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g(x) = \(\frac{5x -6}7\)

g(x) = 0

\(\therefore\) \(\frac{5x -6}7\) = 0

\(\therefore\) x = 6/5

Answer : (c) 1 – x³ 

Let y = f (x) = (1 – x) ^1/3 

⇒ y³ = 1 – x 

⇒ x = 1 – y³ 

⇒ f ^–1(x) = 1 – x³

\(\left(\frac{f(2+h)-f(2)}h\right)\)

\(=\frac{(2+h)^2-3(2+h)+1-(2^2-3(2)+1)}h\) 

 = \(\frac{h^2+h}h\) 

 = h + 1

Given as

f(x) = x² – 3x + 4.

Let us find the x satisfying f (x) = f (2x + 1).

Here, we have,

f(2x + 1) = (2x + 1)² – 3(2x + 1) + 4

= (2x) ² + 2(2x) (1) + 1² – 6x – 3 + 4

= 4x² + 4x + 1 – 6x + 1

= 4x² – 2x + 2

Then, f (x) = f (2x + 1)

x² – 3x + 4 = 4x² – 2x + 2

4x² – 2x + 2 – x² + 3x – 4 = 0
3x² + x – 2 = 0

3x² + 3x – 2x – 2 = 0

3x(x + 1) – 2(x + 1) = 0

(x + 1)(3x – 2) = 0

x + 1 = 0 or 3x – 2 = 0

x = –1 or 3x = 2

x = –1 or 2/3

Thus, the values of x are –1 and 2/3.

f(x) = e^{x – [x]}

0 ≤x-[x]<1 

e⁰ ≤ e^x-[x] <e¹ 

1 ≤ e^x-[x] < e

Therefore, 

R(f) = [1, e)

It is given that

f(x) = (2x- 7)/ 4

We know that

y = (2x- 7)/ 4

It can be written as

4y = 2x – 7

So we get

4y + 7 = 2x

Where x = (4y + 7)/2

So, f ^-1 = (4y + 7)/2 for all y ∈ R.

It is given that

f(x) = 10x + 3

Consider y = 10x + 3

It can be written as

y – 3 = 10x

On further calculation

x = (y – 3)/ 10

So we get

f ^-1 (y) = (y – 3)/ 10

char c; c = cin.get();

We know that

f(x₁) = f(x₂)

It can be written as

3x₁ – 4 = 3x₂ – 4

So we get

3x₁ = 3x₂

Where x₁ = x₂

f is one-one.

Take y = 3x – 4

It can be written as

y + 4 = 3x

So we get

x = (y + 4)/ 3

If y ∈ R there exists x = (y + 4)/ 3 ∈ R

We know that f (x) = f([y + 4]/ 3) = 3 ([y + 4]/ 3) – 4 = y

f is onto

Here, f is one-one onto and invertible.

Take y = f(x)

So we get

y = 4x – 3

It can be written as

x = (y + 4)/ 3

So f ^-1 (y) = (y + 4)/ 3

Hence, we define f ^-1: R → R: f ^-1(y) = (y + 4)/ 3 for all y ∈ R

We know that

f(x₁) = f(x₂)

It can be written as

½ (3x₁ + 1) = ½ (3x₂ + 1)

So we get

3x₁ + 1 = 3x₂ + 1

On further calculation

3x₁ = 3x₂ where x₁ = x₂

f is one-one.

Take y = ½ (3x + 1)

It can be written as

2y = 3x + 1

We get

2y – 1 = 3x

So x = (2y – 1)/3

If y ∈ R there exists x = (2y – 1)/3 ∈ R

We know that

f(x) = f([2y – 1]/ 3) = ½ (3([2y – 1]/ 3) + 1) = y

f is onto

Here, f is one-one and invertible.

Take y = f(x)

So y = (3x + 1)/2

It can be written as

x = (2y – 1)/ 3

Hence, we define f ^-1: R → R: f ^-1(y) = (2y – 1)/ 3 for all y ∈ R

Given,

f(x) = (x-2)/(2-x)

For function to be defined,

2 - x ≠ 0 

x ≠ 2 

Therefore,

D(f) = R-{2}.

Let y = \(\frac{x-2}{2-x}\)

y = -1 

Therefore, 

R(f) = {-1}.

It writes a single character into the stream.

It reads a single character from the stream.

Answer: (b) f is odd g is even

\(f(x) = \frac{1}{x^3+2x|x|}\) 

∴  f(-x) = \(f(x) = \frac{1}{(-x)^3+2(-x)|-x|}\) 

= \(f(x) = \frac{1}{-x^3-2x|x|}\) (∴  | – x | = | x |)

= \(f(x) =- \frac{1}{x^3+2x|x|}\)  = - f(x) ⇒ f is odd.

g(x) = \(\sqrt{x^4+\frac{3}{x^2}}\) 

∴  g(- x) = \(\sqrt{(-x)^4+\frac{3}{(-x)^2}} = \sqrt{x^4+\frac{3}{x^2}}\) 

= g(x) 

⇒ g is even.

The strrev() function is used to reverse the string.

Option : (B)

Given,

f(x) = cos [x], for -π/2< x <π/2

For -π/2< x <-1,

[x]= -2 

f(x)= cos[x]= cos(-2) 

= cos2 

Because,

cos(-x) = cos(x) 

For-1 ≤x<0 

[x] = -1 

f(x) = cos[x] 

=cos(-1) 

= cos1 

For 0 ≤x< 1,

[x] = 0 

f(x) = cos 0 

= 1

For 1≤ x <π/2,

[x] = 1 

f(x) = cos 1 

Therefore, 

R(f) = {1, cos 1,cos 2} 

Option B is correct.

Option : (B)

f(x) = 2x + |x|

f(2x) = 2(2x)+|2x| = 4x+2|x| 

f(-x) = 2(-x)+|-x| 

f(2x)+f(-x )- f(x) 

= 4x+2|x|-2x+|-x|-(2x+|x|) 

= 4x+2|x|-2x+|x|-2x-|x|

= 2|x|

Answer : (a) an odd function

f(x) = sin \(\big[\,log\,\big(x+\sqrt {x^2+1}\big)\big]\) 

∴  f(- x) = sin \(\big[\,log\,\big(-x+\sqrt {(-x)^2+1}\big)\big]\)

= \(sin\big[\,log\,(\sqrt{1+x^2}-x)\big]\) 

= \(sin\big[\,log\,(\sqrt{1+x^2}-x)\frac{(\sqrt{1+x^2}+x)}{(\sqrt{1+x^2}+x)}\big]\) 

= \(sin\big[\log\,\big(\frac{(1+x^2)-x^2}{\sqrt{1+x^2}+x}\big)\big]\) 

= \(sin \big[\,log\,(\sqrt{1+x^2}+x)^{-1}\big]\)

= \(sin\big[\,-log\,(\sqrt{1+x^2}+x)\big]\)   (∴  log a ^–1 = – log a)

= \(-sin[\,log\,(\sqrt{1+x^2}+x)]\)  (∴  sin (– x) = – sin x)

= – f (x) 

⇒ f is an odd function.

Option : (C)

Let y = (x²-x)/(x²+2x)

y(x²+2x) = x²-x 

yx(x+2) = x(x-1) 

y(x+2) = x-1

x(y-1) = -(1+2y)

x = \(-\frac{(1+2y)}{y-1}\) 

Value of x can’t be zero or it cannot be not defined.. 

y≠1, -1/2 

So, 

Range = R-{-1/2, 1}

Option : (D)

f(x) = |x-1| 

f(x²) = |x²-1| 

f(x)² = (x-1)² 

= x²+1-2x 

So, 

f(x²) ≠ [f(x)]² 

f(x + y) = |x+y-1| 

f(x)f(y) = (x-1)(y-1) 

So, 

f(x + y) ≠ f(x)f(y) 

f(|x|) = ||x|-1| 

Therefore, 

Option D is correct.

Answer: (c) = 0

f (x) = x² + kx + 1 

Given, f (x) is an even function 

⇒ f (x) = f (– x) 

⇒ x² + kx + 1 = (– x) ² + k (– x) + 1 

⇒ x² + kx + 1 = x² – kx + 1 

⇒ 2kx = 0 

⇒ k = 0.

Answer : (b) an even function

F(x) = \(\big[\frac{g(x)-g(-x)}{f(x)+f(-x)}\big]^m\) 

∴  F(-x) = \(\big[\frac{g(-x)-g(x)}{f(-x)+f(x)}\big]^m =\big[\frac{-g(x)-g(-x)}{f(-x)+f(x)}\big]^m\)

= \(\big[-\big(\frac{g(x)-g(-x)}{f(-x)+f(x)}\big)\big]^{2n}\)

= \(\big[\frac{g(x)-g(-x)}{f(-x)+f(x)}\big]\) (∴ 2n ⇒ even power)

= F (x) 

⇒ F is an even function.

Option : (C)

Given,

f(x) = x/|x|

We know that,

|x| = -x in (-∞, 0) and 

|x| = x in [0, ∞)

So, 

f(x) = \(\frac{x}{-x}\)

= -1 in (-∞, 0) 

And f(x) = \(\frac{x}{-x}\) in (0, ∞) 

As clearly shown above f(x) has only two values 1 and -1

So, 

Range of f(x) = {-1, 1}

The function strcmp() is used to compare two strings.

The strcpy() is used to copy a string into another string.

Option : (B)

Given,

f(x) = |x – 1|

A modulus function always gives a positive value..

R(f) = [0,∞)

Option : (A)

f(x) = (x+2)/|x+2|

When x>-2,

We have,

f(x) = \(\frac{x+2}{x+2}\) 

= 1 

When x<-2, 

We have,

f(x) = \(\frac{x+2}{-(x+2)}\) 

= -1 

R(f) = {-1, 1}

Answer: (b) =  ϕ  

f (x) = x² + 1 

Let y = f (x) = x² + 1 

⇒ y – 1 = x² 

⇒ x = ± \(\sqrt{y-1}\) 

⇒ f ^–1(x) = ± \(\sqrt{x-1}\)

∴  f ^–1(– 5) = ±  \(\sqrt{-5-1}\)  ± \(\sqrt{-6}\) ∉  R 

∴  f ^–1(– 5) is the null set.

The strcat() function is used to combine the two strings.

Answer: (d) = none of these 

f (x) = sin x + cos x 

f (– x) = sin (– x) + cos (– x) = – sin x + cos x ≠ f (x) or – f (x) 

∴  f (x) is neither even nor odd nor constant.

Answer: (d) = \(\frac{x}{1-x}\)

f (x) = x – x² + x³ – x⁴ + ..... ∞, | x | < 1 

This is an infinite G.P. with a = x, r = – x. 

∴  f (x) = S_∞ = \(\frac{a}{1-r}\) = \(\frac{x}{1+x}\) 

Let y = f(x) = \(\frac{x}{1+x}\)

⇒ y (1 + x) = x 

⇒ y + xy = x 

⇒ y = x (1 – y) 

⇒ y = \(\frac{y}{1-y}\) 

⇒ \(f^{-1}(x) = \frac{x}{1-x}\)

The strlen() function is used to find the length of a string.

Answer: (c) = 1

Given,

f (x) = \(\frac{4^x}{4^x+2}\)

f (1 – x) = \(\frac{4^{1-x}}{4^{1-x}+2}\) = \(\frac{4.4^{-x}}{4^1.4^{-x}+2}\) 

= \(\frac{4.\frac{1}{4^x}}{4.\frac{1}{4^x}+2}\) 

= \(\frac{\frac{4}{4^x}}{\frac{4+2.4^x}{4^x}}\) = \(\frac{4}{4+2.4^x}=\frac{2}{2+4^x}\) 

Now f (x) + f (1 – x) = \(\frac{4^x}{4^x+2}\) + \(\frac{2}{2+4^x}\) 

= \(\frac{4^x+2}{4^x+2}\) 

= 1 

Answer : (d) = 0 

Given, f (x) = cos (log x) 

⇒ f (x²) = cos (log x²) = cos (2 log x) 

f (y²) = cos (log y²) = cos (2 log y)

\(f\big(\frac{x^2}{y^2}\big)\) = \(cos\big(log\frac{x^2}{y^2}\big)\) = cos (log x² - log y² )

= cos (2 log x – 2 log y) 

f (x²y²) = cos (log x²y²) = cos (log x² + log y²) = cos (2 log x + 2 log y)

Now, f(x²). f(y²) - \(\frac{1}{2}\) \(\big[f\big(\frac{x^2}{y^2}\big) +f (x^2y^2)\big]\) 

= cos (2 log x) . cos (2 log y) – \(\frac{1}{2}\) [cos (2 log x – 2 log y) + cos (2 log x + 2 log y)]

= cos (2 log x) . cos (2 log y) - 

\(\frac{1}{2}\) \(\big[ 2\,cos\big(\frac{2\,log\,x-2\,log\,y+2\,log\,x+2\,log\,y}{2}\big) cos \big( \frac{2\,log\,x+2\,log\,y-2\,log\,x+2\,log\,y}{2}\big)\big]\)

\(\big(\because cos\,C + cos\,D = 2 \,cos\big(\frac{C+D}{2}\big)cos\,\big(\frac{D-C}{2}\big)\big)\) 

= cos (2 log x) . cos (2 log y) – \(\frac{1}{2}\)[2 cos (2 log x) . cos (2 log y)] = 0.

The function write() is used for string output.

f(x) = 4x – x²

f(a+1)-f(a-1) = [4(a+1) -(a+1)²] - [4(a-1)-(a-1)²] 

= 4[(a+1) -(a-1)]- [ (a+1)² -(a+1)²] 

= 4(2)-[(a+1+a-1) (a+1-a+1)] 

Using : a² - b²= (a + b)(a-b) 

f(a+1)-f(a-1)=4(2)-2a(2) 

= 4(2-a)

Answer: (c) = \(\frac{2}{5}\) 

Given, f (x) = \(\frac{1}{1+\frac{1}{x}}\) and g(x) = \(\frac{1}{1+\frac{1}{f(x)}}\)

∴  f(x) = \(\frac{1}{\frac{x+1}{x}} = \frac{x}{x+1}\) 

⇒ g(x) = \(\frac{1}{1+ \frac{1}{\frac{x}{x+1}}}\) 

= \(\frac{1}{1+ \frac{x+1}{x}}\) = \(\frac{x}{2x+1}\) 

⇒ g(2) = \(\frac{2}{2\times 2+1}\) 

= \(\frac{2}{5}\)

To Show: that f is invertible

To Find: Inverse of f

[NOTE: Any functions is invertible if and only if it is bijective functions (i.e. one-one and onto)]

one-one function: A function f : A B is said to be a one-one function or injective mapping if different

elements of A have different images in B. Thus for x₁, x₂ ∈ A & f(x₁), f(x₂) ∈ B, f(x₁) = f(x₂) ↔ x₁= x₂ or x₁ ≠ x₂ ↔ f(x₁) ≠  f(x₂)

onto function: If range = co-domain then f(x) is onto functions.

So, We need to prove that the given function is one-one and onto.

Let x₁, x₂ ∈ Q and f(x) = 3x-4.So f(x₁) = f(x₂) → 3x₁ - 4 = 3x₂ - 4 → x₁=x₂

So f(x₁) = f(x₂) ↔ x₁= x₂, f(x) is one-one

Given co-domain of f(x) is Q.

Let y = f(x) = 3x- 4 , So x = [Range of f(x) = Domain of y]

So Domain of y is Q = Range of f(x)

Hence, Range of f(x) = co-domain of f(x) = Q

So, f(x) is onto function

As it is bijective function. So it is invertible

Invers of f(x) is f -1(y) =\(\frac{y+4}{3}\)

When the length of the string is 0 it is called a null string.

Answer : (d) = ϕ (a) + ϕ (b)

ϕ (x) = \(\frac{b(x-a)}{b-a} + \frac{a(x-b)}{a-b}\) 

Then, ϕ (a + b) = \(\frac{b(a+b-a)}{b-a} + \frac{a(a+b-b)}{a-b}\) 

= \(\frac{b^2}{b-a} +\frac{a^2}{a-b} = \frac{b^2-a^2}{b-a} =b+a\) ...(i)

Now ϕ (a) = \(\frac{b\times 0}{b-a}\) + \(\frac{a(a-b)}{a-b}\) = a ...(ii)

ϕ (b) = \(\frac{b(b-a)}{b-a}\) + \(\frac{a\times 0}{a-b}\)  ...(iii)

(i), (ii) and (iii) ⇒ ϕ a + b) = ϕ(a) + ϕ(b).

Option : (D)

f(x) = x, g(x) = 1/x; 

h(x) = f(x) g(x)

h(x) = 1 

f(x)g(x) = 1

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

x ≠ 0

Given: f: X → R, f(x) = x³ + 1 

Here, X = {-1, 0, 3, 7, 9} 

For x = -1 

f(-1) = (-1)3 + 1 = -1 + 1 = 0 

For x = 0 

f(0) = (0)³ + 1 = 0 + 1 = 1 

For x = 3 

f(3) = (3)³ + 1 = 27 + 1 = 28 

For x = 7 

f(7) = (7)³ + 1 = 343 + 1 = 344 

For x = 9 

f(9) = (9)³ + 1 = 729 + 1 = 730 

∴ the ordered pairs are (-1, 0), (0, 1), (3, 28), (7, 344), (9, 730)