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Relations And Functions

Question
CBSEENMA12036049

If g is the inverse of a function f and f'(x) = fraction numerator 1 over denominator 1 plus straight x to the power of 5 end fraction comma then g'(x) is equal to 

  • 1+ x6

  • 5x4

  • fraction numerator 1 over denominator 1 plus space left curly bracket straight g left parenthesis straight x right parenthesis right square bracket to the power of 5 end fraction
  • 1+{g(x)}5

Solution

D.

1+{g(x)}5

Here 'g' is the inverse of f(x)
⇒ fog (x) =x
On differentiating w.r.t x, we get
f'{g(x)} x g'(x) =1
space straight g apostrophe left parenthesis straight x right parenthesis space equals space fraction numerator 1 over denominator straight f apostrophe left parenthesis straight g left parenthesis straight x right parenthesis right parenthesis end fraction space equals space fraction numerator 1 over denominator begin display style fraction numerator 1 over denominator 1 plus left curly bracket straight g left parenthesis straight x right parenthesis to the power of 5 right curly bracket end fraction end style end fraction
open square brackets because space straight f apostrophe left parenthesis straight x right parenthesis space equals space fraction numerator 1 over denominator 1 plus straight x to the power of 5 end fraction close square brackets
straight g apostrophe left parenthesis straight x right parenthesis space equals space 1 space plus space left curly bracket straight g space left parenthesis straight x right parenthesis right curly bracket to the power of 5

Some More Questions From Relations and Functions Chapter

Let L be the set of all lines in a plane and R be the relation in L defined as R = {(L1, L2) : L1 is perpendicular to L2}. Show that R is symmetric but neither reflexive nor transitive.

 Determine whether each of the following relations are reflexive, symmetric and transitive :

(i) Relation R in the set A = {1, 2, 3,....., 13, 14} defined as

R = {(x, y) : 3 x – y = 0}

(ii) Relation R in the set N of natural numbers defined as R = {(x, y) : y = x + 5 and x < 4} (iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as R = {(x,y) : y is divisible by x} (iv) Relation R in the set Z of all integers defined as R = {(x,y) : x – y is an integer}

(v) Relation R in the set A of human beings in a town at a particular time given by
(a)    R = {(x, y) : x and y work at the same place}
(b)    R = {(x,y) : x and y live in the same locality}
(c)    R = {(x, y) : x is exactly 7 cm taller than y}
(d)    R = {(x, y) : x is wife of y}
(e)    R = {(x,y) : x is father of y}

Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b) : b = a + 1} is reflexive, symmetric or transitive.

Show that the relation R in R defined as R = {(a, b) : a ≤ b}, is reflexive and transitive but not symmetric.

Check whether the relation R in R defined by R = {(a,b) : a ≤ b3} is refleive, symmetric or transitive.

If R and R’ arc reflexive relations on a set then so are R ∪ R’ and R ∩ R’. 

If R and R’ are symmetric relations on a set A, then R ∩ R’ is also a sysmetric relation on A.

Show that the union of two symmetric relations on a set is again a symmetric relation on that set.

Let A = {1. 2. 3}. Then show that the number of relations containing (1,2) and (2. 3) which are reflexive and transitive but not symmetric is four.