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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