Chapter 4: Problem 28
\(f(x)+2 f(-x)=-\sin x\) Put \(x=-x\) \(f(-x)+2 f(x)=-\sin x\) Solve, (I) \& (II), we get \(f(x)=-\sin x\) \(f^{\prime}(x)=-\cos x\) \(\mathrm{f}^{\prime}\left(\frac{\pi}{4}\right)=-\frac{1}{\sqrt{2}}\)
Chapter 4: Problem 28
\(f(x)+2 f(-x)=-\sin x\) Put \(x=-x\) \(f(-x)+2 f(x)=-\sin x\) Solve, (I) \& (II), we get \(f(x)=-\sin x\) \(f^{\prime}(x)=-\cos x\) \(\mathrm{f}^{\prime}\left(\frac{\pi}{4}\right)=-\frac{1}{\sqrt{2}}\)
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(A) If \(|g(x)-g(y)| \leq|x-y|^{99},(n \in N)\) then \(g^{\prime}(x)\) equals
(B) Differential coefficient of $\sin ^{-1} \frac{2
\mathrm{x}}{1+\mathrm{x}^{2}}\( with respect to \)\cos ^{-1}
\frac{1-x^{2}}{1+x^{2}}\( in the domain of \)f(x)=\frac{1}{\sqrt{1-x^{2}}}$ is
(C) If \(y=\cot ^{-1}(\tan x)+\tan ^{-1}(\cot x), \pi / 2
\(f(x)=\frac{x}{1+e^{1 / x}}\) \(f(x)\left(1+e^{1 / x}\right)-x=0\) \(f^{\prime}(x)\left(1+e^{1 / x}\right)-\frac{f(x) e^{l / x}}{x^{2}}-1=0\) \(x^{2} f^{\prime}(x) \frac{(x)}{f(x)}-f(x) e^{1 / x}-x^{2}=0\)
If for some differentiable function \(\mathrm{f}, \mathrm{f}(\alpha)=0\) and \(\mathrm{f}^{\prime}(\alpha)=0\). Assertion (A) : The sign of \(\mathrm{f}(\mathrm{x})\) does not change in the neighbourhood of \(\mathrm{x}=\alpha\). Reason \((\mathbf{R}): \alpha\) is repeated root of \(\mathrm{f}(\mathrm{x})=0\)
The value of $\frac{\mathrm{f}(\mathrm{t})}{\mathrm{f}^{\prime}(\mathrm{t})} \cdot \frac{\mathrm{f}^{\prime \prime}(-\mathrm{t})}{\mathrm{f}^{\prime}(-\mathrm{t})}+\frac{\mathrm{f}(-\mathrm{t})}{\mathrm{f}^{\prime}(-\mathrm{t})} \cdot \frac{\mathrm{f}^{\prime \prime}(\mathrm{t})}{\mathrm{f}^{\prime}(\mathrm{t})}$ \(\forall \mathrm{t} \in \mathrm{R}\), is equal to (A) \(-2\) (B) 2 (C) \(-4\) (D) 4
As $\mathrm{f}(\mathrm{x}), \mathrm{f}^{\prime}(\mathrm{x}), \mathrm{f}^{\prime \prime}(\mathrm{x})\( are all \)+\mathrm{ve} \forall \mathrm{x} \in[0,7]$ \(\Rightarrow \mathrm{f}(\mathrm{x})\) is increasing function and concave up. and so is \(\mathrm{f}^{-1}(\mathrm{x})\). $\Rightarrow f^{-1}(5)+4 f^{-1}\left(\frac{2 g}{5}\right)\( is always \)+v e$.
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