第12类RD Sharma解决方案–第2章功能–练习2.3

问题1.如果有雾和雾，请查找

(i)f(x)= e ^x ， $g (x) = \log_ex$

解决方案：

Let f: R → (0, ∞); and g: (0, ∞) → R

Clearly, the range of g is a subset of the domain of f.

So, fog: (0, ∞) → R and we know, (fog)(x) = f(g(x))

$= f(\log_ex)$

$= e^{log_ex}$

(fog)(x) = x

Clearly, the range of f is a subset of the domain of g.

⇒ fog: R→ R

(gof)(x) = g (f (x))

= g(e^x)

$= log_ee^x$

(gof)(x) = x

(ii)f(x)= x ² ，g(x)= cos x

解决方案：

f: R→ [0, ∞) ; g: R→[−1, 1]

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(cosx)

(fog)(x) = cos²x

Clearly, the range of f is a subset of the domain of g.

⇒ fog: R→R

(gof)(x) = g(f (x))

= g (x²)

(gof)(x) = cos x²

(iii)f(x)= | x |，g(x)= sin x

解决方案：

f: R → (0, ∞) ; g : R→[−1, 1]

Clearly, the range of g is a subset of the domain of f.

⇒ fog: R→R

(fog)(x) = f (g (x))

= f (sin x)

(fog)(x) = |sin x|

Clearly, the range of f is a subset of the domain of g.

⇒ fog : R→ R

(gof)(x) = g (f (x))

= g (|x|)

(gof)(x) = sin |x|

(iv)f(x)= x + 1，g(x)= e ^x

解决方案：

f: R→R ; g: R → [ 1, ∞)

Clearly, range of g is a subset of domain of f.

⇒ fog: R→R

(fog)(x) = f (g (x))

= f(e^x)

(fog)(x) = e^x + 1

Clearly, range of f is a subset of domain of g.

⇒ fog: R→R

(gof)(x) = g(f (x))

= g(x+1)

(gof)(x) = e^x+1

(v)f(x)= sin ^-1 x，g(x)= x ²

解决方案：

f: [−1,1]→ [(-π)/2 ,π/2]; g : R → [0, ∞)

Domain (fog) = {x: x ∈ R and x ∈ [−1, 1]}

So, Domain of (fog) = [−1, 1]

fog: [−1,1] → R

(fog)(x) = f (g (x))

= f(x²)

(fog)(x) = sin⁻¹(x²)

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)

(gof)(x) = (sin⁻¹x)²

(vi)f(x)= x + 1，g(x)= sinx

解决方案：

f: R→R ; g: R→[−1, 1]

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(sinx)

(fog)(x) = sin x + 1

Now we have to compute 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)

(gof)(x) = sin(x+1)

(vii)f(x)= x + 1，g(x)= 2x + 3

解决方案：

f: R→R ; g: R → R

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

(fog)(x) = 2x + 4

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

(gof)(x) = 2x + 5

(viii)f(x)= c，g(x)= sin x ²

解决方案：

f: R → {c} ; g: R→ [ 0, 1 ]

Clearly, the range of g is a subset of the domain of f.

fog: R→R

(fog)(x) = f(g(x))

= f(sinx²)

(fog)(x) = c

Clearly, the range of f is a subset of the domain of g.

⇒ fog: R→ R

(gof)(x) = g (f (x))

= g(c)

(gof)(x) = sinc²

(ix)f(x)= x ² + 2并且$g(x) = 1 - \frac{1}{1 - x}$

解决方案：

f: R → [2, ∞)

For domain of g: 1− x ≠ 0

⇒ x ≠ 1

⇒ Domain of g = R − {1}

$g(x) = 1 - \frac{1}{1 - x}$

=$\frac{(1 - x - 1)}{(1 - x)}$

$= \frac{-x}{(1 - x)}$

Range of g = R − {1}

So, g: R − {1} → R − {1}

Clearly, the range of g is a subset of the domain of f.

⇒ fog: R − {1} → R

(fog) (x) = f (g (x))

$= f (\frac{-x}{1 - x})$

$(fog)(x) = (\frac{-x}{1 - x})^2 + 2$

Clearly, the range of f is a subset of the domain of g.

⇒ gof: R→R

(gof)(x) = g (f (x))

= g(x² + 2)

$(gof)(x) = \frac{x^2 + 2}{x^2 + 1}$

问题2。令f(x)= x ² + x + 1且g(x)= sin x。显示雾≠gof。

解决方案：

Given f(x) = x² + x + 1 and g(x) = sin x

Now we have to prove fog ≠ gof

(fog)(x) = f(g(x))

= f(sin x)

(fog)(x) = sin²x + sin x + 1 …..(1)

And (gof)(x) = g (f (x))

= g (x²+ x + 1)

(gof)(x) = sin (x2+ x + 1) ….(2)

From (1) and (2), we get

fog ≠ gof.

问题3.如果f(x)= | x |，则证明fof = f。

解决方案：

Given f(x) = |x|,

Now we have to prove that fof = f.

Consider (fof)(x) = f (f(x))

= f(|x|)

= ||x||

= |x|

= f(x)

So, (fof) (x) = f (x), ∀x ∈ R

Hence, fof = f.

问题4.如果f(x)= 2x + 5和g(x)= x ² +1是两个实函数，则描述以下每个函数：

(i)雾

解决方案：

f(x) and g(x) are polynomials.

⇒ 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)高尔夫球

解决方案：

(gof)(x) = g (f (x))

= g (2x +5)

= (2x + 5)² + 1

= 4x² + 20x + 26

(iii)fof

解决方案：

(fof)(x) = f (f (x))

= f (2x +5)

= 2 (2x + 5) + 5

= 4x + 10 + 5

= 4x + 15

(iv)f ² (x)

解决方案：

f²(x) = f(x) x f(x)

= (2x + 5)(2x + 5)

= (2x + 5)²

= 4x² + 20x +25

问题5.如果f(x)= sin x和g(x)= 2x是两个实函数，则描述gof和fog。这些功能相等吗?

解决方案：

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 not equal functions.

问题6.令f，g，h是由f(x)= sin x，g(x)= 2x和h(x)= cos x给出的实函数。证明雾= go(fh)。

解决方案：

Given that f(x) = sin x, g (x) = 2x and h (x) = cos x

Now, fog(x) = f(g(x))

= f(2x)

fog(x) = sin2x ….(1)

And (go (f h)) (x) = g ((f(x). h(x))

= g (sin x cos x)

= 2sin x cos x

= sin (2x) ….(2)

From (1) and (2), fog(x) = go(fh) (x).

问题7.令f为任何实函数，令g为g(x)= 2x给出的函数。证明：gof = f + f。

解决方案：

We know, (gof)(x) = g(f(x))

= 2(f(x))

= f(x) + f(x)

= f + f.

Hence proved.

问题8。 $f(x) = \sqrt{1 - x}$和$g(x) = log_ex$有两个实函数，即雾和果岭。

解决方案：

Clearly the domain of f and g are R.

Now, fog(x) = f(g(x))

$= f(log_ex)$

fog(x)$= \sqrt{1 - log_ex}$

(gof)(x) = g(f(x))

$= g(\sqrt{1 - x})$

(gof)(x)$= log_e\sqrt{1 - x}$

问题9.如果f(x)= tan x并且$g(x) = \sqrt{1 - x^2}$ ，找到雾气和自由度。

解决方案：

fog(x) = f(g(x))

$= f(\sqrt{1 - x^2})$

$fog(x)= tan\sqrt{1 - x^2}$

(gof)(x) = g(f(x))

= g(tan x)

$(gof)(x) = \sqrt{1 - tan^2x}$

问题10.如果$f(x) = \sqrt{x + 3}$和g(x)= x ² +1是两个实函数，求雾和gof。

解决方案：

fog(x) = f(g(x))

= f(x² + 1)

$fog(x) = \sqrt{x^2 + 4}$

(gof)(x) = g(f(x))

$= g(\sqrt{x + 3})$

$= (\sqrt{x + 3})^2 + 1$

(gof)(x) = x + 4

问题11：设F是由下式给出一个真正的函数$f(x) = \sqrt{x - 2}$ 。找：

(i)fof

解决方案：

fof(x) = f(f(x))

$= f(\sqrt{x - 2})$

$fof(x) = \sqrt{\sqrt{x - 2} - 2}$

(ii)fofof

解决方案：

We know, fof(x) = f(f(x))

$= f(\sqrt{x - 2})$

Thus,$fof(x) = \sqrt{\sqrt{x - 2} - 2}$

Now, fofof(x) = fof(f(x))

$= fof(\sqrt{x - 2})$

$fofof(x) = \sqrt{\sqrt{\sqrt{x - 2} - 2} - 2}$

(iii)f ²

解决方案：

f²(x) = f(x).f(x)

=$(\sqrt{x - 2})^2$

f²⁽x) = x – 2

问题12 $f(x)= \begin{cases}1+x,0\lex\le2\\3-x\ \ \ ,2\lex\le3\end{cases}$找到fof。

解决方案：

Range of f = [0,3]

fof(x) = f(f(x))

$= f{\begin{cases}1+x,0\lex\le2\\3-x\ \ \ ,2\lex\le3\end{cases}}$

$fof(x) = \begin{cases}2+x\ \ ,0\lex\le1\\2-x\ \ \ \ \ ,1\lex\le2\\4-x\ \ \ \ \ ,2\lex\le3\end{cases}$