Question

Match $\text{List I}$ with the $\text{List II}$ and select the correct answer using the code given below the lists :

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& \text{Lines with direction ratios}(1,–c,–b),(–c,1,–a)\text{and}(–b,–a,1)& \text{(P)}& -1\\ & \text{are coplanar. Then}{a}^2+b^2+c^2+2abc\text{is}& & \\ \text{(B)}& \text{If the lines}x=ay+1,z=by+2\text{and}x=cy+3,z=dy+4& \text{(Q)}& 1\\ & \text{are perpendicular, then}ac+bd\text{is equal to}& & \\ \text{(C)}& \text{If}(a,b,c)\text{lies on a plane which forms}△ABC\text{with coordinate axes}& \text{(R)}& 3\\ & \text{whose centroid lies on}(\alpha ,\beta ,\gamma ),\text{then}\frac{a}{\alpha }+\frac{b}{\beta }+\frac{c}{\gamma }\text{is equal to}& & \\ \text{(D)}& \text{Let}\left[x\right]\text{denote the greatest integer less than or equal to}x.\text{Then}& \text{(S)}& 0\\ & f\left(x\right)=\left[x\sin \pi x\right]\text{is not differentiable when}x\text{is equal to}& & \\ & & \text{(T)}& 2\end{array}$

Which of the following is the only CORRECT combination?

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

Answer 1

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

Let $A(x_1,0,0),B(0,y_1,0),C(0,0,z_1)$
Centroid $(\alpha ,\beta ,\gamma )=(\frac{x_1}{3},\frac{y_1}{3},\frac{z_1}{3})$
The equation of plane $ABC$ is
$\frac{x}{x_1}+\frac{y}{y_1}+\frac{z}{z_1}=1$
$\Rightarrow \frac{x}{\alpha }+\frac{y}{\beta }+\frac{z}{\gamma }=3$
$(a,b,c)$ lies on the plane.
So, $\frac{a}{\alpha }+\frac{b}{\beta }+\frac{c}{\gamma }=3$

$f\left(x\right)=\left[x\sin \pi x\right]$
We know, greatest integer function is discontinuous where it takes integral values.
Clearly, $f\left(x\right)$ is discontinuous at $\dots ,-3,-2,-1,1,2,3,\dots$ and hence not differentiable.

At $x=0,$ $f\left(x\right)$ is continuous.
If $-1<x<0,$ then $-1\le \sin \pi x<0$
$\Rightarrow 0<x\sin \pi x<1$
$\Rightarrow \left[x\sin \pi x\right]=0$

If $0<x<1,$ then $0<\sin \pi x\le 1$
$\Rightarrow 0<x\sin \pi x<1$
$\Rightarrow \left[x\sin \pi x\right]=0$

$\therefore f\left(x\right)=0$ for all $x\in (-1,1)$
So, $f\left(x\right)$ is continuous and differentiable at $x=0$