Question

Match the elements from $\text{Column-I}$ to $\text{Column-II}$.
$\begin{array}{cccc}& \text{Column-I}& & \text{Column-II}\\ \text{(A)}& \text{Let}f\left(x\right)\text{be a continuous function, where}f\left(1\right)=3& \text{(P)}& 1\\ & \text{and}F\left(x\right)\text{is defined as}F\left(x\right)=\underset{0}{\overset{x}{\int }}(t^2\cdot \underset{1}{\overset{t}{\int }}f\left(u\right)du)dt\text{.}& & \\ & \text{Then the value of}{F}^{\prime \prime }\left(1\right)\text{is}& & \\ \text{(B)}& f_a,f_b\text{and}{f}_c\text{denote the lengths of the interior angle}& \text{(Q)}& 10\\ & \text{bisector in a triangle of side lengths}a,b,c\text{and area}T.& & \\ & \text{If}\frac{f_a\cdot f_b\cdot f_c}{abc}=\frac{\lambda T(a+b+c)}{(a+b)(b+c)(c+a)}\text{, then the value}& & \\ & \text{of}\lambda \text{is}& & \\ \text{(C)}& \text{Let}{a}_n\text{be the}{n}^{\text{th}}\text{term of an A.P. Let}{S}_n\text{be the sum}& \text{(R)}& 3\\ & \text{of the first}n\text{terms of the A.P. where}{a}_1=1\text{and}& & \\ & a_3=3a_8.\text{If}{S}_n\text{is maximum, then the value of}n\text{is}& & \\ \text{(D)}& \text{If}x={\tan }^{-1}\left(t\right)\text{is substituted in the differential}& \text{(S)}& 4\\ & \text{equation}\frac{d^{2}y}{d{x}^2}+xy\frac{dy}{dx}+{\sec }^{2}x=0,\text{it becomes}& & \\ & (1+t^2)\frac{d^{2}y}{d{t}^2}+(2t+y{\tan }^{-1}(t\left)\right)\frac{dy}{dt}=k.\text{Then}& & \\ & \text{the value of}k\text{is}& & \\ & & \text{(T)}& -1\end{array}$

Which of the following is correct combination?

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

Answer 1

The correct option is A $\left(C\right)\to \left(Q\right),\left(D\right)\to \left(T\right)$

$\left(C\right)$
$a_1=1$
Let the common difference be $d$
Now, $a_3=3a_8$
$\Rightarrow 1+2d=3(1+7d)\Rightarrow d=-\frac{2}{19}$
Therefore, $S_n=\frac{n}{2}[2+(n-1\left)d\right]$
$\Rightarrow S_n=\frac{n}{2}[2+(n-1\left)(-\frac{2}{19})\right]\Rightarrow S_n=\frac{-n^2+20n}{19}\Rightarrow S_n=\frac{100-(n-10{)}^2}{19}$
So, the maximum value of $S_n$ occurs at $n=10.$
$\left(C\right)\to \left(Q\right)$

$\left(D\right)$
Given D.E. is
$\frac{d^{2}y}{d{x}^2}+xy\frac{dy}{dx}+{\sec }^{2}x=0$
Substituting $x={\tan }^{-1}t$
$\Rightarrow \frac{dx}{dt}=\frac{1}{1+t^2}\Rightarrow \frac{dy}{dx}=\frac{dy}{dt}\cdot \frac{dt}{dx}=\frac{dy}{dt}(1+t^2)\Rightarrow \frac{d^{2}y}{d{t}^2}=\frac{d}{dt}\left[\frac{dy}{dt}\right(1+t^2\left)\right]\cdot \frac{dt}{dx}\Rightarrow \frac{d^{2}y}{d{t}^2}=[2t\frac{dy}{dt}+(1+t^2\left)\frac{d^{2}y}{d{t}^2}\right](1+t^2)$

Putting it in the given D.E., we get
$(1+t^2)[2t\frac{dy}{dt}+(1+t^2\left)\frac{d^{2}y}{d{t}^2}\right]+y{\tan }^{-1}t\left[\frac{dy}{dx}\right(1+t^2\left)\right]+(1+t^2)=0\Rightarrow (1+t^2)\frac{d^{2}y}{d{t}^2}+(2t+y{\tan }^{-1}t)\frac{dy}{dt}=-1\therefore k=-1$
$\left(D\right)\to \left(T\right)$