Limits and Derivatives

JEE Main 2024 (Online) 5th April Evening ShiftNumericalHard📈 Topic in 14% papers+4

Ques. 1JEE Main 2024, 5 April (Evening Shift)

Let a > 0 be a root of the equation 2x2 + x - 2 = 0. If x 1a 16 (1 - (2 + x - 2x2 ) )(1 - ax)2 = + 17, where , Z, then + is equal to .

Answer

170

Solution

Since 2a2 + a - 2 = 0, dividing by a2 gives 2 + 1a - 2a2 = 0, so 1a is a root of 2 + x - 2x2 = 0; its other root is -a.

Hence 2 + x - 2x2 = -2 (x - 1a )(x + a) and (1 - ax)2 = a2 (x - 1a )2.

Using 0 1 - 2 = 12 with = 2 + x - 2x2 0:

Limit = x 1a 16 12 4 (x - 1a )2(x + a)2a2 (x - 1a )2 = 32a2 ( 1a + a )2

a = 17 - 14 and 1a = 17 + 14, so 1a + a = 172 and 1a2 = 9 + 178

Limit = 32 9 + 178 174 = 153 + 17 17

+ = 153 + 17 = 170

Answer: 170

Keep practising — Limits and Derivatives

Ques. 2JEE Main 2024, 4 April (Morning Shift)

If x 1 (5x+1)1/3-(x+5)1/3(2x+3)1/2-(x+4)1/2= m 5n(2n)2/3, where (m, n)=1, then 8m+12n is equal to .

Ques. 3JEE Main 2024, 4 April (Evening Shift)

Let f(x) = 0x (t + (1 - et ) )dt, x R. Then, x 0 f(x)x3 is equal to

16
- 16
23
- 23

Ques. 4JEE Main 2024, 1 February (Morning Shift)

Let \x\ denote the fractional part of x and f(x)= -1 (1-\x\2 ) -1(1-\x\)\x\-\x\3, x 0. If L and R respectively denotes the left hand limit and the right hand limit of f(x) at x=0, then 32 2 (L2+R2 ) is equal to .

Ques. 5JEE Main 2024, 1 February (Evening Shift)

Let f(x)= casesx-1, & x is even,\ 2x, & x is odd, cases x . If for some a , f(f(f(a)))=21, then x - \ |x|3a- [ xa ] \, where [t] denotes the greatest integer less than or equal to t, is equal to :

Ques. 6JEE Main 2024, 31 January (Morning Shift)

x 0 e2| x|-2| x|-1x2