(i) Given a*b = 3a + b²
\(\Rightarrow 4*3=3(4)+(3)^{ 2 }\)
= 12 + 9
4*3 = 21
(ii) Given a*b = 3a + b²
\(\Rightarrow 5*2=3(5)+2^{ 2 }\)
\(\Rightarrow 5*2=15\)
\(\therefore 5*2=19\)

We have 2*(3*x) = 33
\(\Rightarrow 2*(3+x+3x)=33\)
\(\Rightarrow 2*(3+4x)=33\)
\(\Rightarrow 2+(3+4x)+2(3+4x)=33\)
\(\Rightarrow 2+3+4x+6+8x=33\)
\(\Rightarrow 12x+11=33\)
\(\Rightarrow 12x=22\Rightarrow x=\frac { 11 }{ 6 } \)

\(f(x)=x^{ 2 }+1,g(x)=2x+1\)
\(fog=f[(g(x)\} =f(2x+1)\)
\(=4x^{ 2 }+4x+1+1\)
\(=4x^{ 2 }+4x+2\)
\(gof=g(f(x)=g(x^{ 2 }+1)\)
\(=2(x^{ 2 }+1)+1\)
\(=2x^{ 2 }+2+1\)
\(=2x^{ 2 }+3\)

\(f(x)=x^{ 2 },x\in Ra\)
\(f(1.1)=(1.1)^{ 2 }z\)
\(=1.21\)
\(f(1)=(1)^{ 2 }=1\)
\(\frac { f(1.1)-f(1) }{ 1.1-1 } =\frac { 1.21 }{ 1.1-1 } =\frac { 0.21 }{ 0.1 } \)
\(=2.1\)

Symmetric Relation : A relation R on a set A is called symmetric relation if aRb implies bRa, for every a,b \(\in a\) i.,e if (a,b)  \(\in R\) \(\Rightarrow \) (b,a)  \(\in R\)For every a,b \(\in A\)
Example 
A = (1,2,3)
A x A =(1,2) (2,1) (1,1) (2,2) (3,3) (1,3) (2,3)  (3,1) (3,2 )  \(\in R\) 
since (a,b)  \(\in R\)  (b,a)  \(\in R\) for every a, b  \(\in A\) 
Relation is said to be symmetric 

\(\cos \frac{\pi}{3}=\frac{1}{2} \text {, also } \cos \frac{5 \pi}{3}=\frac{1}{2} \text {, not one-one }\)

No, as \(f(-2)=(-2)^2=4\) and \(f(2)=(2)^2=4\)
i.e \(x_1 \neq x_2 \Rightarrow f\left(x_1\right)=f\left(x_2\right)\).
Not one-one

Let \(y=\frac{3 x-4}{5} \Rightarrow 5 y=3 x-4\)
\(\Rightarrow x=f^{-1}(y)=\frac{5 y+4}{3} \Rightarrow f^{-1}(x)=\frac{5 x+4}{3}\)

\( a * b=a+3 b^2 \\ 2 * 4=2+3(4)^2 \\ =2+3 \times 16=50 \)

Assume, \(f(x)=x^{ 2 }+4x+7\) 
= \((x+2)^{ 2 }+3\)
\(\Rightarrow y-3=(x+2)^{ 2 }\)
\(\Rightarrow x+2=\pm \sqrt { y-3 } \)
\(\Rightarrow x=\sqrt { y-3 } -2\quad y>3\)
\(g:s\rightarrow N\)
\(g(y)=\sqrt { y-3 } -2\)
\(gof(x)=g[f(x)]=g(x^{ 2 }+4x+7)\)
\(=g[(x+2)^{ 2 }+3]\)
\(=\sqrt { (x+3)^{ 2 }+3-3-2 } \)
\(=x+2-2=x\)
\(fog(y)=f[g(y)]\)
\( =f\sqrt { y-3 } -2\)
\(=(\sqrt { y-3 } -2+2)^{ 3 }=y\)
\( \Rightarrow gof=I_{ N }\)
\(gog=I_{ s }\)
\(\Rightarrow f^{ -1 }=g=\sqrt { x-3 } -2\)
Yes, interest & knowledge have bijective relation Value Interest leads to knowledge

Here \(a*b=\frac { 3ab }{ 5 } ;a,b,\in Q\) is a binary operation:
(I) Commutativity.
For \(a,b,\in Q\),
\(a*b=\frac { 3ab }{ 5 } =\frac { 3ba }{ 5 } .\)
[\(\because \) Rational numbers are commutative under multiplication]
= b * a
Hence,'* is commutative on Q.
(II) Associativity.
For \(a,b,c\in Q,\)
\((a*b)*c=\frac { 3ab }{ 5 } *c\)
\(=\frac { 3\frac { 3ab }{ 5 } c }{ 5 } =\frac { 9abc }{ 25 } \)
And  \(=\frac { 3a\frac { 3bc }{ 5 } }{ 5 } =\frac { 9abc }{ 25 } \)
Thus \(a*(b*c)=a*(b*c)\).
Hence, ' * ' is associative on Q.
(III) Let 'e' be the identity element.
Then \(a*e=a\Rightarrow e*a\Rightarrow \frac { 3ae }{ 5 } \)
\(\Rightarrow \) \(e=\frac { 5 }{ 3 } \).
Hence, the identity element = \(\frac { 5 }{ 3 } \). 

Let 'x' and 'y' the number of packets of food P and Q respectively.
We have:  \(x\ge 0\)  ...(1)
\(y\ge 0\)  ....(2)
\(12x+3y\ge 240\quad i.e.4x+y\ge 80\) .....(3)
\(4x+20y\ge 460\quad i.e.x+5y\ge 115\) .....(4)
\(6x+4y\le 300\quad i.e.3x+2y\le 150\) .....(5)
The mathematical problem is as below:
Minimize Z=6x+3y subject to (1)-(5)
Draw the lines
x = 0, 4x + y = 80, x + 5y = 80, x + 5y = 115 and 3x + 2y = 150.
The lines 3x + 2y = 150 and 4x + y = 80 meet at L (2, 72)
The lines 4x + y = 80 and x + 3y = 115 meet at M (15, 20)
The lines 3x + 2y = 150 and x + 5y = 115 meet at N(40,15).

The shaded portion represents the feasible region, which is bounded.
Apply Corner Point Method, we have:

Hence, minimum amount of vitamin A = 150 units.

Let 'x' hectares land be allocated to crop X and 'y' hectares to crop Y.
We have: \(x\ge 0\) .....(1)
\(y\ge 0\) .....(2)
\(x+y\le 50\) .....(3)
\(20x+10y\le 800\)
\(\therefore \) i.e.  \(2x+y\le 80\)....(4)
The mathematical formulation of the problem is as be low:
Maximize Z = 10500x + 9000y subject to the constraints (1)-(4).
For solution set, we draw the lines:
x = 0, y = 0, x + y = 50 and 2x + y = 80.
The lines x + y = 50 and 2x + y = 80 meet at E (30, 20).

The shaded portion represents the feasible region, which is bounded.
Applying Corner Point Method, we have:

Hence, the maximum profits is Rs. 4,95,000 by allocating 30 hectares for crop X and 20 hectares for crop Y.

Let \(z\in C\) be the given arbitrary element.
Then there exists pre-image y of z  under g such that
\(g(y)=z.\)  \(\left[ \because \quad g\quad is\quad onto \right] \)
Again for \(y\in B\) , there exists pre-image x of y under  f such that \(f(x)=y\).
\(\left[ \because \quad f\quad is\quad onto \right] \)
\(\because \)   \(gof(x)=g(f(x)=g(y)=z\)
Hence, gof is onto.  

No