We have:
\(f(x)=2x^{ 2 }-3x\)
\( \therefore f(x)=4x-3\)
(a) For \(f(x)\)to be strictly decreasing function of x. 
\(f'(x)>0\)
\(\Rightarrow 4x-3>0\)
\(\Rightarrow x>\frac { 3 }{ 4 } \)
Hence \('f'\) is strictly decreasing function of x
\(f'(x)<0\Rightarrow 4x-3<0\)
\(\Rightarrow x<\frac { 3 }{ 4 } \)
Hence  \('f'\) is strictly decreasing in  \(\left( \frac { 3 }{ 4 } \infty \right) \)
(b) For \(f(x)\) to be strictly decreasing function of x.
\(f(x)<0\Rightarrow 4x-3<0\)
\(\Rightarrow x<\frac { 3 }{ 4 } \)
Hence \('f'\) is strictly decreasing in \(\left( -\infty ,\frac { 3 }{ 4 } \right) \)

\(Let\quad y={ e }^{ x^{ 3 } }\)
\(\frac { dy }{ dx } ={ e }^{ x^{ 3 } }\frac { d }{ dx } ({ x }^{ 3 })\)
\(={ e }^{ x^{ 3 } }.{ 3 }x^{ 2 }={ 3 }x^{ 2 }{ e }^{ x^{ 3 } }\)

\(A=\left[\begin{array}{rr} 3 & 1 \\ -1 & 2 \end{array}\right] \)
\(A^{2}=A \cdot A=\left[\begin{array}{rr} 3 & 1 \\ -1 & 2 \end{array}\right]\left[\begin{array}{rr} 3 & 1 \\ -1 & 2 \end{array}\right]=\left[\begin{array}{cc} 9-1 & 3+2 \\ -3-2 & -1+4 \end{array}\right]=\left[\begin{array}{rr} 8 & 5 \\ -5 & 3 \end{array}\right] \)
\(\therefore A^{2}-5 A+7 I \)
\(=\left[\begin{array}{rr} 8 & 5 \\ -5 & 3 \end{array}\right]-5\left[\begin{array}{rr} 3 & 1 \\ -1 & 2 \end{array}\right]+7\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \)
\(=\left[\begin{array}{rr} 8 & 5 \\ -5 & 3 \end{array}\right]-\left[\begin{array}{ll} 15 & 5 \\ -5 & 10 \end{array}\right]+\left[\begin{array}{ll} 7 & 0 \\ 0 & 7 \end{array}\right] \)
\(=\left[\begin{array}{ll} -7 & 0 \\ 0 & -7 \end{array}\right]+\left[\begin{array}{ll} 7 & 0 \\ 0 & 7 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \)
\(\text { Hence, } A^{2}-5 A+7 I=O .\)
\(\therefore A \cdot A-5 A=-7 I \)
\(\Rightarrow A \cdot A\left(A^{-1}\right)-5 A A^{-1}=-7 I A^{-1} \quad\left[\text { Post-multiplying by } A^{-1} \text { as }|A| \neq 0\right] \)
\(\Rightarrow A\left(A A^{-1}\right)-5 I=-7 A^{-1} \)
\(\Rightarrow A I-5 I=-7 A^{-1} \)
\(\Rightarrow A^{-1}=-\frac{1}{7}(A-5 I) \)
\(\Rightarrow A^{-1}=\frac{1}{7}(5 I-A) \)
\(=\frac{1}{7}\left(\left[\begin{array}{ll} 5 & 0 \\ 0 & 5 \end{array}\right]-\left[\begin{array}{rr} 3 & 1 \\ -1 & 2 \end{array}\right]\right)=\frac{1}{7}\left[\begin{array}{lr} 2 & -1 \\ 1 & 3 \end{array}\right] \)
\(\therefore A^{-1}=\frac{1}{7}\left[\begin{array}{rr} 2 & -1 \\ 1 & 3 \end{array}\right]\)

Observe that the function is defined for every real number. The function f may be thought of as a composition g o h of the two functions g and h, where g(x) = sin x and h(x) = x². Since both g and h are continuous functions, by Theorem 2, it can be deduced that f is a continuous function

\(LHS={ tan }^{ -1 }\frac { 1 }{ 2 } +{ tan }^{ -1 }\frac { 2 }{ 11 } \)
\(={ tan }^{ -1 }\frac { \frac { 1 }{ 2 } +\frac { 2 }{ 11 } }{ 1-\frac { 1 }{ 2 } .\frac { 2 }{ 11 } } ={ tan }^{ -1 }\frac { 15 }{ 20 } ={ tan }^{ -1 }\frac { 3 }{ 4 } =RHS.\)

Let A be foot of ladder x m away from the wall and if y m be height of ladder on wall
\(y=\sqrt{169-x^{2}} \Rightarrow \frac{d y}{d t}=\frac{-x}{\sqrt{169-x^{2}}} \cdot \frac{d x}{d t} \)
\(\left.\frac{d y}{d t}\right]_{x=5}=-\frac{5}{12} \times 2=-\frac{5}{6} \mathrm{~cm} / \mathrm{s} \quad\left[\frac{d x}{d t}=2 \mathrm{~cm} / \mathrm{s}\right] \)
Hence,height  of the wall is decreasing at \(5\over6\)cm/sec

(i) \(\text { Let } y=\left(x^{5}-5 x+8\right)\left(x^{3}+7 x+9\right)\)
\(\text { Let } x^{2}-5 x+8=u \text { and } x^{3}+7 x+9=v\)
\(\therefore y=u v\)
\(\Rightarrow \frac{d y}{d x}=\frac{d u}{d x} \cdot v+u \cdot \frac{d v}{d x}\)
\(\text { (By using product rule) }\)
\(\Rightarrow \frac{d y}{d x}=\frac{d}{d x}\left(x^{2}-5 x+8\right) \cdot\left(x^{3}+7 x+9\right)+\left(x^{2}-5 x+8\right) \cdot \frac{d}{d x}\left(x^{3}+7 x+9\right)\)
\(\Rightarrow \frac{d y}{d x}=(2 x-5)\left(x^{3}+7 x+9\right)+\left(x^{2}-5 x+8\right)\left(3 x^{2}+7\right)\)
\(\Rightarrow \frac{d y}{d x}=2 x\left(x^{3}+7 x+9\right)-5\left(x^{3}+7 x+9\right)+x^{2}\left(3 x^{2}+7\right)-5 x\left(3 x^{2}+7\right)+8\left(3 x^{2}+7\right)\)
\(\Rightarrow \frac{d y}{d x}=\left(2 x^{4}+14 x^{2}+18 x\right)-5 x^{3}-35 x-45+\left(3 x^{4}+7 x^{2}\right)-15 x^{3}-35 x+24 x^{2}+56\)
\(\therefore \frac{d y}{d x}=5 x^{4}-20 x^{3}+45 x^{2}-52 x+11 \)
(ii) \(y=\left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right) \)
\(= x^{2}\left(x^{3}+7 x+9\right)-5 x\left(x^{3}+7 x+9\right)+8\left(x^{3}+7 x+9\right) \)
\(= x^{5}+7 x^{3}+9 x^{2}-5 x^{4}-35 x^{2}-45 x+8 x^{3}+56 x+72 \)
\(= x^{5}-5 x^{4}+15 x^{3}-26 x^{2}+11 x+72 \)
\(\therefore \frac{d y}{d x}=\frac{d}{d x}\left(x^{5}-5 x^{4}+15 x^{3}-26 x^{2}+11 x+72\right) \)
\(=\frac{d}{d x}\left(x^{5}\right)-5 \frac{d}{d x}\left(x^{4}\right)+15 \frac{d}{d x}\left(x^{3}\right)-26 \frac{d}{d x}\left(x^{2}\right)+11 \frac{d}{d x}(x)+\frac{d}{d x}(72) \)
\(=5 x^{4}-5 \times 4 x^{3}+15 \times 3 x^{2}-26 \times 2 x+11 \times 1+0 \)
\(=5 x^{4}-20 x^{3}+45 x^{2}-52 x+11 \)
(iii) \(y=\left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right)\)
Taking logarithm on both the sides, we obtain
\(\log y=\log \left(x^{2}-5 x+8\right)+\log \left(x^{3}+7 x+9\right)\)
Differentiating both sides with respect to x, we obtain
\(\frac{1}{y} \frac{d y}{d x}=\frac{d}{d x} \log \left(x^{2}-5 x+8\right)+\frac{d}{d x} \log \left(x^{3}+7 x+9\right) \)
\(\Rightarrow \frac{1}{y} \frac{d y}{d x}=\frac{1}{x^{2}-5 x+8} \cdot \frac{d}{d x}\left(x^{2}-5 x+8\right)+\frac{1}{x^{3}+7 x+9} \cdot \frac{d}{d x}\left(x^{3}+7 x+9\right) \)
\(\Rightarrow \frac{d y}{d x}=y\left[\frac{1}{x^{2}-5 x+8} \times(2 x-5)+\frac{1}{x^{3}+7 x+9} \times\left(3 x^{2}+7\right)\right] \)
\(\Rightarrow \frac{d y}{d x}=\left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right)\left[\frac{2 x-5}{x^{2}-5 x+8}+\frac{3 x^{2}+7}{x^{3}+7 x+9}\right] \)
\(\Rightarrow \frac{d y}{d x}=\left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right)\left[\frac{(2 x-5)\left(x^{3}+7 x+9\right)+\left(3 x^{2}+7\right)\left(x^{2}-5 x+8\right)}{\left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right)}\right] \)
\(\Rightarrow \frac{d y}{d x}=2 x\left(x^{3}+7 x+9\right)-5\left(x^{3}+7 x+9\right)+3 x^{2}\left(x^{2}-5 x+8\right)+7\left(x^{2}-5 x+8\right) \)
\(\Rightarrow \frac{d y}{d x}=\left(2 x^{4}+14 x^{2}+18 x\right)-5 x^{3}-35 x-45+\left(3 x^{4}-15 x^{3}+24 x^{2}\right)+\left(7 x^{2}-35 x+56\right) \)
\(\Rightarrow \frac{d y}{d x}=5 x^{4}-20 x^{3}+45 x^{2}-52 x+11 \)
From the above three observations, it can be concluded that all the results of \(\frac{d y}{d x}\) are same

\(f(x)=\left\{\begin{array}{cl} \frac{|x|}{x}, & \text { if } x \neq 0 \\ 0, & \text { if } x=0 \end{array}\right.\)
The given function f is defined at all the points of the real line.
Let c be a point on the real line.
Case I: \(\text { If } c<0, \text { then } f(c)=-1 \)
\(\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(-1)=-1 \)
\(\therefore \lim _{x \rightarrow c} f(x)=f(c) \)
Therefore, f is continuous at all points x < 0
Case II:
If c = 0, then the left hand limit of f at x = 0 is,
\(\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}(-1)=-1\)
The right hand limit of f at x = 0 is,
\(\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}(1)=1\)
It is observed that the left and right hand limit of f at x = 0 do not coincide.
Therefore, f is not continuous at x = 0
Case III: \(\text { If } c>0, \text { then } f(c)=1 \)
\(\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(1)=1 \)
\(\therefore \lim _{x \rightarrow c} f(x)=f(c) \)
Therefore, f is continuous at all points x, such that x > 0
Hence, x = 0 is the only point of discontinuity of f.

We have
f(x) = sin x + cos x,
or f'(x) = cos x – sin x

Now f'(x) = 0 gives sin x = cos x which gives that \(x=\frac{\pi}{4}, \frac{5 \pi}{4} \text { as } 0 \leq x \leq 2 \pi\)
The points \(x=\frac{\pi}{4} \text { and } x=\frac{5 \pi}{4}\) divide the interval [0, 2p] into three disjoint intervals namely,
\(\left[0, \frac{\pi}{4}\right),\left(\frac{\pi}{4}, \frac{5 \pi}{4}\right) \text { and }\left(\frac{5 \pi}{4}, 2 \pi\right]\)
\(\text {Note that } f^{\prime}(x)>0 \text {if } x \in\left[0, \frac{\pi}{4}\right) \cup\left(\frac{5 \pi}{4}, 2 \pi\right]\)
or f is increasing in the intervals \(\left[0, \frac{\pi}{4}\right) \text { and }\left(\frac{5 \pi}{4}, 2 \pi\right]\)
\(\text {Also }f^{\prime}(x)<0 \text { if } x \in\left(\frac{\pi}{4}, \frac{5 \pi}{4}\right)\)
or is plecreasing in \( \left(\frac{\pi}{4}, \frac{5 \pi}{4}\right)\)

the required trapezium is as given in figure. Draw perpendiculars DP and CQ on AB. Let AP = x cm. Note that \(\Delta \)APD \(\cong \) \(\Delta \) BQC.
Therefore, QB = x cm. Also, by pythagoras theorem DP = QC = \(\sqrt { 100-{ x }^{ 2 } } \)
Let A be the area of the trapezium.

Then, A \(\equiv \) A(x)
\(=\frac { 1 }{ 2 } (sum\ of\ parallel\ sides)\times (height)\)
\(=\frac { 1 }{ 2 } (2x+10+10)\sqrt { 100-{ x }^{ 2 } } \)
= (x + 10)\(\sqrt { 100-{ x }^{ 2 } } \)
\(or\ A'(x)=(x+10)\frac { (-2x) }{ 2\sqrt { 100-{ x }^{ 2 } } } +(\sqrt { 100-{ x }^{ 2 }) } \)
\(=\frac { -2{ x }^{ 2 }-10x+100 }{ \sqrt { 100-{ x }^{ 2 } } } \)
Now, A'(x) = 0 gives 2x² + 10x - 100 = 0,
i.e., x = 5 and x = -10
So, x = 5.
Now, A''(x) = \(\frac { \sqrt { 100-{ x }^{ 2 } } (-4x-10)-(-2x^{ 2 }-10x+100)\frac { (-2x) }{ 2\sqrt { 100-{ x }^{ 2 } } } }{ 100-{ x }^{ 2 } } \)
\(=\frac { { 2x }^{ 3 }-300x-1000 }{ { (100-{ x }^{ 2 }) }^{ \frac { 1 }{ 2 } } } \)
(on simplification)
or \(A''(5)=\frac { { 2(5) }^{ 3 }-300(5)-1,000 }{ (100-(5{ ) }^{ 2 })^{ \frac { 3 }{ 2 } } } \)
\(=\frac { -2,250 }{ 75\sqrt { 75 } } =\frac { -30 }{ \sqrt { 75 } } <0\)
Thus, area of trapezium is maximum at x = 5 and the maximum area is given by
A(5) = (5 + 10)\(\sqrt { 100-{ (5) }^{ 2 } } \)
= 15\(\sqrt { 75 } =75\sqrt { 3 } { cm }^{ 2 }\)

\({ A }^{ 2 }=\left[ \begin{matrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{matrix} \right] \left[ \begin{matrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 4+1+1 & -2-2-1 & 2+1+2 \\ -2-2-1 & 1+4+1 & -1-2-2 \\ 2+1+2 & -1-2-2 & 1-1-4 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{matrix} \right] \)
\({ A }^{ 3 }=\left[ \begin{matrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{matrix} \right] \left[ \begin{matrix} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{matrix} \right] \left[ \because { A }^{ 3 }=A.{ A }^{ 2 } \right] \)
\(=\left[ \begin{matrix} 12+5+5 & -10-6-5 & 10+5+6 \\ -6-10-5 & 5+12+5 & -5-10-6 \\ 6+5+10 & -5-6-10 & 5+5+12 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22 \end{matrix} \right] \)
Now, A³- 6A² + 9A - 4I
\(=\left[ \begin{matrix} 22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22 \end{matrix} \right] -6\left[ \begin{matrix} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{matrix} \right] +9\left[ \begin{matrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{matrix} \right] -4\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] I\)
\(=\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] =0\)(zero matrix)
\(\therefore\) A³ - 6A² + 9A - 4I = 0
Pre-multiplying by A^-1, we get
A^-1A.A² - 6A^-1A.A + 9A^-1A. ​​​​​​- 4A^-1I = 0
A² - 6A + 9I - 4A^-1 = 0
\(\Rightarrow\) 4A^-1 = A²- 6A + 9I
\({ A }^{ -1 }=\frac { 1 }{ 4 } \left[ \begin{matrix} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{matrix} \right] \)
\(=\left[ \begin{matrix} \frac { 3 }{ 4 } & \frac { 1 }{ 4 } & -\frac { 1 }{ 4 } \\ \frac { 1 }{ 4 } & \frac { 3 }{ 4 } & \frac { 1 }{ 4 } \\ -\frac { 1 }{ 4 } & \frac { 1 }{ 4 } & \frac { 3 }{ 4 } \end{matrix} \right] \)

Let ABCD be a rectangle inscribed in a given circle with centre at 0 and radius a.
Let AB = 2x and BC = 2y

Then, OA² = OM² + AM² 
\(\Rightarrow a^2=y^2+x^2\)
\(\Rightarrow y=\sqrt{a^2-x^2}\)
Let A be the area of the rectangle.
\(\therefore A=4xy=4x\sqrt{x^2-x^2}\)
\(\Rightarrow \ \ \frac{dA}{dx}=4\{\frac{a^2-2x^2}{\sqrt{a^2-x^2}}\}\)
For maximum or minimum value of A,
\(\frac{dA}{dx}=0\)
\(=4\{\frac{a^2-2x^2}{\sqrt{a^2-x^2}}\}=0\Rightarrow\ \ x=\frac{a}{\sqrt{2}}\)
Now, \(\frac{d^2A}{dx^2}=4\frac{d}{dx}\{(a^2-2x^2)(a^2-x^2)^{-1/2}\}\)
\(\Rightarrow \frac{d^2A}{dx^2}=4[-4x(a^2-x^2)^{-1/2}+(a^2-2x^2)\times(-1/2)(a^2-x^2)^(-3/2)(-2x)]\)
\(=[\frac{-4x}{\sqrt{a^2-x^2}}+\frac{x(a^2-2x^2)}{(a^2-x^2)^{3/2}}]\)
\(\therefore\ (\frac{d^2A}{dx^2})_{x=\frac{a}{\sqrt2}}=-16<0\)
Thus A is maximum when \(x=\frac{a}{\sqrt2}\)
putting \(x=\frac{a}{\sqrt2}\) in (i) \(y=\frac{a}{\sqrt2}\)
Therefore \(x=y=\frac{a}{\sqrt2}\)
Hence area is maximum when x = y ⇒ 2x = 2y
i.e., the rectangle is a square.

As in the example we find that f is continuous at all real numbers x \(\ne\) 1. The left hand limit of f at x = 1 is
\(\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}(x+2)=1+2=3\)
The right hand limit of f at x = 1 is
\(\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(x-2)=1-2=-1\)
Since, the left and right hand limits of f at x = 1 do not coincide, f is not continuous at x = 1. Hence x = 1 is the only point of discontinuity of f.

Given, 2A - 3B + 5C = O
\(\Rightarrow 2 A=3 B-5 C\)
\(\Rightarrow 2 A=3\left[\begin{array}{rrr} -2 & 2 & 0 \\ 3 & 1 & 4 \end{array}\right]-5\left[\begin{array}{rrr} 2 & 0 & -2 \\ 7 & 1 & 6 \end{array}\right]\)
\(\Rightarrow 2 A=\left[\begin{array}{rrr} -6 & 6 & 0 \\ 9 & 3 & 12 \end{array}\right]-\left[\begin{array}{rrr} 10 & 0 & -10 \\ 35 & 5 & 30 \end{array}\right]\)
\(\Rightarrow 2 A=\left[\begin{array}{rrr} -6-10 & 6-0 & 0-(-10) \\ 9-35 & 3-5 & 12-30 \end{array}\right]\)
\( \Rightarrow A=\frac{1}{2}\left[\begin{array}{rrr} -16 & 6 & 10 \\ -26 & -2 & -18 \end{array}\right] \Rightarrow A=\left[\begin{array}{rrr} -8 & 3 & 5 \\ -13 & -1 & -9 \end{array}\right]\)

A = {1, 2, 3 ...13, 14}
R = {(x, y) : 3x - y = 0}
∴ R = {(1, 3), (2, 6), (3, 9),(4, 12)}
R is not reflexive since (1, 2), (2, 2)...(14, 14) ∉ R
Also, R is not symmetric as (1, 3) ∈R, but (3, 1) ∉ R.[3(3) - 1 ≠ 0]
Also, R is not transitive as (1, 3), (3, 9) ∈R, but (1, 9) ∉ R.
[3(1) - 9 ≠ 0]
Hence R is neither reflexive, nor symmetric, nor transitive. 

The area of a circle (A) with radius (r) is given by.
Therefore, the rate of change of area (A) with respect to time (t) is given by,
[By chain rule]
It is given that.
Thus, when r = 8 cm,

Hence, when the radius of the circular wave is 8 cm, the enclosed area is increasing at the rate of 80 π cm²/s.

Note that
f '(x) = 3x² – 6x + 4
= 3(x² – 2x + 1) + 1
= 3(x – 1)² + 1 > 0, in every interval of R
Therefore, the function f is increasing on R.

Given, \(f(x)=\left\{\begin{array}{l} a x+1, \text { if } x \leq 3 \\ b x+3, \text { if } x>3 \end{array}\right.\) 
and f(x) is continuous at x = 3.
Here, a and b are two unknowns
Now, at x = 3, f(3) = a(3) + 1 = 3a + 1
\( \mathrm{LHL} =\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{-}}(a x+1) \)
\(=\lim _{h \rightarrow 0}\{a(3-h)+1\} \text { [put } x=3- \text { ; when } x \rightarrow 3^{-} \text {then } \left.h \rightarrow 0\right]\)
\(=\lim _{h \rightarrow 0}(3 a-a h+1)=3 a+1 \) 
\( \text {and } \mathrm{RHL} =\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3^{+}}\{b x+3\} \)
\(=\lim _{h \rightarrow 0}\{b(3+h)+3\} \text { [put } x=3+h ; \text { when } x \rightarrow 3^{+}, \text {then } \left.h \rightarrow 0\right]\)
\(=\lim _{h \rightarrow 0}\{3 b+b h+3\}=3 b+3 \) 
Since, f(x) is continuous at x = 3.
At x = 3, f(3) = LHL = RHL
\(\text { i.e. } \quad 3 a+1=3 a+1=3 b+3 \)
\(\Rightarrow 3 a+1=3 b+3 \)
\(\Rightarrow 3 a=3 b+3-1 \)
\(\Rightarrow a=\frac{(3 b+2)}{3} \Rightarrow a=b+\frac{2}{3} \) 
which is the required relation between a and b

\(\left\{ \frac { cosx }{ 2\sqrt { sinx } \left( 1+sinx \right) } \right\} \)