Subject

Mathematics

Class

JEE Class 12

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JEE Mathematics 2009 Exam Questions

Multiple Choice Questions

11. integral subscript 0 superscript straight pi[cot x]dx, where [.] denotes the greatest integer function, is equal to
  • π/2

  • 1

  • -1

  • – π/2 


D.

– π/2 

Let space straight I space space equals space integral subscript 0 superscript straight pi space left square bracket space cot space straight x right square bracket space dx space space... space left parenthesis 1 right parenthesis
space equals space integral subscript 0 superscript straight pi space left square bracket space cot space left parenthesis straight pi minus straight x right parenthesis right square bracket space dx space equals space integral subscript 0 superscript straight x left square bracket negative cot space straight x right square bracket space dx space space... space left parenthesis 2 right parenthesis
adding space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis
2 straight I space equals space integral subscript 0 superscript straight pi space left square bracket space cot space straight x right square bracket space dx space space plus space integral subscript 0 superscript straight pi left square bracket negative cot space straight x right square bracket space dx space space equals integral subscript 0 superscript straight pi space left parenthesis negative 1 right parenthesis dx
equals space left square bracket negative straight x right square bracket subscript 0 superscript straight pi space equals space minus space straight pi
therefore comma space straight I space equals space straight pi over 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 open vertical bar table row cell space space space straight a end cell cell straight a plus 1 end cell cell space space straight a minus 1 end cell row cell negative straight b end cell cell straight b plus 1 end cell cell space space straight b minus 1 end cell row cell space space space straight c end cell cell space straight c minus 1 end cell cell space space straight c plus 1 end cell end table close vertical bar plus open vertical bar table row cell straight a plus 1 end cell cell straight b plus 1 end cell blank row cell straight a minus 1 end cell cell straight b minus 1 end cell cell straight c plus 1 end cell row cell left parenthesis negative 1 right parenthesis to the power of straight n plus 2 end exponent straight a end cell cell left parenthesis negative 1 right parenthesis to the power of straight n plus 1 end exponent straight b end cell cell left parenthesis negative 1 right parenthesis to the power of straight n space straight c end cell end table close vertical bar space equals space 0,then the value of 'n' is 

  • 0

  • any even integer

  • any odd integer

  • any integer


C.

any odd integer

open vertical bar table row straight a cell negative straight b end cell straight c row cell straight a plus 1 end cell cell straight b plus 1 end cell cell straight c minus 1 end cell row cell straight a minus 1 end cell cell straight b minus 1 end cell cell straight c plus 1 end cell end table close vertical bar open vertical bar table row cell left parenthesis negative 1 right parenthesis to the power of straight n plus 2 end exponent straight a end cell cell left parenthesis negative 1 right parenthesis to the power of straight n plus 1 end exponent straight b end cell cell left parenthesis negative 1 right parenthesis to the power of straight n space straight c end cell row cell straight a plus 1 end cell cell straight b plus 1 end cell cell straight c minus 1 end cell row cell straight a minus 1 end cell cell straight b minus 1 end cell cell straight c plus 1 end cell end table close vertical bar

space equals space open vertical bar table row cell straight a left parenthesis 1 plus left parenthesis negative 1 right parenthesis to the power of straight n right parenthesis end cell cell left parenthesis negative straight b right parenthesis left parenthesis 1 plus left parenthesis negative 1 right parenthesis to the power of straight n right parenthesis end cell cell straight c left parenthesis 1 plus left parenthesis negative 1 right parenthesis to the power of straight n end cell row cell straight a plus 1 end cell cell straight b plus 1 end cell cell straight c minus 1 end cell row cell straight a minus 1 end cell cell straight b minus 1 end cell cell straight c plus 1 end cell end table close vertical bar space equals space 0
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.

straight g left parenthesis straight f left parenthesis straight x right parenthesis right parenthesis space equals space sin space left parenthesis straight f left parenthesis straight x right parenthesis right parenthesis space equals space open curly brackets table attributes columnalign left end attributes row cell sin space straight x squared space comma space space straight x greater or equal than end cell row cell negative sin space straight x squared comma space straight x less than 0 end cell end table close
left parenthesis straight g left parenthesis straight f left parenthesis straight x right parenthesis right parenthesis right parenthesis apostrophe space equals space open curly brackets table attributes columnalign left end attributes row cell 2 straight x space cos space straight x squared comma space space straight x greater or equal than 0 end cell row cell negative 2 straight x space cos space straight x squared comma space straight x less than 0 end cell end table close
straight R. straight H. straight D space of space left parenthesis straight g left parenthesis straight f left parenthesis 0 right parenthesis right parenthesis right parenthesis apostrophe space equals space limit as straight h rightwards arrow 0 to the power of plus of space fraction numerator 2 straight h space cos space straight h squared over denominator straight h end fraction space equals space 2
straight L. straight H. straight D space of space left parenthesis straight g space left parenthesis straight f left parenthesis 0 right parenthesis right parenthesis right parenthesis apostrophe space equals space limit as straight h rightwards arrow 0 to the power of plus of space fraction numerator 2 straight h space cosh squared over denominator negative straight h end fraction space equals space minus 2
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 fraction numerator straight n squared minus 1 over denominator 4 end fraction
Statement 2: The sum of first n natural numbers is fraction numerator straight n space left parenthesis straight n plus 1 right parenthesis over denominator 2 end fraction and the sum of squares of first n natural numbers is fraction numerator straight n left parenthesis straight n plus 1 right parenthesis left parenthesis 2 straight n plus 2 right parenthesis over denominator 6 end fraction

  • 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
straight x to the power of straight x space left parenthesis 1 space plus space lnx right parenthesis space equals space left parenthesis negative cosec squared straight y space minus cosecy space cot space straight y right parenthesis space dy over dx
When space straight x space equals 1 space and space straight y space equals straight pi over 2
dy over dx space equals space minus space 1

179 Views

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