For some a, b, let f(x)=a+sin⁡xx1ba1+sin⁡xxba1b+sin⁡xx, x≠0, limx→0 f(x)=λ+μa+vb. Then (λ+μ+v)2 is equal to :

limx→0 f(x)=a+11ba1+1ba1b+1=(a+1)(2(b+1)−b)+1(ab−a(b+1))+ba=(a+1)(b+2)−a+ab=b+a+2=λ+μa+vbλ=2,μ=1,v=1⇒(λ+μ+v)2=16

For some a, b, let f(x)=a+sin⁡xx1ba1+sin⁡xxba1b+sin⁡xx, x≠0, limx→0 f(x)=λ+μa+vb. Then (λ+μ+v)2 is equal to :

limx→0 f(x)=a+11ba1+1ba1b+1=(a+1)(2(b+1)−b)+1(ab−a(b+1))+ba=(a+1)(b+2)−a+ab=b+a+2=λ+μa+vbλ=2,μ=1,v=1⇒(λ+μ+v)2=16

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For some a, b, let $f (x) = |\begin{array}{ccc} a + \frac{\sin x}{x} & 1 & b \\ a & 1 + \frac{\sin x}{x} & b \\ a & 1 & b + \frac{\sin x}{x} \end{array}|, x \neq 0,$ $lim_{x \to 0} f (x) = λ + μa + vb . Then (λ + μ + v)^{2}$ is equal to :

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Detailed Solution

$lim_{x \to 0} f (x) = |\begin{array}{ccc} a + 1 & 1 & b \\ a & 1 + 1 & b \\ a & 1 & b + 1 \end{array}|$
$\begin{array}{l} = (a + 1) (2 (b + 1) - b) + 1 (ab - a (b + 1)) + ba \\ = (a + 1) (b + 2) - a + ab \\ = b + a + 2 = λ + μa + vb \\ λ = 2, μ = 1, v = 1 \Rightarrow (λ + μ + v)^{2} = 16 \end{array}$