﻿ The population p(t) at time t of a certain mouse species satisfies the differential equation . if p (0) = 850, then the  time at which the population becomes zero is from Mathematics Class 12 JEE Year 2012 Free Solved Previous Year Papers

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# JEE Mathematics 2012 Exam Questions

#### Multiple Choice Questions

21.

If g(x) =  then g(x +π) equals

• g(x)/g(π)

• g(x) +g(π)

• g (x) - g(π)

• g(x). g(π)

B.

g(x) +g(π)

C.

g (x) - g(π)

Integral
To find g(x+π) in terms of g(x) of g(π)

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22.

Let ABCD be a parallelogram such that  and ∠BAD be an acute angle. If   is the vector that coincides with the altitude directed from the vertex B the side AD, then  is given byLet ABCD be a parallelogram such that AB = q,AD = p and ∠BAD be an acute angle. If r is the vector that coincides with the altitude directed from the vertex B to the side AD, then r is given by (1)

D.

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23.

Consider the function f(x) = |x – 2| + |x – 5|, x ∈ R.
Statement 1: f′(4) = 0
Statement 2: f is continuous in [2, 5], differentiable in (2, 5) and f(2) = f(5).

• Statement 1 is false, statement 2 is true

• 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

C.

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

f(x) = 7 – 2x; x < 2
= 3; 2 ≤ x ≤ 5
= 2x – 7; x > 5
f(x) is constant function in [2, 5]
f is continuous in [2, 5] and differentiable in (2, 5) and f(2) = f(5)
by Rolle’s theorem f′(4) = 0
∴ Statement 2 and statement 1 both are true and statement 2 is correct explanation for statement 1.

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24.

Let  be  two unit vectors. If the vectors  and  are perpendicular to each other, then the angle between  is

• π/6

• π/2

• π/3

• π/4

C.

π/3

(iii) c and d are perpendicular to each other, i.e., c.d =0
To find Angle between a and b
Now, c.d = 0
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25.

If: R →R is a function defined by  where [x] denotes the greatest integer function, then f is

• continuous for every real x

• discontinous only at x = 0

• discontinuous only at non-zero integral values of x

• continuous only at x =0

A.

continuous for every real x

therefore, [x] sin π x is continuous for every real x.
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26.

If the integral   then an equal to

• -1

• -2

• 1

• 2

D.

2

Now, let us assume that I,

Multiplying by cos x in numerator and denominator, we get

This special integration requires special substitution of type
N' = A (D') +
⇒ Let 5 sin x = (A + 2B) sin x + (B- 2A) cos x
Comapring the coefficients of sin x and cosx,
we get
A +2B =5 and B- 2A = 0
Solving the above two equations in A and B
we get
A = 1 an B= 2
⇒ 5 sin x = ( sin x - 2 cos x)+ 2 (cos x + 2 sin x)

where k is the constant of integration. Now, by comparing the value of l in eq. (i) and (ii) we get a = 2
 ⇒
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27.

Let P and Q be 3 × 3 matrices with P ≠ Q. If P3= Qand P2Q = Q2P, then determinant of(P2+ Q2) is equal to

• -2

• 1

• 0

• -1

C.

0

P3= Q3
P3– P2Q = Q3– Q2P
P2(P – Q) = Q2(Q – P)
P2(P – Q) + Q2(P – Q) = O
(P2+ Q2)(P – Q) = O
⇒ |P2+ Q2| = 0

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28.

The population p(t) at time t of a certain mouse species satisfies the differential equation . if p (0) = 850, then the  time at which the population becomes zero is

• 2 log 18

• log 9

• log 18

A.

2 log 18

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29.

Let a, b ∈ R be such that the function f given by f(x) = ln |x| + bx
2+ ax, x ≠ 0 has extreme values at x = –1 and x = 2.
Statement 1: f has local maximum at x = –1 and at x = 2.
Statement 2:

• Statement 1 is false, statement 2 is true

• 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

B.

Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1

(i) A function f, such that f(x)= log |x| +bx2 +ax, x≠0
(ii) The function 'f' has extrema at x = -1 and x =2 i.e, f'(1) = f'(2) = 0 and f''(-1) ≠ 0≠f''(2)
Now, given function f is given by
f(x) = log |x| +bx2 +ax

Since 'f' has extrema at x = - 1 and x =2
Hence, f'(-1) = 0 =f'(2)
f'(-1) = 0
⇒ a-2b =1 ..... (i)
and f'(2) = 0
⇒ a+ 4b = -1/2
solving eq. (i) and (ii), we get
a =1/2 and b = -1/4

⇒ f'' has local maxima at both x = - 1 and x =2
Thus, a statement I is correct. Also, while solving for the statement I, we found values of a and b, which justify that statement 2 is also correct.

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30.

Assuming the balls to be identical except for difference in colours, the number of ways in which one or more balls can be selected from 10 white, 9 green and 7 black balls is

• 880

• 629

• 630

• 879

D.

879

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