\(I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { dx }{ 1+\sqrt { tanx } } } \)
\(=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sqrt { cosx } dx }{ \sqrt { cosx } +\sqrt { sinx } } } \) ....(i)
Apply the property
\(\int _{ 0 }^{ a }{ f(x)dx } =\int _{ 0 }^{ a }{ f(a-x)dx, } \)
\(I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sqrt { cos\left( \frac { \pi }{ 2 } -x \right) } }{ \sqrt { cos\left( \frac { \pi }{ 2 } -x \right) +\sqrt { sin\left( \frac { \pi }{ 2 } -x \right) } } } } \)
\(\Rightarrow I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sqrt { sinx } }{ \sqrt { sinx } +\sqrt { sinx } } } dx\) ..(ii)
by adding eqn. (i) and (ii),
\(\Rightarrow 2I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sqrt { cosx } +\sqrt { sinx } }{ \sqrt { cosx } +\sqrt { sinx } } } dx\)
\(\Rightarrow 2I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ dx } \Rightarrow \quad \left[ x \right] ^{ \frac { \pi }{ 2 } }_{ 0 }\)
\(\Rightarrow 2I=\left( \frac { \pi }{ 2 } -0 \right) =\frac { \pi }{ 2 } \)
\(\Rightarrow I=\frac { \pi }{ 4 } \)

\( \int_0^1 \frac{d x}{1+x^2}=\left[\tan ^{-1} x\right]_0^1 \)
\( =\tan ^{-1} 1-\tan ^{-1} 0 \\ \)
\( =\frac{\pi}{4}-0=\frac{\pi}{4} \)

\(\int _{ 0 }^{ 2 }{ \sqrt { 4-{ x }^{ 2 } } } dx\)
\(=\int _{ 0 }^{ 2 }{ \sqrt { { 2 }^{ 2 }-{ x }^{ 2 } } } dx\)
\(=\left[ \frac { x }{ 2 } \sqrt { 4-{ x }^{ 2 } } +\frac { 4 }{ 2 } { sin }^{ -1 }\left( \frac { x }{ 2 } \right) \right] ^{ 2 }_{ 0 }\)
\(=\left[ 0+2sin^{ -1 }(1) \right] -\left[ 0+2{ sin }^{ -1 }0 \right] \)
\(=2\times \frac { \pi }{ 2 } -(0++2\times 0)=\pi \)

Given integral \(=\int { \left( \frac { 1 }{ 1+{ x }^{ 2 } } -\frac { 2x }{ (1+{ x }^{ 2 })^{ 2 } } \right) } { e }^{ x }dx\)
\(=\frac { 1 }{ 1+{ x }^{ 2 } } { e }^{ x }+C \left[ as\frac { d }{ dx } \left( \frac { 1 }{ 1+{ x }^{ 2 } } \right) =\frac { -2x }{ (1+{ x }^{ 2 })^{ 2 } } \right] \)

Given, \(y=x^2-ax+7\)
\(\Rightarrow \frac{dy}{dx}=2x-a\)
\(m_1=2x-a\)
Line  \(2x-y+9=0\)
\(\Rightarrow 2-\frac{dy}{dx}=0\)
\(\Rightarrow \frac{dy}{dx}=2=m_2\)
for parallel, \(m_1=m_2\)
\(\therefore 2x-a=2\)
\(\Rightarrow 2(-1)-a=2\)
\(\Rightarrow -2-2=a\)
\(\Rightarrow a=-4\)

Given, \(x^2+2y+y^2=0\)
\(2x+2\frac{dy}{dx}+2y\frac{dy}{dx}=0\)
\(\frac{dy}{dx}(2+2y)=-2x\)
\(\frac{dy}{dx}=\frac{-2x}{2(1+y)}=-\frac{x}{1+y}\)
Slope of tangent at (-1,2)
\(\frac{-1(-1)}{1+2}=\frac{1}{3}\)
Slope of normal at (-1,2)
\(=-\frac{3}{1}=-3\)

\(\int { \left[ \frac { 2sinxcosx }{ 2 } \right] ^{ 2 } } dx\)
\(=\frac { 1 }{ 4 } \int { (sin2x)^{ 2 } } dx\)
\(=\frac { 1 }{ 4 } \int { { sin }^{ 2 }2x } dx\)
\(\frac { 1 }{ 4 } \int { \frac { 1-cos4x }{ 2 } } dx=\frac { 1 }{ 8 } \int { 1dx } -\frac { 1 }{ 8 } \int { cos4xdx } \)
 \(=\frac { x }{ 8 } -\frac { 1 }{ 8 } \frac { sin4x }{ 4 } \)
\(=\frac { 1 }{ 8 } \left( x-\frac { 1 }{ 4 } sin4x \right) +c\)

\(\int { sin^{ 2 }x } dx=\int { \left( \frac { 1-cos2x }{ 2 } \right) dx } \)
\([\therefore cos2x=1-2sin^{ 2 }x]\)
\(=\frac { 1 }{ 2 } \int { dx } -\frac { 1 }{ 2 } \int { cos2x } dx\)
\(=\frac { 1 }{ 2 } x-\frac { 1 }{ 4 } sin2x+C\)

\(cosC.sinD=\frac { 1 }{ 2 } \left[ sin(C+D)-sin(C-D) \right] \)
\(=\frac { 1 }{ 2 } \int { \left[ sin(3x+2x)-sin(3x-2x) \right] } dx\)
\(=\frac { 1 }{ 2 } \int { (sin5x-sinx) } dx\)
\(=\frac { 1 }{ 2 } \left( -\frac { cos5x }{ 5 } \right) -\frac { 1 }{ 2 } \left[ -cosx \right] \)
\(=\frac { 1 }{ 2 } \left[ cosx-\frac { 1 }{ 5 } cos5x \right] +C\)

Let A denote the area of rectangle at instant t
∴  A = xy  (area of rectangle),
\(\frac { dx }{ dt } =-5cm/min\)
\(\frac { dx }{ dt } =4cm/min\)
\(\frac { dx }{ dt } =x\frac{dy}{dt}+y\frac{dx}{dt}\)
\(\Rightarrow \frac{dA}{dt}=5 \times 4+8\times -5\)
\(\Rightarrow \frac{dA}{dt}=20-40\)
\(\Rightarrow \frac{dA}{dt}=-20\ cm^2/min\)
(-) ve sign shows that area is decreasing at the rate of 20cm²/min.

\(y'\ne 0,\)for any x

0.75

\(\int \sqrt{\tan x} \cdot \sec ^{2} x d x=\int \sqrt{t} d t=\frac{2}{3} t^{\frac{3}{2}}+C=\frac{2}{3}(\tan x)^{\frac{3}{2}}+C \)

\(\int \frac{1+\tan x}{1-\tan x} d x =\int \frac{\cos x+\sin x}{\cos x-\sin x} d x \)
\(=-\int \frac{1}{t} d t \)
\(=-\log |t|+C \)
\(=-\log |\cos x-\sin x|+C\)

\(\int \frac{x^{2}+4 x}{x^{3}+6 x^{2}+5} d x =\frac{1}{3} \int \frac{1}{t} d t \)
\(=\frac{1}{3} \log |t|+C=\frac{1}{3} \log \left|x^{3}+6 x^{2}+5\right|+C\)

\(\int \frac{x^{2}}{1+x^{3}} d x =\frac{1}{3} \int \frac{1}{t} d t=\frac{1}{3} \log |t|+C \)
\(=\frac{1}{3} \log \left|1+x^{3}\right|+C\)

30.255 Concern for environment;Responsibility for pollution free envirnment     

Let \(I= \int { \sin ^{ -1 }{ \sqrt { \frac { x }{ a+x } } } } dx.\)
\(Put\ x=a\tan ^{ 2 }{ \theta } \)
so that \(dx=2a \tan { \theta } \sec ^{ 2 }{ \theta } d\theta .\)
\(\therefore I= \int { \sin ^{ -1 }{ \sqrt { \frac { a\tan ^{ 2 }{ \theta } }{ a(1+\tan ^{ 2 }{ \theta ) } } } } } (2a\tan { \theta } \sec ^{ 2 }{ \theta } )\)
\(=2a\int { \sin ^{ -1 }{ \sqrt { \frac { \tan ^{ 2 }{ \theta } }{ \sec ^{ 2 }{ \theta } } } } } (\tan { \theta } \sec ^{ 2 }{ \theta } )d\theta \)
\(=2a\int { \sin ^{ -1 }{ (\sin { \theta } ) } } (\tan { \theta } \sec ^{ 2 }{ \theta } )d\theta \)
\(=2a\int { \theta . } (\tan { \theta } \sec ^{ 2 }{ \theta } )d\theta \)
\(=2a\left[ \theta .\frac { \tan ^{ 2 }{ \theta } }{ 2 } -\int { (1) } \frac { \tan ^{ 2 }{ \theta } }{ 2 } d\theta \right] \)
[Integrating by Parts]
\(=a\theta \tan ^{ 2 }{ \theta } -a\int { (\sec ^{ 2 }{ \theta } -1) } d\theta \)
\(=a\theta \tan ^{ 2 }{ \theta } -a(\tan { \theta } -\theta )+c\)
\(=a\theta \tan ^{ 2 }{ \theta } -a\tan { \theta } +a\theta +c\)
\(=a\left( \tan ^{ -1 }{ \sqrt { \frac { x }{ a } } } \right) \left( \frac { x }{ a } \right) -a\sqrt { \frac { x }{ a } } +\tan ^{ -1 }{ \sqrt { \frac { x }{ a } } } +c.\)
\(\left[ \because x=a\tan ^{ 2 }{ \theta } \Rightarrow \tan ^{ 2 }{ \theta } =\frac { x }{ a } \Rightarrow \tan ^{ 2 }{ \theta } =\sqrt { \frac { x }{ a } } \Rightarrow \theta =\tan ^{ -1 }{ \sqrt { \frac { x }{ a } } } +c \right]\)
\(=x\tan ^{ -1 }{ \sqrt { \frac { x }{ a } } } -\sqrt { ax } +a\tan ^{ -1 }{ \sqrt { \frac { x }{ a } } } +c\)​​​​​​​

We have :
\((i)f(x)=(2x-1)^{ 2 }+3\)
Since \((2x-1)^{ 2 }\ge 0\)
Min value of \((2x-1)^{ 2 }\)= 0
Min value of f(x) = 3
Max value of f(x) does not exist.
(iii) We  have \(f(x)=-(x-1)^{ 2 }+10\)
Max value of -\((x-1)^{ 2 }=0\)
Min value of f(x) does not exist 
We have :\(f(x)=x^{ 3 }+1\)
Since \(x\longrightarrow \infty \Rightarrow f(x)\longrightarrow \infty \)
\(x\longrightarrow -\infty \Rightarrow f(x)\longrightarrow -\infty \)
Moreover \(f'(x)=3x^{ 2 }+1=+ve\forall x\in R\)
\(\Rightarrow f(x)\) in an increasing function on R.
Hence there is no maximum or minimum value.

We have:\(f(x)=2x^{ 3 }-3x^{ 2 }-36x+7\quad \)
\(\therefore f'(x)=6x^{ 2 }-6x-36\)
(a) For \(f(x)\) to be strictly increasing function of x:
\(f'(x)>0\quad 6x^{ 2 }-6x-36>0\)
\(\Rightarrow x^{ 2 }-x-6>0\)
\(\Rightarrow (x-3)(x+2)>0\)
\(\Rightarrow (x-(-2))(x-3)>0\)
\(\Rightarrow x\in (-\infty ,-2)\cup (3,\infty )\)
Hence \('f'\)is strictly decreasing function of x,
\(f'(x)<0\quad 6x^{ 2 }-6x-36<0\)
\(\Rightarrow x^{ 2 }-x-6<0\)
\(\Rightarrow (x-3)(x+2)<0\)
\(\Rightarrow (x-(-2))(x-3)<0\)
\(\Rightarrow x\in (-2,3).\)
Hence \('f'\)is strictly decreasing in (-2,3).

\(\int { { \left( \frac { { x }^{ 3 }+{ 3x }+4 }{ { x }^{ 2 } } \right) } } dx\)
\(=\int { { \left( \frac { { x }^{ 3 } }{ \sqrt { x } } +3\frac { x }{ \sqrt { x } } +\frac { 4 }{ \sqrt { x } } \right) } } dx\)
\(=\int { { x }^{ \frac { 5 }{ 2 } }dx+3 } \int { { x }^{ \frac { 1 }{ 2 } }dx+4 } \int { { x }^{ \frac { 1 }{ 2 } }dx } \)
\(=\frac { { x }^{ \frac { 5 }{ 2 } +1 } }{ \frac { 5 }{ 2 } +1 } +3\frac { { x }^{ \frac { 1 }{ 2 } +1 } }{ \frac { 1 }{ 2 } +1 } +4\frac { { x }^{ -\frac { 1 }{ 2 } +1 } }{ -\frac { 1 }{ 2 } +1 } +C\)
\(=\frac { 2 }{ 7 } { x }^{ \frac { 7 }{ 2 } }+{ 2x }^{ \frac { 3 }{ 2 } }+8{ x }^{ \frac { 1 }{ 2 } }+C.\)

The anti derivative of (ax + b)³ is the function of x whose derivative is (ax + b)³
It is known that,
\(\frac { d }{ dx } \left( (ax+{ b) }^{ 3 } \right) =3a(ax+{ b) }^{ 2 }\)
\(\Rightarrow \frac { d }{ dx } \left( \frac { 1 }{ 3a } (ax+{ b) }^{ 3 } \right) =(ax+{ b) }^{ 2 }\)
Hence,  \(\int { (ax+{ b) }^{ 2 } } dx=\frac { 1 }{ 3a } (ax+{ b) }^{ 3 }+C.\)

Note that
\(\int \sqrt{x^{2}+2 x+5} d x=\int \sqrt{(x+1)^{2}+4} d x\)
Put x + 1 = y, so that dx = dy. Then
\(\int \sqrt{x^{2}+2 x+5} d x =\int \sqrt{y^{2}+2^{2}} d y \)
\(=\frac{1}{2} y \sqrt{y^{2}+4}+\frac{4}{2} \log \left|y+\sqrt{y^{2}+4}\right|+C \quad[\text { using } 7.6 .2 \text { (ii) }] \)
\(=\frac{1}{2}(x+1) \sqrt{x^{2}+2 x+5}+2 \log \left|x+1+\sqrt{x^{2}+2 x+5}\right|+C \)

Let 'r' be the radius of spherical ball of salt.
By the question.
\(\frac { dV }{ dt } =-ks\)
\(\Rightarrow \frac { d }{ dt } \left( \frac { 4 }{ 3 } \pi r^{ 3 } \right) =-k(4\pi r^{ 2 })\)
\(\Rightarrow \frac { 4 }{ 3 } \pi (3r^{ 2 })\frac { dr }{ dt } =-k(4\pi r^{ 2 })\)
\(\Rightarrow 4\pi r^{ 2 }\frac { dr }{ dt } =-k(4\pi r^{ 2 })\)
\(\Rightarrow \frac { dt }{ dt } =-k(-ve\quad constant)\)
Hence, the radius is decreasing at a constant rate.

We have:
\(f(x)=x^{ 3 }-3x^{ 2 }+3x-100\quad \)
\(\therefore f(x)=3x^{ 2 }-6x+3=3(x-1)^{ 2 }\)
Which is for all \(x\in R\)
Hence,f(x) is increasing function on R.

Let 'r' be the radius of the circle.
Then A,the area of the circle = \(\pi r^{ 2 }\).
\(\therefore \)  \(\frac { dA }{ dr } =2\pi r\)
Hence, the rate of change of the area of the circle = \(2\pi r\)
(a) When r = 3 cm, then the rate of change of the area of the circle = \(2\pi (3)=6\pi cm^{ 2 }/s.\)
(b) When r = 4 cm, then the rate of change of the area of the circle = \(2\pi (4)=8\pi cm^{ 2 }/s.\)

Let 'r' be the radius of the circular disc.
Then  \(A=\pi r^{ 2 }\)
\(\therefore \frac { da }{ dt } =2\pi r\frac { dr }{ dt } \)
Now approximate rate of increase of radius \(=d r=\frac{d r}{d t} \Delta t=0.05 \mathrm{~cm} / \mathrm{s}\)
Therefore, the approximate rate of increase in area is given by
\( d \mathrm{~A} =\frac{d \mathrm{~A}}{d t}(\Delta t)=2 \pi r\left(\frac{d r}{d t} \Delta t\right) \)
\(=2 \pi(3.2)(0.05)=0.320 \pi \mathrm{cm}^2 / \mathrm{s} \quad(r=3.2 \mathrm{~cm}) \)

\(y=x^{ 2 }-3x.\)
\( \therefore \frac { dy }{ dx } =2x-3.\)
Now \(\frac { dy }{ dx } =0\)
\(\Rightarrow 2x-3=0\)
\(\Rightarrow x=\frac { 3 }{ 2 } \)
Now \(y]_{ x=0 }=0,\quad y]_{ x=\frac { 3 }{ 2 } }=\left( \frac { 9 }{ 4 } \right) -3\left( \frac { 3 }{ 2 } \right) =\frac { 9 }{ 4 } -\frac { 9 }{ 2 } =\frac { -9 }{ 4 } \)
and \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
Hence,absolute minimum value\(=\frac { -9 }{ 4 } \)

\(f^{\prime}(x) =2 \cos 2 x-1, \)
\(f^{\prime}(x) =0 \Rightarrow \cos 2 x=\frac{1}{2}=\cos \frac{\pi}{3} \)
\(\Rightarrow 2 x =\pm \frac{\pi}{3} \Rightarrow x=\pm \frac{\pi}{6} \)
\(f^{\prime \prime}(x) =-4 \sin 2 x \)
\(\text { when } x =\frac{\pi}{6}, f^{\prime \prime}\left(\frac{\pi}{6}\right)=-4 \sin \frac{\pi}{3}<0\)
\(\therefore \text { function attains local maximum at }\)
\(x=\frac{\pi}{6}\)
\(\text { Local maximum value }=f\left(\frac{\pi}{6}\right)\)
\(=\sin \frac{\pi}{3}-\frac{\pi}{6}=\frac{\sqrt{3}}{2}-\frac{\pi}{6}\)
\(\text { when } x=\frac{-\pi}{6} \text { , }\)
\(f^{\prime \prime}\left(\frac{-\pi}{6}\right)=-4 \sin \left(\frac{-\pi}{3}\right)=4 \times \frac{\sqrt{3}}{2}>0\)
\(\therefore \text { function attains local minimum at }\)
\(x=\frac{\pi}{6}\)
\(\text { Local minimum value }\)
\(=f\left(\frac{-\pi}{6}\right)=\sin \left(\frac{-\pi}{3}\right)+\frac{\pi}{6}=-\frac{\sqrt{3}}{2}+\frac{\pi}{6}\)

Let the length and breadth of the tank be x and y m, respectively
Then, volume = 75m³
\(\Rightarrow 3 x y=75 \quad[\because \text { depth of tank }=3 \mathrm{~m}]\)
\(\Rightarrow y=\frac{25}{x} \) 
Let C be the cost of the tank.
Then, \(C^{\prime}=100 x y+50(3 \times 2 x+3 \times 2 y)\) 
= 100xy + 300x + 300y
\( =100 x \times \frac{25}{x}+300 x+300 \times \frac{25}{x} \quad\left[\because y=\frac{25}{x}\right] \)
\(\Rightarrow C =2500+300 x+\frac{7500}{ } \) 
On differentiating twice w.r.t. x, we get
\( \frac{d C}{d x}=300-\frac{7500}{x^{2}} \)
\(\text { and } \ \frac{d^{2} C}{d x^{2}}=\frac{15000}{x^{3}}>0 \) 
For minimum value put \(\frac{d C}{d x}=0\) 
\( \Rightarrow 300-\frac{7500}{x^{2}}=0 \Rightarrow x^{2}=25\)
\(\Rightarrow x=5 \quad[\because \text { length cannot be negative }]\)
\(\text { At } x=5, \frac{d^{2} C}{d x^{2}}=\frac{15000}{5^{3}}=120>0 \) 
So, e is minimum.
When x = 5, then e = 2500 + 1500 + 1500 = 5500
Hence, the cost of least expensive tank is Rs. 5500.

\({x\over a}+{y\over b}=1\)

Radius (say r) of sphere 
\(=\frac { 1 }{ 2 } (diameter)\)
\(=\frac { 1 }{ 2 } ,\frac { 3 }{ 2 } (2x+3)\)
\(=\frac { 3 }{ 4 } (2x+3)\)
Let V be the volume of the sphere 
Then\(V=\frac { 4 }{ 3 } \pi r^{ 3 }=\frac { 4 }{ 3 } \pi \left( \frac { 3 }{ 4 } (2x+3 \right) ^{ 3 }\)
\(=\frac { 9 }{ 16 } \pi (2x+3)^{ 3 }\)
\(\therefore \) Rate of change of volume w,r.t.x
\(=\frac { dv }{ dx } =\frac { 9 }{ 16 } \pi .3(2x+3)^{ 2 }.2\)
\(=\frac { 27 }{ 8 } \pi (2x+3)^{ 2 }\)

Let f be continuous at a critical point c in I. Then
(i) If f ′(x) changes sign from positive to negative as x increases through c, i.e., if f ′(x) > 0 at every point sufficiently close to and to the left of c, and f ′(x) < 0 at every point sufficiently close to and to the right of c, then c is a point of local maxima.
(ii) If f ′(x) changes sign from negative to positive as x increases through c, i.e., if f ′(x) < 0 at every point sufficiently close to and to the left of c, and f ′(x) > 0 at every point sufficiently close to and to the right of c, then c is a point of local minima.
(iii) If f ′(x) does not change sign as x increases through c, then c is neither a point of local maxima nor a point of local minima. Infact, such a point is called point of inflection

Let r be the radius, h be the height and V be the volume of the sand cone
Also given that, \(\frac{d V}{d t}=12 \mathrm{~cm}^{3} / \mathrm{s}, h=\frac{1}{6} r\) 
\(\Rightarrow r=6 h \text { and } h=4 \mathrm{~cm}\) 
Volume of sand cone,
\(V=\frac{1}{3} \pi r^{2} h\) 
\( \Rightarrow V=\frac{1}{3} \pi(6 h)^{2} h \)
\(\Rightarrow V=\frac{1}{3} \pi \times 36 h^{2} \times h=12 \pi h^{3} \) 
On differentiating both sides w.r.t. t, we get
\( \frac{d V}{d t}=12 \pi \times 3 h^{2} \frac{d h}{d t}=36 \pi h^{2} \frac{d h}{d t} \)
\(\Rightarrow 12=36 \pi(4)^{2} \frac{d h}{d t} \)
\(\Rightarrow \frac{d h}{d t}=\frac{12}{36 \pi \times 16}=\frac{1}{48 \pi} \mathrm{cm} / \mathrm{s} \) 
Hence, the height of the sand cone is increasing at the rate of \(\frac{1}{48 \pi} \mathrm{cm} / \mathrm{s}\). when the height is 4 cm,

Here, f(x) = x² + 3, a = 0, b = 2 and nh = b - a = 2
\(\int _{ 0 }^{ 2 }{ \left( { x }^{ 2 }+1 \right) dx } =\int _{ a }^{ b }{ f\left( x \right) dx } \)
\(={ lim }_{ h\rightarrow 0 }\) h[f(a) + f(a + h) + f(a + 2h) + ... + f(a + (n - 1)h)]
\(={ lim }_{ h\rightarrow 0 }\) h[3 + 1²h² + 3 + 2²h² + 3 + ... + (n - 1)²h² + 3]
\(={ lim }_{ h\rightarrow 0 }\) h[3n + h² {1² + 2² + 3² + ... (n - 1)²}]
\(={ lim }_{ h\rightarrow 0 }h\left[ 3n+{ h }^{ 2 }\left\{ \frac { \left( n-1 \right) n\left( 2n-1 \right) }{ 6 } \right\} \right] \)
\(={ lim }_{ h\rightarrow 0 }h\left[ 3nh+\left\{ \frac { \left( nh-h \right) nh\left( 2nh-h \right) }{ 6 } \right\} \right] \)
\(={ lim }_{ h\rightarrow 0 }h\left[ 3\times 2+\left\{ \frac { \left( 2-h \right) 2\left( 4-h \right) }{ 6 } \right\} \right] \)
\(=6+\frac { 16 }{ 6 } ,i.e.,\frac { 26 }{ 3 } \)

Slope of tangent = \(\frac{dy}{dx}=2x-2\)
(i) Tangent parallel to \(2x-y+9=0\)
Slope of line = m₁(say)
ஃ m₁ = 2
ஃ  They are parallel
\(\therefore \frac{dy}{dx}=m_1\)
\(\therefore\ 2x-2=2\)
\(x=2,\ y=7\)
Equation of tangent through (2, 7) and parallel to the given line is \(y-7=2(x-2)\Rightarrow y=2x+3\)
(ii) Tangent perpendicular to \(5y-15x=13\)
Slope of line = m₂(say)
ஃ m₂ = 3
ஃ  They are perpendicular
\(\therefore \frac{dy}{dx}=-\frac{1}{m}\)
\(\therefore (2x-2)\cdot3=-1\)
\(\therefore x=\frac{5}{6},y=\frac{217}{36}\)
Equation of tangent through (\(\frac{5}{6},\frac{217}{36}\)) and perpendicular to the line is \(y=-\frac{217}{36}=-\frac{1}{3}(x-\frac{5}{6})\)
\(\Rightarrow y=\frac{-x}{3}+\frac{227}{36}\)
\(12x+36y=227\)