RBSE Solutions for Class 12 Maths Chapter 1 संयुक्त फलत Ex 1.1 is part of RBSE Solutions for Class 12 Maths. Here we have given Rajasthan Board RBSE Class 12 Maths Chapter 1 संयुक्त फलत Exercise 1.1.
Rajasthan Board RBSE Class 12 Maths Chapter 1 संयुक्त फलत Ex 1.1
प्रश्न 1.
यदि f : R→R तथा g : R→R दो फलन निम्न प्रकार से परिभाषित हों, तो (fog)(x) तथा (gof)(x) ज्ञात कीजिए :
(i) f(x) = 2x + 3, g(x) = x² + 5
(ii) f(x) = x² + 8, g(x) = 3x3 + 1
(iii) f(x) = x, g(x) = | x |
(iv) f(x) = x² + 2x + 3, g(x) = 3x – 4
हल :
(i) f(x) = 2x +3
g(x) = x² + 5.
(fog)(x) = f(g(x))
= f(x² + 5)
= 2(x² + 5) + 3
= 2x² + 10 + 3
= 2x² + 13
(gof)(x) = g(f(x))
= g(2x + 3)
= (2x + 3)² + 5
= 4x² + 9 + 12x + 5
= 4x² + 12x + 14
(ii) f(x) = x² + 8
g(x) = 3x3 + 1
(fog)(x) = f(g(x))
= f(3x3 + 1)
= (3x3 + 1)² + 8
= 9x6 + 6x3 + 1 + 8
= 9x6 + 6x + 9
(gof)(x) = g(f(x))
= g(x² + 8)
= 3(x² + 8)3 + 1
= 3(x6 + 3x4 × 8 + 32 × 64 + 83) + 1
= 3x6 + 727x4 + 576x2 + 15136
(iii) f(x) = x
g(x) = | x |
(fog)(x) = f(g(x))
= f(| x |) = | x |
(gof)(x) = g(f(x))
= g(x) = | x |
(iv) f(x) = x² + 2x + 3
g(r) = 3x – 4
(fog)(x) = f(g(x))
= f(3x – 4)
= (3x – 4)² + 2(3x – 4) + 3
= 9x² – 24x + 16 + 6 – 8 + 3
= 9x² – 18x + 11
(gof)(x) = g(f(x)) = g(x² + 2x + 3)
= 3(x² + 2 + 3) – 4
= 3x² + 6x + 9 – 4
= 3x² + 6x + 5.
प्रश्न 2.
यदि A = {a, b, c}, B = {u, v, w} यदि f: A→B तथा g: B→A निम्न प्रकार परिभाषित हों कि
f= {(a, v), (b, u), (c, w)}
g= {(u, b), (v, a), (w, c)} तो (fog) तथा (gof) ज्ञात कीजिए।
हल :
दिया है, f = {(a, v), (b, u), (c, w)}
g = {(u, b), (v, a), (w, c}}
∴ f(a) = v तथा g(u) = b
f(b) = u तथा g(v) = a
f(c) = w तथा g(w) = c
अत: fog(x) = f[g(x)] से
fog(u) = f[g(u)] = f(b) =u
fog(v) = f[g(v)] = f(a) = v
fog(w) = f[g(w)] =f(c) = w
अत: fog= {(u, u), (v, v), (w, w)}
gof(a) = g[f(a)] = g(v) = a
gof(b) = g[f(b)] = g(u) = b
gof(c) = g[f(c)] = g(w) = c
∴ gof = {(a, a), (b, b), (c, c)}.
प्रश्न 3.
यदि f : R+ → R+ तथा g : R+ → R+ निम्न प्रकार परिभाषित हों कि
f(x) = x² तथा g(x) = √x तो
gof तथा fog ज्ञात कीजिए। क्या ये तुल्य फलन है?
हल :
दिया है, f: R+ → R+,f(x) = x²
g: R+ → R+, g(x) = √x
(gof)(x) = g[f(x)] = g(x)² = √x² = x
(fog)(x) = f[g(x)] = f(√x) = (√x)² = x
उपरोक्त से स्पष्ट है कि
(fog)(x) = (gof) (x) = x, ∀x ∈ R+
अत: (fog) तथा (gof) तुल्य फलन है।
प्रश्न 4.
यदि f: R → R तथा g : R → R दो ऐसे फलन हैं कि f(x) = 3x + 4 तथा g(x) = 
हल :
दिये गये फलन हैं,
f: R → R, f(x) = 3x + 4
g: R→ R, g(x) =
∴ (fog)(x) = f(g(x))
प्रश्न 5.
यदि f, g, h तीन फलन R से R पर इस प्रकार परिभाषित हैं कि f(x) = x², g(x) = cos x एवं h(x) = 2x + 3, तो {ho(gof)} √2π का मान लिखिये।
हल :
दिया है : f(x) = x², g(x) = cos x, h(x) = 2x + 3
∴{ho(gof)}(x) = hog{{(x)} = h[g{f(x)}]
= h[g(x²)] = h(cos x²)
= 2 cos x² + 3
∴{ho(gof}}√2π = 2 cos (√2π)² + 3
= 2 cos 2π + 3
= 2 × 1 + 3 = 5
प्रश्न 6.
यदि f तथा g निम्न प्रकार परिभाषित हों, तो (gof) (x) ज्ञात कीजिए
(i) f : R → R, f(x) = 2x + x-2
(ii) g : R → R, g(x) = x4 + 2x + 4
हल :
दिया है,f : R → R
f(x) = 2x + x-2
g : R → R, g(x) = x4 + 2x + 4
(gof)(x) = g{f(x)}
= g{2x + x-2}
= (2x + x-2)4 + 2(2x + x-2) + 4
प्रश्न 7.
यदि A = {1, 2, 3, 4},f: R→ R, f(x) = x² + 3x + 1 g: R → R, g(x) = 2x – 3, तब तब ज्ञात कीजिए :
(i) (fog)(x)
(ii) (gof)(x)
(iii) (fof)(x)
(iv) (gog)(x)
हल :
दिया है :
f: R → R, (x) = x² + 3x + 1
g : R – R, g(x) = 2x – 3
(i) ∴ (fog)(x) = f{g(x)}
=f{2x – 3}
= (2x – 3)² + 3(2x – 3) + 1
= 4x² – 12x + 9 + 6x – 9 + 1
= 4x² – 6 + 1
(ii) (gof)(x) = g{(x)}
= g(x² + 3x + 1)
= 2(x² + 3x + 1) – 3
= 2x² + 6x + 2 – 3,
= 2x² + 6x – 1.
(iii) (fof)(x) = f{f(x)}
= f(x² + 3x + 1)
= (x² + 3x + 1)² + 3(x² + 3x + 1) + 1
= x4 + 9x² + 1 + 6x3 + 6x + 2x² + 3x² + 9x + 3 + 1
= x4 + 6x3 + 14x² + 15x + 15
(iv) (gog)(x) = g{g(x)}
= g(2x – 3)
= 2(2x – 3) – 3
= 4x – 6 – 3
= 4x – 9
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