Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

If $ f(x) = \left\{ \begin{array}{l l} \large\frac{x^3+x^2-16x+20}{(x-2)^2} & \quad when \quad x \neq 3 \\ k & \quad when \quad x = 2 \end{array} \right. $ and $f(x)$ is continuous at $x=3$ find the value of k.

$\begin{array}{1 1}(A)\;7 \\(B)\;6 \\(C)\;8 \\(D)\;0 \end{array}$

Can you answer this question?

1 Answer

0 votes
Given $f(3)= k$----(1)
Now $\lim \limits_{x \to 3-0} f(x) =\lim \limits_{3+0} f(x) =\lim \limits_{x \to 3 } \large\frac{x^3+x^2-16 x +20}{(x-2)^2}$
=>$\lim \limits_{x \to 3} \large\frac{x^3-2x^2+3x^2-6x-10x+20}{(x-2)^2}$
=>$\lim \limits_{x \to 3} \large\frac{(x-2)(x^2+3x-10)}{(x-2)^2}$
=>$\lim \limits_{x \to 3} \large\frac{(x-2)(x-2)(x+5)}{(x-2)^2}$
=>$\lim \limits_{x \to 3} (x+5)$
=> $3+5$
=> $8$
$f(x) $ is continous at $x=3$
$\lim \limits_{x \to 3-0} f(x) =\lim \limits_{x \to 3+0} f(x)=f(3)$
So $k=8$
Hence C is the correct answer.
answered Apr 22, 2014 by meena.p

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App