Given, \(y=x^2-7x+18\)
\(\frac{dy}{dx}=2x-7\)
\(\frac{dy}{dx} \ at\ (3,0)=2(3)-7 \)
\(=6-7=-1\)
\(\frac{dy}{dx}\ at\ (4,0)=2(4)-7\)
\(=8-7=1\)
\(m_1=-1, m_2=1\)
\(m_1\times m_2=-1\times 1=-1\)
Hence tangent are at right angles to each other,

\(y^2 =8x + 3\)
\(2y\frac{dy}{dt}=8\frac{dx}{dt}\)
\(\frac{dy}{dt}=8\frac{dx}{dt}\)
\(2y\cdot8\frac{dx}{dt}=8\frac{dx}{dt}\)
\(\Rightarrow y=\frac{8}{16}=\frac{1}{2}\)
\(y=\frac{1}{2},y^2=8x+3\)
\((\frac{1}{2})^2=8x+3\)
\(\frac{1}{4}-3=8x\)
\(\Rightarrow x=\frac{1-12}{4}\times \frac{1}{8}=-\frac{11}{32}\)
Points  \((-\frac{11}{32},\frac{1}{2})\)

2\(\pi\) x 5 x 0.1=\(\pi\)sq units

(i)\(y=\pm 5.\)  \(\therefore\) points are:\((0,\pm 5)\)
(ii)\(x=\pm2\)  \(\therefore\) points are:\((\pm2,0)\)

Curve is \(\frac{x^2}{9}+\frac{y^2}{9}=1 \Rightarrow \frac{2 x}{9}+\frac{2 y}{4} \cdot y^{\prime}=0\)
\(\Rightarrow \frac{y}{2} y^{\prime}=\frac{-2 x}{9} \Rightarrow y^{\prime}=\frac{-4 x}{9 y}\)
For tangent to be parallel to the y-axis, slope is not defined.
⇒ 9y = 0 ⇒ y = 0
Substituting in curve, we get \(\frac{x^2}{9}+0=1 \Rightarrow x= \pm 3\)
Points are (± 3, 0)

y = x³ + 6 ⇒ y′ = 3x²
\(\left.\left.y^{\prime}\right]_{(-1,5)}=3(-1)^2=3 \text { and } y^{\prime}\right]_{(1,7)}=3(1)^2=3\)
As slope at these points are equal,hence tangents are parallel.

a=-2

\(y^{\prime}=-\frac{3}{x^2}<0, \text { for } x \in R, x \neq 0\). Hence, function is strictly decreasing

m < 0

Marginal revenue (MR) = dR/dx = \(\frac{d}{dx} (3x^2 + 36x + 5)\)
= 6x + 36
when x = 5
Marginal revenue (MR) = 6 x 5 + 36 = 66

1368 Concern for children health and nutrient food for every child.

30.255 Concern for environment;Responsibility for pollution free envirnment     

The given curves are:
\(x=y^{ 2 }\) ..(1)
and xy=k ....(2)
From(2), using (1) we get:
\(y^{ 2 }y=k\Rightarrow y^{ 3 }=k\Rightarrow y=k^{ 1/3 }\)
Putting in (1),\(x=k^{ 1/3 }\)
Thus  the given curves intersect at P(\(k^{ 2/3 },k^{ 1/3 })\)
Diff(1)w.r.t.x,
\(1=2y\frac { dy }{ dx } \Rightarrow \frac { dy }{ dx } =\frac { 1 }{ 2y } \)
\(\therefore m_{ 1 }=\left[ \frac { dy }{ dx } \right] _{ P }=\frac { 1 }{ 2k^{ 1/3 } } \)
Diff(2)w.r.t.x,x,
\(x\frac { dy }{ dx } \)
\(\Rightarrow \frac { dy }{ dx } =-\frac { y }{ x } \)
\( \therefore m_{ 2 }=\left[ \frac { dy }{ dx } \right] _{ P }=\frac { k^{ 1/3 } }{ k^{ 2/3 } } =\frac { 1 }{ k^{ 1/3 } } \)
Hence the given curves cut at right angles at P
\(m_{ 1 },m_{ 2 }=-1\)
\(\Rightarrow \left( \frac { 1 }{ 2k^{ 1/3 } } \right) \left( \frac { 1 }{ k^{ 1/3 } } \right) =-1\)
\(\Rightarrow 2k^{ 1/3 }=1\Rightarrow 8k^{ 2 }=1,\) 
Which is line. 

We have:
\(f(x)=\sqrt { 3 } sinx-cosx-2ax+b\)
\(\therefore f(x)=\sqrt { 3 } cosx-sinx-2a\)
\(=2\left( \frac { \sqrt { 3 } }{ 2 } cosx+\frac { 1 }{ 2 } sinx \right) -2a\)
\(=2cos\left( x-\frac { \pi }{ 6 } \right) -2a=2\left\lfloor cos\left( x-\frac { \pi }{ 6 } \right) -a \right\rfloor \)
\(= \ge 0.\)

Marginal cost is the rate of change of total cost with respect to output.
Now \(C(x)=0.007x^{ 3 }-0.003x^{ 2 }+15x+4000\)
\(\therefore \) Marginal Cost (MC)=\(\frac { dC }{ dx } \) 
=\((0.007)(3x^{ 2 })-0.003(2x)+15\)
When  x = 17, MC = (0.007) (3\(17)^{ 2 })\)- 0.003 (2(17)) + 15
= 6.069 - 0.102 + 15 = 20.967
Hence the reqd marginal cost is Rs. 21 (nearly).

Let 'r' be the radius of the circle.
\(\frac { dr }{ dt } =0.7cm/s\)
Now   \(C=2\pi r\).
\(\therefore \frac { dc }{ dt } =2\pi \left( \frac { dr }{ dt } \right) \)
\(\Rightarrow \frac { dC }{ dt } =2\pi (0.7)=\frac { 14\pi }{ 10 } =\frac { 7\pi }{ 5 } cm/s.\)

The slope of the tangent at x = 2 is given by
\(\left.\left.\frac{d y}{d x}\right]_{x=2}=3 x^{2}-1\right]_{x=2}=11\)

True. Here \(f(x)=x^{ 2 },\) which has no maximum value.[\(x^{ 2 }\) goes on increasing]

We have
\(y=\frac{4 \sin \theta}{(2+\cos \theta)}-\theta \)
\(\therefore \frac{d y}{d x} =\frac{(2+\cos \theta)(4 \cos \theta)-4 \sin \theta(-\sin \theta)}{(2+\cos \theta)^{2}}-1 \)
\(=\frac{8 \cos \theta+4 \cos ^{2} \theta+4 \sin ^{2} \theta}{(2+\cos \theta)^{2}}-1 \)
\(=\frac{8 \cos \theta+4}{(2+\cos \theta)^{2}}-1 \)
\(\text { Now, } \frac{d y}{d x}=0 .\)
\(\Rightarrow \frac{8 \cos \theta+4}{(2+\cos \theta)^{2}}=1\)
\(\Rightarrow 8 \cos \theta+4=4+\cos ^{2} \theta+4 \cos \theta \)
\(\Rightarrow \cos ^{2} \theta-4 \cos \theta=0\)
\(\Rightarrow \cos \theta(\cos \theta-4)=0 \)
\(\Rightarrow \cos \theta=0 \text { or } \cos \theta=4\)
\(\text { Since } \cos \theta \neq 4, \cos \theta=0\)
\(\cos \theta=0 \Rightarrow \theta=\frac{\pi}{2}\)
\(\text { Now, }\frac{d y}{d x}=\frac{8 \cos \theta+4-\left(4+\cos ^{2} \theta+4 \cos \theta\right)}{(2+\cos \theta)^{2}}=\frac{4 \cos \theta-\cos ^{2} \theta}{(2+\cos \theta)^{2}}=\frac{\cos \theta(4-\cos \theta)}{(2+\cos \theta)^{2}}\)
\(\text { In interval }\left(0, \frac{\pi}{2}\right), \text { we have } \cos \theta>0 . \text { Also, } 4>\cos \theta \Rightarrow 4-\cos \theta>0 \text { . }\)
\(\therefore \cos \theta(4-\cos \theta)>0 \text { and also }(2+\cos \theta)^{2}>0 \)
\(\Rightarrow \frac{\cos \theta(4-\cos \theta)}{(2+\cos \theta)^{2}}>0 \)
\(\Rightarrow \frac{d y}{d x}>0 \)
 Therefore, y is strictly increasing in interval \( \left(0, \frac{\pi}{2}\right) \text { . }\)
Also, the given function is continuous at  \( x=0 \text { and } x=\frac{\pi}{2} \text { . }\)
Hence, y is increasing in interval  \( \left[0, \frac{\pi}{2}\right] \text { . }\)

Let f be twice differentiable at c. Then
(i) x = c is a point of local maxima if f ′(c) = 0 and f ″(c) < 0
The value f (c) is local maximum value of f .
(ii) x = c is a point of local minima if f '(c) = 0 and f ″(c) > 0
In this case, f (c) is local minimum value of f .
(iii) The test fails if f ′(c) = 0 and f ″(c) = 0.
In this case, we go back to the first derivative test and find whether c is a point of local maxima, local minima or a point of inflexion.

Since marginal cost is the rate of change of total cost with respect to the output, we have
Marginal \(\operatorname{cost}(\mathrm{MC})=\frac{d C}{d x}=0.005\left(3 x^{2}\right)-0.02(2 x)+30\)
When x = 3, MC = 0.015(3² ) − 0.04(3) + 30
= 0.135 – 0.12 + 30 = 30.015
Hence, the required marginal cost is Rs. 30.02 (nearly).

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
\(=\frac { dv }{ dx } =\frac { 9 }{ 16 } \pi .3(2x+3)^{ 2 }.2\)
\(=\frac { 27 }{ 8 } \pi (2x+3)^{ 2 }\)

\(x^{x^2-3}+(x-3)^{x^2}, \text { for } x>3\)
Let \(y=x^{x^2-3}+(x-3)^{x^2}\)
And let \(u=x^{x^2-3}, v=(x-3)^{x^2}\)
\(\boldsymbol{y}=\boldsymbol{u}+\boldsymbol{v}\)
Differentiating both sides w.r.t. x
\(\frac{d y}{d x}=\frac{d(u+v)}{d x} \)
\( \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\)
Differentiating both sides with respect to x, we obtain
\( \frac{1}{u} \frac{d u}{d x}=\log x \cdot \frac{d}{d x}\left(x^2-3\right)+\left(x^2-3\right) \cdot \frac{d}{d x}(\log x) \)
\( \Rightarrow \frac{1}{u} \frac{d u}{d x}=\log x \cdot 2 x+\left(x^2-3\right) \cdot \frac{1}{x} \)
\( \Rightarrow \frac{d u}{d x}=x^{x^2-1}\left[\frac{x^2-3}{x}+2 x \log x\right]\)
Now, v=(x-3)^x
Taking logarithm on both the sides, we obtain
\(\log v =\log (x-3)^{x^2} =x^2 \log (x-3)\)
Differentiating both sides with respect to x, we obtain
\(\frac{1}{v} \frac{d v}{d x}=\log (x-3) \cdot \frac{d}{d x}\left(x^2\right)+\left(x^2\right) \cdot \frac{d}{d x}[\log (x-3)] \)
\(\Rightarrow \frac{1}{v} \frac{d v}{d x}=\log (x-3) \cdot 2 x+x^2 \cdot \frac{1}{x-3} \cdot \frac{d}{d x}(x-3) \)
\( \Rightarrow \frac{d v}{d x}=v\left[2 x \log (x-3)+\frac{x^2}{x-3} \cdot 1\right] \)
\( \Rightarrow \frac{d v}{d x}=(x-3)^{x^2}\left[\frac{x^2}{x-3}+2 x \log (x-3)\right]\)
From (1), (2), and (3), we obtain

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 \(y=-\frac{217}{36}=-\frac{1}{3}(x-\frac{5}{6})\)
\(\Rightarrow y=\frac{-x}{3}+\frac{227}{36}\)
\(12x+36y=227\)