Let y = 3x -5.
\(\Rightarrow y+5=3x\Rightarrow \frac { y+5 }{ 3 } =x\)
Let g(y) = \(\frac { y+5 }{ 3 } \)
\(gof(x)=g(f(x))=g(3x-5)=\frac { 3x-5+5 }{ 3 } =\frac { 3x }{ 3 } =y\)
Also f o g(y) = f(g(y)) = \(f\left( \frac { y+5 }{ 3 } \right) =3\left( \frac { y+5 }{ 3 } \right) -5=y+5-5=y\)
Thus g o f = Ix and fog = Iy.
This implies that f and g are bijections and inverses to each other.
Hence f is a bijection and f^-1(y) = \(\frac { y+5 }{ 3 } \)
Replacing y by x we get, f^-1(x) = \(\frac { x+5 }{ 3 } \)

f(-3) = (-3)² - 3    \(\left[ \therefore \ f(x)={ x }^{ 2 }-3\quad when\ x=-3 \right] \)
= 9 - 3 = 6
f(5) = 5² + 3(5)-2   \(\left[ \therefore f(x)={ x }^{ 2 }+3x-2\quad when\quad x=5 \right] \)
= 25 + 15 - 2
= 38
f(2) = 2² - 3
= 4 - 3 = 1   \(\left[ \therefore \ f(x)={ x }^{ 2 }-3\ when\ x=2 \right] \)
f(-1) = (-1)² + (-1) -5   \(\left[ \therefore \ f(x)={ x }^{ 2 }+x-5\ when\ x=-1 \right] \)
= 1-1-5 = -5
f(0) = 0²-3 = -3  \(\left[ \therefore \ f(x)={ x }^{ 2 }-3\ when\ x=0 \right] \)
\(\therefore\) f(-3) = 6, f(5) = 38, f(2) = 1, f(-1) = -5, f(0) = -3

(i) y = -x

Graph of y = - x is the reflection of the graph of y = x about the X - axis.
(ii) y = 2x

The graph of y = 2x compresses towards the Y-axis that is moves away from the X-axis since the multiplying factor is 2, which is greater than 1.
(iii) y = x + 1

The graph of y = x + 1, causes the shift to the upward for one unit.
(iv)  \(y={1\over 2}x+1\)

The graph of y = \(\frac{1}{2}\) x + 1, stretches towards the X-axis since the multiplying factor is \(\frac{1}{2}\) which is less than one and shift to the upward for one unit.
(v) ⇒ y = -2x - 3

The graph of y = -2x - 3, stretches towards the X-axis since the multiplying factor is - 2 which is less than one and causes the shifts to the downward for 3 units.

(i) y = sin (-x)

Let y = sin x.
Then sin(-x) is the reflection of the graph of sin x, about y-axis.
(ii) y = -sin(-x)

-sin(-x) is the reflection of the graph of sin(-x) about the x-axis.
(iii) \(y=sin\left( {\pi\over 2}+x\right)\)
Let y = sinx.
Then \(sin\left( {\pi\over 2}+n\right)\)causes the shift to the left for \(\pi\over 2\) unit to the sin x curve.
(iv) \(y=sin\left({\pi\over 2}-x \right)\)

Let y = sin x. Then \(sin\left( {\pi\over 2}-n\right)\) causes the shift to the left for \(\pi\over 2\) unit to the sin (-x) curve.

Given f(x) = 3x - 4
Let y = 3x - 4 ⇒ y + 4 = 3x
\(⇒ x={y+4\over 3}\)
Let g(y) = \(y+4\over 3\)
Now gof(n) = g(f(n)) = g(3\(\times\) -4) = \({3x-4+4\over 3}={3x\over 3}=x\)
and fog(y) = f(g(y)) = \(f\left(y+4\over 4\right)=3\left(y+4\over 3\right)-4=y+4-4=y\)
Thus, gof(x) = I_x and fog (y) = I_y
This implies that f and g are bijections and inverses to each other
Hence f is bijection and \(f^{-1} (x)={y+4\over 3}\)
Replacing y by x, we get f^-1 (x) = \(\frac { x+4 }{ 3 } \)

Hence, the graph of y = f^-1(x) is the reflection of the graph of f in y = x

Given A(x) = 30,000 + 0.04x
S(x) = 25,000 + 0.05x
∴ (A + S)(x) = 30,000 + 0.04x + 25,000 + 0.05x
= 55,000+0.09x
Given x = Rs. 1,50,00,000
Then (A+S)(x) = 55000 + 0.09(1,50,00,000)
= 55000 + 1,350,000
= 1,405,000
Hence total family income = Rs. 14,05,000

Clearly
\(-1\le\cos x\le 1 \\ \Rightarrow -3\ge -3\cos x \ge-3\)
\(\Rightarrow -3\le-3\cos x\le3 \\ \Rightarrow 1-3\le 1-3\ \cos x\le 1+3\)
Thus we get \(-2\le 1-3\ \cos\ x\) and \(1-3\cos\ x\le 4.\)
By taking reciprocals, we get \({1 \over 1-3\cos x}\le-{1\over2}\) and \({1\over 1-3\cos\ x}\ge{1\over 4}.\)
Hence the range of f is \(\left( -\infty, {1\over 2} \right]\cup[{1\over4},\infty).\)

Range of cosine function is -1 \(\le \)cos x \(\le \) 1
\(\Rightarrow\) -2 \(\le \) 2 cos x \(\le \) 2        (Multiplied by 2)
\(\Rightarrow\) -2 -1 \(\le \) 2 cos x -1 \(\le \) 2-1
\(\Rightarrow\) -3 \(\le \) 2 cos x-1 \(\le \) 1
\(\Rightarrow \frac { -1 }{ 3 } >\frac { 1 }{ 2cosx-1 } >\frac { 1 }{ 1 } \)
\(\Rightarrow \frac { -1 }{ 3 } f(x)>1\)
\(\therefore \ Range \) \(=\left(-\infty,-\frac{1}{3}\right] \cup[1, \infty)\)

(i) f( m) = f( n) \(\Rightarrow\) m² = n² \(\Rightarrow\) m = n since \(m,\ n\in N.\) Thus f is one-to-one. But, non-perfect square elements in the co-domain do not have pre-images and hence not onto.
(ii) Two different elements in the domain have same images and hence f is not one-to-one. Clearly the range of f is a proper subset of R. Thus it is not onto.

As m - m = 0 and 0 = 0 \(\times\)12, hence mRm proving that R is reflexive
Let mRn; Then m -n = 12k for some integer k, thus n - m = 12 (-k) and hence nRm. This shows that R is symmetric.
Let mRn and nRp; then m - n =12k and n - p = 12l for some integers k and l.
So m - p = 12 (k + l) and hence mRp. This shows that R is transitive.
Thus R is an equivalence relation.