If 3 f tan x -2 f x - x -3- x -0- -2- then 15 f -3-correct answer - 1 - wrong answer -0-25 A- 4 -1B- 4 -7C- 2 -3D- 2 -9

The correct option is D 2π+93f(tan x)+2f( x)=x+3 ⋯(1) Changing x to π2 x, we get 3f( x)+2f(tan x)=π2 x+3 ⋯(2) From eqn(1) and (2), we have f(tan x)=x+3 π5 ...

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Byju's Answer

Standard XII

Mathematics

Napier’s Analogy

If 3 f tan x ...

Question

If $3 f (tan x) + 2 f (cot x) = x + 3$ , $x \in (0, \frac{π}{2})$ , then $15 f (\sqrt{3}) =$
(correct answer + 1, wrong answer - 0.25)

4 π + 1

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4 π + 7

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2 π + 3

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2 π + 9

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Solution

The correct option is D $2 π + 9$
$3 f (tan x) + 2 f (cot x) = x + 3 \dots (1)$
Changing $x$ to $\frac{π}{2} - x$ , we get
$3 f (cot x) + 2 f (tan x) = \frac{π}{2} - x + 3 \dots (2)$
From eqn $(1)$ and $(2),$ we have
$f (tan x) = x + \frac{3 - π}{5}$
Now, at $x = \frac{π}{3}$
$f (\sqrt{3}) = \frac{π}{3} + \frac{3 - π}{5}$
$\Rightarrow 15 f (\sqrt{3}) = 5 π + 9 - 3 π = 2 π + 9$

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If 3f(tanx)+2f(cotx)=x+3, x∈(0,π2), then 15f(√3)= (correct answer + 1, wrong answer - 0.25)

The correct option is D 2π+93f(tanx)+2f(cotx)=x+3 ⋯(1) Changing x to π2−x, we get 3f(cotx)+2f(tanx)=π2−x+3 ⋯(2) From eqn(1) and (2), we have f(tanx)=x+3−π5 Now, at x=π3 f(√3)=π3+3−π5 ⇒15f(√3)=5π+9−3π=2π+9

If $3 f (tan x) + 2 f (cot x) = x + 3$ , $x \in (0, \frac{π}{2})$ , then $15 f (\sqrt{3}) =$
(correct answer + 1, wrong answer - 0.25)