If f1x-2x-f2x-3 sin x-xcos x- then for x-0-2

The correct option is B f_1(x)> f_2(x) Let f(x) = f_1(x) f_2(x) f(x)= 2x 3sin x + xcos x ⇒f'(x)=2 2cos x xsin x ⇒f”(x)=sin x xcos x= ...

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Standard XII

Mathematics

Variable Separable Method

If f1x=2x,f...

Question

If $f_{1} (x) = 2 x, f_{2} (x) = 3 sin x - x cos x,$ then for $x \in (0, \frac{π}{2})$

f_{1} (x) < f_{2} (x)

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f_{1} (x) > f_{2} (x)

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f_{1} (| x |) < f_{2} (| x |)

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none of these

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Solution

The correct option is B $f_{1} (x) > f_{2} (x)$
Let $f (x) = f_{1} (x) - f_{2} (x)$
$f (x) = 2 x - 3 sin x + x cos x$
$\Rightarrow f^{'} (x) = 2 - 2 cos x - x sin x$
$\Rightarrow f^{''} (x) = sin x - x cos x = cos x (tan x - x)$
$\Rightarrow f^{''} (x) > 0 \forall x \in (0, \frac{π}{2}) .$
Thus $f^{'} (x)$ is increasing in $(0, \frac{π}{2}); f^{'} (0) = 0$
$\Rightarrow f^{'} (x) > 0 \forall x \in (0, \frac{π}{2}) .$
$\Rightarrow f (0) = 0$
$\Rightarrow f (x) > 0 \forall x \in (0, \frac{π}{2}) .$
$\Rightarrow f_{1} (x) > f_{2} (x)$
$\Rightarrow f (x)$ is increasing in $(0, \frac{π}{2})$

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If f1(x)=2x,f2(x)=3sinx−xcosx, then for x∈(0,π2)

If $f_{1} (x) = 2 x, f_{2} (x) = 3 sin x - x cos x,$ then for $x \in (0, \frac{π}{2})$