For real x- the maximum value of esin x-ecos x is

F(x)=e^sin x/e^cos x=e^sin x cos x nbsp; nbsp;...(i) f'(x)=e^sin x cos x[cos x+sin x] amp; nbsp; f”(x)=e^sin x cos[(cos x+sin x)^2+(cos x sin x)] For ...

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Standard XII

Mathematics

Evaluation of a Determinant

For real x, t...

Question

For real $x$ , the maximum value of $\frac{e^{s i n x}}{e^{c o s x}}$ is

1

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e^{\sqrt{2}}

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Solution

The correct option is C $e^{\sqrt{2}}$
$f (x) = \frac{e^{s i n x}}{e^{c o s x}} = e^{s i n x - c o s x}$ ...(i)
$f^{'} (x) = e^{s i n x - c o s x} [c o s x + s i n x]$ &
$f^{''} (x) = e^{s i n x - c o s} [(c o s x + s i n x)^{2} + (c o s x - s i n x)]$
For critical point $f^{'} (x) = 0$
$\Rightarrow c o s x + s i n x = 0$
$t a n x = - 1$
$x = \frac{- π}{4}$ & $\frac{3 π}{4}$
$∵ f^{''} (\frac{3 π}{4}) < 0 \Rightarrow$ maximum at $x = \frac{3 π}{4}$
Maximum value $f (\frac{3 π}{4}) = [e^{s i n x - c o s x}]_{x = \frac{3 π}{4}} = e^{\sqrt{2}}$

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For real x, the maximum value of esinxecosx is

For real $x$ , the maximum value of $\frac{e^{s i n x}}{e^{c o s x}}$ is