Question

If the function $f(x) = \dfrac{{3{x^2} + ax + a + 3}}{{{x^2} + x - 2}}$ is continuous at x=-2, then the value of f (-2) is
a. 0
b. -1
c. 1
d. 2

Answer 1

Hint: A continuous function is a function in which there is no discontinuity or abrupt change throughout its entire path.
 A function is said to be continuous at point $x = a$, if
$\mathop {\lim }\limits_{x \to a} f(x)$ Exists
$\mathop {\lim }\limits_{x \to a} f(x) = f(a)$
It implies that if left hand limit, right hand limit and the value of the function $x = a$ exists and these three are equal in parameter with each other, then the function f(x) is said to be continuous at $x = a$

Complete step-by-step answer: $f(x) = \dfrac{{3{x^2} + ax + a + 3}}{{{x^2} + x - 2}}$
Appling formula $\mathop {\lim }\limits_{x \to a} f(x) = f(a)$
$ \Rightarrow f( - 2) = \mathop {\lim }\limits_{x \to - 2} f(x)$
$\mathop {\lim }\limits_{x \to - 2} \dfrac{{3{x^2} + ax + a + 3}}{{{x^2} + x - 2}}$ (Putting function inside the limit)
$ \Rightarrow \dfrac{{3{{( - 2)}^2} + a( - 2) + a + 3}}{{{{( - 2)}^2} + ( - 2) - 2}}$ (Putting value of limit)
$ \Rightarrow \dfrac{{12 + - 2a + a + 3}}{{4 - 2 - 2}} = \dfrac{{15 - a}}{0}$ (Opening the brackets and then solving)
This limit will only exists when $15 - a = 0$

15 - a = 0
therefore a = 15

Putting the value of ‘a’ into the function and then finding out the functions value at -2
$ \Rightarrow \mathop {\lim }\limits_{x \to - 2} \dfrac{{3{x^2} + 15x + 15 + 3}}{{{x^2} + x - 2}}$
$ \Rightarrow \mathop {\lim }\limits_{x \to - 2} \dfrac{{3{x^2} + 15x + 18}}{{{x^2} + x - 2}}$
$ \Rightarrow \mathop {\lim }\limits_{x \to - 2} \dfrac{{3({x^2} + 5x + 6)}}{{{x^2} + x - 2}}$ (Taking 3 common from the numerator)
$ \Rightarrow \mathop {\lim }\limits_{x \to - 2} \dfrac{{3(x + 2)(x + 3)}}{{(x + 2)(x - 1)}}$
 (Doing Factors of the quadratic equation using sridharacharya formula i.e. $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$)
$ \Rightarrow \mathop {\lim }\limits_{x \to - 2} \dfrac{{3(x + 3)}}{{(x - 1)}}$ [Cancelling $(x + 2)$ from numerator and denominator]
$ \Rightarrow \dfrac{{3( - 2 + 3)}}{{( - 2 - 1)}}$ (Putting limit value)
$ \Rightarrow \dfrac{{3 \times (1)}}{{( - 3)}} = - 1$
$\therefore $$f( - 2) = - 1$
So the correct option is (B)
Additional note: - A function f(x) is said to be continuous in the closed interval [a,b] if it satisfies three conditions: -
1. f(x) should be continuous in the open interval (a,b)
2. f(x) is continuous at the point a from right hand side if $\mathop {\lim }\limits_{x \to a} f(x) = f(a)$
3. f(x) is continuous at the point b from left hand side if $\mathop {\lim }\limits_{x \to b} f(x) = f(b)$

Note: Putting limits can sometimes become confusing so make sure you are doing it cautiously. Also when there are quadratic equations in both numerator and denominator then making factors can make the calculation part smooth and fast.