JEE Main202331 Jan 2023Evening ShiftMathematicsDefinite IntegrationActual
If ϕ ( x ) = 1 x ∫ π 4 x 4 2 sin t - 3 ϕ ' ( t ) d t , x > 0 then ϕ ' π 4 is equal to
Options
- A4 6 + π
- B8 6 + π
- C8 π
- D4 6 - π
Correct answer
B. 8 6 + π
Step-by-step solution
Given, ϕ ( x ) = 1 x ∫ π / 4 x 4 2 sin t - 3 ϕ ' ( t ) d t Differentiate both sides w.r.t. x we get, ϕ ' ( x ) = - 1 2 x 3 / 2 ∫ π / 4 x 4 2 sin t - 3 ϕ ' ( t ) d t + 1 x 4 2 sin x - 3 ϕ ' ( x ) Put x = π 4 ϕ ' ( π 4 ) = - 1 2 π 4 3 / 2 × 0 + 4 π 4 2 × 1 2 - 3 ϕ ' ( π 4 ) ⇒ π 4 ϕ ' ( π 4 ) + 3 ϕ ' ( π 4 ) = 4 ⇒ ϕ ' π 4 3 + π 4 = 4 ⇒