Question

$\text{List I}$ has four entries and $\text{List II}$ has four entries. Each entry of $\text{List I}$ has a unique match in $\text{List II}.$

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& \text{Let}f\text{be a real-valued differentiable}& \text{(P)}& 18\\ & \text{function on}R\text{such that}{f}^{\prime }\left(1\right)=6& & \\ & \text{and}{f}^{\prime }\left(2\right)=2.\text{Then}& & \\ & \underset{h\to 0}{lim}\frac{f(3\cos h+4\sin h-2)-f\left(1\right)}{f(3e^h-5\sec h+4)-f\left(2\right)}& & \\ & \text{is equal to}& & \\ \text{(B)}& \text{If}(0.5{)}^{{\log }_3{\log }_{1/5}(x^2-\frac{4}{5})}>1,\text{then the least}& \text{(Q)}& 4\\ & \text{positive integral value of}x\text{is}& & \\ \text{(C)}& \text{The value of}\underset{x\to 1}{lim}\frac{{\sin }^2(x^3+x^2+x-3)}{1-\cos (x^2-4x+3)}\text{is}& \text{(R)}& 5\\ \text{(D)}& \text{Number of points of non-differentiability of}& \text{(S)}& 2\\ & f\left(x\right)=min(|\sin x|,|\cos x|,\frac{1}{4})\text{in}(0,\pi )\text{is}& & \end{array}$

Which of the following is the only CORRECT combination?

• A. $\text{(C)}\to \text{(Q)}$
• B. $\text{(D)}\to \text{(Q)}$
• C. $\text{(C)}\to \text{(P)}$
• D. $\text{(D)}\to \text{(S)}$

Answer 1

The correct option is C $\text{(C)}\to \text{(P)}$

$\text{(C)}$
$\underset{x\to 1}{lim}\frac{{\sin }^2(x^3+x^2+x-3)}{1-\cos (x^2-4x+3)}$
$=\underset{x\to 1}{lim}\frac{{\sin }^2(x^3+x^2+x-3)}{{(x^3+x^2+x-3)}^2}\cdot \frac{{(x^3+x^2+x-3)}^2}{1-\cos (x^2-4x+3)}$
$=\underset{x\to 1}{lim}1\cdot \frac{{(x^2-4x+3)}^2}{1-\cos (x^2-4x+3)}\cdot \frac{{(x^3+x^2+x-3)}^2}{{(x^2-4x+3)}^2}$
$=\underset{x\to 1}{lim}1\cdot 2\cdot {\left(\frac{x^3+x^2+x-3}{x^2-4x+3}\right)}^2$
$=2l^2$, where $l=\underset{x\to 1}{lim}\frac{3x^2+2x+1}{2x-4}=\frac{6}{-2}=-3$
$\therefore$ Limit $=2(-3{)}^2=18$

$\text{(D)}$


Hence, graph of $min(|\sin x|,|\cos x|,\frac{1}{4})$ can be plotted as


Clearly, from the graph, total number of non-differentiable points of $f\left(x\right)$ in the interval $(0,\pi )$ is $5.$