⇒Application Of Derivative

Correct Option is (B)

$\begin{array}{rl}\mathrm{f}(x)=& {\sin }^{4}x+{\cos }^{4}x\\ \therefore {\mathrm{f}}^{\prime }(x)& =4{\sin }^{3}x\cos x-4{\cos }^{3}x\sin x\\ & =4\sin x\cos x({\sin }^{2}x-{\cos }^{2}x)\\ & =-2\sin 2x\cos 2x\\ & =-\sin 4x\end{array}$

$\therefore$ If $\mathrm{f}(x)$ is increasing, then ${\mathrm{f}}^{\prime }(x)>0$

$\begin{array}{rl}\text{ i.e., }-\sin 4x>0& \Rightarrow \pi <4x<2\pi \\ & \Rightarrow \frac{\pi }{4}<x<\frac{\pi }{2}\end{array}$

Correct Option is (C)

$f(x)=x^3-3(a-2)x^2+3ax+7$

As $\mathrm{f}(x)$ is increasing in $(0,1]$ and decreasing in $[1,5)$, we get that $\mathrm{f}(x)$ has critical point at $x=1$

$\begin{array}{rl}& \Rightarrow {\mathrm{f}}^{\prime }(1)=0\\ & {\mathrm{f}}^{\prime }(x)=3x^2-6(\mathrm{a}-2)x+3\mathrm{a}\\ & \therefore 3(1{)}^2-6(\mathrm{a}-2)+3\mathrm{a}=0\\ & \therefore \mathrm{a}=5\\ & \therefore \frac{\mathrm{f}(x)-14}{(x-1{)}^2}=\frac{x^3-9x^2+15x-7}{(x-1{)}^2}\\ & =\frac{(x-1{)}^2(x-7)}{(x-1{)}^2}\\ & =x-7\end{array}$

$\therefore$ The required root is 7.

Correct Option is (A)

$\begin{array}{rl}\mathrm{f}(x)& =x{\mathrm{e}}^{x(1-x)}\\ \therefore {\mathrm{f}}^{\prime }(x)& =x{\mathrm{e}}^{x(1-x)}[x(-1)+(1-x)]+{\mathrm{e}}^{x(1-x)}\\ & ={\mathrm{e}}^{x(1-x)}(x-2x^2+1)\end{array}$

For $\mathrm{f}(x)$ to be increasing, ${\mathrm{f}}^{\prime }(x)\ge 0$

$\begin{array}{rl}& \Rightarrow {\mathrm{e}}^{x(1-x)}(x-2x^2+1)\ge 0\\ & \Rightarrow x-2x^2+1\ge 0\\ & \Rightarrow 2x^2-x-1\le 0\\ & \Rightarrow (2x+1)(x-1)\le 0\\ & \Rightarrow x\in [-\frac{1}{2},1]\end{array}$

For $\mathrm{f}(x)$ to be decreasing, ${\mathrm{f}}^{\prime }(x)\le 0$

$\begin{array}{rl}& \Rightarrow (2x+1)(x-1)\ge 0\\ & \Rightarrow x\in (-\mathrm{\infty },-\frac{1}{2}]\cup [1,\mathrm{\infty })\end{array}$

Correct Option is (C)

$\begin{array}{ll}& \text{ Given equation }{x}^3+x-1=0\\ & \text{ Let }\mathrm{f}(x)=x^3+x-1\\ \therefore & {\mathrm{f}}^{\prime }(x)=3x^2+1\\ & \Rightarrow {\mathrm{f}}^{\prime }(x)>0\mathrm{\forall }x\in \mathrm{R}\\ & \Rightarrow \mathrm{f}(x)\text{ is increasing }\\ \therefore & \text{ for }{x}_2>x_1,\mathrm{f}\left(x_2\right)>\mathrm{f}\left(x_1\right)\\ & \text{ Now, }\mathrm{f}(0)=-1\text{ and }\mathrm{f}(1)=1\\ \therefore & \mathrm{f}(x)=0\text{ for some }x\in (0,1)\\ \therefore & \text{ Equation has one real root. }\end{array}$

Correct Option is (A)

$\begin{array}{rl}f(x)& =x^3+6x^2-36x+7\\ f^{\prime }(x)& =3x^2+12x-36\\ & =3(x^2+4x-12)\end{array}$

For $\mathrm{f}(x)$ to be increasing,

$\begin{array}{rl}& {\mathrm{f}}^{\prime }(x)>0\\ & \Rightarrow 3(x^2+4x-12)>0\\ & \Rightarrow x^2+4x-12>0\\ & \Rightarrow (x+6)(x-2)>0\\ & \Rightarrow x\in (-\mathrm{\infty },-6)\cup (2,\mathrm{\infty })\end{array}$