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Continuity — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsContinuity2 Marks
MCQ
If $f ( x )$ is continuous at $x=0$, where$f(x)=\left\{\begin{array}{ll}x^2+a, & x \geq 0 \\2 \sqrt{x^2+1}+b, & x<0\end{array}\right.$and $f\left(\frac{1}{2}\right)=2$, then the values of $a$ and $b$ are respectively
  • A
    $\frac{7}{4}, \frac{1}{4}$
  • ✓
    $\frac{7}{4},-\frac{1}{4}$
  • C
    $\frac{-1}{4}, \frac{7}{4}$
  • D
    $-\frac{7}{4},-\frac{1}{4}$
✓

Answer

Correct option: B.
$\frac{7}{4},-\frac{1}{4}$
(B)
$f \left(\frac{1}{2}\right)=\left(\frac{1}{2}\right)^2+ a$
$\Rightarrow 2=\frac{1}{4}+a \Rightarrow a=\frac{7}{4}\quad\ldots(i)$
Since $f (x)$ is continuous at $x=0$
$\therefore \lim _{x \rightarrow 0^{-}} f (x)=\lim _{x \rightarrow 0^{+}} f (x)$
$\Rightarrow \lim _{x \rightarrow 0^{-}}\left(2 \sqrt{x^2+1}+ b \right)=\lim _{x \rightarrow 0^{+}}\left(x^2+ a \right)$
$\Rightarrow 2 \sqrt{0+1}+b=0+a$
$\Rightarrow 2+b=\frac{7}{4} \quad \ldots[From (i)]$
$\Rightarrow b =-\frac{1}{4}$

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