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CBSE

Subject

Mathematics

Class

JEE Class 12

JEE Mathematics 2011 Exam Questions

Multiple Choice Questions

1.

Consider 5 independent Bernoulliís trials each with a probability of success p. If the probability of at least one failure is greater than or equal to 31/32, then p lies in the interval

  • (1/2, 3/4]

  • (3/4, 11/12]

  • [0, 1/2]

  • [11/12,1 ]


C.

[0, 1/2]

1 minus straight P to the power of 5 space greater or equal than space 31 over 32
straight P to the power of 5 space less or equal than space 1 over 32
straight P less or equal than space 1 half
straight P space element of space open square brackets 0 comma space 1 half close square brackets
250 Views

2.

The value of p and q for which the function f(x) =open curly brackets table attributes columnalign left end attributes row cell table attributes columnalign left columnspacing 1.4ex end attributes row cell fraction numerator sin space left parenthesis straight p space plus 1 right parenthesis space straight x space plus space sin space straight x over denominator straight x end fraction comma end cell cell straight x less than 0 end cell row cell straight q comma end cell cell straight x space equals space 0 space is space continuous space for space all space straight x space in space straight R comma space are colon end cell end table
fraction numerator square root of straight x space plus straight x squared end root minus square root of straight x over denominator straight x to the power of 3 divided by 2 end exponent end fraction space comma space straight x greater than space 0 space end cell row space end table close

  • p = 1/2. q = -3/2

  • p = 5/2, q = 1/2

  • p = - 3/2, q = 1/2

  • p = 1/2, q = 3/2


C.

p = - 3/2, q = 1/2

f(0) = q
straight f left parenthesis 0 to the power of plus right parenthesis space equals stack space lim with straight x space rightwards arrow 0 to the power of plus below space fraction numerator left parenthesis 1 space plus straight x right parenthesis to the power of 1 divided by 2 end exponent minus 1 space over denominator straight x end fraction
space equals space stack space lim with straight x space rightwards arrow 0 to the power of minus below space fraction numerator left parenthesis 1 space plus begin display style 1 half end style straight x...... negative 1 space over denominator straight x end fraction space equals space 1 half
straight f left parenthesis 0 to the power of minus right parenthesis space equals space stack space lim with straight x space rightwards arrow 0 to the power of minus below fraction numerator sin space left parenthesis straight p plus 1 right parenthesis space straight x space plus space sin space straight x over denominator straight x end fraction
straight f left parenthesis 0 to the power of minus right parenthesis space space equals space stack space lim with straight x space rightwards arrow 0 to the power of minus below space fraction numerator left parenthesis cos space left parenthesis straight p space plus 1 right parenthesis space straight x right parenthesis space left parenthesis straight p plus 1 right parenthesis space space plus space left parenthesis cos space straight x right parenthesis over denominator 1 end fraction
space equals space left parenthesis straight p space plus 1 right parenthesis space space plus 1 space equals space straight p space plus 2
therefore space straight p space plus 2 space equals space straight q space equals space 1 half
rightwards double arrow space straight p space equals space minus 3 divided by 2 comma space straight q space equals space 1 divided by 2

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3.

Consider the following statements
P: Suman is brilliant
Q: Suman is rich
R: Suman is honest. The negation of the statement ì Suman is brilliant and dishonest if and only if Suman is richî can be ex- pressed as

  •  ~ P ^ (Q ↔ ~ R)

  • ~ (Q ↔ (P ^ ~R)

  • ~ Q ↔ ~ P ^ R

  • ~ (P ^ ~ R)↔ Q


B.

~ (Q ↔ (P ^ ~R)

Negation of (PΛ~ R) ↔ Q is ~ ↔(PΛ ~ R)↔Q)
It may also be written as ~ (Q ↔ (PΛ ~ R))

183 Views

4.

If ω(≠1) is a cube root of unity, and (1 + ω)7 = A + Bω.Then (A, B) equals

  • (0,1)

  • (1,1)

  • (1,0)

  • (-1,1)


B.

(1,1)

(1 + ω)7 = A + Bω
(-ω2)7 = A + Bω
- ω14 = A + Bω

2 = A + Bω
1 + ω = A + Bω
therefore,
(A, B) = (1, 1)

159 Views

5.

Let R be the set of real numbers.
Statement-1 : A = {(x, y) ∈R × R : y - x is an integer} is an equivalence relation on R.
Statement-2 : B = {(x, y) ∈ R × R : x = αy for some rational number α} is an equivalence relation on R.

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

  • (2) 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. 

Statement - 1 :
(i) x - x is an integer ∀ x ∈ R so A is a reflexive relation.
(ii) y - x∈ I ⇒ x - y ∈ I so A is a symmetric relation. (iii) y - x ∈ I and z - y ∈ I ⇒ y - x + z - y ∈ ⇒
z - x ∈ I so A is a transitive relation. ThereforeAis equivalence relation.
Statement - 2 :
(i) x = αx when α = 1 ⇒ B is reflexive relation
(ii) for x = 0 and y = 2, we have 0 = α(2) for α = 0
But 2 = α(0) for no α so B is not symmetric so not equivalence.

171 Views

6.

The coefficient of x7 in the expansion of (1 - x - x2 +x3)6 is :

  • 144

  • -132

  • -144

  • 132


C.

-144

left parenthesis 1 minus straight x minus straight x squared plus straight x cubed right parenthesis to the power of 6
left parenthesis 1 minus straight x right parenthesis to the power of 6 left parenthesis 1 minus straight x squared right parenthesis to the power of 6
left parenthesis to the power of 6 straight C subscript 0 minus to the power of 6 straight C subscript 1 straight x to the power of 1 plus to the power of 6 straight C subscript 2 straight x squared minus to the power of 6 straight C subscript 3 straight x cubed plus to the power of 6 straight C subscript 4 straight x to the power of 4 minus space to the power of 6 straight C subscript 5 straight x to the power of 5 space plus to the power of 6 straight C subscript 6 straight x to the power of 6 right parenthesis
left parenthesis to the power of 6 straight C subscript 0 minus to the power of 6 straight C subscript 1 straight x squared space plus to the power of 6 straight C subscript 2 straight x to the power of 4 minus to the power of 6 straight C subscript 3 straight x squared space plus to the power of 6 straight C subscript 4 straight x to the power of 8 space plus...... plus to the power of 6 straight C subscript 6 straight x to the power of 12 right parenthesis
Now space coefficient space of space straight x to the power of 7 space equals space to the power of 6 straight C subscript 1 to the power of 6 straight C subscript 3 minus to the power of 6 straight C subscript 3 to the power of 6 straight C subscript 2 space plus to the power of 6 straight C subscript 5 to the power of 6 straight C subscript 1
space equals space 6 space straight x space 20 space minus space 20 space space straight x space 15 space plus 36
space equals space 120 minus 300 plus 36
space equals space 156 minus 300
space equals space minus 144 space space
174 Views

7.

If the mean deviation about the median of the numbers a, 2a, ....., 50a is 50, then |a| equals 

  • 2

  • 3

  • 4

  • 5


C.

4

Median = 25.5 a
Mean deviation about median = 50

rightwards double arrow space fraction numerator straight capital sigma space vertical line straight x subscript straight i space minus 25.5 straight a vertical line over denominator 50 end fraction space equals space 50
⇒24.5 a + 23.5a + ..... + 0.5a + 0.5a + .... + 24.5a = 2500
⇒ a + 3a + 5a + ..... + 49a = 2500
⇒ 25/2 (50a) = 2500 ⇒ a = 4

314 Views

8.

Let α,  β be real and z be a complex number. If z2 + αz + β = 0 has two distinct roots on the line Re z = 1, then it is necessary that

  • β ∈(0, 1)

  • β ∈(-1, 0)

  • |β| = 1

  • β ∈ (1, ∞)


D.

β ∈ (1, ∞)

Let roots be p + iq and p - iq p, q ∈ R
root lie on line Re(z) = 1
⇒ p = 1
product of roots = p2 + q2 =  β = 1 + q2
⇒ β∈ (1, ∞), (q ≠ 0, ∵ roots are distinct)

286 Views

9. fraction numerator straight d squared straight x over denominator dy squared end fraction equal to
  • open parentheses fraction numerator straight d squared straight x over denominator dy squared end fraction close parentheses to the power of negative 1 end exponent
  • negative open parentheses fraction numerator straight d squared straight y over denominator dx squared end fraction close parentheses to the power of negative 1 end exponent open parentheses dy over dx close parentheses to the power of negative 3 end exponent
  • open parentheses fraction numerator straight d squared straight y over denominator dx squared end fraction close parentheses open parentheses dy over dx close parentheses to the power of negative 2 end exponent
  • negative open parentheses fraction numerator straight d squared straight y over denominator dx squared end fraction close parentheses open parentheses dy over dx close parentheses to the power of negative 3 end exponent

D.

negative open parentheses fraction numerator straight d squared straight y over denominator dx squared end fraction close parentheses open parentheses dy over dx close parentheses to the power of negative 3 end exponent
dy over dx space equals space fraction numerator 1 over denominator begin display style dx over dy end style end fraction

fraction numerator straight d squared straight y over denominator dx squared end fraction space equals space straight d over dx open parentheses fraction numerator 1 over denominator dx divided by dy end fraction close parentheses space
equals space fraction numerator begin display style straight d end style over denominator begin display style dx end style end fraction open parentheses fraction numerator begin display style 1 end style over denominator begin display style dx divided by dy end style end fraction close parentheses. fraction numerator begin display style straight d end style over denominator begin display style dx end style end fraction
equals space minus space space fraction numerator begin display style 1 end style over denominator begin display style open parentheses dx over dy close parentheses squared end style end fraction. fraction numerator begin display style fraction numerator straight d squared straight x over denominator dy squared end fraction end style over denominator begin display style dx over dy end style end fraction
space equals space fraction numerator begin display style negative fraction numerator straight d squared straight x over denominator dy squared end fraction end style over denominator begin display style open parentheses dx over dy close parentheses cubed end style end fraction
equals negative open parentheses fraction numerator begin display style straight d squared straight x end style over denominator begin display style dy squared end style end fraction close parentheses open parentheses fraction numerator begin display style dy end style over denominator begin display style dx end style end fraction close parentheses cubed
167 Views

10.

A man saves Rs. 200 in each of the first three months of his service. In each of the subsequent months his saving increases by Rs. 40 more than the saving of immediately previous month. His total saving from the start of service will be Rs. 11040 after

  • 18 Months

  • 19 Months

  • 20 Months

  • 21 Months


D.

21 Months

a = Rs. 200
d = Rs. 40
savings in first two months = Rs. 400
remained savings = 200 + 240 + 280 + ..... upto n terms =
n/2[400 + (n -1)40] = 11040 - 400
200n + 20n2 - 20n = 10640
20n2 + 180 n - 10640 = 0
n2 + 9n - 532 = 0
(n + 28) (n - 19) = 0
n = 19
∴ no. of months = 19 + 2 = 21

140 Views

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