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Let f : R\rightarrow R defined as f(x)=\left\{\begin{matrix} x^5\sin \left ( \frac{1}{x}+5x^2 \right ),x<0\\0,x=0 \\x^5\cos\left ( \frac{1}{x}+\lambda z^2,x<0 \right ) \end{matrix}\right. The value of \lambda for which f"(x) exists is.
Option: 1 2
Option: 2 3
Option: 3 5
Option: 4 6

Answers (1)

best_answer

f'(x)=\left\{\begin{matrix} 5x^4\sin \left ( \frac{1}{x} \right )-x^3\cos\left ( \frac{1}{x} \right )+10x,x<0\\0,\qquad\qquad x=0 \\5x^4\cos\left ( \frac{1}{x} \right )+x^3\sin\left ( \frac{1}{x} \right )+2\lambda ,x>0 \end{matrix}\right.

f''(x)=\left\{\begin{matrix} \text{term having x in muliplication}+10,x<0\\0,\qquad\qquad x=0 \\\text{term having x in muliplication}+2\lambda ,x>0 \end{matrix}\right.

L.H.L = R.H.L.

\\ 2\lambda =10\\ \lambda =5

Correct Answer : 5

Posted by

himanshu.meshram

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