SATHEE: Limit Continuity and Differentiability 4 Question 6

Limit Continuity and Differentiability 4 Question 6

6. $lim_{x \to 0} \frac{\int_{0}^{x^{2}} \cos^{2} t d t}{x \sin x} = \dots$

(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,

$f (x) = \begin{array}{ccc} 5 & if & x \leq 1 \\ a + b x & if & 1 < x < 3 \\ b + 5 x & if & 3 \leq x < 5 \end{array}$

Clearly, for $f (x)$ to be continuous, it has to be continuous at $x = 1, x = 3$ and $x = 5$

[ $∵$ In rest portion it is continuous everywhere]

$∴ lim_{x \to 1^{+}} (a + b x) = a + b = 5$

$[∵ lim_{x \to 1^{-}} f (x) = lim_{x \to 1^{+}} f (x) = f (1)]$

$\begin{array}{r} lim_{x \to 5^{-}} (b + 5 x) = b + 25 = 30 \\ [∵ lim_{x \to 5^{-}} f (x) = lim_{x \to 5^{+}} f (x) = f (5)] \end{array}$

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,

$\begin{aligned} lim_{x \to 3^{-}} (a + b x) = a + 3 b = 15 \\ lim_{x \to 3^{+}} (b + 5 x) = b + 15 = 20 \end{aligned}$

and

Hence, for $a = 0$ and $b = 5, f (x)$ is not continuous at $x = 3$ $∴ f (x)$ cannot be continuous for any values of $a$ and $b$ .

Limit Continuity and Differentiability 4 Question 6

6. $lim_{x \to 0} \frac{\int_{0}^{x^{2}} \cos^{2} t d t}{x \sin x} = \dots$

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Limit Continuity and Differentiability 4 Question 6

6. limx→0∫0x2cos2⁡tdtxsin⁡x=…

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6. $lim_{x \to 0} \frac{\int_{0}^{x^{2}} \cos^{2} t d t}{x \sin x} = \dots$