Projection of \(\vec { a } \) on \(\vec { b } \) 
(i) Here \(\vec { a } =\hat { i } -\hat { j } \) ;\(\vec { a } .\vec { b } =\left( \hat { i } -\hat { j } \right) \) 
So projection of \(\hat { i } -\hat { j } \) on Z-axis = O.
(ii) \(\vec { a } =\hat { i } +2\hat { j } -2\hat { k } \) ;\(\vec { b } =2\hat { i } -\hat { j } +5\hat { k } \) 
\(\vec { a } .\vec { b } =2-2-10=-10\) ;\(\left| \vec { b } \right| =\sqrt { 4+1+25 } =\sqrt { 30 } \) 
ஃ Projection of \(\vec { a } \) on \(\vec { b } \) =\(\cfrac { -10 }{ \sqrt { 3 } } \) units.
(iii) \(\vec { a } =3\hat { i } +\hat { j } \) ;\(\vec { b } =4\hat { i } -\hat { j } +2\hat { k } \) 
\(\vec { a } .\vec { b } =12-1-2=9\) ;\(\left| \vec { b } \right| =\sqrt { 16+1+4 } =\sqrt { 21 } \) 
∴ Projection of \(\vec { a } \vec { b } =\cfrac { 9 }{ \sqrt { 21 } } \)

\(A=\left[ \begin{matrix} 1 & 2 \\ 2 & 0 \end{matrix} \right] ,B=\left[ \begin{matrix} 3 & -1 \\ 1 & 0 \end{matrix} \right] \) 
\(\therefore A+B=\left[ \begin{matrix} 1 & 2 \\ 2 & 0 \end{matrix} \right] +\left[ \begin{matrix} 3 & -1 \\ 1 & 0 \end{matrix} \right] =\left[ \begin{matrix} 4 & 1 \\ 3 & 0 \end{matrix} \right] \)   ...(1)
\(A-B=\left[ \begin{matrix} 1 & 2 \\ 2 & 0 \end{matrix} \right] -\left[ \begin{matrix} 3 & -1 \\ 1 & 0 \end{matrix} \right] \) 
= \(\left[ \begin{matrix} 1 & 2 \\ 2 & 0 \end{matrix} \right] +\left[ \begin{matrix} -3 & 1 \\ -1 & 0 \end{matrix} \right] =\left[ \begin{matrix} -2 & 3 \\ 1 & 0 \end{matrix} \right] \)     ...(2)


= \(\left[ \begin{matrix} 16+3 & 4+0 \\ 12+0 & 3+0 \end{matrix} \right] =\left[ \begin{matrix} 19 & 4 \\ 12 & 3 \end{matrix} \right] \)   ...(7)

= \(\left[ \begin{matrix} 4+3 & -6+0 \\ -2+0 & 3+0 \end{matrix} \right] =\left[ \begin{matrix} 7 & -6 \\ -2 & 3 \end{matrix} \right] \) 
(i) \(LHS=\left( A+B \right) ^{ 2 }=\left[ \begin{matrix} 19 & 4 \\ 12 & 3 \end{matrix} \right] [from7]\) 
RHS = A² + AB + BA + B²
= \(\left[ \begin{matrix} 5 & 2 \\ 2 & 4 \end{matrix} \right] +\left[ \begin{matrix} 5 & -1 \\ 6 & -2 \end{matrix} \right] +\left[ \begin{matrix} 1 & 6 \\ 1 & 2 \end{matrix} \right] +\left[ \begin{matrix} 8 & -3 \\ 3 & -1 \end{matrix} \right] \)  (from (3), (5), (6) and (4))
= \(\left[ \begin{matrix} 5+5+1+8 & 2-1+6-3 \\ 2+6+1+3 & 4-2+2-1 \end{matrix} \right] =\left[ \begin{matrix} 19 & 4 \\ 12 & 3 \end{matrix} \right] \) 
LHS = RHS ⇒ (A + B)² = A² + AB + BA + B²
(ii) LHS .\({ \left( A-B \right) }^{ 2 }=\left[ \begin{matrix} 7 & -6 \\ -2 & 3 \end{matrix} \right] \)  from (8)
RHS. \({ A }^{ 2 }-AB-BA+{ B }^{ 2 }=\left[ \begin{matrix} 5 & 2 \\ 2 & 4 \end{matrix} \right] -\left[ \begin{matrix} 5 & -1 \\ 6 & -2 \end{matrix} \right] -\left[ \begin{matrix} 1 & 6 \\ 1 & 2 \end{matrix} \right] +\left[ \begin{matrix} 8 & -3 \\ 3 & -1 \end{matrix} \right] \) 
= \(\left[ \begin{matrix} 5 & 2 \\ 2 & 4 \end{matrix} \right] +\left[ \begin{matrix} -5 & 1 \\ -6 & 2 \end{matrix} \right] +\left[ \begin{matrix} -1 & -6 \\ -1 & -2 \end{matrix} \right] +\left[ \begin{matrix} 8 & -3 \\ 3 & -1 \end{matrix} \right] \) 
= \(\left[ \begin{matrix} 5-51+8 & 2+1-6-3 \\ 2-6-1+3 & 4+2-2-1 \end{matrix} \right] =\left[ \begin{matrix} 7 & -6 \\ -2 & 3 \end{matrix} \right] \) 
LHS = RHS ⇒ (A - B)² = A² - AB - BA + B²

(i) \(AB=\left[ \begin{matrix} 1 & 8 \\ 4 & 3 \end{matrix} \right] \left[ \begin{matrix} 1 & 3 \\ 7 & 4 \end{matrix} \right] =\left[ \begin{matrix} (1)(1)+(8)(7) & (1)(3)+(8)(4) \\ (4)(1)+(3)(7) & (4)(3)+(3)(4) \end{matrix} \right] \) 
= \(\left[ \begin{matrix} 1+56 & 3+32 \\ 4+21 & 12+12 \end{matrix} \right] =\left[ \begin{matrix} 57 & 35 \\ 25 & 24 \end{matrix} \right] \) ...(1)
\(BA=\left[ \begin{matrix} 1 & 3 \\ 7 & 4 \end{matrix} \right] \left[ \begin{matrix} 1 & 8 \\ 4 & 3 \end{matrix} \right] =\left[ \begin{matrix} (1)(1)+(3)(4) & (1)(8)+(3)(3) \\ (7)(1)+(4)(4) & (7)(8)+(4)(3) \end{matrix} \right] \) 
= \(\left[ \begin{matrix} 1+12 & 8+9 \\ 7+16 & 56+12 \end{matrix} \right] =\left[ \begin{matrix} 13 & 17 \\ 23 & 68 \end{matrix} \right] \)    ...(2)
From (1) and (2) we have AB ≠ BA
(ii) \(\\ (AB)C=\left[ \begin{matrix} 57 & 35 \\ 25 & 24 \end{matrix} \right] \left[ \begin{matrix} -4 & 6 \\ 3 & -5 \end{matrix} \right] \) 
= \(\left[ \begin{matrix} (57)(-4)+(35)(3) & (57)(6)+(35)(-5) \\ (25)(-4)+(24)(3) & (25)(6)+(24)(-5) \end{matrix} \right] \) 
= \(\left[ \begin{matrix} -228+105 & 342-175 \\ -100+72 & 150-120 \end{matrix} \right] \) 
\(\therefore (AB)C=\left[ \begin{matrix} -123 & 167 \\ -28 & 30 \end{matrix} \right] \)     ..(3)
= \(BC=\left[ \begin{matrix} 1 & 3 \\ 7 & 4 \end{matrix} \right] \left[ \begin{matrix} -4 & 6 \\ 3 & -5 \end{matrix} \right] \) 
= \(\left[ \begin{matrix} (1)(-4)+(3)(3) & (1)(6)+(3)(-5) \\ (7)(-4)+(4)(3) & (7)(6)+(4)(-5) \end{matrix} \right] =\left[ \begin{matrix} -4-9 & 6-15 \\ -28+12 & 42-20 \end{matrix} \right] \) 
\(BC=\left[ \begin{matrix} 5 & -9 \\ -16 & 22 \end{matrix} \right] \)    ...(1)
\(A(BC)\left[ \begin{matrix} 1 & 8 \\ 4 & 3 \end{matrix} \right] =\left[ \begin{matrix} 5 & -9 \\ -16 & 22 \end{matrix} \right] \) 
= \(\left[ \begin{matrix} (1)(5)+(8)(-16) & (1)(-9)+(8)(22) \\ (4)(5)+(3)(-16) & (4)(-9)+(3)(22) \end{matrix} \right] =\left[ \begin{matrix} 5-128 & -9+176 \\ 20-48 & -36+66 \end{matrix} \right] \) 
\(A(BC)=\left[ \begin{matrix} 1-123 & 167 \\ -28 & 30 \end{matrix} \right] \)   ..(4)
From (3) and (4) we have, (AB)C = A (BC)
(iii) 
\(B+C=\left[ \begin{matrix} 1 & 3 \\ 7 & 4 \end{matrix} \right] +\left[ \begin{matrix} -4 & 6 \\ 3 & -5 \end{matrix} \right] =\left[ \begin{matrix} 1-4 & 3+6 \\ 7+3 & 4-5 \end{matrix} \right] =\left[ \begin{matrix} -3 & 9 \\ 10 & -1 \end{matrix} \right] \) 
\(A(B+C)=\left[ \begin{matrix} 1 & 8 \\ 4 & 3 \end{matrix} \right] \left[ \begin{matrix} -3 & 9 \\ 10 & -1 \end{matrix} \right] =\left[ \begin{matrix} -3+80 & 9-8 \\ -12+30 & 36-3 \end{matrix} \right] \) 
\(A(B+C)=\left[ \begin{matrix} 77 & 1 \\ 18 & 33 \end{matrix} \right] \)   .(5)
\(AB=\left[ \begin{matrix} 57 & 35 \\ 25 & 24 \end{matrix} \right] \)     ..
\(AC=\left[ \begin{matrix} 1 & 8 \\ 4 & 3 \end{matrix} \right] \left[ \begin{matrix} -4 & 6 \\ 3 & -5 \end{matrix} \right] =\left[ \begin{matrix} -4+24 & 6-40 \\ -16+9 & 24-15 \end{matrix} \right] =\left[ \begin{matrix} 20 & -34 \\ -7 & 9 \end{matrix} \right] \)
 \(AB+AC=\left[ \begin{matrix} 57 & 35 \\ 25 & 24 \end{matrix} \right] +\left[ \begin{matrix} 20 & -34 \\ -7 & 9 \end{matrix} \right] =\left[ \begin{matrix} 57+20 & 35-34 \\ 25-7 & 24+9 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 77 & 1 \\ 18 & 33 \end{matrix} \right] \)     ...(6)
From equations(5) and (6) we haveA(B + C) = AB + AC
(iv) Sinceorder of A is 2 \(\times\) 2, take \(I=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)
\(AI=\left[ \begin{matrix} 1 & 8 \\ 4 & 3 \end{matrix} \right] \left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] =\left[ \begin{matrix} 1(1)+8(0) & 1(0)+8(1) \\ 4(1)+3(3) & 4(0)+3(1) \end{matrix} \right] =\left[ \begin{matrix} 1+0 & 0+8 \\ 4+0 & 0+3 \end{matrix} \right] \) 
= \(\left[ \begin{matrix} 1 & 8 \\ 4 & 3 \end{matrix} \right] =A\)      ...(7)
\(IA=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \left[ \begin{matrix} 1 & 8 \\ 4 & 3 \end{matrix} \right] =\left[ \begin{matrix} 1(1)+0(4) & 1(8)+0(3) \\ 0(1)+1(4) & 0(8)+1(3) \end{matrix} \right] =\left[ \begin{matrix} 1+0 & 8+0 \\ 0+4 & 0+3 \end{matrix} \right] \) 
= \(\left[ \begin{matrix} 1 & 8 \\ 4 & 3 \end{matrix} \right] =A\)    ...(8)
∴From (7) and (8) AI = IA = A
 

Consider the following events.
E_1: I purse is chosen
E2: II purse is chosen
A: Coin pulled out is silver
\(\Rightarrow \quad \therefore P(E_{ 1 })=P({ E }_{ 2 })=\frac { 1 }{ 2 }\)
There are 3 silver and 4 copper coins in I purse
\(\Rightarrow \quad P(A/{ E }_{ 1 })=\frac { 3 }{ 7 } \)
There are 4 silver and 3 copper coins in II Purse
\(\therefore\) P(A/E₁) =  \(\frac { 4 }{ 7 } \)
By the theorem of total probability
P(A) = P(E₁).P(A/E₁)+P(E₂).P(A/E₂) = \(\frac { 1 }{ 2 } \times \frac { 3 }{ 7 } +\frac { 1 }{ 2 } \times \frac { 4 }{ 7 } =\frac { 3 }{ 14 } +\frac { 4 }{ 17 } =\frac { 7 }{ 14 } =\frac { 1 }{ 2 } \)

Let A be the event of getting award for design of railway bridge and B be the event of getting award for the efficient use of materials
Then, \(P(A)=0.48, P(B)=0.36, P(A \cap B)=0.2\)
(i) P (atleast one of the two awards)
\(=P(A \cup B) \)
\(=P(A)+P(B)-P(A \cap B) \)
\(=0.18+0.36-0.2=0.64\)
(ii) P (will get only one of the award)
\(=P(A \cap \bar{B})+P(\bar{A} \cap B) \)
\(=P(A)-P(A \cap B)+P(B)-P(A \cap B) \)
\(=0.48-0.20+0.36-0.20 \)
\(=0.44\)

Let \(\frac{x+1}{(x+2)(x+3)}=\frac{A}{x+2}+\frac{B}{x+3}\)
Multiplying both sides by (x + 2)(x + 3)
x + 1 =  A(x + 3) + B(x + 2) ............(1)
Putting x = -2 in (1)
-2 + 1 = A(-2 + 3) + 0
-1 = A(1) ⇒ A = -1
Putting x = -3 in (1)
-3 +1 = 0 + B(-3 + 2)
-2 = B(-1) ⇒ B = 2
\(\therefore \frac{x+1}{(x+2)(x+3)}=\frac{-1}{x+2}+\frac{2}{x+3}\)
\(\therefore \int \frac{x+1}{(x+2)(x+3)} d x=\int\left[\frac{-1}{x+2}+\frac{2}{x+3}\right] d x\)
\( =-\int \frac{1}{x+2} d x+2 \int \frac{1}{x+3} d x\)
\(=-\log |x+2|+2 \log |x+3|+c \)
\(=2 \log |x+3|-\log |x+2|+c \)

We have x⁴ + y⁴ = 16.
Differentiating implicitly, 4x³ +4y³y' = 0
Solving for y' gives
\(y'=-{x^3\over y^3}\)
To find y'' we differentiate this expression for y' using the quotient rule and remembering that y is a function of x.
\( y^{\prime \prime}=\frac{d}{d x}\left(\frac{-x^3}{y^3}\right) =\frac{-\left[y^3 \frac{d}{d x}\left(x^3\right)-x^3 \frac{d}{d x}\left(y^3\right)\right]}{\left.\left(y^3\right)^2\right]} \\ \)
\(=-\frac{\left[y^3 \cdot 3 x^2-x^3\left(3 y^2 y^{\prime}\right)\right]}{y^6} \)
\( =-\frac{3 x^2 y^3-3 x^3 y^2\left(-\frac{x^3}{y^3}\right)}{y^6} \)
\( =-\frac{3\left(x^2 y^4+x^6\right)}{y^7}=\frac{-3 x^2\left[x^4+y^4\right]}{y^7} \)
\( = \frac{-3 x^2(16)}{y^7}=\frac{-48 x^2}{y^7} . \)

\(\frac{\sin x}{|x|}= \begin{cases}\frac{\sin (x)}{-x} & \text { if }-1<x<0 \\ \frac{\sin x}{x} & \text { if } 0<x<1\end{cases}\)
Therefore, \(lim_{x\rightarrow 0^-}f(x)=-1\)
\(lim_{x\rightarrow 0^+}f(x)=1\)
Hence the limit does not exist.

Let ABC be a triangle and let O be the origin.
Let D and E be the midpoints of AB and AC\(\overrightarrow{OE}={\overrightarrow{OA}+\overrightarrow{OC}\over 2}​={\overrightarrow{a}+\overrightarrow{c}\over 2}​\)

\( \text { Let } \overrightarrow{O A}=\vec{a}, \overrightarrow{O B}=\vec{b}, \overrightarrow{O C}=\vec{c} \)
\(\overrightarrow{O D}= \frac{\vec{a}+\vec{b}}{2}, \quad \overrightarrow{O E}=\frac{\vec{a}+\vec{c}}{2} \)
\(\left.\therefore \overrightarrow{D E}=\overrightarrow{O E}-\overrightarrow{O D}=\left(\frac{\vec{a}+\vec{c}}{2}\right)-\left(\frac{\vec{a}+\vec{b}}{2}\right)=\frac{\vec{a}+\vec{c}-\vec{a}-\vec{b}}{2}\right) \)
\(=\frac{\vec{c}-\vec{b}}{2}=\frac{(\overrightarrow{O C}-\overrightarrow{O B})}{2}\)
\( \overrightarrow{D E}=\frac{\overrightarrow{B C}}{2}=\frac{1}{2} \overrightarrow{B C} \)
\(\therefore D E \| B C . \)
\(\text {Also } \overrightarrow{D E}=\frac{1}{2} \overrightarrow{B C} \Rightarrow|\overrightarrow{D E}|=\frac{1}{2}|\overrightarrow{B C}| \)
\(D E=\frac{1}{2} B C \)
Hence, \(D E \| B C \text { and } D E=\frac{1}{2} B C \text {. }\) 

The straight lines joining the origin and the points of intersection of given equations is a second degree homogeneous equation.
Following steps show, the way of homogenizing the
3x² + 5xy - 2y² + 4x + 5y = 0 with 3x - 2y + 2 = 0
3x² + 5xy - 2y² + (4x + 5y)(1) = 0 and \(\frac { (3x-2y) }{ -2 } \) = 1
3x² + 5xy - 2y² + (4x + 5y)\(\left( \frac { 3x-2y }{ -2 } \right) \)= 0
(-2) (3x² + 5xy - 2y²) + (4x + 5y) (3x - 2y) = 0
On simplification,
We get, 2x² + xy - 2y² = 0 ⇒ a = 2, b = -2 ⇒ a + b = 0
Since sum of the coefficient of x² and y² is equal to zero, the lines are at right angles.

Let, P(n): = (r(cos ፀ + i sin ፀ)n = rⁿ (cos nፀ+i sin nፀ)
Substituting the value of n = 1, in the statement we get,
P(1) =(r(cos ፀ + i sin ፀ))¹ = r(cos ፀ+i sin ፀ)
Hence, P(1) is true.
Let us assume that the statement is true for n = k. Then
(r(cos ፀ + i sin ፀ))^k = r^k (cos kፀ + i sin kፀ)
We need to show that P(k + 1) is true. Consider,
P(k+1) = ((r(cos ፀ + i sin ፀ))^k+1
=(r(cos ፀ + i sin ፀ))^k+1
=(r(cos ፀ + i sin ፀ))^k \(\times\) r(cosፀ+ isinፀ)
=r^k(coskፀ+i sin kፀ) \(\times\) r(cosፀ+ i sinፀ)
=r^k+1 x (cos kፀ cos ፀ + i² sin kፀ sin) + i(sin kፀ cos ፀ + cos kፀ sin ፀ)
=r^k+1 x (cos(k+1)ፀ + i sin(k+1)ፀ).
This implies that P(k + 1) is true. The validity of P(k + 1) follows from that of P(k).
Therefore by the principle of mathematical induction, for any natural number n,
(r(cos ፀ + i sin ፀ))ⁿ = rⁿ(cos(nፀ) + i sin(nፀ))

Given \({a\over r}.a.ar=5832\)
⇒ a³ = 5832 =  18³
⇒ a =18
Also given \({a\over r},a+6, ar+9\) form an A.P
\(∴ a+6-{a\over r}=ar+9-a-6\)
\(⇒\ 18 + 6 --{18\over r}=18r + 9 - 18 - 6\ \ \ [∵ a = 18 ]\)
\(⇒\ 24-{18\over r}=18r-15\)
\(⇒ 24 + 15=18r+{18\over r}\)
\(⇒ 39 ={18r^2+18\over r}\)

⇒ 39r =  18r² + 18
⇒ (2r- 3) (3r- 2) =0
⇒ \(r={3\over2},{2\over 3}\)
Case (i) When \(a=18,r={3\over 2},\)the numbers m G.P. are \({18\over {3\over 2}},18,18\left(2\over 3\right)⇒27,18,12\)
Case (ii) When \(a=18,r={2\over 3},\) the numbers m G.P. are \({18\over {2\over 3}},18,18\left(2\over 3\right)⇒27,18,12\)

Fourth root of 623 = \(\left( \frac { -2 }{ 625 } \right) \)
\(\left( \frac { -2 }{ 625 } \right) \)
\(={ \left[ 625\left( 1-\frac { 2 }{ 625 } \right) \right] }^{ \frac { 1 }{ 4 } }=5{ \left[ 1+\left( -\frac { 2 }{ 625 } \right) \right] }^{ \frac { 1 }{ 4 } }\)
\(={ 5\left[ 1+\frac { 1 }{ 4 } \left( \frac { -2 }{ 625 } \right) +\frac { \frac { 1 }{ 4 } \left( -\frac { 3 }{ 4 } \right) }{ 1.2 } { \left( \frac { -2 }{ 625 } \right) }^{ 2 } \right] }\)
Other terms will have more than seven zeroes after the decimal]
= 5[1-0.0008 - 0.0000009]
\(\sqrt [ 4 ]{ 623 } =4.9959955\)

Given a = 4, b = 6 and c = 8.
By cosine formula, \(\cos { B } =\frac { { a }^{ 2 }+{ c }^{ 2 }-{ b }^{ 2 } }{ 2ac } \)
\(\Rightarrow \cos { B } =\frac { 16+64-36 }{ 2(4)(8) } =\frac { 80-36 }{ 64 } =\frac { 44 }{ 64 } =\frac { 11 }{ 16 } ....(1)\)
Also \(\cos { C } =\frac { { a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 } }{ 2ab } =\frac { 16+36-64 }{ 2(4)(6) } \)
\(=\frac { 52-64 }{ 48 } =\frac { -12 }{ 48 } =\frac { -1 }{ 4 }.....(2)\)
Now, LHS = \(4\cos { B } +3\cos { C } \)
\(=4\left( \frac { 11 }{ 16 } \right) +3\left( \frac { -1 }{ 4 } \right) \)      [ From (1) and (2) ]
= \(\frac { 11 }{ 4 } -\frac { 3 }{ 4 } =\frac { 11-3 }{ 4 } =\frac { 8 }{ 4 } =2=RHS\)   
Hence Proved.

Since the numerator's degree is equal to the denominators degree, let us divide the numerator by the denominator.

\(∴\ {2x^3+5x+11\over x^2+2x-3}=2+{x-5\over x^3+2x-3}\)
Consider \({x-5\over x^2+2x-3}={x-5\over (x+3)(x-1)}{A\over x+3}+{B\over x-1}\)
∴ x-5 = A(x-1) + B(x+3)
Putting x = 1 in (2) we get
-4 = B(4) ⇒ B = 1
Putting x = -3 in (2) we get
-8 = A(-4) ⇒ A = 2

\(∴\ {x-5\over x^2+2x-3}={2\over x+3}-{1\over x-1}\)
Substituting this in (1) we get
\({{2x^2+5x-11}\over{x^2+2x-3}}=2+{{2}\over{x+3}}-{{1}\over{x-1}}\)

If 2x + 3y = 6

x + 4y = 4

x > y > 0 represents the area in the 1 quadrant.
 
All points bounded between x = 0, y = 0, x + 4y = 4 and 2x + 3y = 6 is required region. Darkly shaded area will represents the solution set of the given linear inequalities.

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

LHS = \({{x}\over{a+x}}+{{y}\over{b+y}}+{{z}\over{c+z}}=1\)
\({{ax}\over{a^2+ax}}+{{by}\over{b^2+by}}-{{cz}\over{c^2+cz}}\) [ Multiplying the numerator and denominator be, a, b, c respectively]
\(={{ax}\over{by+cz+ax}}+{{by}\over{cz+ax+by}}+{{cz}\over{ax+by+cz}}\)   \([\therefore a^2=by+cz;b^2=cz+ax;c^2=ax+by]\)
\(\frac { ax+by+cz }{ ax+by+cz } =1\) = RHS
Hence proved.

Given functions are f(x) = x³ and g(x) = \(x^{ ^{ \frac { 1 }{ 3 } } }\)
Now, f o g(x) = f(g(x)
= \(f\left( { x }^{ \frac { 1 }{ 3 } } \right) \)
= \(\left( { x }^{ 3 } \right) ^{ \frac { 1 }{ 3 } }=x\)

Since f o g(x) = x is symmetric about the line y = x, g(x) is the inverse of f(x)
∴ g(x) = f^-1(x).
where f o g(x) = g o f(x) = x so that
(i) f o g is bijective.
(ii) both f(x) and g(x) also bijective.
(iii) f and g are symmetrical about y = x

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