Question ID: #409
For some $a, b$, let $f(x) = \begin{vmatrix} a+\frac{\sin x}{x} & 1 & b \\ a & 1+\frac{\sin x}{x} & b \\ a & 1 & b+\frac{\sin x}{x} \end{vmatrix}, x \neq 0$. If $\lim_{x\rightarrow0}f(x)=\lambda+\mu a+\nu b$. Then $(\lambda+\mu+\nu)^{2}$ is equal to:
- (1) 25
- (2) 9
- (3) 36
- (4) 16
Solution:
$$ \lim_{x \to 0} f(x) = \begin{vmatrix} a+1 & 1 & b \\ a & 1+1 & b \\ a & 1 & b+1 \end{vmatrix} $$
Applying $R_1 \to R_1 – R_2$ and $R_2 \to R_2 – R_3$:
$$ = \begin{vmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ a & 1 & b+1 \end{vmatrix} $$
$$ = 1(b+1+1) – (-1)(a) + 0 $$
$$ = b+2+a = \lambda + \mu a + \nu b $$
Comparing coefficients:
$$ \lambda = 2, \quad \mu = 1, \quad \nu = 1 $$
$$ (\lambda + \mu + \nu)^2 = (2+1+1)^2 = 16 $$
Ans. (4)
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