The least integral value of k for fx-e-x-k-8x2 is decreasing for all x- is

F(x)=e^ x√(k+8x^2) ⇒ f'(x)=√(k+8x^2)( e^ x) e^ x16x2√(k+8x^2)(k+8x^2) =(k+8x^2)( e^ x) e^ x8x(k+8x^2)^3/2 = e^ x(8x^2+8x+k)(k+8x^2)^3/2 for f(x) always to be d ...

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Byju's Answer

Standard XII

Mathematics

Function Defintion

The least int...

Question

The least integral value of $k$ for $f (x) = \frac{e^{- x}}{\sqrt{k + 8 x^{2}}}$ is decreasing for all $x \in R$ is

2.0

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2.00

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Solution

$f (x) = \frac{e^{- x}}{\sqrt{k + 8 x^{2}}}$
$\Rightarrow f^{'} (x) = \frac{\sqrt{k + 8 x^{2}} (- e^{- x}) - e^{- x} \frac{16 x}{2 \sqrt{k + 8 x^{2}}}}{(k + 8 x^{2})}$
$= \frac{(k + 8 x^{2}) (- e^{- x}) - e^{- x} 8 x}{(k + 8 x^{2})^{3 / 2}}$
$= \frac{- e^{- x} (8 x^{2} + 8 x + k)}{(k + 8 x^{2})^{3 / 2}}$

for $f (x)$ always to be decreasing $f^{'} (x) \leq 0 \forall x \in R$
$\Rightarrow 8 x^{2} + 8 x + k \geq 0 \forall x \in R$
$\Rightarrow D \leq 0 \Rightarrow 64 - 32 k \leq 0 \Rightarrow k \geq 2$
$∴$ Least integral value of $k$ is $2$

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The least integral value of k for f(x)=e−x√k+8x2 is decreasing for all x∈R is

f(x)=e−x√k+8x2 ⇒f′(x)=√k+8x2(−e−x)−e−x16x2√k+8x2(k+8x2) =(k+8x2)(−e−x)−e−x8x(k+8x2)3/2 =−e−x(8x2+8x+k)(k+8x2)3/2 for f(x) always to be decreasing f′(x)≤0 ∀ x∈R ⇒ 8x2+8x+k≥0 ∀x∈R ⇒D≤0 ⇒64−32k≤0⇒k≥2 ∴ Least integral value of k is 2

The least integral value of $k$ for $f (x) = \frac{e^{- x}}{\sqrt{k + 8 x^{2}}}$ is decreasing for all $x \in R$ is