Chapter 4: Problem 3
\(f(x)=x^{2} \ln g(x)\) \(f^{\prime}(x)=2 x \ln g(x)+\frac{x^{2} g^{\prime}(x)}{g(x)}\) \(f^{\prime}(2)=4 \ln 3-\frac{16}{3}\)
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\(f(x)=\ln \sin x\) \(f^{\prime}(x)=\frac{1}{\ln \sin x} \times \frac{1}{\sin x} \times \cos x\) \(f^{\prime}\left(\frac{\pi}{6}\right)=\frac{-1}{\ln 2} \times \sqrt{3}\)
$\begin{aligned}
&f(x)=\left|x^{2}-3\right| x|+2| \\
&f(x)= \begin{cases}x^{2}-3 x+2, & 0
Assume that \(f\) is differentiable for all \(x\). The sign of \(f^{\prime}\) is as follows: \(\mathrm{f}^{\prime}(\mathrm{x})>0\) on \((-\infty,-4)\) \(\mathrm{f}^{\prime}(\mathrm{x})<0\) on \((-4,6)\) \(\mathrm{f}^{\prime}(\mathrm{x})>0\) on \((6, \infty)\) Let \(g(x)=f(10-2 x)\). The value of \(g^{\prime}(L)\) is (A) positive (B) negative (C) zero (D) the function \(g\) is not differentiable at \(x=5\)
$\frac{f(2 x+2 y)-f(2 x-2 y)}{f(2 x+2 y)+f(2 x-2 y)}=\frac{\cos x \sin y}{\sin x \cos y}$ Applying C \& D, $\frac{\mathrm{f}(2 \mathrm{x}+2 \mathrm{y})}{\mathrm{f}(2 \mathrm{x}-2 \mathrm{y})}=\frac{\sin (\mathrm{x}+\mathrm{y})}{\sin (\mathrm{x}-\mathrm{y})}$ \(\frac{\mathrm{f}(2 \mathrm{x}+2 \mathrm{y})}{\sin (\mathrm{x}+\mathrm{y})}\) is constant \(\Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{C} \sin \frac{\mathrm{x}}{2}\) \(\mathrm{C}=1\) using \(\mathrm{f}^{\prime}(0)=\frac{1}{2}\) \(\Rightarrow \mathrm{f}(\mathrm{x})=\sin \frac{\mathrm{x}}{2}\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2} \cos \frac{\mathrm{x}}{2}\) $\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{-1}{4} \sin \frac{\mathrm{x}}{2}$ $\Rightarrow 4 \mathrm{f}^{\prime \prime}(\mathrm{x})+\mathrm{f}(\mathrm{x})=0$
Assertion (A) : Let \(f: R \rightarrow R\) be a smooth function such that \(\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{f}(1-\mathrm{x})\) for all \(\mathrm{x}\) and \(\mathrm{f}(0)=1 .\) Then \(\mathrm{f}(\mathrm{x})=\cos\) \(x+(\sec 1+\tan 1) \sin x\) Reason (R) : Differentiating the given equation gives $f^{\prime \prime}(x)=-f(x)\(. This has solution of the form \)f(x)=A \cos x+$ \(\mathrm{B} \sin \mathrm{x}\), when \(\mathrm{A} \& \mathrm{~B}\) are determined by the boundary conditions.
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