JEE Mains · Maths · STD 12 - 5. continuity and differentiation
Let \(f: R \rightarrow R\) be defined as \(f(x)=\left\{\begin{array}{ccc}x^{5} \sin \left(\frac{1}{x}\right)+5 x^{2}& , & x<0 \\ 0 & , & x=0 \\ x^{5} \cos \left(\frac{1}{x}\right)+\lambda x^{2} & , & x>0\end{array} .\right.\) The value of \(\lambda\) for which \(f^{\prime \prime}(0)\) exists, is
- A \(5\)
- B \(10\)
- C \(15\)
- D \(20\)
Answer & Solution
Correct Answer
(A) \(5\)
Step-by-step Solution
Detailed explanation
\(f(x)=x^{5} \cdot \sin \frac{1}{x}+5 x^{2} \quad\) if \(x<0\) \(f(x)=0\) if \(x=0\) \(f(x)=x^{5} \cdot \cos \frac{1}{x}+\lambda x^{2} \quad\) if \(x>0\) LHD of \(f^{\prime}(x)\) at \(x=0\) is 10 \(RHD\) of \(f^{\prime}( x )\) at \(x =0\) is \(2 \lambda\) if…
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