﻿ One ticket is selected at random from 50 tickets numbered 00, 01, 02, ... , 49. Then the probability that the sum of the digits on the selected ticket is 8, given that the product of these digits is zero, equals from Mathematics Class 12 JEE Year 2009 Free Solved Previous Year Papers

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JEE Mathematics 2009 Exam Questions

Multiple Choice Questions

11. [cot x]dx, where [.] denotes the greatest integer function, is equal to
• π/2

• 1

• -1

• – π/2

D.

– π/2

104 Views

12.

For real x, let f(x) = x3+ 5x + 1, then

• f is one–one but not onto R

• f is onto R but not one–one

• f is one–one and onto R

• f is neither one–one nor onto R

C.

f is one–one and onto R

f(x) = x3+ 5x + 1
f'(x )3x2 +5>0
⇒ f is one–one
therefore, f is cubic
⇒ f is onto
‘f’ is one–one and onto.

102 Views

13.

Let a, b, c be such that 0 (a +c) ≠ . If ,then the value of 'n' is

• 0

• any even integer

• any odd integer

• any integer

C.

any odd integer

167 Views

14.

Let f(x) = (x + 1)2– 1, x ≥ – 1
Statement – 1: The set {x : f(x) = f–1(x)} = {0, –1}.
Statement – 2: f is a bijection.

• 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

(x + 1)2 – 1 = x

(x + 1)2= x + 1
⇒ x = 0, −1
Since co–domain of function is not given.So if we assume function
(a) as onto then A is correct
(b) as not onto then none of the answer is correct.

116 Views

15.

Let f(x) = x|x| and g(x) = sinx

Statement 1 : gof is differentiable at x = 0 and its derivative is continuous atthat point
Statement 2: gof is twice differentiable at x = 0

• 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.

Clearly gof is twice differentiable at x = 0 hence it is differentiable at x = 0 and its derivative is continuous at x = 0
106 Views

16.

One ticket is selected at random from 50 tickets numbered 00, 01, 02, ... , 49. Then the probability that the sum of the digits on the selected ticket is 8, given that the product of these digits is zero, equals

•  1/14

• 1/7

• 5/14

• 1/50

A.

1/14

A = Events that sum of the digits on selected ticket is 8
= {08, 17, 26, 35, 44}
n(A) = 5
Event that product of digits is zero
= {00, 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 20,30, 40}
⇒ n(B) = 14

=P(A/B) = (5/14)

168 Views

17.

Statement 1: ~ (p ↔ ~ q) is equivalent to p ↔ q
Statement 2 : ~ (p ↔ ~ q) is a tautology

• 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.

 p q ~q (p ↔ ~q) ~ (p ↔ ~q) p ↔ q T T F F T T T F T T F F F T F T F F F F T F T T

Clearly, ~ (p ↔ ~q) is not a tautology because it does not contain T in the column of its truth table. Also, ~(p ↔ ~ q) & p ↔ q have the same truth value
135 Views

18.

Given P(x) = x4+ ax3 + cx + d such that x = 0 is the only real root of P′ (x) = 0. If P(–1) < P(1),then in the interval [–1, 1].

• P(–1) is the minimum and P(1) is the maximum of P

• P(–1) is not minimum but P(1) is the maximum of P

• P(–1) is the minimum but P(1) is not the maximum of P

• neither P(–1) is the minimum nor P(1) is the maximum of P

B.

P(–1) is not minimum but P(1) is the maximum of P

P(x) = x4+ ax3+ bx2+ cx + d
P′(x) = 4x3+ 3ax2+ 2bx + c
As P′(x) = 0 has only root x = 0
⇒ c = 0
P′(x) = x(4x2+ 3ax + 2b)
⇒ 4x3+ 3ax + 2b = 0 has non real root.
and 4x2+ 3ax + 2b > 0 ∀ x ∈ [−1, 1].

As P(−1) < P(1) ⇒ P(1) is the max. of P(x) in [−1, 1]

141 Views

19.

Statement 1: The variance of first n even natural numbers is
Statement 2: The sum of first n natural numbers is  and the sum of squares of first n natural numbers is

• 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

D.

Statement–1 is false, Statement–2 is true

265 Views

20.

Let y be an implicit function of x defined by x2x – 2xxcoty – 1 = 0. Then y′ (1) equals

• -1

• 1

• log 2

• -log 2

A.

-1

When x = 1, y=π/2
=(xx– cot y)2= cosec2y
xx = cot y + |cosec y|
when x = 1, y=π/2

⇒ xx = cot y + cosec y
diff. w.r.t. to x

179 Views