Let gx-ftan x-f x for x-2- If f-x-0 - x-2- then

G(x)=f(tan x)+f( x) ∀ x∈(π2,π) g'(x)=f'(tan x)^2x f'( x)(cosec^2x) For increasing, g'(x) gt;0 f”(x) lt;0⇒ f'(x) is decreasing ∵tan x lt; x ∀ x∈(π2,3π4) ...

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Byju's Answer

Standard XII

Mathematics

Domain and Range of Basic Inverse Trigonometric Functions

Let gx=ftan x...

Question

Let $g (x) = f (tan x) + f (cot x)$ for $x \in (\frac{π}{2}, π)$ . If $f^{''} (x) < 0 \forall x \in (\frac{π}{2}, π),$ then

g (x)

is increasing in

(\frac{π}{2}, \frac{3 π}{4})

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g (x)

is increasing in

(\frac{3 π}{4}, π)

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g (x)

is decreasing in

(\frac{3 π}{4}, π)

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g (x)

has local maximum at

x = \frac{3 π}{4}

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Solution

The correct option is D $g (x)$ has local maximum at $x = \frac{3 π}{4}$
$g (x) = f (tan x) + f (cot x) \forall x \in (\frac{π}{2}, π)$
$g^{'} (x) = f^{'} (tan x) {sec}^{2} x - f^{'} (cot x) ({cosec}^{2} x)$
For increasing, $g^{'} (x) > 0$
$f^{''} (x) < 0 \Rightarrow f^{'} (x)$ is decreasing
$∵ tan x < cot x \forall x \in (\frac{π}{2}, \frac{3 π}{4}) ∴ f^{'} (tan x) > f^{'} (cot x)$
Also, ${sec}^{2} x > {cosec}^{2} \forall x \in (\frac{π}{2}, \frac{3 π}{4})$
$∴ g^{'} (x) > 0 \Rightarrow g (x)$ is increasing in
$(\frac{π}{2}, \frac{3 π}{4})$

Similarly, $g (x)$ is decreasing in $(\frac{3 π}{4}, π)$
Also, $g^{'} (\frac{3 π}{4}) = 0$
At $x = \frac{3 π}{4},$ sign of $g^{'} (x)$ changes from positive to negative
So, $g (x)$ has local maximum at $x = \frac{3 π}{4}$

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Similar questions

Q. Let

f (x)

be defined on

[0, π]

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} x + a \sqrt{2} sin x & , 0 \leq x \leq π / 4 2 x cot x + b & , \frac{π}{4} < x \leq \frac{π}{2} a cos 2 x - b sin x & , \frac{π}{2} < x < π \end{matrix}

. If

f

is continuous on

[0, π]

then

Q. Let

f (x) = \frac{1 - tan x}{4 x - π}

x \neq π / 4

x \in [0, \frac{π}{2}]

. If

f (x)

is continuous in

[0, \frac{π}{2}]

then

f (\frac{π}{4})

is?

Q. Let

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} - 2 sin x, - π \leq x \leq - \frac{π}{2} a sin x + b, - \frac{π}{2} < x < \frac{π}{2} cos x, \frac{π}{2} \leq x \leq π \end{matrix}

f (x)

is continuous on

[- π, π]

, then

Q. If

f'' (x) > 0, \forall x \in R, f' (3) = 0

and

g (x) = f ({tan}^{2} x - 2 tan x + 4), 0 < x < \frac{π}{2}

, then

g (x)

is increasing in

Q. Let

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} - 2 sin x f o r - π \leq x \leq - \frac{π}{2} a sin x + b f o r - \frac{π}{2} < x < \frac{π}{2} cos x f o r \frac{π}{2} \leq x < π \end{matrix}

, If f is continuous on

[- π, π]

then find the values of a & b.

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Let g(x)=f(tanx)+f(cotx) for x∈(π2,π). If f′′(x)<0 ∀ x∈(π2,π), then

Let $g (x) = f (tan x) + f (cot x)$ for $x \in (\frac{π}{2}, π)$ . If $f^{''} (x) < 0 \forall x \in (\frac{π}{2}, π),$ then