f(x)=x^2-7 x+10 x^2+2 x-8 , for x [-6,-3] - Vidyadip

MCQ

$f(x)=\frac{x^2-7 x+10}{x^2+2 x-8}$, for $x \in[-6,-3]$

A
$f$ is discontinuous at $x=2$
✓
$f$ is discontinuous at $x=-4$
C
$f$ is discontinuous at $x=0$
D
$f$ is discontinuous at $x=2$ and $x=-4$

✓

Answer

Correct option: B.

$f$ is discontinuous at $x=-4$

(B) $f$ is discontinuous at $x =-4$
Hint:
$f(x)=\frac{x^2-7 x+10}{x^2+2 x-8}, \text { for } x \in[-6,-3] $
$=\frac{x^2-7 x+10}{(x+4)(x-2)}$
Here $f(x)$ is a rational function and is continuous everywhere except at the points Where denominator becomes zero.
Here, denominator becomes zero when $x=-4$ or $x=2$
But $x=2$ does not lie in the given interval.
$\therefore x =-4$ is the point of discontinuity.

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