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Continuity — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsContinuity1 Mark
MCQ
If $\text{f(x)}=\begin{cases}\frac{\sin(\text{a}+1)}{\text{x}},&\text{x}<0\\\text{c},&\text{x}=0\\\frac{\sqrt{\text{x+bx}^2}-\sqrt{\text{x}}}{\text{bx}\sqrt{\text{x}}},&\text{x}>0&\end{cases}$ is continuouse at $x = 0,$ then:
  • A
    $\text{a}=-\frac{3}{2},\text{b}=0,\text{c}=\frac{1}{2}$
  • B
    $\text{a}=-\frac{3}{2},\text{b}=1,\text{c}=-\frac{1}{2}$
  • ✓
    $\text{a}=-\frac{3}{2},\text{b}\in\text{R}-\{0\},\text{c}=\frac{1}{2}$
  • D
    $\text{None of these}.$
✓

Answer

Correct option: C.
$\text{a}=-\frac{3}{2},\text{b}\in\text{R}-\{0\},\text{c}=\frac{1}{2}$
$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{\text{x+bx}^2}-\sqrt{\text{x}}}{\text{bx}\sqrt{\text{x}}}=\text{c}$
$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{\text{x+bx}^2}-\sqrt{\text{x}}}{\text{bx}\sqrt{\text{x}}}\times\frac{\sqrt{\text{x+bx}^2}+\sqrt{\text{x}}}{\sqrt{\text{x+bx}^2}+\sqrt{\text{x}}}=\text{c }$
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x+bx}^2-\text{x}}{\text{bx}\sqrt{\text{x}}}\times\frac{1}{\sqrt{\text{x+bx}^2}+\sqrt{\text{x}}}=\text{c}$
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{bx}^2}{\text{bx}\sqrt{\text{x}}}\times\frac{1}{\sqrt{\text{x+bx}^2}+\sqrt{\text{x}}}=\text{c}$
$\lim\limits_{\text{x}\rightarrow0}\sqrt{\text{x}}\times\frac{1}{\sqrt{\text{x+bx}^2}+\sqrt{\text{x}}}=\text{c}$
$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{\text{x}}}{\sqrt{\text{x}\Big(1+\text{bx}^\frac{3}{2}\Big)}+\sqrt{\text{x}}}=\text{c}$
$\lim\limits_{\text{x}\rightarrow0}{\frac{\sqrt{\text{x}}}{\sqrt{\text{x}}\Bigg[\sqrt{\Big(1+\text{b}^\frac{3}{2}}\Big)+1\Bigg]}}=\text{c}$
$\lim\limits_{\text{x}\rightarrow0}\frac{1}{\Bigg[\sqrt{\Big(1+\text{bx}^{\frac{3}{2}}\Big)}+1\Bigg]}=\text{c}$
$\text{c}=\frac{1}{2}$
$\lim\limits_{\text{x}\rightarrow0}\frac{\sin(\text{a+1})\text{x}+\sin\text{x}}{\text{x}}=\text{c}$
$\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\bigg[\frac{(\text{a+1)x+x}}{2}\bigg]\cos\bigg(\frac{\text{ax}}{2}\bigg)}{\text{x}}=\frac{1}{2}$
$\lim\limits_{\text{x}\rightarrow0}\frac{\sin\Big[\frac{(\text{a+2)x}}{2}\Big]}{\frac{(\text{a+2)x}}{2}}\times\frac{(\text{a}+2)}{2}\cos\Big(\frac{\text{ax}}{2}\Big)=\frac{1}{4}$
$1\times\frac{(\text{a+2)}}{2}\times1=\frac{1}{4}$
$\text{a}+2=\frac{1}{2}$
$\text{a}=\frac{-3}{2}$
$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{\text{x+bx}^2}-\sqrt{\text{x}}}{\text{bx}\sqrt{\text{x}}}$ exist if $\text{b}\neq0$
$\text{b }\in\text{ R}-(0)$

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