Limits and Derivatives · Question 1

JEE Advanced 2025 Paper 1NumericalMedium📈 Topic in 17% papers+4

Let and be the real numbers such that x 0 1x3 ( 2 0x 11 - t2 dt + x x ) = 2. Then the value of + is .

Answer

2.35 to 2.45

Solution

For small t: 11 - t2 = 1 + t2 + 0x dt1 - t2 = x + x33 +

x x = x - x32 +

Bracket = ( 2 + )x + ( 6 - 2 )x3 +

Finite limit: 2 + = 0; limit value: 6 - 2 = 2.

= - 2 6 + 4 = 2 = 245, = - 125.

+ = 125 = 2.4.

Answer: 2.4

JEE Advanced 2025 Paper 1

Keep practising — Limits and Derivatives

Let x0 be the real number such that ex0+x0=0. For a given real number , define g(x)= 3xex+3x- ex- x3(ex+1) for all real numbers x. Then which one of the following statements is TRUE?

For =2, x x0 | g(x)+ex0x-x0 |=0
For =2, x x0 | g(x)+ex0x-x0 |=1
For =3, x x0 | g(x)+ex0x-x0 |=0
For =3, x x0 | g(x)+ex0x-x0 |= 23

JEE Advanced 2025 Paper 2

Let k R. If x 0+ ( ( kx)+ x+x ) 2x=e6, then the value of k is

JEE Advanced 2024 Paper 2

Let S be the set of all ( , ) R R such that x (x2)( e x) ( 1x2 )x ( e(1+x)) =0. Then which of the following is (are) correct?

(-1,3) S
(-1,1) S
(1,-1) S
(1,-2) S

JEE Advanced 2024 Paper 2

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