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JEE Mathematics 2012 previous year questions with solutions

114 JEE Mathematics questions from 2012 in the Quantrex bank. Featured questions below open unique solution pages. Browse by chapter on JEE Mathematics.

114 questions2012
1

Let f(x)=(1-x)² ² x+x² for all x I R and let g(x)= ₁^ x ( 2(t-1) t+1 - t ) f(t) d t for all x (1, ) . Question: Which of the following is true?

JEE Advanced 2012 Application of Derivatives Solution
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2

If f(x)= ₀^ x e^ t² (t-2)(t-3) d t for all x (0, ) , then

JEE Advanced 2012 Application of Derivatives Solution
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3

Let f: I R I R be defined as f(x)=|x|+ |x²-1 | . The total number of points at which f attains either a local maximum or a local minimum is

JEE Advanced 2012 Application of Derivatives Solution
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4

Let p(x) be a real polynomial of least degree which has a local maximum at x=1 and a local minimum at x=3 . If p(1)=6 and p(3)=2 , then p^ (0) is

JEE Advanced 2012 Application of Derivatives Solution
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5

Let S be the area of the region enclosed by y=e^ -x² , y=0, x=0 and x=1 ; then

JEE Advanced 2012 Area Under Curves Solution
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6

The area bounded by the parabola y^2=4 x and the line 2 x-3 y+4=0 , in square unit, is

JEE Main 2012 Area Under Curves Solution
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7

The area of the region bounded by the curve y=x^3 , and the lines, y=8 , and x=0 , is

JEE Main 2012 Area Under Curves Solution
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8

The area enclosed by the curves y=x^2, y=x^3 , x=0 and x=p , where p>1 , is 1 / 6 . The p equals

JEE Main 2012 Area Under Curves Solution
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9

If a straight line y-x=2 divides the region x^2+y^2 4 into two parts, then the ratio of the area of the smaller part to the area of the greater part is

JEE Main 2012 Area Under Curves Solution
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10

The parabola y^2=x divides the circle x^2+y^2=2 into two parts whose areas are in the ratio

JEE Main 2012 Area Under Curves Solution
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11

The area bounded between the parabolas x^2= y 4 and x^2=9 y , and the straight line y=2 is

JEE Main 2012 Area Under Curves Solution
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12

A tangent P T is drawn to the circle x²+y²=4 at the point P( 3 , 1) . A straight line L , perpendicular to P T is a tangent to the circle (x-3)²+y²-1 . Question

JEE Advanced 2012 Circle Solution
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13

A tangent P T is drawn to the circle x²+y²=4 at the point P( 3 , 1) . A straight line L , perpendicular to P T is a tangent to the circle (x-3)²+y²-1 . Question

JEE Advanced 2012 Circle Solution
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14

The locus of the mid-point of the chord of contact of tangents drawn from points lying on the straight line 4 x-5 y=20 to the circle x²+y²=9 is

JEE Advanced 2012 Circle Solution
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15

The number of common tangents of the circles given by x^2+y^2-8 x-2 y+1=0 and x^2+y^2+6 x+8 y=0 is

JEE Main 2012 Circle Solution
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16

If the line y=m x+1 meets the circle x^2+y^2+3 x=0 in two points equidistant from and on opposite sides of x -axis, then

JEE Main 2012 Circle Solution
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17

The equation of the circle passing through the point (1,2) and through the points of intersection of x^2+y^2-4 x-6 y-21=0 and 3 x+4 y+5=0 is given by

JEE Main 2012 Circle Solution
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18

The length of the diameter of the circle which touches the x -axis at the point (1,0) and passes through the point (2,3) is

JEE Main 2012 Circle Solution
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19

Let z be a complex number such that the imaginary part of z is non-zero and a=z²+z+1 is real. Then a cannot take the value

JEE Advanced 2012 Complex Number Solution
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20

For every integer n , let a_ n and b_ n be real numbers. Let function f: I R I R be given by f(x)= array lll a_ n + x, & for & x [2 n, 2 n+1] b_ n + x, & for &

JEE Advanced 2012 Continuity and Differentiability Solution
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21

Let (x)= array r x² | x |, x 0 0, x=0 array , x R . then is

JEE Advanced 2012 Continuity and Differentiability Solution
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22

If f(x)=a| x|+b e^ |x| +c|x|^3 , where a, b, c R , is differentiable at x=0 , then

JEE Main 2012 Continuity and Differentiability Solution
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23

Let f:[1,3] R be a function satisfying x [x] f(x) 6-x , for all x 2 and f(2)=1 , where R is the set of all real numbers and [x] denotes the largest integer less

JEE Main 2012 Continuity and Differentiability Solution
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24

Statement 1: A function f: R R is continuous at x₀ if and only if _ x x₀ f(x) exists and _ x x₀ f(x)=f (x₀ ) Statement 2: A function f: R R is discontinuous at

JEE Main 2012 Continuity and Differentiability Solution
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