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f(x) = (x2 − 4) \left|x^2-5 x+6\right|+\cos (|x|) is non-differentiable at

 

Option: 1

x = 0


Option: 2

x = 2


Option: 3

x = −2


Option: 4

x = 3

 


Answers (1)

best_answer

f(x) = (x2 – 4) |(x – 2) (x – 3)| + cosx


f(x)=\left\{\begin{array}{lr} \left(x^2-4\right)\left(x^2-5 x+6\right)+\cos x, & x \leq 2 \text { or } x \geq 3 \\ \left(4-x^2\right)\left(x^2-5 x+6\right), & x \in(2,3) \end{array}\right.

only points where f(x) may be non–differentiable are x = 2 and x = 3

\begin{aligned} & \quad\left\{\begin{array}{lr} \left(x^2-4\right)(2 x-5)+2 x\left(x^2-5 x+6\right)-\sin x, & x<2 \text { or } x>3 \\ \left(4-x^2\right)(2 x-5)-\left(x^2-5 x+6\right)(2 x)-\sin x, & x \in(2,3) \end{array}\right. \\ & f^{\prime}(x)= \\ & f^{\prime}(2-0)=-\sin 2, \\ & f^{\prime}(2+0)=-\sin 2, \\ & f^{\prime}(3-0)=-5-\sin 3 \\ & f^{\prime}(3+0)=5-\sin 3 \end{aligned}Thus, f(x) is differentiable at x = 2 but not at x = 3.

 

Posted by

Ajit Kumar Dubey

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