SATHEE: Limit Continuity and Differentiability 5 Question 2

Limit Continuity and Differentiability 5 Question 2

2. If $f (x) =$

$\begin{array}{cc} q, & x = 0 \\ \frac{\sqrt{x + x^{2}} - \sqrt{x}}{x^{3 / 2}}, & x > 0 \end{array}$

is continuous at $x = 0$ , then the ordered pair $(p, q)$ is equal to

(2019 Main, 10 April I)

(a) $- \frac{3}{2}, - \frac{1}{2}$

(b) $- \frac{1}{2}, \frac{3}{2}$

(d) $- \frac{3}{2}, \frac{1}{2}$

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Answer:

Correct Answer: 2. (c)

Solution:

$g \circ f (x) =$

$f (x) + 1, if f (x) < 0$

$Extra close brace or missing open brace$

$\begin{aligned} = (x + a - 1)^{2} + b, if - a \leq x < 0 \\ (| x - 1 | - 1)^{2} + b, if x \geq 0 \end{aligned}$

As $g \circ f (x)$ is continuous at $x = - a$

$\begin{array}{r} gof (- a) = gof (- a^{+}) = gof (- a^{-}) \\ 1 + b = 1 + b = 1 \Rightarrow b = 0 \end{array}$

$\Rightarrow 1 + b = 1 + b = 1 \Rightarrow b = 0$

Also, $g \circ f (x)$ is continuous at $x = 0$

$\begin{aligned} \Rightarrow gof (0) = gof (0^{+}) = gof (0^{-}) \\ \Rightarrow b = b = (a - 1)^{2} + b \Rightarrow a = 1 \\ Hence, gof (x) = \begin{array}{cl} x + 2, & if x < - 1 \\ x^{2}, & if - 1 \leq x < 0 \\ (| x - 1 | - 1)^{2}, & if x \geq 0 \end{array} \end{aligned}$

In the neighbourhood of $x = 0$ , $g \circ f (x) = x^{2}$ , which is differentiable at $x = 0$ .

Limit Continuity and Differentiability 5 Question 2

2. If $f (x) =$

Answer:

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Limit Continuity and Differentiability 5 Question 2

2. If f(x)=

Answer:

Engineering Preparation

Medical Preparation

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Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

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NCERT Exercise Video Solution

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2. If $f (x) =$