Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
99 Percentile Qs Bank for JEE MainMathematicsLimits

If f: R R is defined by f(x)= array cc 3 x- x x^2 , & for x 0 & , for x=0 array . and if f is continuous at x=0 , then is equal to

Options

  1. A-2
  2. B-4
  3. C-6
  4. D-8

Correct answer

B. -4

Step-by-step solution

Given that, f(x)= array cc 3 x- x x^2 , & for x 0 , & for x=0 array . Now, aligned LHL & = _ x 0⁻ f(x) & = _ x 0⁻ 3 x- x x^2 & = _ h 0 3(0-h)- (0-h) (0-h)^2 & = _ h 0 3 h- h h^2 & = _ h 0 -3 3 h+ h 2 h & = _ h 0 -9 3 h+ h 2 & = -9+1 2 =-4 aligned (using L' Hospital's rule) Since, f(x) is continuous at x=0 array ll & _ x 0⁻ f(x)=f(0) & -4= =-4 array

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 99 Percentile Qs Bank for JEE Main PYQs