RD Chapter 2 Functions Ex 2.3 Solutions
Question - 1 : - Find fog and gof, if
(i) f (x) = ex, g(x) = loge x
(ii) f (x) = x2, g(x) = cos x
(iii) f (x) = |x|, g (x) = sin x
(iv) f (x) = x+1, g(x) = ex
(v) f (x) = sin−1 x, g(x) = x2
(vi) f (x) = x+1, g (x) = sin x
(vii) f(x)= x + 1, g (x) = 2x + 3
(viii) f(x) = c, c ∈ R, g(x) = sin x2
(ix) f(x) = x2 + 2 , g (x) = 1 − 1/ (1-x)
Answer - 1 : -
(i) Given f (x) =ex, g(x) = loge x
Letf: R → (0, ∞); and g: (0, ∞) → R
Now we have tocalculate fog,
Clearly, the range of g is a subset of the domain of f.
fog: ( 0, ∞) → R
(fog) (x) = f (g (x))
= f (loge x)
= loge ex
= x
Now we have tocalculate gof,
Clearly, the range of f is a subset of the domain of g.
⇒ fog: R→ R
(gof) (x) = g (f (x))
= g (ex)
= loge ex
= x
(ii)f (x) = x2, g(x) = cos x
f: R→ [0, ∞) ; g: R→[−1, 1]
Now we have tocalculate fog,
Clearly, the range of g is not a subset of the domain of f.
⇒ Domain (fog) = {x: x∈domain of g and g (x) ∈ domain of f}
⇒ Domain (fog) = x: x ∈ R and cos x ∈ R}
⇒ Domain of (fog) = R
(fog): R→ R
(fog) (x) = f (g (x))
= f (cos x)
= cos2 x
Now we have tocalculate gof,
Clearly, the range of f is a subset of the domain of g.
⇒ fog: R→R
(gof) (x) = g (f (x))
= g (x2)
= cos x2
(iii) Givenf (x) = |x|, g(x) = sin x
f: R → (0, ∞) ; g : R→[−1, 1]
Now we have to calculatefog,
Clearly, the range of g is a subset of the domain of f.
⇒ fog: R→R
(fog) (x) = f (g (x))
= f (sin x)
= |sin x|
Now we have tocalculate gof,
Clearly, the range of f is a subset of the domain of g.
⇒ fog : R→ R
(gof) (x) = g (f (x))
= g (|x|)
= sin |x|
(iv) Givenf (x) = x + 1, g(x) = ex
f: R→R ; g: R → [ 1, ∞)
Now we have calculatefog:
Clearly, range of g is a subset of domain of f.
⇒ fog: R→R
(fog) (x) = f (g (x))
= f (ex)
= ex +1
Now we have to computegof,
Clearly, range of f is a subset of domain of g.
⇒ fog: R→R
(gof) (x) = g (f (x))
= g (x+1)
= ex+1
(v) Givenf (x) = sin −1 x, g(x) = x2
f: [−1,1]→[(-π)/2 ,π/2]; g : R → [0, ∞)
Now we have tocompute fog:
Clearly, the range of g is not a subset of the domain of f.
Domain (fog) = {x: x ∈ domain of g and g (x) ∈ domain of f}
Domain (fog) ={x: x ∈ R and x2 ∈ [−1, 1]}
Domain (fog) ={x: x ∈ R and x ∈ [−1, 1]}
Domain of (fog)= [−1, 1]
fog: [−1,1] → R
(fog) (x) = f (g (x))
= f (x2)
= sin−1 (x2)
Now we have tocompute gof:
Clearly, the range of f is a subset of the domain of g.
fog: [−1,1] → R
(gof) (x) = g (f (x))
= g (sin−1 x)
= (sin−1 x)2
(vi) Givenf(x) = x+1, g(x) = sin x
f: R→R ; g: R→[−1, 1]
Now we have tocompute fog
Clearly, the range of g is a subset of the domain of f.
Set of the domain of f.
⇒ fog: R→ R
(fog) (x) = f (g (x))
= f (sin x)
= sin x + 1
Now we have to computegof,
Clearly, the range of f is a subset of the domain of g.
⇒ fog: R → R
(gof) (x) = g (f (x))
= g (x+1)
= sin (x+1)
(vii) Givenf (x) = x+1, g (x) = 2x + 3
f: R→R ; g: R → R
Now we have tocompute fog
Clearly, the range of g is a subset of the domain of f.
⇒ fog: R→ R
(fog) (x) = f (g (x))
= f (2x+3)
= 2x + 3 + 1
= 2x + 4
Now we have tocompute gof
Clearly, the range of f is a subset of the domain of g.
⇒ fog: R → R
(gof) (x) = g (f (x))
= g (x+1)
= 2 (x + 1) + 3
= 2x + 5
(viii) Givenf (x) = c, g (x) = sin x2
f: R → {c} ; g: R→ [ 0, 1 ]
Now we have tocompute fog
Clearly, the range of g is a subset of the domain of f.
fog: R→R
(fog) (x) = f (g (x))
= f (sin x2)
= c
Now we have to computegof,
Clearly, the range of f is a subset of the domain of g.
⇒ fog: R→ R
(gof) (x) = g (f (x))
= g (c)
= sin c2
(ix) Givenf (x) = x2+ 2 and g (x) = 1 – 1 /(1 – x)
f: R → [ 2, ∞ )
For domain of g: 1− x ≠ 0
⇒ x ≠ 1
⇒ Domain of g= R − {1}
g (x )= 1 –[1/(1 – x)] = (1 – x – 1)/ (1 – x) = -x/(1 – x)
For range of g
y = (- x)/ (1 – x)
⇒ y – xy = − x
⇒ y = xy − x
⇒ y = x (y−1)
⇒ x = y/(y – 1)
Range of g =R − {1}
So, g: R −{1} → R − {1}
Now we have tocompute fog
Clearly, the range of g is a subset of the domain of f.
⇒ fog: R − {1}→ R
(fog) (x) = f (g (x))
= f (-x/ (1 – x))
= ((-x)/ (1 – x))2 +2
= (x2 +2x2 + 2 – 4x) / (1 – x)2
= (3x2 –4x + 2)/ (1 – x)2
Now we have tocompute gof
Clearly, the range of f is a subset of the domain of g.
⇒ gof: R→R
(gof) (x) = g (f (x))
= g (x2 + 2)
= 1 – 1 / (1 – (x2 +2))
= – 1/ (1 – (x2 +2))
= (x2 +2)/ (x2 + 1)
Question - 2 : - Let f(x) = x2 + x + 1 and g(x) =sin x. Show that fog ≠ gof.
Answer - 2 : -
Given f(x) = x2 + x +1 and g(x) = sin x
Now we have to provefog ≠ gof
(fog) (x) = f (g (x))
= f(sin x)
= sin2 x + sin x + 1
And (gof) (x) = g (f (x))
= g (x2+ x + 1)
= sin (x2+ x + 1)
So, fog ≠ gof.
Question - 3 : - If f(x) = |x|, prove that fof = f.
Answer - 3 : -
Given f(x) = |x|,
Now we have to provethat fof = f.
Consider(fof) (x) = f (f (x))
= f (|x|)
= ||x||
= |x|
= f (x)
So,
(fof) (x) = f (x), ∀x ∈ R
Hence, fof = f
Question - 4 : - If f(x) = 2x + 5 and g(x) = x2 + 1 betwo real functions, then describe each of the following functions:
(i) fog
(ii) gof
(iii) fof
(iv) f2
Also, show that fof ≠ f2
Answer - 4 : -
f(x) and g(x) arepolynomials.
⇒f: R → R and g: R → R.
So, fog: R → R and gof: R → R.
(i) (fog) (x) = f (g(x))
= f (x2 +1)
= 2 (x2 +1) + 5
=2x2 +2 + 5
= 2x2 +7
(ii) (gof) (x) = g (f(x))
= g (2x +5)
= (2x + 5)2 +1
= 4x2 +20x + 26
(iii) (fof) (x) = f (f(x))
= f (2x +5)
= 2 (2x + 5) + 5
= 4x + 10 + 5
= 4x + 15
(iv) f2 (x)= f (x) x f (x)
= (2x + 5) (2x +5)
= (2x + 5)2
= 4x2 +20x +25
Hence, from (iii) and(iv) clearly fof ≠ f2
Question - 5 : - If f(x) = sin x and g(x) = 2x be two real functions,then describe gof and fog. Are these equal functions?
Answer - 5 : -
Given f(x) =sin x and g(x) = 2x
We know that
f: R→ [−1, 1] and g: R→ R
Clearly, the range of f is a subset of the domain of g.
gof: R→ R
(gof) (x) = g (f (x))
= g (sin x)
= 2 sin x
Clearly, the range of g is a subset of the domain of f.
fog: R → R
So, (fog) (x) = f (g (x))
= f (2x)
= sin (2x)
Clearly, fog ≠gof
Hence they are notequal functions.
Question - 6 : - Let f, g, h be real functions given by f(x) =sin x, g (x) = 2x and h (x) = cos x. Provethat fog = go (f h).
Answer - 6 : -
Given that f(x) =sin x, g (x) = 2x and h (x) = cos x
We know that f: R→[−1, 1] and g: R→ R
Clearly, the range of g is a subset of the domain of f.
fog: R → R
Now, (fh) (x) = f (x)h (x) = (sin x) (cos x) = ½ sin (2x)
Domain of fh is R.
Since range of sin x is [-1,1], −1 ≤ sin 2x ≤ 1
⇒ -1/2 ≤ sin x/2 ≤ 1/2
Range of fh = [-1/2, 1/2]
So, (fh): R → [(-1)/2, 1/2]
Clearly, range of fh is a subset of g.
⇒ go (fh): R → R
⇒Domains of fog and go (fh) are the same.
So, (fog) (x)= f (g (x))
= f (2x)
= sin (2x)
And (go (fh)) (x) = g ((f(x). h(x))
= g (sinx cos x)
= 2sin x cos x
= sin (2x)
⇒ (fog) (x) = (go(f h)) (x), ∀x ∈ R
Hence, fog = go (fh)
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