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JEE Mathematics 2014 previous year questions with solutions

118 JEE Mathematics questions from 2014 in the Quantrex bank. Featured questions below open unique solution pages. Browse by chapter on JEE Mathematics.

118 questions2014
1

Let f : 0 , ∞ → R be given by f x = ∫ 1 x x e - t + 1 t d t t , then

JEE Advanced 2014 Application of Derivatives Solution
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2

Let a ∈ R and let f : R → R be given by f x = x 5 - 5 x + a , then

JEE Advanced 2014 Application of Derivatives Solution
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3

For every pair of continuous functions f , g : 0 , 1 → R such that max ⁡ f x : x ∈ 0 , 1 = max ⁡ g x : x ∈ 0 , 1 , then the correc

JEE Advanced 2014 Application of Derivatives Solution
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4

The slope of the tangent to the curve y - x 5 2 = x 1 + x 2 2 at the point (1, 3) is ________.

JEE Advanced 2014 Application of Derivatives Solution
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5

The area of the region ( in square units ) above the x -axis bounded by the curve y = tan x , 0 ≤ x ≤ π 2 and the tangent to the curve at x = &

JEE Main 2014 Area Under Curves Solution
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6

Let A = x , y : y 2 ≤ 4 x , y - 2 x ≥ - 4 . The area of the region A in square units is

JEE Main 2014 Area Under Curves Solution
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7

The area (in sq. unit) of the region described by A = x , y : x 2 + y 2 ≤ 1 and y 2 ≤ 1 - x is

JEE Main 2014 Area Under Curves Solution
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8

Coefficient of x 11 in the expansion of 1 + x 2 4 1 + x 3 7 1 + x 4 12 is

JEE Advanced 2014 Binomial Theorem Solution
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9

A circle S passes through the point (0, 1) and is orthogonal to the circles x - 1 2 + y 2 = 16 and x ⁡ 2 + y ⁡ 2 = 1 . Then

JEE Advanced 2014 Circle Solution
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10

The equation of the circle described on the chord 3 x ⁡ + y ⁡ + 5 = 0 of the circle x ⁡ 2 + y ⁡ 2 = 1 6 as the diameter is

JEE Main 2014 Circle Solution
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11

For the two circles x^2+y^2=16 and x^2+y^2-2 y=0 , there is/are

JEE Main 2014 Circle Solution
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12

The set of all real values of for which exactly two common tangents can be drawn to the circles x^2+y^2-4 x-4 y+6=0 and x ^2+ y ^2-10 x -10 y + =0 is the interv

JEE Main 2014 Circle Solution
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13

Let L ₁ be the length of the common chord of the curves x^2+y^2=9 and y^2=8 x , and L₂ be the length of the latus rectum of y^2=8 x , then:

JEE Main 2014 Circle Solution
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14

If the point 1 , 4 lies inside the circle x 2 + y 2 - 6 x + 10 y + p = 0 and the circle does not touch or intersect the coordinate axes, then the set of all pos

JEE Main 2014 Circle Solution
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15

Let C be the circle with center at 1 , 1 and radius = 1 . If T is the circle centered at 0 , y , passing through the origin and touching the circle C externally

JEE Main 2014 Circle Solution
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16

Let z k = cos ⁡ 2 k π 10 + i sin ⁡ 2 k π 10 ; k = 1 , 2 , … , 9 List – I List – II (A) For each z k there exists a z j s

JEE Advanced 2014 Complex Number Solution
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17

Let f : a , b → 1 , ∞ be a continuous function and let g : R → R be defined as g x = 0 i f x < a , ∫ a x f t d t i f a ≤ x &#8

JEE Advanced 2014 Continuity and Differentiability Solution
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18

Let f : R → R and g : R → R be respectively given by f x = x + 1 and g x = x 2 + 1 . Define h : R → R by h x = max ⁡ f x , g x i f x &#8

JEE Advanced 2014 Continuity and Differentiability Solution
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19

If the function f x = 2 + cos x - 1 π - x 2 , x ≠ π k , x = π is continuous at x = π , then k equals

JEE Main 2014 Continuity and Differentiability Solution
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20

Let f : R → R be a function such that f x ≤ x 2 , for all x ∈ R . Then, at x = 0 , f is

JEE Main 2014 Continuity and Differentiability Solution
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21

Let f , g : R R be two functions defined by f(x)= cases x ( 1 x ) & , x 0 0, & , x=0 cases and g(x)=x f(x) Statement I: f is a continuous function at x =0 . Sta

JEE Main 2014 Continuity and Differentiability Solution
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22

Let f(x)=x|x|, g(x)= x and h(x)=(g f)(x) . Then

JEE Main 2014 Continuity and Differentiability Solution
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23

The following integral ∫ π 4 π 2 2 c o s e c x 17 d x is equal to

JEE Advanced 2014 Definite Integration Solution
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24

Paragraph: Given that for each a (0,1) , _ h 0⁺ _ h ^ 1-h t^ -a (1-t)^ a-1 d t exists. Let this limit be g(a) . In addition, it is given that the function g(a)

JEE Advanced 2014 Definite Integration Solution
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