If fx=1+tanx1+tanπ4-x and gx is a function with domain R, then ∫01x3gof(x)dx=

fx=1+tanx1+tanπ4-x =1+tanx1+1-tan x1+tan x =1+tanx1+tanx+1-tanx1+tanx=2∫01x3(gof)(x)dx=∫01x3g(f(x))dx=∫01x3g(2)dx=g(2)x4401=g(2)4

If fx=1+tanx1+tanπ4-x and gx is a function with domain R, then ∫01x3gof(x)dx=

fx=1+tanx1+tanπ4-x =1+tanx1+1-tan x1+tan x =1+tanx1+tanx+1-tanx1+tanx=2∫01x3(gof)(x)dx=∫01x3g(f(x))dx=∫01x3g(2)dx=g(2)x4401=g(2)4

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

If $f (x) = (1 + \tan x)$ $(1 + \tan (\frac{π}{4} - x))$ and $g (x)$ is a function with domain R, then $\int_{0}^{1} x^{3} g o f (x) d x =$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{1}{2} g (\frac{π}{4})$

$\frac{1}{4} g (2)$

$\frac{1}{4} g (1)$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$f (x) = (1 + \tan x) (1 + \tan (\frac{π}{4} - x)) = (1 + \tan x) (1 + \frac{1 - \tan x}{1 + \tan x}) = (1 + \tan x) \frac{(1 + \tan x + 1 - \tan x)}{(1 + \tan x)} = 2$

$\begin{array}{l} \int_{0}^{1} x^{3} (g o f) (x) d x \\ = \int_{0}^{1} x^{3} g (f (x)) d x \\ = \int_{0}^{1} x^{3} g (2) d x \\ = g (2) {(\frac{x^{4}}{4})}_{0}^{1} \\ = \frac{g (2)}{4} \end{array}$