$\mathrm{f}(\mathrm{x})=\mathrm{x} \cos \frac{1}{\mathrm{x}}, \mathrm{x} \geq 1$

$\mathrm{f}^{\prime}(\mathrm{x})=\cos \frac{1}{\mathrm{x}}+\frac{1}{\mathrm{x}} \sin \frac{1}{\mathrm{x}}$

$\lim _{x \rightarrow \infty} \mathrm{f}^{\prime}(\mathrm{x})=1$

$\mathrm{Now} \quad \mathrm{f}^{\prime \prime}(\mathrm{x})=-\sin \frac{1}{\mathrm{x}} \times-\frac{1}{\mathrm{x}^{2}}-\frac{1}{\mathrm{x}^{2}} \sin \frac{1}{\mathrm{x}}+\frac{1}{\mathrm{x}} \cos \frac{1}{\mathrm{x}} \times-\frac{1}{\mathrm{x}^{2}}$

$\therefore \mathrm{f}^{\prime}(\mathrm{x}+2)<\mathrm{f}^{\prime}(\mathrm{x})$

$\therefore \mathrm{f}^{\prime}(\mathrm{x}+2)<\mathrm{f}^{\prime}(\mathrm{x})$

Also $\lim _{\mathrm{x} \rightarrow \infty} \mathrm{f}(\mathrm{x}+2)-\mathrm{f}(\mathrm{x})$

$=\lim _{x \rightarrow \infty}\left[(x+2) \cos \frac{1}{(x+2)}-x \cos \frac{1}{x}\right]=2$

$\therefore f(x+2)-f(x)>2 \quad \forall x \geq 1$

$\mathrm{f}(\mathrm{x})=\ln \mathrm{x}+\int_{0}^{\mathrm{x}} \sqrt{1+\sin \mathrm{t}} \mathrm{d} \mathrm{t}$

$\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\mathrm{x}}+\sqrt{1+\sin \mathrm{x}}$

$\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\mathrm{x}}+\sqrt{2}\left|\cos \left(\frac{\mathrm{x}}{2}-\frac{\pi}{4}\right)\right|$

$\because\left|\cos \left(\frac{\mathrm{x}}{2}-\frac{\pi}{4}\right)\right|$ is non-derivable

$\therefore \mathrm{f}^{\prime}(\mathrm{x})$ is non-derivable but continuous.

hence option (A) is incorrect \& option (B) is correct.

For option $\mathrm{C}$

$\mathrm{f}(\mathrm{x})=(\ell \mathrm{nx})+\int_{0}^{\mathrm{x}}(\sqrt{1+\sin \mathrm{x}}) \mathrm{d} \mathrm{x}$

since $\mathrm{f}(\mathrm{x})$ is positive increasing function for all $\mathrm{x}>1$

$\Rightarrow|\mathrm{f}(\mathrm{x})|=\mathrm{f}(\mathrm{x}) \&\left|\mathrm{f}^{\prime}(\mathrm{x})\right|=\mathrm{f}^{\prime}(\mathrm{x})$

Let $\mathrm{f}(\mathrm{x})=\mathrm{y}$

$\mathrm{f}^{\prime}(\mathrm{x})-\mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{x}}-\ell \mathrm{nx}+\sqrt{1+\sin \mathrm{x}}-\int_{0}^{\mathrm{x}} \sqrt{1+\sin \mathrm{t}} \mathrm{d} \mathrm{t}$

$\mathrm{f}^{(\mathrm{x})}-\mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{x}}-\ln \mathrm{x}+\sqrt{1+\sin \mathrm{x}}-\sqrt{2} \int_{0}^{\mathrm{x}}\left(\frac{\mathrm{t}}{2}-\frac{\pi}{4}\right) \| \mathrm{dt}$

$\frac{1}{\mathrm{x}}-\ln \mathrm{x}<0 ;$ when $\alpha>\mathrm{e}$

$0 \leq \sqrt{1+\sin \mathrm{x}} \leq \sqrt{2}$

$\int_{0}^{\mathrm{x}}\left|\cos \left(\frac{\mathrm{t}}{2}-\frac{\pi}{4}\right)\right| \mathrm{dt}>\sqrt{2} \forall \alpha>\frac{3 \pi}{2}$

$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})-\mathrm{f}(\mathrm{x})<0 \quad \forall \quad \alpha>\frac{3 \pi}{2}>1$

Hence option (C) is correct.

For option (D) $|\mathrm{f}(\mathrm{x})|+\left|\mathrm{f}^{\prime}(\mathrm{x})\right| \rightarrow \infty$

when $\mathrm{x} \rightarrow \infty$

Therefore option (D) is incorrect.

Alternate :

$\mathrm{f}(\mathrm{x})=\ln \mathrm{x}+\int_{0}^{\mathrm{x}} \sqrt{1+\sin \mathrm{t}} \mathrm{dt}$

$\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\mathrm{x}}+\sqrt{1+\sin \mathrm{x}}$ ……..(i)

for $\mathrm{x}>1$

$\frac{1}{\mathrm{x}}+\sqrt{1+\sin \mathrm{x}}<1+\sqrt{2}$

but $\ell \mathrm{nx}+\int_{0}^{\mathrm{x}} \sqrt{1+\sin \mathrm{t}} \mathrm{dt}$ will always be more than $1+\sqrt{2}$ for some $\alpha>1$

$\because \int_{0}^{x} \sqrt{1+\sin t}>0 \quad \& \quad \ln x$ is increasing in $(1, \infty)$

$\Rightarrow \quad \mathrm{f}(\mathrm{x})>\mathrm{f}^{\prime}(\mathrm{x}) \forall \alpha>1$

$\therefore(\mathrm{C})$ is correct

$\mathrm{f}^{\prime \prime}(\mathrm{x})=-\frac{1}{\mathrm{x}^{2}}+\frac{\cos \mathrm{x}}{2 \sqrt{1+\sin \mathrm{x}}}$

$\Rightarrow \mathrm{f}$ is not derivable on $(0, \infty)$

at $\frac{3 \pi}{2}, \frac{7 \pi}{2}$

$\therefore$ (B) is also correct

$\mathrm{f}(\mathrm{x})$ is unbounded near $\mathrm{x}=0$ in $(0,1)$ hence $|\mathrm{f}(\mathrm{x})|$ can never be made less

than a finite number hence $|\mathrm{f}(\mathrm{x})|+\left|\mathrm{f}^{\prime}(\mathrm{x})\right|$ can never be less than $\beta$

$f:(0,1) \rightarrow \mathrm{R}$

$f(\mathrm{x})=\frac{\mathrm{b}-\mathrm{x}}{1-\mathrm{bx}}$

$\mathrm{b} \in(0,1)$

$\Rightarrow \quad f^{\prime}(x)=\frac{b^{2}-1}{(b x-1)^{2}}$

$\Rightarrow f^{\prime}(x)<0 \forall x \in(0,1)$

hence $f(x)$ is decreasing function

hence its range $(-1, b)$

$\Rightarrow$ co-domain $\neq$ range

$\Rightarrow f(x)$ is non-invertible function

Let $f(x)=x^{4}-4 x^{3}+12 x^{2}+x-1$

$f^{\prime}(x)=4 x^{3}-12 x^{2}+24 x$

$f^{\prime \prime}(x)=12 x^{2}-24 x+24$

$=12\left(x^{2}-2 x+2\right)>0$

$\Rightarrow f^{\prime}(x)$ is strictly increasing function

$\because f^{\prime}(x)$ is cubic polynomial

hence number of roots of $f^{\prime}(x)=0$ is 1

$\Rightarrow$ Number of maximum roots of $f(x)=0$ are 2

Now $f(0)=-1, f(1)=9, f(-1)=15$

$\Rightarrow f(x)$ has exactly 2 distinct real roots.

$f(x)=(1-x)^{2} \sin ^{2} x+x^{2}$

$P: f(x)+2 x=2\left(1+x^{2}\right)$

$\Rightarrow(1-x)^{2} \sin ^{2} x+x^{2}+2 x=2+2 x^{2}$

$\Rightarrow(1-x)^{2} \sin ^{2} x-x^{2}+2 x-2=0$

$(1-x)^{2} \cos ^{2} x+1=0$

which is not possible.

$\therefore \mathrm{P}$ is false.

$\mathrm{Q}: 2 f(\mathrm{x})+1=2 \mathrm{x}(1+\mathrm{x})$

$2 \mathrm{x}^{2}+2(1-\mathrm{x})^{2} \sin ^{2} \mathrm{x}+1=2 \mathrm{x}^{2}+2 \mathrm{x}$

$2(1-\mathrm{x})^{2} \sin ^{2} \mathrm{x}-2 \mathrm{x}+1=0$

Let $\mathrm{h}(\mathrm{x})=2(1-\mathrm{x})^{2} \sin ^{2} \mathrm{x}-2 \mathrm{x}+1$

clearly $\mathrm{h}(1)=-1$

and $\mathrm{h}(\mathrm{x})=2\left(\mathrm{x}^{2}-2 \mathrm{x}+1\right) \sin ^{2} \mathrm{x}-2 \mathrm{x}+1$

$=\mathrm{x}^{2}\left[2\left(1-\frac{2}{\mathrm{x}}+\frac{1}{\mathrm{x}^{2}}\right) \cdot \sin ^{2} \mathrm{x}-\frac{2}{\mathrm{x}}+\frac{1}{\mathrm{x}^{2}}\right]$

$\therefore \mathrm{h}(\mathrm{x}) \rightarrow \infty$ as $\mathrm{x} \rightarrow \infty$

$\therefore$ By intermediate value theorem

$\mathrm{h}(\mathrm{x})=0$ has a root which is greater than 1

Hence $\mathrm{Q}$ is true.

$\mathrm{g}(\mathrm{x})=\int_{1}^{\mathrm{x}}\left(\frac{2(\mathrm{t}-1)}{(\mathrm{t}+1)}-\ell \mathrm{nt}\right) f(\mathrm{t}) \mathrm{dt}$

$g^{\prime}(x)=\left(\frac{2(x-1)}{x+1}-\ln x\right) f(x)$

$f(\mathrm{x})>0 \quad \forall \mathrm{x} \in \mathrm{R}$

Suppose.

$h(x)=\frac{2(x-1)}{x+1}-\ln x$

$h(x)=2-\left(\frac{4}{x+1}+\ln x\right)$

$h^{\prime}(x)=\frac{4}{(x+1)^{2}}-\frac{1}{x}$

$h^{\prime}(x)=-\frac{(x-1)^{2}}{x(x+1)^{2}}$

$h^{\prime}(x)<0$

So h(x) is decreasing

so h(x) is decreasing

so h(x) $<0 \quad$ (1). $\quad \forall x>1$

h(x) $<0 \quad \forall x>1$

$\{h(1)=0\}$

So $g^{\prime}(x)=h(x) f(x)$

$g^{\prime}(x)<0 \quad \forall x>1$

$\mathrm{g}(\mathrm{x})$ is decreasing in $(1, \infty)$

$f(\mathrm{x})=\int_{0}^{\mathrm{x}} e^{\mathrm{t}^{2}}(\mathrm{t}-2)(\mathrm{t}-3) \mathrm{dt}$

$\Rightarrow f^{\prime}(x)=e^{x^{2}}(x-2)(x-3)$

$\therefore f^{\prime}(2)=f^{\prime}(3)=0$

$\Rightarrow f^{\prime \prime}(\mathrm{c})=0$ for same $\mathrm{c} \in(2,3)$ (by Rolle’s theorem)

$f(x)=x^{2}-x \sin x-\cos x$

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

$\quad=x(2-\cos x)$

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

$\tan \pi \mathrm{x}=-\pi \mathrm{x}$

$\mathrm{y}=\tan \mathrm{x} \pi \& \mathrm{y}=-\pi \mathrm{x}$

Let $F(x)=f(x)-3 g(x)$

$\therefore F(-1)=3 ; F(0)=3 \& F(2)=3$

$\therefore F^{\prime}(x)$ will vanish at least twice in $(-1,0) \cup(0,2)$

$\because F^{\prime \prime}(x)>0$ or $<0 \forall x \in(-1,0) \cup(0,2)$

so there will be exactly one solution in $(-1,0)$ and one in $(0,2)$

Alter

Let $\mathrm{H}(\mathrm{x})=\mathrm{f}(\mathrm{x})-3 \mathrm{g}(\mathrm{x})$

$\mathrm{H}(-1)=\mathrm{H}(0)=\mathrm{H}(2)=3$

applying Rolle’s theorem in interval $[-1,0]$

$\mathrm{H}(\mathrm{x})=\mathrm{f}(\mathrm{x})-3 \mathrm{g}^{\prime}(\mathrm{x})=0$ for at least one c $\mathrm{c} \in(-1,0)$

As $\mathrm{H}^{\prime \prime}(\mathrm{x})$ never vanises in the interval

$\Rightarrow$ exactly one $\mathrm{c} \in(-1,0)$ for which $\mathrm{H}^{\prime}(\mathrm{x})=0$

similarily apply rolle’s theorems in the interval $[0,2]$

$\mathrm{H}^{\prime}(\mathrm{x})=0$ has exactly one solution in $(0,2)$

Let $f(\mathrm{x})=\int_{0}^{\mathrm{x}} \frac{\mathrm{t}^{2}}{1+\mathrm{t}^{4}} \mathrm{dt}-2 \mathrm{x}+1$

$\Rightarrow f^{\prime}(x)=\frac{x^{2}}{1+x^{4}}-2$

as $\quad \frac{1+x^{4}}{x^{2}} \geq 2$

$\Rightarrow \frac{\mathrm{x}^{2}}{1+\mathrm{x}^{4}} \leq \frac{1}{2}$

$\Rightarrow f^{\prime}(x) \leq-\frac{3}{2}$

$\Rightarrow f(\mathrm{x})$ is continuous and decreasing

$f(0)=1$ and $f(1)=\int_{0}^{1} \frac{t^{2}}{1+t^{4}} d t-2 \leq-\frac{3}{2}$

by IVT $f(x)=0$ possesses exactly one solution in $[0,1]$

For option (A),

Let $g(x)=e^{x}-\int_{0}^{x} f(t) \sin t d t$

$\therefore g^{\prime}(x)=e^{x}-(f(x) \cdot \sin x)>0 \forall x \in(0,1)$

$\Rightarrow g(x)=e^{x}-(f(x) \cdot \sin x)>0 \forall x \in(0,1)$

$\Rightarrow g(x)$ is strictly incrasing function.

Also, $g(0)=1$

$\Rightarrow \mathrm{g}(\mathrm{x})>1 \forall \mathrm{x} \in(0,1)$

$\therefore$ option $(\mathrm{A})$ is not possible.

For option $(\mathrm{B}),$ let

$\mathrm{k}(\mathrm{x})=\mathrm{x}^{9}-f(\mathrm{x})$

Now, $\mathrm{k}(0)=-f(0)<0(\mathrm{As} f \in(0,1))$

Also, $\mathrm{k}(1)=1-f(1)>0(\mathrm{As} f \in(0,1))$

$\Rightarrow \mathrm{k}(0) . \mathrm{k}(1)<0$

So, option(B) is correct.

For option (C), let

$\mathrm{T}(\mathrm{x})=f(\mathrm{x})+\int_{0}^{\frac{\pi}{2}} f(\mathrm{t}) \cdot \sin \mathrm{t} \mathrm{dt}$

$\Rightarrow \mathrm{T}(\mathrm{x})>0 \forall \mathrm{x} \in(0,1) \text { (As } f \in(0,1))$

so, option(C) is not possible.

For option (D)

Let $\mathrm{M}(\mathrm{x})=\mathrm{x}-\int_{0}^{\frac{\pi}{2}-\mathrm{x}} f(\mathrm{t}) \cos \mathrm{t} \mathrm{dt}$

$\therefore \mathrm{M}(0)=0-\int_{0}^{\pi / 2} f(\mathrm{t}) \cdot \cos \mathrm{t} \mathrm{dt}<0$

Also, $\mathrm{M}(1)=1-\int_{0}^{\frac{\pi}{2}-1} f(\mathrm{t}) \cdot \cos \mathrm{td} \mathrm{t}>0$

$\Rightarrow \mathrm{M}(0) \cdot \mathrm{M}(1)<0$

$\therefore$ option (D) is correct.