KCET · Maths · Differentiation
If \(f(x)=\frac{g(x)+g(-x)}{2}+\frac{2}{[h(x)+h(-x)]^{-1}}\) where \(g\) and \(h\) are differentiable function, then \(\mathrm{f}^{\prime}(0)\)
- A 1
- B \(\frac{1}{2}\)
- C \(\frac{3}{2}\)
- D 0
Answer & Solution
Correct Answer
(D) 0
Step-by-step Solution
Detailed explanation
Given,
\(f(x)=\frac{g(x)+g(-x)}{2}+\frac{2}{[h(x)+h(-x)]^{-1}}\) \(\Rightarrow f(x)=\frac{g(x)+g(-x)}{2}+2[h(x)+h(-x)]\)
On differentiating w.r.t. \(x\), we get
\(f^{\prime}(x)=\frac{g^{\prime}(x)-g^{\prime}(-x)}{2}+2\left[h^{\prime}(x)-h^{\prime}(-x)\right]\) \(\therefore f^{\prime}(0)=\frac{g^{\prime}(0)-g^{\prime}(0)}{2}+2\left[h^{\prime}(0)-h^{\prime}(0)\right]\) \(=0\)
\(f(x)=\frac{g(x)+g(-x)}{2}+\frac{2}{[h(x)+h(-x)]^{-1}}\) \(\Rightarrow f(x)=\frac{g(x)+g(-x)}{2}+2[h(x)+h(-x)]\)
On differentiating w.r.t. \(x\), we get
\(f^{\prime}(x)=\frac{g^{\prime}(x)-g^{\prime}(-x)}{2}+2\left[h^{\prime}(x)-h^{\prime}(-x)\right]\) \(\therefore f^{\prime}(0)=\frac{g^{\prime}(0)-g^{\prime}(0)}{2}+2\left[h^{\prime}(0)-h^{\prime}(0)\right]\) \(=0\)
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