Question

Let $\alpha ,\beta \in R$ be such that $\underset{x\to 0}{lim}\frac{x^2\sin \left(\beta x\right)}{ax-\sin x}=1$. Then $6\left(\alpha +\beta \right)$ equals

• A. $5$
• B. $7$
• C. $8$
• D. $4$

Answer 1

Answer: (A)

$5$

Solution :-

${lim}_{x\to 0}\frac{x^{2}sin\left(\beta x\right)}{ax-sinx}=1$

Apply L Hospital Rule

${lim}_{x\to 0}\frac{2xsin\left(\beta x\right)+x^2\beta cos\left(\beta x\right)}{a-cosx}=1$

Numerator is 0

Denominator needs to be zero

$a=1$

Apply L Hospital rule

${lim}_{x\to 0}\frac{2sin\left(\beta \pi \right)+2xcos\left(\beta x\right)\beta -{\beta }^2{x}^{2}sin\left(\beta x\right)+2x{\beta }^{2}cos\left(\beta x\right)}{sinx}$

Apply again them

$2\beta +2\beta +2\beta =-1$

[only writing terms not containing x and sin $\left(\beta x\right)]$

$\beta =\frac{-1}{6}$

$6(a+\beta )=\frac{6\times 5}{6}=5$

A is correct