Skip to main content

Visit this link for 1 : 1 LIVE Classes.

JEE 2012 Math: ${\lim _{a \to {0^ + }}}\alpha (a)$ & ${\lim _{a \to {0^ + }}}\beta (a)$

Let $\alpha (a)$ and $\beta (a)$ be the roots of the equation

$\left( {\sqrt[3]{{1 + a}} - 1} \right){x^2} + \left( {\sqrt {1 + a}  - 1} \right)x + \left( {\sqrt[6]{{1 + a}} - 1} \right) = 0$ where a>-1.

Then ${\lim _{a \to {0^ + }}}\alpha (a)$ and ${\lim _{a \to {0^ + }}}\beta (a)$ are

(A) $-\frac {5}{2}$ and 1
(B) $-\frac {1}{2}$ and -1
(C) $-\frac {7}{2}$ and 2
(D) $-\frac {9}{2}$ and 3

image

Solution

Looking at the options and noticing that while the sum of the limits is same the product of the limits is different and hence the following quick method can be employed:

Consider,

${\lim _{a \to {0^ + }}}\alpha (a).{\lim _{a \to {0^ + }}}\beta (a)$

$= {\lim _{a \to {0^ + }}}\alpha (a).\beta (a)$ assuming both limits exist

$= {\lim _{a \to {0^ + }}}\frac{{\sqrt[6]{{1 + a}} - 1}}{{\sqrt[3]{{1 + a}} - 1}}$

$= {\lim _{a \to {0^ + }}}\frac{{\sqrt[6]{{1 + a}} - 1}}{{\sqrt[3]{{1 + a}} - 1}} \times \frac{{\sqrt[6]{{1 + a}} + 1}}{{\sqrt[6]{{1 + a}} + 1}}$

$= {\lim _{a \to {0^ + }}}\frac{{\sqrt[3]{{1 + a}} - 1}}{{\sqrt[3]{{1 + a}} - 1}} \times \frac{1}{{\sqrt[6]{{1 + a}} + 1}}$

$= \frac{1}{2}$

Hence, (B).

Popular posts from this blog

${\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$

The range of the function $f(x) = {\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$ is: (A) $[ - 2,2]$ (B) $\left[ {\frac{1}{{\sqrt 5 }},\sqrt 5 } \right]$ (C) $(0,\sqrt 5 )$ (D) $[ 0,2]$ Solution We have, $f(x) = {\log _{\sqrt 5 }}\left( {3 - 2\sin \frac{{3\pi }}{4}\sin x + 2\cos \frac{\pi }{4}\cos x} \right)$ $ \Rightarrow f(x) = {\log _{\sqrt 5 }}\left[ {3 + \sqrt 2 (\cos x - \sin x)} \right]$ Now, $ - \sqrt 2  \le \cos x - \sin x \le \sqrt 2 $ $\therefore - 2 \le \sqrt 2 (\cos x - \sin x) \le 2$ $\therefore 1 \le 3 + \sqrt 2 (\cos x - \sin x) \le 5$ $\therefore{\log _{\sqrt 5 }}1 \le {\log _{\sqrt 5 }}[3 + \sqrt 2 (\cos x - \sin x)] \le {\log _{\sqrt 5 }}5$ $ \Rightarrow 0 \le f(x) \le 2$ Answer: (D)