Let k>0 and λ=limx→0 k1−4k2−x2x2k2−x2 be finite. Then the value of λk, is

A
$λ = 8, k = \frac{1}{2}$
B
$λ = 8, k = \frac{1}{4}$
C
$λ = 4, k = \frac{1}{2}$
D
$6 λ = 4, k = \frac{1}{4}$

Solution:

It is given that

$λ = lim_{x \to 0} \frac{k (1 - 4 \sqrt{k^{2} - x^{2}})}{x^{2} \sqrt{k^{2} - x^{2}}}$ exists

$\begin{array}{l} \Rightarrow lim_{x \to 0} k (1 - 4 \sqrt{k^{2} - x^{2}}) = 0 \Rightarrow 1 - 4 k = 0 \Rightarrow k = \frac{1}{4} \\ ∴ λ = lim_{x \to 0} \frac{1 - \sqrt{1 - 16 x^{2}}}{x^{2} \sqrt{1 - 16 x^{2}}} \end{array}$

$\begin{array}{l} \Rightarrow λ = lim_{x \to 0} \frac{1 - (1 - 16 x^{2})}{x^{2} (1 + \sqrt{1 - 16 x^{2}}) \sqrt{1 - 16 x^{2}}} \\ \Rightarrow λ = 16 lim_{x \to 0} \frac{1}{(1 + \sqrt{1 - 16 x^{2}}) \sqrt{1 - 16 x^{2}}} = 8 \end{array}$

Hence, $λ = 8$ and $k = \frac{1}{4}$

Let $k > 0$ and $λ = lim_{x \to 0} \frac{k (1 - 4 \sqrt{k^{2} - x^{2}})}{x^{2} \sqrt{k^{2} - x^{2}}}$ be finite. Then the value of $λ k,$ is

Solution:

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