Mathematics - Application of Derivative Question with Solution | TestHub

(P) We have

$$\left(e^x+e^{-x}\right)=e^{-y}$$

Differentiate both sides w.r.t x , we get

$$\left({\mathrm{e}}^{\mathrm{x}}-{\mathrm{e}}^{-\mathrm{x}}\right)=-{\mathrm{e}}^{-\mathrm{y}}\cdot \frac{\mathrm{dy}}{\mathrm{dx}}$$

Again differentiate both sides w.r.t x , we get

$$-\left(e^x+e^{-x}\right)=e^{-y}\frac{d^{2}y}{d{x}^2}-e^{-y}{\left(\frac{dy}{dx}\right)}^2$$

$\Rightarrow y^{\prime \prime }-{\left(y^{\prime }\right)}^2+1=0$

$\therefore \mathrm{c}=1$.

(Q) $f(x)=x^2+4\cos x+5,x\in R$

$f^{\prime }(\mathrm{x})=2\mathrm{x}-4\cos \mathrm{x}+0=2(\mathrm{x}-2\sin \mathrm{x})$

Also, $f^{\prime \prime }(\mathrm{x})=2(1-2\cos \mathrm{x})$

Now, put $f^{\prime }(\mathrm{x})=0$, we get

$\mathrm{X}=\alpha \in \left(-\pi ,-\frac{\pi }{2}\right),\mathrm{X}=0\&\mathrm{X}=\beta \in \left(\frac{\pi }{2},\pi \right)$ (using graphically or IVT)

$\therefore \mathrm{x}=0$ (point of local maximum) and $x=\alpha ,\beta$ (point of local minimum)

⇒ Total number of extremum points $=3$

(R) $\mathrm{y}=\frac{{\mathrm{x}}^3}{(\mathrm{x}-\alpha )(\mathrm{x}-\beta )(\mathrm{x}-\gamma )}$

$\therefore \mathrm{y}(\mathrm{x}=2)=\frac{8}{(2-\alpha )(2-\beta )(2-\gamma )}=\frac{8}{4}=2$

(S) Put $|\mathrm{x}|=\mathrm{t}$ As $\mathrm{x}\in [-1,2]\Rightarrow \mathrm{t}\in [0,2]$

let $g(t)=2t^3+3t^2-12t+1$

$\therefore g^{\prime }(t)=6(t+2)(t-1)$

$\therefore g(0)=1$

$$g(1)=-6$$

and $g(2)=5$, is the greatest value.