Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE MainMathematicsLimits

If _ x 4 a - ( x + x)^8 b + 4x = L , where L is a finite non-zero real number, then the value of a + b + L is equal to

Options

  1. A23
  2. B33
  3. C21
  4. D25

Correct answer

D. 25

Step-by-step solution

For the limit to be finite and non-zero, the numerator and denominator must both approach 0 as x 4 . Setting the limit of the numerator to 0: a - ( 4 + 4 )^8 = 0 a - ( 2 )^8 = 0 a = 16 . Setting the limit of the denominator to 0: b + (4 4 ) = 0 b + = 0 b = 1 . Now, substitute a and b back into the limit: L = _ x 4 16 - ( x + x)^8 1 + 4x Let t = x + x . As x 4 , t 2 . Squaring both sides, t^2 = 1 + 2x 2x = t^2 - 1 . The denominator can be written as: 1 + 4x = 2 ^2 2x = 2(1 - ^2 2x) = 2(1 - (t^2 - 1)^2) = 2(2t^2 - t^

Practice Limits on Quantrex Academy →

More from Limits

Let _ x 2 ( (x-2))(rx^2 + (p-2)x - 2p) (x-2)^2 = 5 for some r, p R . If the set of all possible values of q , such that the roots of the equation rx^2 - px + q = 0 lie in (0, 2) , 2026The value of _ x 0 ( x^2 ^2 x x^2 - ^2 x ) is: 2026Let f(x) = _ y 0 (1 - (xy)) (xy) y^3 . Then the number of solutions of the equation f(x) = x , x R is : 2026The product of all possible values of , for which _ x 0 ( 1 - ( x) (( +1)x) (( +2)x) ^2(( +1)x) ) = 2 , is: 2026If _ x 2 (x^3 - 5x^2 + ax + b) ( x-1 - 1) _e(x-1) = m , then a + b + m is equal to : 2026The value of _ x 0 _ e ( (e x) (e² x ) (e¹⁰ x ) ) e²-e^ 2 x is equal to 2026If _ x 0 e ^ ( a -1) x +2 ~b x+( c -2) e ^ -x x x- _ e (1+x) =2 , then a ²+ b ²+ c ² is equal to : 2026Let [ ] denote the greatest integer function and f(x)= _ n 1 n ³ _ k =1 ^ n [ k ² 3^ x ] . Then 12 _ j =1 ^ f( j ) is equal to _ _ _ _ . 2026 Full Limits list All JEE Main PYQs