(5)

$\because f:R\to R\text{ and }f\left(0\right)=0, f\left(1\right)=1, f\left(2\right)=-1, f\left(3\right)=2\text{ and }f\left(4\right)=-2,\text{ then}$

$\therefore f\left(x\right)\text{ has at least 4 real roots.}$

$\text{Then }f\text{'}\left(x\right)\text{ has atleast 3 real roots and }f\text{'}\text{'}\left(x\right)\text{ has atleast 2 real roots.}$

$\frac{d}{dx}(f^3·f\text{'}\text{'})=3f^2·f\text{'}·f\text{'}\text{'}+f^3·f\text{'}\text{'}\text{'}=f^2(3f\text{'}·f\text{'}\text{'}+f·f\text{'}\text{'}\text{'}).$

$\text{Hence, }{f}^3.f\text{'}\text{'}\text{ has at least 6 roots.}$

$\text{Then its differentiation has at least 5 distinct roots.}$

(17)

Given, $f\left(x\right)=[\frac{{\displaystyle x}}{{\displaystyle 2}}+3]-\left[\sqrt{x}\right]=\left[\frac{{\displaystyle x}}{{\displaystyle 2}}\right]-\left[\sqrt{x}\right]+3$

$\left[\frac{{\displaystyle x}}{{\displaystyle 2}}\right]$ $\text{is discontinuous at 2,4,}6,8\text{ and [}\sqrt{x}\text{] discontinuous at }1,4$

$\therefore f\left(x\right)\text{ is discontinuous at }1,2,6,8$                

                                                                               $[\because f(4^{+})=3=f(4^{-}\left)\right]$

$\therefore \sum _{a\in S}a=1+2+6+8=17$

(81)

$\text{Since, }f\text{ is continuous at }x=\frac{\pi }{2}$

$\therefore \underset{x\to \frac{{\pi }^{-}}{2}}{\lim }f\left(x\right)=f\left(\frac{{\displaystyle \pi }}{{\displaystyle 2}}\right)=\underset{x\to \frac{{\pi }^{+}}{2}}{\lim }f\left(x\right)$

$\text{Now, }\underset{x\to \frac{{\pi }^{-}}{2}}{\lim }{\text{(}\frac{8}{7}\text{)}}^{\left(\frac{{\displaystyle \tan 8x}}{{\displaystyle \tan 7x}}\right)}$

$=\underset{h\to 0}{\lim }{\left(\frac{{\displaystyle 8}}{{\displaystyle 7}}\right)}^{\frac{\tan (4\pi -8h)}{\tan {\displaystyle (}{\displaystyle 3}{\displaystyle \pi }{\displaystyle +}{\displaystyle \frac{{\displaystyle \pi }}{{\displaystyle 2}}}{\displaystyle -}{\displaystyle 7}{\displaystyle h}{\displaystyle )}}}=\underset{h\to 0}{\lim }{\left(\frac{{\displaystyle 8}}{{\displaystyle 7}}\right)}^{\frac{\tan (-8h)}{\cot \left(7h\right)}}={\left(\frac{{\displaystyle 8}}{{\displaystyle 7}}\right)}^0=1$

$\Rightarrow a-8=1\Rightarrow a=9$

$\text{Also, }\underset{\pi \to {\frac{\pi }{2}}^{+}}{\lim }f\left(x\right)=\underset{\pi \to \frac{\pi }{2}}{\lim }{(1+|\cot x\left|\right)}^{\frac{b}{a}\left|\tan x\right|}$

$=\underset{h\to 0}{\lim }{(1-\tan h)}^{-\frac{b}{9}\cot h}$

$=\underset{h\to 0}{\lim }{(1-\tan h)}^{\left(\frac{{\displaystyle 1}}{{\displaystyle \tan h}}\right)·\left(\tan h\right)·\left(\frac{{\displaystyle -b}}{{\displaystyle 9}}\cot h\right)}=e^{\frac{b}{9}}=1\Rightarrow b=0$

Hence, $a^2+b^2=81+0=81$

(61)

We have $f\text{'}\left(x\right)=3f\left(x\right)+\alpha$

Let $f\left(x\right)=y$ and $\frac{dy}{dx}=f\text{'}\left(x\right)\Rightarrow \frac{dy}{dx}=3y+\alpha \Rightarrow \frac{dy}{3y+\alpha }=dx$

$\Rightarrow \frac{1}{3}\log (3y+\alpha )=x+c \text{... (i) [On integrating]}$

Now, $f\left(0\right)=1,$ so $y\left(0\right)=1$

$\Rightarrow \frac{1}{3}\log (3+\alpha )=c$

$\Rightarrow \frac{1}{3}\log \left(\frac{3y+\alpha }{3+\alpha }\right)=x$                        [Putting value of $c$ in eq. (i)]

$\Rightarrow \frac{3y+\alpha }{3+\alpha }=e^{3x}\Rightarrow y=\frac{e^{3x}(3+\alpha )-\alpha }{3}=f\left(x\right)$

Now, $\underset{x\to -\infty }{\lim }f\left(x\right)=7$

$\Rightarrow \underset{x\to -\infty }{\lim }\frac{e^{3x}(3+\alpha )-\alpha }{3}=7\Rightarrow \frac{-\alpha }{3}=7$

$\Rightarrow \alpha =-21$

$\therefore f\left(x\right)=7-6e^{3x}$

So, $9f(-{\log }_{e}3)=9(7-6e^{3(-{\log }_{e}3)})$

$=9(7-\frac{{\displaystyle 6}}{{\displaystyle 27}})=63-2=61$

(105)

We have, $y=\frac{(\sqrt{x}+1)(x^2-\sqrt{x})}{x(\sqrt{x}+1)+\sqrt{x}}+\frac{1}{15}(3{\cos }^{2}x-5){\cos }^{3}x$

$=\frac{\sqrt{x}(x-1)(x+1+\sqrt{x})}{\sqrt{x}(x+\sqrt{x}+1)}+\frac{1}{15}(3{\cos }^{2}x-5){\cos }^{3}x$

$=x-1+\frac{1}{5}{\cos }^{5}x-\frac{1}{3}{\cos }^{3}x$

$\Rightarrow y\text{'}=1+\frac{1}{5}·5{\cos }^{4}x·(-\sin x)-\frac{1}{3}·3·{\cos }^{2}x·(-\sin x)$

$=1-\sin x·{\cos }^{4}x+\sin x·{\cos }^{2}x$

$\Rightarrow 96y\text{'}\left(\frac{\pi }{6}\right)=96[1-\frac{1}{2}·\frac{9}{16}+\frac{1}{2}·\frac{3}{4}]=105$

(2890)

$f\left(x\right)-f\left(y\right)\ge {\log }_e\left(\frac{x}{y}\right)+x-y\forall (x,y)\in (0,\infty )$                ...(i)

Now, $f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f(x+h)-f\left(x\right)}{h}=\underset{h\to 0}{\lim }{\log }_e\frac{\left(\frac{x+h}{x}\right)+h}{h}$

$=\underset{h\to 0}{\lim }[\frac{{\log }_e(1+\frac{h}{x})}{h}+1]=\frac{1}{x}+1$                          ..(ii)

Now, $\sum _{n=1}^{20}f\text{'}\left(\frac{1}{n^2}\right)=\sum _{n=1}^{20}\frac{1}{\frac{1}{n^2}}+1$                                [Using (i)]

$=\sum _{n=1}^{20}(n^2+1)={[\frac{n(n+1)(2n+1)}{6}+n]}_{n=20}$

$=\frac{20\times 21\times 41}{6}+20=2870+20=2890$

(202)

$f\left(x\right)=x^3+x^{2}f\text{'}\left(1\right)+xf\text{'}\text{'}\left(2\right)+f\text{'}\text{'}\text{'}\left(3\right), x\in R$

$f\text{'}\left(x\right)=3x^2+2xf\text{'}\left(1\right)+f\text{'}\text{'}\left(2\right),$

$f\text{'}\text{'}\left(x\right)= 6x+2f\text{'}\left(1\right)$

$f\text{'}\text{'}\text{'}\left(x\right)=6\Rightarrow f\text{'}\text{'}\text{'}\left(3\right)=6$

$f\text{'}\text{'}\left(2\right)=6\left(2\right)+2f\text{'}\left(1\right)=12+2f\text{'}\left(1\right),$

             $f\text{'}\left(1\right)=3{\left(1\right)}^2+2\left(1\right)f\text{'}\left(1\right)+f\text{'}\text{'}\left(2\right)=3+2f\text{'}\left(1\right)+f\text{'}\text{'}\left(2\right)$

$\Rightarrow f\text{'}\left(1\right)=3+2f\text{'}\left(1\right)+12+2f\text{'}\left(1\right)$

$\Rightarrow 3f\text{'}\left(1\right)=-15\Rightarrow f\text{'}\left(1\right)=-5$

$\therefore f\text{'}\text{'}\left(2\right)=12+2(-5)=2$

Now, $f\text{'}\left(10\right)=3{\left(10\right)}^2+2\left(10\right)f\text{'}\left(1\right)+f\text{'}\text{'}\left(2\right)$

                       $=300+20(-5)+2=202$

(15)

$f\left(x\right)$$=\{\begin{array}{cc}\frac{1}{\left|x\right|},& \left|x\right|\ge 2\\ a{x}^2+2b,& \left|x\right|<2\end{array}$

Since, $f$ is differentiable so $f$ must be continuity

$\text{⇒R.H.L. at }2=\text{L.H.L. at }2$

$\Rightarrow \frac{1}{2}=4a+2b$                                                                    ...(i)

Also, $f$ is differentiable at $x=2,$

Now, around 2, $f\left(x\right)$$=\{\begin{array}{cc}\frac{1}{x},& x\ge 2\\ a{x}^2+2b,& x<2\end{array}$

$\Rightarrow$$f\text{'}\left(x\right)$$=\{\begin{array}{cc}\frac{-1}{x^2},& x\ge 2\\ 2ax,& x<2\end{array}$

Now, R.H.D. at $x=2=$L.H.D. at $x=2$

$\Rightarrow -\frac{1}{4}=4a\Rightarrow a=-\frac{1}{16}\Rightarrow 2b=\frac{1}{2}+\frac{1}{4}=\frac{3}{4} \text{[Using (i)]}$

$\Rightarrow b=\frac{3}{8}$

So, $48(a+b)=48(\frac{{\displaystyle 3}}{{\displaystyle 8}}-\frac{{\displaystyle 1}}{{\displaystyle 16}})=\frac{48\times 5}{16}=15$