The function f(x) = 2 x^2 + 7 x^3 + 3 x^2 - x - 3 is discontinuou… - Vidyadip

MCQ

The function $f(x) = \frac{{2{x^2} + 7}}{{{x^3} + 3{x^2} - x - 3}}$ is discontinuous for

A
$x = 1$ only
B
$x = 1$ and $x = - 1$ only
✓
$x = 1,x = - 1,x = - 3$ only
D
$x = 1,x = - 1,x = - 3$ and other values of $x$

✓

Answer

Correct option: C.

$x = 1,x = - 1,x = - 3$ only

c
(c) $f(x) = \frac{{2{x^2} + 7}}{{{x^2}(x + 3) - 1(x + 3)}} = \frac{{2{x^2} + 7}}{{({x^2} - 1)(x + 3)}}$

$ = \frac{{2{x^2} + 7}}{{(x - 1)(x + 1)(x + 3)}}$

Hence points of discontinuity are

$x = 1$, $x = - 1$ and $x = - 3$ only.

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