Functions Question 541
Question: The values of a and b such that $ \underset{x\to 0}{\mathop{\lim }},\frac{x(1+a\cos x)-b\sin x}{x^{3}}=1 $ , are
[Roorkee 1996]
Options:
A) $ \frac{5}{2},\ \frac{3}{2} $
B) $ \frac{5}{2},\ -\frac{3}{2} $
C) $ -\frac{5}{2},\ -\frac{3}{2} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ \underset{x\to 0}{\mathop{\lim }},\frac{x,(1+a\cos x)-b,\sin x}{x^{3}}=1 $
Þ $ \underset{x\to 0}{\mathop{\lim }}\frac{x{ 1+a,( 1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\frac{x^{6}}{6!}+… ) }-b,{ x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-… }}{x^{3}}=1 $
Þ $ \underset{x\to 0}{\mathop{\lim }}\frac{(1+a-b)+x^{2},( \frac{b}{3!}-\frac{a}{2!} )+x^{4}( \frac{a}{4!}-\frac{b}{5!} )+…}{x^{2}}=1 $ …..(i)
If $ 1+a-b\ne 0, $ then L.H.S. $ \to \infty $ as $ x\to 0 $ while R.H.S. =1, therefore $ 1+a-b=0. $
Now from (i), $ \underset{x\to 0}{\mathop{\lim }}\frac{x^{2}( \frac{b}{3!}-\frac{a}{2!} )+x^{4},( \frac{a}{4!}-\frac{b}{5!} )+…}{x^{2}}=1 $
$ \Rightarrow \frac{b}{3,!}-\frac{a}{2!}=1,\Rightarrow b-3a=6 $ . Solving $ 1+a-b=0 $ and $ b-3a=6, $ we get $ a=-5/2,b=-3/2 $ .
BETA
