Subject

Mathematics

Class

JEE Class 12

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JEE Mathematics 2010 Exam Questions

Multiple Choice Questions

11.

Let A be a 2 × 2 matrix with non-zero entries and let A2 = I, where I is 2 × 2 identity matrix. Define Tr(A) = sum of diagonal elements of A and |A| = determinant of matrix A.
Statement-1: Tr(A) = 0.
Statement-2: |A| = 1.

  • Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

  • Statement-1 is true, Statement-2 is true; statement-2 is not a correct explanation for Statement-1.

  • Statement-1 is true, Statement-2 is false.

  • Statement-1 is false, Statement-2 is true.


C.

Statement-1 is true, Statement-2 is false.

A satisfies A2 -Tr(A). A + (det A) l = 0
on comparing with A2-I = 0, 
we get 
Tr (A) = 0, |A| = - 1

152 Views

12.

If two tangents drawn from a point P to the parabola y2= 4x are at right angles, then the locus of P is

  • X = 1

  • 2x +1 = 0

  • x = -1

  • 2x-1 = 0


C.

x = -1

We know that the locus of point P from which two perpendicular tangents are drawn to the parabola is the directrix of the parabola.
Hence, the required locus is x = -1

281 Views

13.

let f : (-1, 1) → R be a differentiable function
with f(0) = -1 and f'(0) = 1.
Let g(x) = [f(2f(x) + 2)]2. Then g'(0) =

  • 4

  • -4

  • 0

  • -2


B.

-4

g(x) = (f(2(f(x) + 2))2
g'(x) 2f (2f (x) 2) f '(2f (x) 2) 2f '(x)
g'(0) 2f (2f (0) 2) f '(2f (0) 2) 2f '(0)
= 4f(0) × (f '(0))2– 4

123 Views

14.

The number of 3 × 3 non-singular matrices, with four entries as 1 and all other entries as 0, is 

  • less than 4

  • 5

  • 6

  • atleast 7


D.

atleast 7

178 Views

15.

Let f : R → R be a positive increasing function with limit as infinity space rightwards arrow 0 of space fraction numerator straight f left parenthesis 3 straight x right parenthesis over denominator straight f left parenthesis straight x right parenthesis end fraction space equals 1 space then space limit as infinity space rightwards arrow 0 of space fraction numerator begin display style straight f left parenthesis 2 straight x right parenthesis end style over denominator begin display style straight f left parenthesis straight x right parenthesis end style end fraction space space is space equal space to

  • 1

  • 2/3

  • 3/2

  • 3


A.

1

Since f(x) is a positive increasing function.
⇒ 0< f(x)<f(2x)<f(3x)
⇒ 0<1<fraction numerator straight f left parenthesis 2 straight x right parenthesis over denominator straight f left parenthesis straight x right parenthesis end fraction less than space fraction numerator straight f left parenthesis 3 straight x right parenthesis over denominator straight f left parenthesis straight x right parenthesis end fraction
limit as straight x space rightwards arrow 0 of space less or equal than space stack lim space with straight x rightwards arrow infinity below space fraction numerator straight f space left parenthesis 2 straight x right parenthesis over denominator straight f left parenthesis straight x right parenthesis end fraction space less or equal than space limit as straight x rightwards arrow infinity of space fraction numerator straight f left parenthesis 3 straight x right parenthesis over denominator straight f left parenthesis straight x right parenthesis end fraction
by space sandwich space theorem comma space limit as straight x space rightwards arrow infinity of space fraction numerator straight f left parenthesis 2 straight x right parenthesis over denominator straight f left parenthesis straight x right parenthesis end fraction space equals space 1

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16.

Let f : R → R be defined by
straight f left parenthesis straight x right parenthesis space equals space open curly brackets table attributes columnalign left end attributes row cell straight k minus 2 straight x comma if space straight x space less or equal than space minus 1 end cell row cell 2 straight x space plus 3 comma space if space straight x greater than negative 1 end cell end table close
If f has a local minimum at x = - 1 then a possible value of k is

  • 1

  • 0

  • -1/2

  • -1


C.

-1/2

k – 2x > 1 k + 2 = 1
k > 1 + 2x k = -1
k > 1 + 2(-1)
k > -1

137 Views

17.

The equation of the tangent to the curvestraight y equals straight x space plus space 4 over straight x squared, that is parallel to the x-axis, is

  • y = 0

  • y = 1

  • y = 3

  • y =2


C.

y = 3

space straight y space equals space straight x space plus space 4 over straight x squared
On differentiating w.r.t, we get
dy/dx = 1-8/x3
since the tangent is parallel to X-axis, therefore
⇒ x3 = 8
⇒ x = 2 and y = 3
133 Views

18.

Let f : R → R be a continuous function defined
by f(x) = 1/ex + 2e-x
Statement - 1: f(c) = 1/3, for some c ∈ R.
Statement-2: 0 < f(x)≤ fraction numerator 1 over denominator 2 square root of 2 end fraction, for all x ∈ R.

  • Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

  • Statement-1 is true, Statement-2 is true; statement-2 is not a correct explanation for Statement-1.

  • Statement-1 is true, Statement-2 is false.

  • Statement-1 is false, Statement-2 is true.


A.

Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

because space straight f space left parenthesis straight x right parenthesis space equals space fraction numerator 1 over denominator straight e to the power of straight x space plus space begin display style 2 over straight e to the power of straight x end style end fraction
Using comma space AM space greater or equal than space GM
fraction numerator straight e to the power of straight x space plus space begin display style 2 over straight e to the power of straight x end style over denominator 2 end fraction space greater or equal than space open parentheses straight e to the power of straight x. space 2 over straight e to the power of straight x close parentheses to the power of 1 divided by 2 end exponent comma space as space straight e to the power of straight x greater than 0
rightwards double arrow space straight e to the power of straight x space plus space 2 over straight e to the power of straight x space greater or equal than space 2 square root of 2
rightwards double arrow space 0 less than space fraction numerator 1 over denominator straight e to the power of straight x space plus begin display style 2 over straight e to the power of straight x end style end fraction space less or equal than space fraction numerator 1 over denominator 2 square root of 2 end fraction
therefore space 0 less than straight f left parenthesis straight x right parenthesis space less or equal than space fraction numerator 1 over denominator 2 square root of 2 end fraction comma space for space all space straight x element of straight R

Statement I us true and statement I as for some 'c'
f(c) = 1/3
134 Views

19. Let space straight S subscript 1 space equals space sum from straight j space equals 1 to 10 of space straight j space left parenthesis negative 1 right parenthesis space to the power of 10 straight C subscript straight j space comma space straight S subscript 2 space equals space sum from straight j space equals 1 to 10 of space straight j to the power of 10 straight C subscript straight j space and space straight S subscript 3 space equals sum from straight j space equals 1 to 10 of space straight j squared space to the power of 10 straight C subscript straight j

Statement-1: S3 = 55 × 29.
Statement-2: S1 = 90 × 28 and S2 = 10 × 28.

  • Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

  • Statement-1 is true, Statement-2 is true; statement-2 is not a correct explanation for Statement-1. 

  • Statement-1 is true, Statement-2 is false.

  • Statement-1 is false, Statement-2 is true.


C.

Statement-1 is true, Statement-2 is false.

straight S subscript 1 space equals space straight capital sigma with straight i space equals space 1 below and 10 on top space straight j space left parenthesis straight j minus 1 right parenthesis fraction numerator 10 factorial over denominator straight j space left parenthesis straight j minus 1 right parenthesis left parenthesis straight j minus 2 right parenthesis factorial left parenthesis 10 minus straight j right parenthesis factorial end fraction
space equals space 90 sum from straight j space equals 2 to 10 of space fraction numerator 8 factorial over denominator left parenthesis straight j minus 2 right parenthesis factorial left parenthesis 8 minus left parenthesis straight j minus 2 right parenthesis right parenthesis factorial end fraction space equals space 90.2 to the power of 8
space and space straight S subscript 2 space equals space sum from straight i space equals 1 to 10 of fraction numerator 9 factorial over denominator straight j left parenthesis left parenthesis straight j minus 1 right parenthesis factorial space left parenthesis 9 minus left parenthesis straight j minus 1 right parenthesis right parenthesis factorial end fraction
space equals space 10 sum from straight j space equals 1 to 10 of space fraction numerator 9 factorial over denominator left parenthesis straight j minus 1 right parenthesis factorial left parenthesis 9 minus left parenthesis straight j minus 1 right parenthesis right parenthesis factorial end fraction space equals space 10.2 to the power of 9
Also comma space straight S subscript 3 space equals space 10 sum from straight j space equals 1 to 10 of space left square bracket space straight j space left parenthesis straight j minus 1 right parenthesis space plus straight j right square bracket fraction numerator 10 factorial over denominator straight j factorial left parenthesis 10 minus straight j right parenthesis factorial end fraction
space equals space sum from straight j space equals 1 to 10 of space space straight j space left parenthesis straight j minus 1 right parenthesis to the power of 10 straight C subscript straight i space equals space sum from straight j space equals 1 to 10 of space straight j to the power of 10 straight C subscript straight j
space equals space 90.2 to the power of 8 space plus space 10.2 to the power of 8
space equals space 90.2 to the power of 8 space plus space 20.2 to the power of 8
space equals space 1110.2 to the power of 8 space equals space 56.2 to the power of 8
160 Views

20.

Consider the system of linear equation
x1 + 2x2 + x3 = 3
2x1 + 3x2 + x3 = 3
3x1 + 5x2 + 2x3 = 1
The system has

  • infinite number of solutions

  • exactly 3 solutions

  • a unique solution

  • no solution


D.

no solution

Subtracting the Eq. (ii) – Eq. (i)
We get x1 + x2 = 0
Subtract equations
Eq. (iii) – 2 × eq. (ii)
x1 + x2 = 5

Therefore, no solutions

146 Views

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