Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
NTA Abhyas JEE Main2020MathematicsIndefinite IntegrationPractice

The integral I = ∫ e x cos ⁡ e x x d x = f x + c (where, c is the constant of integration) and f ln ⁡ π 4 2 = 2 . Then, the number of solutions of f x = 2 e ∀ x ∈ R - 0 is equal to

Correct answer

0

Step-by-step solution

Given integral is I = ∫ e x cos ⁡ e x x d x Let e x = t ⇒ e x 2 x d x = d t ∴ I = 2 ∫ cos ⁡ t d t = 2 sin ⁡ t + c = 2 sin ⁡ e x + c ∴ f x = 2 sin ⁡ e x which has range = - 2,2 Hence, f x = 2 e has no solution

Practice Indefinite Integration on Quantrex Academy →

More from Indefinite Integration

Let f(x) = ( 16x + 24 x^2 + 2x - 15 ) dx . If f(4) = 14 _e(3) and f(7) = _e(2^ 3^ ) , , N , then + is equal to: 2026Let f(x)= d x x^ ( 2 3 ) +2 x^ ( 1 2 ) be such that f(0)=-26+24 _ e (2) . If f(1)= a + b _ e (3) , where a , b Z , then a + b is equal to : 2026Let f(t)= ( 1- ( _ e t ) 1- ( _ e t ) ) d t, t>1 . If f (e^ / 2 )=-e^ / 2 and f (e^ / 4 )= e^ / 4 , then equals 2026Let I (x)= 3 d x (4 x+6) ( 4 x²+8 x+3 ) and I (0)= 3 4 +20 . If I ( 1 2 )= a 2 b + c , where a, b, c N , gcd (a, b)=1 , then a+b+c is equal to 2026Let f(x)= (2-x² ) e ^ x ( 1+x )(1-x)^ 3 / 2 ~d x . If f(0)=0 , then f ( 1 2 ) is equal to: 2026If ( x)^ -11 2 ( x)^ -5 2 d x=- p₁ q₁ ( x)^ 9 2 - p₂ q₂ ( x)^ 5 2 - p₃ q₃ ( x)^ 1 2 + p₄ q₄ ( x)^ -3 2 + C , where p_ i and q_ i are positive integers with gcd (p_ i , q_ i )=1 for 2026If aligned ( 1 x + 1 x^3 ) & ( [23] 3 x⁻²⁴+x⁻²⁶ ) d x & =- 3( +1) (3 x^ +x^ )^ +1 +C, x 0, aligned ( , , Z) , where C is the constant of integration, then + + is equal to ________ 2025If ( 1+x^2 +x )¹⁰ ( 1+x^2 -x )^9 d x= 1 m ( ( 1+x^2 +x )^n (n 1+x^2 -x ) )+C where C is the constant of integration and m, n N , then m + n is equal to 2025 Full Indefinite Integration list All NTA Abhyas JEE Main PYQs