MCQ
Let $f: R \rightarrow R$ be a function defined as  $f(x)=\left\{\begin{array}{cl}\frac{\sin (a+1) x+\sin 2 x}{2 x} & , \text { if } x<0 \\ b & , \text { if } x=0 \\ \frac{\sqrt{x+b x^{3}}-\sqrt{x}}{b x^{5 / 2}} & , \text { if } x>0\end{array}\right.$ . If $f$ is continuous at $x=0,$ then the value of $a + b$ is equal to ....... .
  • A
    $-\frac{5}{2}$
  • B
    $-2$
  • C
    $-3$
  • $-\frac{3}{2}$

Answer

Correct option: D.
$-\frac{3}{2}$
d
$f(x)$ is continuous at $x=0$

$\lim _{x \rightarrow 0^{+}} f(x)=f(0)=\lim _{x \rightarrow 0^{-}} f(x)....(1)$

$f(0)= b....(2)$

$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}\left(\frac{\sin (a+1) x}{2 x}+\frac{\sin 2 x}{2 x}\right)$

$=\frac{a+1}{2}+1....(3)$

$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} \frac{\sqrt{x+b x^{3}}-\sqrt{x}}{b x^{5 / 2}}$

$=\lim _{x \rightarrow 0^{+}} \frac{\left(x+b x^{3}-x\right)}{b x^{5 / 2}\left(\sqrt{x+b x^{3}}+\sqrt{x}\right)}$

$=\lim _{x \rightarrow 0^{+}} \frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{1+b x^{2}}+1\right)}=\frac{1}{2} \quad \ldots(4)$

Use $(2),(3)$ And $(4)$ in $(1)$

$\frac{1}{2}=b=\frac{a+1}{2}+1$

$\Rightarrow b =\frac{1}{2}, a =-2$

$a+b=\frac{-3}{2}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free