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a² + b² = 7ab

a² + 2ab + b² = 7ab + 2ab

(a + b)² = 9ab

\(\frac{(a+b)^2}9\) = ab

\((\frac{a+b}3)^2 = ab\) 

Taking log on both sides, we get

log\((\frac{a+b}3)^2=log (ab)\) 

2 log\((\frac{a+b}3)\)  = log a + log b

Dividing throughout by 2, we get

log\((\frac{a+b}3)=\frac12log\,a+\frac12 log\,b\)

Since, f:R → R and g:R → R

fog:R → R and gof:R → R

f (x) = x and g (x) = |x|

Now, gof(x)=g(f(x) =g(x)

⇒ gof(x) =|x|

and, fog(x) = f(g(x)) = f (|x|)
⇒ fog(x)=|x|

Hence, gof(x) = fog(x) = |x|

Since, f:R → R and g:R → R

fog:R → R and gof:R → R

f(x)=x² + 8 and g(x)=3x³ + 1

So, gof(x)= g(f(x))

gof(x)= g(x² + 8)

gof(x)= 3(x² + 8)³ + 1

⇒ gof(x)= 3(x⁶ + 512 + 24x⁴ + 192x²) + 1

⇒ gof(x)= 3x⁶ + 72x⁴ + 576x² + 1537

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

⇒ fog(x)= f(3x³ + 1)

⇒ fog(x)=(3x³ + 1)² + 8

⇒ fog(x)=(9x⁶ + 1 + 6x³) + 8

⇒ fog(x)=9x⁶ + 6x³ + 9

So, gof(x) = 3x⁶ + 72x⁴ + 576x² + 1537 and fog(x) = 9x⁶ + 6x³ + 9

Since, f:R → R and g:R → R

fog:R → R and gof:R → R

f(x) = x² + 2x – 3 and g(x) = 3x – 4

Now, gof(x)=g(f(x))= g(x² + 2x – 3)

gof(x) = 3(x² + 2x–3) – 4

⇒ gof(x)= 3x² + 6x – 9 – 4

⇒ gof(x) = 3x² + 6x – 13

and, fog= f(g(x)) = f(3x – 4)

fog(x) = (3x – 4)² + 2(3x – 4) – 3

= 9x² + 16 – 24x + 6x – 8 – 3

∴ fog(x) = 9x² – 18x + 5

Thus, gof(x) = 3x² + 6x – 13 and fog(x) = 9x² – 18x + 5

Let f = {(3,1), (9,3), (12,4)} and

g = {(1,3), (3,3), (4,9), (5,9)}

Now,

range of f = (1, 3, 4}

domain of f = {3, 9, 12}

range of g = {3,9}

domain of g = (1, 3, 4, 5}

since, range of f ⊂ domain of g

∴ gof is well defined.

Again, the range of g ⊆ domain of f

∴ fog in well defined.

Finally, gof = {(3,3), (9 ,3), (12,9)}

fog = {(1,1) , (3,1), (4,3), (5,3)}

We have,

f = {(1, – 1), (4, – 2) , (9, – 3), (16,4)} and

g = {(– 1, – 2), (– 2, – 4), (– 3, – 6), (4,8)}

Now,

Domain of f = {1,4,9,16}

Range of f = {– 1, – 2, – 3, 4}

Domain of g = (– 1, – 2, – 3,4}

Range of g = (– 2, – 4, – 6, 8}

Clearly range of f = domain of g

∴ gof is defined.

but, range of g ≠ domain of f
So, fog is not defined.

Now,

gof(1) = g(– 1)= – 2

gof(4) = g(– 2) = – 4

gof(9) = g (– 3) = – 6

gof(16) = g(4)= 8

So, gof = {(1, – 2), (4, – 4), (9 , – 6), (16,8)}

Correct Answer is (C) x

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

\(\Rightarrow (fofof)(x)=f(f(f(x)))\)

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

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

f(x) = \(\sqrt[3]{x+1}\) 

f is defined for all real x and the values of f(x) ∈ R

∴ Domain of f = R, Range of f = R

L.H.S. = log(1 + 2 + 3) = log 6

R.H.S. = log 1 + log 2 + log 3

= 0 + log (2 × 3)

= log 6

∴ L.H.S. = R.H.S.

L.H.S. = log \(|\sqrt{x^2+1}+x|+log|\sqrt{x^2+1}-x|\) 

= log|(\(\sqrt{x^2+1}+x\))(\(\sqrt{x^2+1}-x\))|

 = log|x² + 1 - x²|

 = log 1

 = 0

= R.H.S.

f(x) = \(\sqrt{4x+5},x\geq\frac{-5}4\)

Let f(x₁) = f(x₂)

\(\therefore\) \(\sqrt{4x_1+5}=\sqrt{4x_2+5}\)

\(\therefore\) x₁ = x₂

∴ f is a one-one function.

f(x) = \(\sqrt{4x+5}=y, \) say(y) \(\geq0\)

Squaring on both sides, we get

y² = 4x + 5

∴ x = \(\frac{y^2-5}4\) 

∴ For every y we can get x.

∴ f is an onto function.

∴ x = \(\frac{y^2-5}4=f^{-1}(y)\) 

Replacing y by x, we get

f^-1(x) = \(\frac{x^2-5}4\) 

f(x) = \(\frac{6x-7}3\)

Let f(x₁ ) = f(x₂)

\(\therefore\) \(\frac{6x_1-7}3=\frac{6x_2-7}3\)

\(\therefore\) x₁ = x₂

∴ f is a one-one function.

f(x) = \(\frac{6x-7}3\) = y (say)

\(\therefore\) For every y, we can get x

\(\therefore\) f is an onto function.

\(\therefore\) x = \(\frac{3y+7}6\) = f^-1(y)

Replacing y by x, we get

f^-1(x) = \(\frac{3x+7}6\)

1. get() function: 

get() is an input function. It is used to read a single character and it does not ignore the white spaces and newline character.

Syntax is cin.get(variable); 

Eg. char ch; 

cin.get(ch);

2. put() function:

put() is an output function. It is used to print a character. 

Syntax is cout.put(variable); 

Eg. char ch; 

cin.get(ch); 

cout.put(ch);

The program will always respond with, “You must provide TWO command-line 

arguments!!!”, regardless of what you write on the command line

It returns the absolute value of x.

The function getline() is used to read a string.

char name[50]=” Bharath”;

The declaration of string is char array name[size];

The null in a string is the control character ‘\0’ that indicates the end of the string.

(B) 4

2 log₂ x = 4, x > 0

∴ log₂ (x²) = 4

∴ x² = 16

∴ x = ±4

∴ x = 4

(A), (B), (C), (D)

log _x²  16 + log_2x64 = 3

\(\therefore\frac{log16}{logx^2}+\frac{log64}{log2x}=3\)

∴ 4 log 2 [log x + log 2] + (6 log 2) (2 log x)

= 3 (2 log x) (log 2 + log x)

Let log 2 = a, log x = t. Then

∴ 4at + 4a² + 12at = 6at + 6t²

∴ 6t²– 10at – 4a² = 0

∴ 3t² – 5at – 2a² = 0

∴ (3t + a) (t – 2a) = 0

∴ t = \(-\frac13\)a, 2a

∴ log x = log(2)^-1/3, log(2²)

∴ x = 2^-1/3, 4

∴ x = \(\frac{1}{\sqrt[3]2},4\)

Answer : (d) = f is neither one -one nor onto

For all x, y ∈ R, 

f (x) = f (y) 

⇒ x⁴ = y⁴ ⇒ x⁴ – y⁴ = 0 

⇒ (x² – y²) (x² + y²) = 0 ⇒ (x – y) (x + y) (x² + y²) = 0

∴   (x – y) = 0 or (x + y) = 0 or (x² + y²) = 0 

⇒ x = y or x = – y or x = ± y 

∴   x = – y 

⇒ f is a many-one function 

Also let z = f (x) = x⁴ ⇒ x = (z)^1/4 

Now z = – 1 ∈ R has no pre-image in R as (–1)^1/4 ∉ R. 

∴  f : R → R is not an onto function.

Answer : (b) = one - one and onto 

. (f – g) (x) = \(\ f(x) - g(x) =\begin {cases}x\,, &\quad \text{if } x \text{ is rational}\\-x\,, &\quad \text{if } x \text{ is irrational}\end{cases}\) 

Distinct elements of R have distinct images in (f – g). 

Hence (f – g) is one-one. 

Also Range of (f – g) = R 

⇒ f is onto. 

⇒ (f – g) R → R is a one-one onto function

To find: f^-1

Given: f : R → R : f(x) = 10x + 3

We have,

f(x) = 10x + 3

Let f(x) = y such that \(y\in R\)

⇒ y = 10x + 3

⇒ y – 3 = 10x

\(\Rightarrow x=\frac{y-3}{10}\)

\(\Rightarrow f^{-1}=\frac{y-3}{10}\)

\(f^{-1}=\frac{y-3}{10}\) for all \(y\in R\)

To find: (f o g) (7)

Formula used: f o g = f(g(x))

Given: (i) f (x) = x + 7

(ii) g (x) = x – 7

We have,

f o g = f(g(x)) = f(x – 7) = [ (x – 7) + 7 ]

⇒ x

(f o g) (x) = x

(f o g) (7) = 7

Answer is (i) 1 (ii) - 1 (iii) - 1 (iv) - 1

 (i) \(f(\frac{1}{2})\)

Here, x = 1/2,which is rational

∴f(1/2) = 1

 (ii) \(f(\sqrt{2})\)

Here, x = √2,which is irrational

∴f(√2) = - 1

 (iii) \(f(\pi)\)

Here, x = ∏, which is irrational

\(f(\pi)\)= -1

 (iv) \(f(2+\sqrt{3}).\)

Here, x = 2 + √3, which is irrational

∴f(2 + √3) = - 1

(i) gof 

To find: gof 

Formula used: g o f = g(f(x)) 

Given: f = {(1, 4), (2, 1), (3, 3), (4, 2)} and g = {(1, 3), (2, 1), (3, 2), (4, 4)} 

Solution: We have, 

gof(1) = g(f(1)) = g(4) = 4 

gof(2) = g(f(2)) = g(1) = 3 

gof(3) = g(f(3)) = g(3) = 2 

gof(4) = g(f(4)) = g(2) = 1 

g o f = {(1, 4), (2, 3), (3, 2), (4, 1)} 

(ii) fog 

To find: fog 

Formula used: fog = f(g(x)) 

Given: f = {(1, 4), (2, 1), (3, 3), (4, 2)} and g = {(1, 3), (2, 1), (3, 2), (4, 4)} 

Solution: We have, 

fog(1) = f(g(1)) = f(3) = 3 

fog(2) = f(g(2)) = f(1) = 4 

fog(3) = f(g(3)) = f(2) = 1 

fog(4) = f(g(4)) = f(4) = 2 

f o g = {(1, 3), (2, 4), (3, 1), (4, 2)} 

(iii) fof 

To find: fof 

Formula used: f o f = f(f(x)) 

Given: f = {(1, 4), (2, 1), (3, 3), (4, 2)} 

Solution: We have, 

fof(1) = f(f(1)) = f(4) = 2 

fof(2) = f(f(2)) = f(1) = 4 

fof(3) = f(f(3)) = f(3) = 3 

fof(4) = f(f(4)) = f(2) = 1 

fof = {(1, 2), (2, 4), (3, 3), (4, 1)}

To prove: function is many-one and into

Given: \(f:R→R :f(x)=\begin{cases}1,\text{ if x is rational}\\ -1,\text{ if x is irrational}\end{cases}\)

We have,

f(x) = 1 when x is rational

It means that all rational numbers will have same image i.e. 1

⇒ f(2) = 1 = f (3) , As 2 and 3 are rational numbers

Therefore f(x) is many-one

The range of function is [{-1},{1}] but codomain is set of real numbers.

Therefore f(x) is into