If f(x) satisfies the relation f(x)=ex+∫01(y+xex) f(y) dy, then e+f(0) is equal to _______ . [2026]
(2)
f(x)=ex+∫01yf(y)dy+xex∫01f(y)dy
f(x)=ex+A+Bxex
A=∫01yf(y)dy=∫01y(A+ey+Byey)dy
A=A2+0-(-1)+B(e-1)
A2+B(1-e)=1
B=∫01f(y)dy
B=∫01(ey+A+Byey)dy
B=(e-1)+A+B(0-(-1))
B=e-1+A+B⇒A=1-e
f(0)=1+A=1-e+1=2-e
e+f(0)=2
Let I(x)=∫3dx(4x+6)(4x2+8x+3) and I(0)=34+20. If I (12)=a2b+c, where a,b,c∈ℕ, gcd(a,b)=1,then a+b+c is equal to. [2026]
31
30
29
28
(1)
Let 4x+6=1t⇒x=1t-64
4dx=-dtt2, {x+1=1t-24
∫3dx(4x+6)4(x+1)2-1
=∫3(-dt)4t2·1t4(1/t-24)2-1
=-34∫dtt(1-2t)24t2-1
=-34∫dt(2t)t1-4t
=-32∫dt1-4t=-32(1-4t12×-4)+C
=341-4t+C ∵ t=14x+6
=341-4(14x+6)+C
=344x+6-44x+6+C
I(x)=344x+24x+6+C
I(0)=3426+C=34+C
⇒C=20
Hence I(x)=344x+24x+6+20
I(12)=3448+20
=342+20=328+20
a+b+c=3+8+20=31
If ∫(1-5cos2xsin5x cos2x)dx=f(x)+C, where C is the constant of integration, then f(π6)-f(π4) is equal to [2026]
43(8-6)
23(4+6)
13(26+3)
13(26-3)
∫dxsin5xcos2x-5∫dxsin5x
=∫sec2x dxsin5x-5∫dxsin5x
By IBP
=tanxsin5x-∫5sin6xcosxtanx dx-5∫dxsin5x
=tanxsin5x+C
f(x)=tanxsin5x
f(π6)-f(π4)=253-(2)5=42-323
=323-42
=43(8-6)