Chapter 6

Here. \(f(x)=\tan ^{-1}(\sin x+\cos x)\).

$$ f(x)=|x|-|x-1|. $$

$$ f(x)=(x-2)|x-3|. $$

$$ \begin{aligned} &f(x) f^{\prime}(x)<0 \forall x \in R \\ &\text { or } \quad \frac{1}{2} \frac{d}{d x}\left(f^{2}(x)\right)<0 \\ &\text { or } \quad \frac{d}{d x}\left(f^{2}(x)\right)<0 \end{aligned} $$

(a) \(f(x)>0, f^{\prime \prime}(x)<0\) (b) \(f(x)<0, f^{\prime \prime}(x)<0\) (c) \(f(x)>0, f^{\prime \prime}(x)>0\) (d) \(f(x)<0, f^{\prime \prime}(x)>0\)

\(f(x)=\left|x \log _{e} x\right|\) For \(g(x)=x \log _{e} x\) $$ g^{\prime}(x)=x \frac{1}{x}+\log _{e} x=1+\log _{e} x $$

$$ g^{\prime}(x)=\left(f^{\prime}\left((\tan x-1)^{2}+3\right)\right) 2(\tan x-1) \sec ^{2} x $$ Since \(f^{\prime \prime}(x)>0, f^{\prime}(x)\) is increasing.

$$ \begin{aligned} &u=\sqrt{c+1}-\sqrt{c} \\ &\qquad u=\frac{1}{\sqrt{c+1}+\sqrt{c}} \text { and } v=\frac{1}{\sqrt{c-1}+\sqrt{c}} \end{aligned} $$ Clearly, \(u

\(\begin{aligned} f^{\prime}(x) &=-\pi \sin \pi x+10+6 x+3 x^{2} \\\ &=3(x+1)^{2}+7-\pi \sin \pi x>0 \text { for all } x \end{aligned}\).

If \(f(x)\) has an extremum at \(x=\pi / 3\), then \(f^{\prime}(x)=0\) at \(x=\pi / 3\).

We have \(f(x)=\sin ^{4} x+\cos ^{4} x=\frac{3}{4}+\frac{1}{4} \cos 4 x\).

\(\frac{d y}{d x}=5 x^{2}(x-1)(x-3)=0\).

$$ \begin{aligned} f(x) &=\frac{(\sin x+\cos x)^{2}-1}{\frac{1}{\sqrt{2}}(\sin x+\cos x)}=\sqrt{2} \frac{t^{2}-1}{t} \\ &=\phi(t)=\sqrt{2}\left(t-\frac{1}{t}\right) \end{aligned} $$ where \(t=g(x)=\sin x+\cos x=\sqrt{2} \cos \left(\frac{\pi}{4}+x\right), x \in[0, \pi / 2]\)

\(f^{\prime}(x)=-3 x^{2}-6 x-2<0 \forall x>0\).

Equation of the tangent to the ellipse \(\frac{x^{2}}{27}+y^{2}=1\) at \((3 \sqrt{3} \cos \theta\), \(\sin \theta), \theta \in(0, \pi / 2)\), is $$ \frac{\sqrt{3} x \cos \theta}{9}+y \sin \theta=1 $$

Fuel charges \(\propto v^{2}\). Let \(F\) represents fuel charges. Then \(F \propto v^{2}\) or \(F=k v^{2}\).

Let \(H\) be the height of the cone and \(\alpha\) be its semi-vertical angle. Suppose that \(x\) is the radius of the inscribed cylinder and \(h\) is its height. $$ \begin{aligned} &h=Q L=O L-O Q=H-x \cot \alpha \\ &V=\text { Volume of the cylinder }=\pi x^{2}(H-x \cot \alpha) \end{aligned} $$

The dimensions of the box after cutting equal squares of side \(x\) on the comer will be \(21-2 x, 16-2 x\), and height \(x\). $$ \begin{aligned} V &=x(21-2 x)(16-2 x) \\ &=x\left(336-74 x+4 x^{2}\right)=4 x^{3}+336 x-74 x^{2} \end{aligned} $$