Limit Continuity and Differentiability 1 Question 1
2. If $\lim _{x \rightarrow 1} \frac{x^{2}-a x+b}{x-1}=5$, then $a+b$ is equal to
(a) -4
(b) 1
(c) -7
(d) 5
(2019 Main, 10 April II)
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Answer:
Correct Answer: 2. (c)
Solution:
- Let $L=\lim _{x \rightarrow 1^{-}} \frac{\sqrt{\pi}-\sqrt{2 \sin ^{-1} x}}{\sqrt{1-x}}$, then
$L=\lim _{x \rightarrow 1^{-}} \frac{\sqrt{\pi}-\sqrt{2 \sin ^{-1} x}}{\sqrt{1-x}} \times \frac{\sqrt{\pi}+\sqrt{2 \sin ^{-1} x}}{\sqrt{\pi}+\sqrt{2 \sin ^{-1} x}}$
[on rationalization]
$=\lim _{x \rightarrow 1^{-}} \frac{\pi-2 \sin ^{-1} x}{\sqrt{1-x}} \times \frac{1}{\sqrt{\pi}+\sqrt{2 \sin ^{-1} x}}$
$=\lim _{x \rightarrow 1^{-}} \frac{\pi-2 \frac{\pi}{2}-\cos ^{-1} x}{\sqrt{1-x}} \times \frac{1}{\sqrt{\pi}+\sqrt{2 \sin ^{-1} x}}$
$$ \because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2} $$
$=\lim _{x \rightarrow 1^{-}} \frac{2 \cos ^{-1} x}{\sqrt{1-x}} \times \lim _{x \rightarrow 1^{-}} \frac{1}{\sqrt{\pi}+\sqrt{2 \sin ^{-1} x}}$
$=\frac{1}{2 \sqrt{\pi}} \lim _{x \rightarrow 1^{-}} \frac{2 \cos ^{-1} x}{\sqrt{1-x}} \quad \because \lim _{x \rightarrow 1^{-}} \sin ^{-1} x=\frac{\pi}{2}$
Put $x=\cos \theta$, then as $x \rightarrow 1^{-}$, therefore $\theta \rightarrow 0^{+}$
Now, $L=\frac{1}{2 \sqrt{\pi}} \lim _{\theta \rightarrow 0^{+}} \frac{2 \theta}{\sqrt{1-\cos \theta}}$
$=\frac{1}{2 \sqrt{\pi}} \lim _{\theta \rightarrow 0^{+}} \frac{2 \theta}{\sqrt{2} \sin \frac{\theta}{2}} \quad \because 1-\cos \theta=2 \sin ^{2} \frac{\theta}{2}$
$=\frac{1}{2 \sqrt{\pi}} \cdot \sqrt{2} \lim _{\theta \rightarrow 0^{+}} \frac{2 \cdot \frac{\theta}{2}}{\sin \frac{\theta}{2}}$
$=\frac{1}{2 \sqrt{\pi}} 2 \sqrt{2}=\sqrt{\frac{2}{\pi}} \quad \because \lim _{x \rightarrow 0^{+}} \frac{\theta}{\sin \theta}=1$
