Question

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& \text{If}f\left(x\right)=\{\begin{array}{c}x;x\le 1\\ x^2+bx+c;x>1\end{array}\text{and}{f}^{\prime }\left(x\right)\text{exist finitely for}x\in R,& \text{(P)}& -1\\ & {\text{then}}^{\prime }b{c}^{\prime }\text{is equal to}& & \\ \text{(B)}& \text{The number of tangents to the}\text{curve}{y}^2-2x^3-4y+8=0\text{that}& \text{(Q)}& 1\\ & \text{pass through}(1,2)\text{is}& & \\ \text{(C)}& f\left(x\right)=2x^3-3a{x}^2+\frac{4}{3}{a}^{2}x+1(a>0)\text{attains its maxima and}& \text{(R)}& 2\\ & \text{minima at}x=k\text{and}x=k^2\text{respectively,}\text{then}a\text{is}& & \\ \text{(D)}& \text{If}f\left(x\right)=\underset{-1}{\overset{1}{\int }}\frac{\sin x}{1+t^2}dt\text{and}{f}^{\prime }\left(\frac{\pi }{3}\right)=\frac{\pi }{k},{\text{then}}^{\prime }{k}^{\prime }\text{is equal to}& \text{(S)}& 3\\ & & \text{(T)}& 4\\ & & \text{(U)}& 6\end{array}$

Which of the following option is CORRECT ?

• A. $\left(A\right)\to \left(Q\right)$
• B. $\left(B\right)\to \left(U\right)$
• C. $\left(C\right)\to \left(S\right)$
• D. $\left(D\right)\to \left(T\right)$

Answer 1

The correct option is D $\left(D\right)\to \left(T\right)$

$\left(A\right)$
$f\left(x\right)=\{\begin{array}{c}x;x\le 1\\ x^2+bx+c;x>1\end{array}$
So the function is continuous at $x=1$,
$1=1+b+c\Rightarrow b+c=0$
Now,
$f^{\prime }\left(x\right)=\{\begin{array}{c}1;x\le 1\\ 2x+b;x>1\end{array}$
So the function is differentiable at $x=1$,
$1=2+b\Rightarrow b=-1$
$\therefore bc=-1\times 1=-1$
$\left(A\right)\to \left(P\right)$

$\left(B\right)$
$y^2-2x^3-4y+8=0\cdots \left(1\right)\Rightarrow 2y{y}^{\prime }-6x^2-4y^{\prime }=0\Rightarrow y^{\prime }=\frac{3x^2}{y-2}$
Tangent at $(x_1,y_1)$,
$\frac{y-y_1}{x-x_1}=\frac{3x_1^2}{y_1-2}\cdots \left(2\right)$
Equation $\left(1\right)$,
$y^2-2x^3-4y+8=0\Rightarrow (y-2{)}^2=2x^3-4\Rightarrow (y_1-2{)}^2=2x_1^3-4$
Equation $\left(2\right)$,
$\frac{y-y_1}{x-x_1}=\frac{3x_1^2}{y_1-2}$
Putting $(1,2)$,
$\Rightarrow -(y_1-2{)}^2=3x_1^2(1-x_1)\Rightarrow -2x_1^3+4=3x_1^2(1-x_1)\Rightarrow (x_1-2{)}^2(x_1+1)=0\Rightarrow x_1=-1,2$
When $x=2$
$(y_1-2{)}^2=3\times 4(1)\Rightarrow y_1=2\pm 2\sqrt{3}$
When $x=-1$
$(y_1-2{)}^2=-6\Rightarrow y\to \text{non real}$
Therefore, two possible tangents.
$\left(B\right)\to \left(R\right)$

$\left(C\right)$
$f\left(x\right)=2x^3-3a{x}^2+\frac{4}{3}{a}^{2}x+1\Rightarrow f^{\prime }\left(x\right)=6x^2-6ax+\frac{4}{3}{a}^2=0\Rightarrow (3x-a)(3x-2a)=0\Rightarrow x=\frac{a}{3},\frac{2a}{3}$
Now,
$f^{\prime \prime }\left(x\right)=12x-6a{f}^{\prime \prime }\left(\frac{a}{3}\right)<0f^{\prime \prime }\left(\frac{2a}{3}\right)>0\therefore p=\frac{a}{3},p^2=\frac{2a}{3}\Rightarrow \frac{a^2}{9}=\frac{2a}{3}\Rightarrow a=6$
$\left(C\right)\to \left(U\right)$

$\left(D\right)$
$f\left(x\right)=\underset{-1}{\overset{1}{\int }}\frac{\sin x}{1+t^2}dt\Rightarrow f\left(x\right)=\sin x\underset{-1}{\overset{1}{\int }}\frac{1}{1+t^2}dt\Rightarrow f\left(x\right)=\sin x{\left[{\tan }^{-1}t\right]}_{-1}^1\Rightarrow f\left(x\right)=\frac{\pi }{2}\sin x\Rightarrow f^{\prime }\left(x\right)=\frac{\pi }{2}\cos x\Rightarrow f^{\prime }\left(\frac{\pi }{3}\right)=\frac{\pi }{4}=\frac{\pi }{k}$
Therefore, $k=4$
$\left(D\right)\to \left(T\right)$