Question

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& \text{If}\underset{n\to \infty }{lim}(\frac{n^2+1}{n+1}-an)-b=0,\text{then}& & \\ & \text{the value of}b\text{is}& \text{(P)}& 0\\ \text{(B)}& \text{If}{x}^{2}y+y^3=2\text{and the value of}\frac{d^{2}y}{d{x}^2}\text{at}x=1& & \\ & \text{is}-\frac{m}{8}\text{, then the value of}m\text{is}& \text{(Q)}& 1\\ \text{(C)}& \text{If}f\left(x\right)=\{\begin{array}{cc}x,& x\le 1\\ x^2+bx+c,& x>1\end{array}\text{and}{f}^{\prime }\left(x\right)& & \\ & \text{exists for all}x\in R\text{, then the value of}c\text{is}& \text{(R)}& 2\\ \text{(D)}& \text{If}f\left(x\right)=\underset{0}{\overset{x}{\int }}t\sin \frac{1}{t}dt\text{, then the number of}& & \\ & \text{point(s) of discontinuity of}f\left(x\right)\text{in}(0,\pi )\text{is}& \text{(S)}& -1\\ & & \text{(T)}& 4\\ & & \text{(U)}& 3\end{array}$

Which of the following is the only INCORRECT combination?

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

Answer 1

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

$\left(A\right)$
$\underset{n\to \infty }{lim}(\frac{n^2+1}{n+1}-an)-b=0\Rightarrow \underset{n\to \infty }{lim}\left(\frac{(1-a)n^2+1-an}{n+1}\right)-b=0$
For the limit to exist,
$1-a=0\Rightarrow a=1$
$\Rightarrow \underset{n\to \infty }{lim}\left(\frac{1-n}{n+1}\right)-b=0\Rightarrow -1-b=0\Rightarrow b=-1$
$\left(A\right)\to \left(S\right)$

$\left(B\right)$
$x^{2}y+y^3=2$
Differentiating w.r.t. $x$,
$2xy+x^2{y}^{\prime }+3y^2{y}^{\prime }=0$
If $x=1$, then $y^3+y-2=0\Rightarrow y=1$
Putting $x=1,y=1$, we get
$2+y^{\prime }+3y^{\prime }=0\Rightarrow y^{\prime }=-\frac{1}{2}\cdots \left(1\right)$
Again differentiating w.r.t. $x$,
$2y+2x{y}^{\prime }+2x{y}^{\prime }+x^2{y}^{\prime \prime }+6y(y^{\prime }{)}^2+3y^2{y}^{\prime \prime }=0$
Putting $x=1,y=1$ and using equation $\left(1\right)$, we get
$2-1-1+y^{\prime \prime }+\frac{3}{2}+3y^{\prime \prime }=0\Rightarrow y^{\prime \prime }=-\frac{3}{8}\therefore m=3$
$\left(B\right)\to \left(U\right)$

$\left(C\right)$
$f\left(x\right)=\{\begin{array}{cc}x,& x\le 1\\ x^2+bx+c,& x>1\end{array}$

$f^{\prime }\left(x\right)=\{\begin{array}{cc}1,& x\le 1\\ 2x+b,& x>1\end{array}$
The function is differentiable, so it is continuous at $x=1$
$1=1+b+c\Rightarrow b+c=0\cdots \left(2\right)$
Also, $f^{\prime }\left(1^{+}\right)=f^{\prime }\left(1^{-}\right)$
$\Rightarrow 2+b=1\Rightarrow b=-1\therefore c=1\left[\text{From (2)}\right]$
$\left(C\right)\to \left(Q\right)$

$\left(D\right)$
$f\left(x\right)=\underset{0}{\overset{x}{\int }}t\sin \frac{1}{t}dt$
In the interval $(0,\pi ),x\sin \frac{1}{x}$ is continuous.
$\therefore f^{\prime }\left(x\right)=x\sin \frac{1}{x}$
So, $f^{\prime }\left(x\right)$ is finite and exists for all $x\in (0,\pi )$, therefore $f\left(x\right)$ is continuous at all points.
So, the number of points of discontinuity of $f\left(x\right)$ in $(0,\pi )$ is $0$.
$\left(D\right)\to \left(P\right)$