Relations and Functions

More Topic from Mathematics

Question 1

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

  • 1

  • -1

  • – π/2 

Solution

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
Question 2

A function is matched below against an interval where it is supposed to be increasing. Which of the following pairs is incorrectly matched?

  • Interval Function
    (-∞, ∞) x3 – 3x2 + 3x + 3
  • Interval Function
    [2, ∞) 2x3 – 3x2 – 12x + 6
  • Interval Function
    (-∞, 1/3] 3x2 – 2x + 1
  • Interval Function
    (- ∞, -4] x3 + 6x2 + 6

Solution

C.

Interval Function
(-∞, 1/3] 3x2 – 2x + 1

Clearly function f(x) = 3x2 – 2x + 1 is increasing when
f′(x) = 6x – 2 ≥ 0 ⇒ x∈[1/3, ∞)

Question 3

A real valued function f(x) satisfies the functional equation f(x – y) = f(x) f(y) – f(a – x) f(a + y) where a is a given constant and f(0) = 1, f(2a – x) is equal to

  • –f(x)

  • f(x)

  • f(a) + f(a – x)

  • f(-x)

Solution

A.

–f(x)

f(a – (x – a)) = f(a) f(x – a) – f(0) f(x) 
 = - f(x) [ ∵ x = 0, y= 0, f(0) = f2 (0)-f2(a) = 0 ⇒ f(a) = 0]

Question 4

Consider the following relations:
R = {(x, y)| x, y are real numbers and x = wy for some rational number w}; S = {(m/p, p/q)| m, n, p and q are integers such that n, q ≠ 0 and qm = pn}. Then

  • R is an equivalence relation but S is not an equivalence relation

  • neither R nor S is an equivalence relation

  • S is an equivalence relation but R is not an equivalence relation

  • R and S both are equivalence relations

Solution

D.

R and S both are equivalence relations

Question 5

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 

Solution

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.