Question
The function $f(x) = \frac{{4 - {x^2}}}{{4x - {x^3}}}$ is
- (a) discontinuous at only one point
- (b) discontinuous at exactly two points
- (c) discontinuous at exactly three points ✓ Correct
- (d) None of the above
Continuity and Differentiability — Class 12 Maths Solution
Step-by-step Solution
We have, $f(x) = \frac{{4 - {x^2}}}{{4x - {x^3}}} = \frac{{\left( {4 - {x^2}} \right)}}{{x\left( {4 - {x^2}} \right)}}$
$= \frac{{\left( {4 - {x^2}} \right)}}{{x\left( {{2^2} - {x^2}} \right)}} = \frac{{4 - {x^2}}}{{x(2 + x)(2 - x)}}$
Clearly, $f(x)$ is discontinuous at exactly three points $x = 0,x = - 2$ and $x = 2$.
NCERT & Exemplar solution for CBSE Class 12 Mathematics, Continuity and Differentiability. Curated by Sachin Sharma. Free for all students.