SATHEE: Limit Continuity and Differentiability 1 Question 1

Limit Continuity and Differentiability 1 Question 1

2. If $lim_{x \to 1} \frac{x^{2} - a x + b}{x - 1} = 5$ , then $a + b$ is equal to

(a) -4

(b) 1

(d) 5

(2019 Main, 10 April II)

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Answer:

Correct Answer: 2. (c)

Solution:

Let $L = lim_{x \to 1^{-}} \frac{\sqrt{π} - \sqrt{2 \sin^{- 1} x}}{\sqrt{1 - x}}$ , then

$L = lim_{x \to 1^{-}} \frac{\sqrt{π} - \sqrt{2 \sin^{- 1} x}}{\sqrt{1 - x}} \times \frac{\sqrt{π} + \sqrt{2 \sin^{- 1} x}}{\sqrt{π} + \sqrt{2 \sin^{- 1} x}}$

[on rationalization]

$= lim_{x \to 1^{-}} \frac{π - 2 \sin^{- 1} x}{\sqrt{1 - x}} \times \frac{1}{\sqrt{π} + \sqrt{2 \sin^{- 1} x}}$

$= lim_{x \to 1^{-}} \frac{π - 2 \frac{π}{2} - \cos^{- 1} x}{\sqrt{1 - x}} \times \frac{1}{\sqrt{π} + \sqrt{2 \sin^{- 1} x}}$

$∵ \sin^{- 1} x + \cos^{- 1} x = \frac{π}{2}$

$= lim_{x \to 1^{-}} \frac{2 \cos^{- 1} x}{\sqrt{1 - x}} \times lim_{x \to 1^{-}} \frac{1}{\sqrt{π} + \sqrt{2 \sin^{- 1} x}}$

$= \frac{1}{2 \sqrt{π}} lim_{x \to 1^{-}} \frac{2 \cos^{- 1} x}{\sqrt{1 - x}} ∵ lim_{x \to 1^{-}} \sin^{- 1} x = \frac{π}{2}$

Put $x = \cos θ$ , then as $x \to 1^{-}$ , therefore $θ \to 0^{+}$

Now, $L = \frac{1}{2 \sqrt{π}} lim_{θ \to 0^{+}} \frac{2 θ}{\sqrt{1 - \cos θ}}$

$= \frac{1}{2 \sqrt{π}} lim_{θ \to 0^{+}} \frac{2 θ}{\sqrt{2} \sin \frac{θ}{2}} ∵ 1 - \cos θ = 2 \sin^{2} \frac{θ}{2}$

$= \frac{1}{2 \sqrt{π}} \cdot \sqrt{2} lim_{θ \to 0^{+}} \frac{2 \cdot \frac{θ}{2}}{\sin \frac{θ}{2}}$

$= \frac{1}{2 \sqrt{π}} 2 \sqrt{2} = \sqrt{\frac{2}{π}} ∵ lim_{x \to 0^{+}} \frac{θ}{\sin θ} = 1$

Limit Continuity and Differentiability 1 Question 1

2. If $lim_{x \to 1} \frac{x^{2} - a x + b}{x - 1} = 5$ , then $a + b$ is equal to

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Limit Continuity and Differentiability 1 Question 1

2. If limx→1x2−ax+bx−1=5, then a+b is equal to

Answer:

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2. If $lim_{x \to 1} \frac{x^{2} - a x + b}{x - 1} = 5$ , then $a + b$ is equal to