For some, a, b, let . Then is equal to : [2025]

Question

For some, a, b, let $f (x) =$ $| \begin{matrix} a + \frac{\sin x}{x} & 1 & b \\ a & 1 + \frac{\sin x}{x} & b \\ a & 1 & b + \frac{\sin x}{x} \end{matrix} |$ $, x \neq 0, \lim_{x \to 0} f (x) = λ + μ a + ν b$ . Then ${(λ + μ + ν)}^{2}$ is equal to : [2025]

Smart Achievers · Accepted Answer

(3)

$f (x) =$ $| \begin{matrix} a + \frac{\sin x}{x} & 1 & b \\ a & 1 + \frac{\sin x}{x} & b \\ a & 1 & b + \frac{\sin x}{x} \end{matrix} |$ $, x \neq 0$

Since, $\lim_{x \to 0} f (x) = λ + μ a + ν b$

$\Rightarrow λ + μ a + ν b = \lim_{x \to 0}$ $| \begin{matrix} a + \frac{\sin x}{x} & 1 & b \\ a & 1 + \frac{\sin x}{x} & b \\ a & 1 & b + \frac{\sin x}{x} \end{matrix} |$

$=$ $| \begin{matrix} a + 1 & 1 & b \\ a & 1 + 1 & b \\ a & 1 & b + 1 \end{matrix} |$ $=$ $| \begin{matrix} a + 1 & 1 & b \\ a & 2 & b \\ a & 1 & b + 1 \end{matrix} |$

$= (a + 1) [2 (b + 1) - b] - 1 [a (b + 1) - a b] + b \times (a - 2 a)$

$= (a + 1) (b + 2) - a - a b = a + b + 2$

On comparing, we get

$λ = 2, μ = 1 and ν = 1$

Hence, ${(λ + μ + ν)}^{2} = {(2 + 1 + 1)}^{2} = 16$ .

Solution

Q.
For some, a, b, let $f (x) =$ $| \begin{matrix} a + \frac{\sin x}{x} & 1 & b \\ a & 1 + \frac{\sin x}{x} & b \\ a & 1 & b + \frac{\sin x}{x} \end{matrix} |$ $, x \neq 0, \lim_{x \to 0} f (x) = λ + μ a + ν b$ . Then ${(λ + μ + ν)}^{2}$ is equal to : [2025]

1 25

2 9

3 16

4 36

Want Us to Email you About Special Offers & Updates?