Download App

STD 12 - 5. continuity and differentiation — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 5. continuity and differentiation4 Marks
MCQ
If $f(x) = \left\{ \begin{array}{l}\frac{{{x^2} - 4x + 3}}{{{x^2} - 1}},\;{\rm{for}}\;x \ne 1\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;2,\;{\rm{for\, }}x = 1\end{array} \right.$, then
  • A
    $\mathop {\lim }\limits_{x \to 1 + } f(x) = 2$
  • B
    $\mathop {\lim }\limits_{x \to 1 - } f(x) = 3$
  • C
    $f(x)$ is discontinuous at $x = 1$
  • D
    None of these
✓

Answer

$f(x) = \left\{ {\frac{{{x^2} - 4x + 3}}{{{x^2} - 1}}} \right\}$, for $x \ne 1$
$ = 2$, for $x = 1$
$f(1) = 2,f(1 + ) = \mathop {\lim }\limits_{x \to 1 + } \frac{{{x^2} - 4x + 3}}{{{x^2} - 1}} = \mathop {\lim }\limits_{x \to 1 + } \frac{{(x - 3)}}{{(x + 1)}} = - 1$
$f(1 - ) = \mathop {\lim }\limits_{x \to 1 - } \frac{{{x^2} - 4x + 3}}{{{x^2} - 1}} = - 1 $
$\Rightarrow f(1) \ne f(1 - )$
Hence the function is discontinuous at $x = 1.$

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

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

If twice the $11^{th}$ term of an $A.P.$ is equal to $7$ times of its $21^{st}$ term, then its $25^{th}$ term is equal to
→
Find the coefficient of $x^{49}$ in the expansion of $(2x + 1) (2x + 3) (2x + 5)----- (2x + 99)$
→
If $\tan \theta = - \frac{1}{{\sqrt {10} }}$ and $\theta $ lies in the fourth quadrant, then $\cos \theta = $
→
If $i{z^4} + 1 = 0$, then $z$ can take the value
→
The value of $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x}{1+e^x} d x$ is equal to
→
Let $f:( - 1,1) \to B$, be a function defined by $f(x) = {\tan ^{ - 1}}\frac{{2x}}{{1 - {x^2}}},$ then $f$ is both one- one and onto when $B$ is the interval
→
If $f:R \to R$ is a differentiable function and $f\left( 2 \right) = 6$, then $\mathop {\lim }\limits_{x \to 2} \int\limits_6^{f\left( x \right)} {\frac{{2\,tdt}}{{\left( {x - 2} \right)}}} $ is
→
Given that the slope of the tangent to a curve $y = y(x)$ at any point $(x, y)$ is $\frac{{2y}}{{{x^2}}}$. If the curve passes through the centre of the circle $x^2 + y^2 - 2x - 2y = 0$, then its equation is
→
$\int_{}^{} {\frac{{dx}}{{2\sqrt x (1 + x)}} = } $
→
Let $f(x)=\max \{|x+1|,|x+2|, \ldots,|x+5|\}$. Then $\int_{-6}^{0} f(x) d x$ is equal to
→