Limit Continuity and Differentiability 4 Question 6
6. $\lim _{x \rightarrow 0} \frac{\int _0^{x^{2}} \cos ^{2} t d t}{x \sin x}=\ldots$
(1997C, 2M)
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Answer:
Correct Answer: 6. (b)
Solution:
- Key Idea A function is said to be continuous if it is continuous at each point of the domain.
We have,
$$ f(x)=\begin{array}{ccc} 5 & \text { if } & x \leq 1 \\ a+b x & \text { if } & 1<x<3 \\ b+5 x & \text { if } & 3 \leq x<5 \end{array} $$
Clearly, for $f(x)$ to be continuous, it has to be continuous at $x=1, x=3$ and $x=5$
[ $\because$ In rest portion it is continuous everywhere]
$\therefore \quad \lim _{x \rightarrow 1^{+}}(a+b x)=a+b=5$
$$ \left[\because \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=f(1)\right] $$
$$ \begin{array}{r} \lim _{x \rightarrow 5^{-}}(b+5 x)=b+25=30 \\ {\left[\because \lim _{x \rightarrow 5^{-}} f(x)=\lim _{x \rightarrow 5^{+}} f(x)=f(5)\right]} \end{array} $$
On solving Eqs. (i) and (ii), we get $b=5$ and $a=0$
Now, let us check the continuity of $f(x)$ at $x=3$.
Here,
$$ \begin{aligned} & \lim _{x \rightarrow 3^{-}}(a+b x)=a+3 b=15 \\ & \lim _{x \rightarrow 3^{+}}(b+5 x)=b+15=20 \end{aligned} $$
and
Hence, for $a=0$ and $b=5, f(x)$ is not continuous at $x=3$ $\therefore f(x)$ cannot be continuous for any values of $a$ and $b$.
