Download App

Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability1 Mark
MCQ
If $\text{f(x)}=\begin{cases}\frac{1-\sin^2\text{x}}{3\cos^2\text{x}},&\text{if}\text{ x}<\frac{\pi}{2}\\\text{a},&\text{if}\text{ x}=\frac{\pi}{2}\\\frac{\text{b}(1-\sin\text{x})}{(\pi-2\text{x})^2},&\text{if}\text{ x }>\frac{\pi}{2}\end{cases}$ Then $f(x)$ is continuous at $\text{x}=\frac{\pi}{2},$ if :
  • A
    $\text{a}=\frac{1}{3},\text{ b}=2$
  • ✓
    $\text{a}=\frac{1}{3},\text{ b}=\frac{8}{3}$
  • C
    $\text{a}=\frac{2}{3},\text{ b}=\frac{8}{3}$
  • D
    None of these
✓

Answer

Correct option: B.
$\text{a}=\frac{1}{3},\text{ b}=\frac{8}{3}$
Given, $\text{f(x)}=\begin{cases}\frac{1-\sin^2\text{x}}{3\cos^2\text{x}},&\text{if}\text{ x} < \frac{\pi}{2}\\\text{a},&\text{if}\text{ x}=\frac{\pi}{2}\\\frac{\text{b}(1-\sin\text{x})}{(\pi-2\text{x})^2},&\text{if}\text{ x } > \frac{\pi}{2}\end{cases}$
We have
$\Big(\text{LHL at x}=\frac{\pi}{2}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}\Big(\frac{\pi}{2}-\text{h}\Big)$
$=\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{1-\sin^2\big(\frac{\pi}{2}-\text{h}\big)}{3\cos^2\big(\frac{\pi}{2}-\text{h}\big)}\Bigg)$
$=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{1-\cos^2\text{h}}{3\sin^2\text{h}}\Big)$
$=\frac{1}{3}\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\sin^2\text{h}}{\sin^2\text{h}}\Big)$
$=\frac{1}{3}$
$\Big(\text{RHL at x}=\frac{\pi}{2}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}\Big(\frac{\pi}{2}+\text{h}\Big)$
$=\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{\text{b}\big[1-\sin\big(\frac{\pi}{2}+\text{h})\big]}{\big[\text{x}-2\big(\frac{\pi}{2}+\text{h}\big)\big]^2}\Bigg)$
$=\lim\limits_{\text{h}\rightarrow0}\bigg(\frac{\text{b}(1-\cos\text{h})}{[-2\text{h}]^2}\bigg)$
$=\lim\limits_{\text{h}\rightarrow0}\bigg(\frac{2\text{b}\sin^{2}\frac{\text{h}}{2}}{4\text{h}^2}\bigg)$
$=\lim\limits_{\text{h}\rightarrow0}\bigg(\frac{2\text{b}\sin^2\frac{\text{h}}{2}}{16\frac{\text{h}^2}{4}}\bigg)$
$=\frac{\text{b}}{8}\lim\limits_{\text{h}\rightarrow0}\bigg(\frac{\sin\frac{\text{h}}{2}}{\frac{\text{h}}{2}}\bigg)^2$
$=\frac{\text{b}}{8}\times1$
$=\frac{\text{b}}{8}$
Also, $\text{f}\big(\frac{\pi}{2}\big)=\text{a}$
If $f(x)$ is continuous at $\text{x}=\frac{\pi}{2},$ then
$\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^+}\text{f(x)}=\text{f}\big(\frac{\pi}{2}\big)$
$\Rightarrow\frac{1}{3}=\frac{\text{b}}{8}=\text{a}$
$\Rightarrow\text{a}=\frac{1}{3}$ and $\text{b}=\frac{8}{3}$

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

Explore more

More "M.C.Q (1 Marks)" from Continuity and Differentiability→Continuity and Differentiability — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Area of the region bounded by the curve $y^2=4 x, y-$ axis and the line $y = 3$, is :
→
Choose the correct answer from the given four options:The area of the region bounded by the $y-$axis, $\text{y}=\cos\text{x}$ and $\text{y}=\sin\text{x},0\leq\text{x}\leq\frac{\pi}{2}$ is:
→
The slope of the tangent at $(x, y)$ to a curve passing through a point $(2, 1)$ is $\frac{{{x^2} + {y^2}}}{{2xy}}$, then the equation of the curve is
→
Choose the correct answer from the given four options. Let $f : R \rightarrow R$ be defined by $\text{f}(\text{x})=\begin{cases}2\text{x}:\text{x}>3\\\text{x}^2:1<\text{x}\leq3\\3\text{x}:\text{x}\leq1\end{cases}$ Then $f(-1) + f(2) + f(4)$ is:
→
Let $N$ be the set of natural numbers and two functions $f$ and $g$ be defined as $f$, $g : N \to N$ such that $f\left( n \right) = \left\{ \begin{gathered}
  \frac{{n + 1}}{2}\,\,\,\,\,\,\,\,\,\,\,{\text{if n is odd}} \hfill \\
  \frac{n}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{if n is even}} \hfill \\ 
\end{gathered}  \right.$ and $g(n) = n - (-1)^n$. Then $fog$ is
→
If the equation of a straight line passing through a point $ a$ and parallel to the vector $ b$ is $r = a + t\,b$, where $ t $ is a parameter, then what is the perpendicular distance from this line to a point $ c$ ?
→
If $x\left[\begin{array}{l}1 \\ 2\end{array}\right]+y\left[\begin{array}{l}2 \\ 5\end{array}\right]=\left[\begin{array}{l}4 \\ 9\end{array}\right],$ then
→
The position vectors of four points  $ P, Q, R, S$  are $2a + 4c,\,$ $5a + 3\sqrt 3 \,b + 4c,$ $ - 2\sqrt 3 b + c$ and $2a + c$ respectively, then
→
The value of $I = \int_{\,0}^{\,1} {\,x\,\left| {x - \frac{1}{2}} \right|\,dx} $ is
→
Let $f(x+y)=f(x).f(y)$ for all $x, y$ where $ f(0) \ne  0$. If $f(5) = 2$ and $f '(0) = 3,$ then $f '(5)$ is equal to
→