Question

JEE ADVANCED MATHS QUESTION SEPTEMBER 2020

Answer 1

AnswEr:-

$\sf \implies F(x) = \int\limits^x_0 {f(t)} \, dt$$\\\\\longrightarrow F'(x)= f(x)$$</p><p>\implies I = \int\limits^\pi_0 {f'(x).cosx} \, dx + \int\limits^\pi_0 {F(x)cos(x)} \, dx =2...(1)$

$\sf Let \ \ I = \int\limits^\pi_0 {f'(x).cosx} \, dx$

Solve by using by parts:

$\sf \implies I_1 = (cosx.f(x))^\pi_0 = \int\limits^\pi_0 {sinx.f(x)} \, dx \\\\\\\implies I_1 = 6-f(0)+\int\limits^\pi_0 {sinx.F'(x)} \, dx \\\\\\\implies I_1 = 6- f(0)+ I_2---(2)$

And ,

$\sf \implies I_2 = \int\limits^\pi_0 {sinx.F'(x).} \, dx$

Using by parts :

$\sf \implies I_2 = (sinx.F(x))^\pi_0 - \int\limits^\pi_0 {cosx.F(x)} \, dx \\\\\\\implies I_2 = - \int\limits^\pi_0 {cosx.F(x)} \, dx$

From eqn (2) :

$\sf \implies I_1 = 6 -f(0)-\int\limits^\pi_0 {cosx.F(x)} \, dx$

From eqn (1) :

$\sf \implies I = 6 - f(0) = 2\\\\\\\implies \boxed{\pink{ f(0) = 4 }}$

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Answer 2

Let f : R → R be a differentiable function such that its derivatives f' is continuous and f(π) = -6

if F : [0, π] → R is defined by F(x) = $\int\limits^x_0{f(t)}\,dt$ and if $\int\limits^{\pi}_0{(f'(x)+F(x))cosx}\,dx=2$

Then the value of f(0) is ...

solution : here F(x) = $\int\limits^x_0{f(t)}\,dt$

differentiating with respect to x we get,

F'(x) = f(x) ......(1)

now again, I = $\int\limits^{\pi}_0{(f'(x)+F(x))cosx}\,dx=2$

= $\int\limits^{\pi}_0{f'(x)cosx}\,dx+\int\limits^{\pi}_0{F(x)cosx}\,dx=2$.....(2)

let I₁ = $\int\limits^{\pi}_0{f'(x)cosx}\,dx$

using integration by parts,

I.e., $\int{udv}=uv-\int{vdu}$

now, I₁ = $[cosx f(x)]^{\pi}_0-\left[\int\limits^{\pi}_0{-sinx f(x)}\,dx$

= $cos(π)f(π)-cos0f(0)+\int\limits^{\pi}_0{sinx F'(x)}\,dx$

= 6 - f(0) + $\int\limits^{\pi}_0{sinxF'(x)}\,dx$ [ from equation (1), f(x) = F'(x)]

now Let I₂ = $\int\limits^{\pi}_0{F'(x)sinx}\,dx$

using integration by parts,

= $[sinxF(x)]^{\pi}_0-\int\limits^{\pi}_0{cosxF(x)}\,dx$

= $-\int\limits^{\pi}_0{cosxF(x)}\,dx$

now putting I₁ and I₂ in equation (2) we get,

6 - f(0) = 2

⇒f(0) = 4

Therefore the value of f(0) is 4