JEE mathematics

Sponsor Area

Question
CBSEENMA11015599

Suppose a population A has 100 observations 101, 102, … , 200, and another population B has 100 observations 151, 152, … , 250. If VA and VB represent the variances of the two populations, respectively, then VA/VB is 

  • 1

  • 9/4

  • 4/9

  • 2/3

Solution

A.

1

straight sigma subscript straight x superscript 2 space equals space fraction numerator sum from space to space of straight d subscript straight i superscript 2 over denominator straight n end fraction(Here deviations are taken from the mean) Since A and B both have 100 consecutive integers, therefore both have same standard deviation and hence the variance. 
therefore straight V subscript straight A over straight V subscript straight B space equals space 1 space left parenthesis As space begin inline style sum from space to space of end style space straight d subscript straight i superscript 2 space is space same space in space both space the space cases right parenthesis

Sponsor Area

Question
CBSEENMA11015600

If the roots of the quadratic equation x2 + px + q = 0 are tan30° and tan15°, respectively then the value of 2 + q − p is

  • 2

  • 3

  • 0

  • 1

Solution

B.

3

x2 + px + q = 0
tan 30° + tan 15° = − p
tan 30° ⋅ tan 15° = q
tan space 45 to the power of straight o space equals space fraction numerator tan space 30 to the power of straight o space plus space tan space 15 to the power of straight o over denominator 1 minus tan space 30 to the power of straight o space tan space 15 to the power of straight o end fraction space
equals space fraction numerator negative straight p over denominator 1 minus straight q end fraction space equals 1
⇒ − p = 1 − q
⇒ q − p = 1
∴ 2 + q − p = 3

Question
CBSEENMA11015601

Let W denote the words in the English dictionary. Define the relation R by :
R = {(x, y) ∈ W × W | the words x and y have at least one letter in common}. Then R is

  • not reflexive, symmetric and transitive

  • reflexive, symmetric and not transitive

  • reflexive, symmetric and transitive

  • reflexive, not symmetric and transitive

Solution

B.

reflexive, symmetric and not transitive

Clearly (x, x) ∈ R ∀ x ∈ W. So, R is reflexive. Let (x, y) ∈ R,
then (y, x) ∈ R as x and y have at least one letter in common. So, R is symmetric. But R is not transitive for example
Let x = DELHI, y = DWARKA and z = PARK then
(x, y) ∈ R and (y, z) ∈ R but (x, z) ∉ R

Question
CBSEENMA11015602

The value of sum from straight k equals 1 to 10 of space open parentheses sin space fraction numerator 2 kπ over denominator 11 end fraction plus space straight i space cos space fraction numerator 2 kπ over denominator 11 end fraction close parentheses space is

  • i

  • 1

  • -i

  • -1

Solution

C.

-i

sum from straight k space equals 1 to 10 of space open parentheses sin space fraction numerator 2 kπ over denominator 11 end fraction space plus space straight i space cos space fraction numerator 2 kπ over denominator 11 end fraction close parentheses
space equals space sum from straight k space equals 1 to 10 of space sin space fraction numerator 2 kπ over denominator 11 end fraction space plus space straight i space sum from straight k space equals 1 to 10 of space cos space fraction numerator 2 kπ over denominator 11 end fraction
space equals space 0 space plus space straight i space left parenthesis negative 1 right parenthesis space equals space minus space straight i

Question
CBSEENMA11015603

All the values of m for which both roots of the equations x2 − 2mx + m2 − 1 = 0 are greater than −2 but less than 4, lie in the interval

  • −2 < m < 0

  • m > 3

  • −1 < m < 3 

  • 1 < m < 4 

Solution

C.

−1 < m < 3 

Equation x2 − 2mx + m2 − 1 = 0
(x − m)2 − 1 = 0
(x − m + 1) (x − m − 1) = 0
x = m − 1, m + 1 − 2 < m − 1 and m + 1 < 4
m > − 1 and m < 3 − 1 < m < 3.