer from the options given below:

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The maximum and the minimum value of the function f(x)=−∣∣x+3∣−5∣+2f(x) = -||x+3|-5| + 2f(x)=−∣∣x+3∣−5∣+2 are :

Solution

Related questions:

Keyboard Shortcuts

CUET Mathematics Past Year Questions (Topic-Wise)

The maximum and the minimum value of the function $f(x) = -||x+3|-5| + 2$ are :