1.0349

\(R_{1}=R_{2}=\frac{C}{2}\)

-1

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

Consider function f(x) = x³ - 3x² + 6x - 100
f'(x) 3x²- 6x + 6 = 3(x² - 2x + 2)
3[(x 1)²+1] > 0,for all x ∈ R
Hence, f is increasing on R.

\(f^{\prime}(x)=\frac{1}{\cos x} \cdot(-\sin x)=-\tan x, \tan x>0 \text { for }\left(0, \frac{\pi}{2}\right)\)
f′(x) < 0. Hence, function is strictly decreasing

f'(x) 3x²- 6x + 3 = 3(x² - 2x + 1)
= 3 (x - 1)² > 0
Hence, function is increasing in R.

30.255 Concern for environment;Responsibility for pollution free envirnment     

0.0608

We have,\((3.968)^{3 / 2}\) 
Let \(y=f(x)=x^{3 / 2}\) 
On differentiating both sides W.r.t. x , we get
\(f^{\prime}(x)=\frac{3}{2} x^{1 / 2}\) 
Let x = 4 and \(x+\Delta x=3.968\) 
Then, \(\Delta x=-0.032\) 
Now, \(f(x+\Delta x) \approx f(x)+f^{\prime}(x) \cdot \Delta x\) 
\(\therefore \ (x+\Delta x)^{3 / 2} \approx(x)^{3 / 2}+\frac{3}{2}(x)^{1 / 2}(-0.032) \)
\(\Rightarrow(4-0.032)^{3 / 2} \approx(4)^{3 / 2}+\frac{3}{2}(4)^{1 / 2}(-0.032) \)
\(\Rightarrow (3.968)^{3 / 2} \approx 8+\frac{3}{2} \cdot 2 \cdot(-0.032)=8-0.096 \)
\(\Rightarrow (3.968)^{3 / 2} \approx 7.904 \) 
Hence, the approximate value of \((3.968)^{3 / 2} \text { is } 7.904\)

We have
\(f(x)=\log \sin x \)
\(\therefore f^{\prime}(x)=\frac{1}{\sin x} \cos x=\cot x \)
In interval \( \left(0, \frac{\pi}{2}\right), f^{\prime}(x)=\cot x>0\)
\(\therefore f \text { is strictly increasing in }\left(0, \frac{\pi}{2}\right) \text { . }\)
In interval \( \left(\frac{\pi}{2}, \pi\right), f^{\prime}(x)=\cot x<0\)
\(\therefore f \text { is strictly decreasing in }\left(\frac{\pi}{2}, \pi\right) \text { . }\)

Let a piece of length l be cut from the given wire to make a square.
Then, the other piece of wire to be made into a circle is of length (28 − l) m.
Now, side of square =.\(\frac{l}{4}\)
Let r be the radius of the circle. Then, \(2 \pi r=28-l \Rightarrow r=\frac{1}{2 \pi}(28-l)\)
The combined areas of the square and the circle (A) is given by,
\(A=(\text { side of the square })^{2}+\pi r^{2} \)
\(=\frac{l^{2}}{16}+\pi\left[\frac{1}{2 \pi}(28-l)\right]^{2} \)
\(=\frac{l^{2}}{16}+\frac{1}{4 \pi}(28-l)^{2} \)
\(\therefore \frac{d A}{d l}=\frac{2 l}{16}+\frac{2}{4 \pi}(28-l)(-1)=\frac{l}{8}-\frac{1}{2 \pi}(28-l) \)
\(\frac{d^{2} A}{d l^{2}}=\frac{1}{8}+\frac{1}{2 \pi}>0 \)
\(\text { Now }, \frac{d A}{d l}=0 \Rightarrow \frac{l}{8}-\frac{1}{2 \pi}(28-l)=0 \)
\(\Rightarrow \frac{\pi l-4(28-l)}{8 \pi}=0 \)
\(\Rightarrow(\pi+4) l-112=0 \)
\(\Rightarrow l=\frac{112}{\pi+4} \)
Thus, when \(l=\frac{112}{\pi+4}, \frac{d^{2} \mathrm{~A}}{d l^{2}}>0 .\)
By second derivative test, the area (A) is the minimum when.\(l=\frac{112}{\pi+4}\)
Hence, the combined area is the minimum when the length of the wire in making the square is \(\frac{112}{\pi+4}\)m while the length of the wire in making the circle is
28 − 112π + 4 = 28π + 4m

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

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,

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 \( \left(\frac{\pi}{4}, \frac{5 \pi}{4}\right)\)

\(r=l \sin \frac{\pi}{6}=\frac{l}{2} \text { and } h=l \cos \frac{\pi}{6}=\frac{l \sqrt{3}}{2}\)
 
\( V=\frac{\sqrt{3}}{24} \pi l^{3} \\ \Rightarrow \quad \frac{d V}{d t}=-1=\frac{\sqrt{3}}{8} \pi l^{2} \frac{d l}{d t} \)

Let the sides of rt. ΔABC be x and y.
\(\therefore\ x^2+y^2=4r^2\)
and A = area of \(\Delta=\frac{1}{2}xy\)
Let, \(S=A^2 =\frac{1}{4}x^2y^2\)
\(=\frac{1}{4}x^2(4r^2-x^2)\)
\(=\frac{1}{4}(4r^2x^2-x^4)\)
\(\therefore\ \frac{dS}{dx}=\frac{1}{4}[8r^2x-4x^3]\)
\(\Rightarrow \ \frac{dS}{dx}=0\Rightarrow x^2=2r^2 \ or\ x=\sqrt{2r}\)
\(y^2=4r^2-2r^2=2r^2\Rightarrow y=\sqrt2r\)
i.e. x = y and  \(\frac{d^2S}{dx^2}=(2r^2-3x^2)=2r^2-6r^2<0\)
⇒ Area is maximum when Δ is isosceles 

\(3x^2-y^2=8\)
Diff. w.r.t. 'x'
\(6x-2y\frac{dy}{dx}=0\)
\(\Rightarrow \frac{dy}{dx}=\frac{3x}{y}\)
\((\frac{dy}{dx})_{x_1.y_1}=\frac{3x_1}{y_1}\)
ஃ The equation of tangent at \((x_1,y_1)\) is
\(y-y_1=\frac{3x_1}{y_1}(x-x_1)\)
Tangent passes through the point \((\frac{4}{3},0)\)
\(\therefore \ 0-y_1=\frac{3x_1}{y_1}(\frac{4}{3}-x_1)\)
\(\Rightarrow -y_1^2=3x_1(\frac{4}{3}-x_1)\)
\(\Rightarrow -y_1^2=4x_1-3x_1^2\)
Using equation (i), \(8-3x_1^2=4x_1-3x_1^2\)
\(\Rightarrow x_1=2\)
\(\therefore\ y=3x-4\)
\(\Rightarrow 12-y_1^2=8\)
\(\Rightarrow y_1^2=4 \)
\(\Rightarrow y_1=\pm2\)
ஃ  The points are (2, 2) and (2, - 2).
ஃ The equation of tangent at (2, 2) is :
\(y - 2 = 3(x - 2)\)
\(\Rightarrow y =3x-4\)
The equation of tangent at (2, - 2) is :
\(y+2 =- 3(x - 2)\)
\(\Rightarrow y =-3x+4\) 

As y’ = 3x² – 3, for a point of absolute maximum or minimum y’=0 ⇒ x = ± 1.
y]_x=0 = 2,
y]_x=1 = 1 – 3 + 2 = 0,
y]_x=-1 = -1 +3+ 2 = 4,
y]_x=2 = 8 – 6 + 2 = 4

As \(\frac { dy }{ dx } ={ 2e }^{ 2x }\Rightarrow \frac { dy }{ dx } \)]_(0,1) = 2e² = 2
Equation of tangent is y-1=2(x-0)
⇒ 2x-y+1=0, if it meets the x-axis, then y = 0 ⇒ x = -\(\frac12\), point is (\(-\frac12\), 0)

As 2y \(\frac{dy}{dx}\) = 1
⇒\(\frac{dy}{dx}=\frac{1}{2y}\) (slope of tangent)
⇒ \(\frac{1}{2y}\) = tan(-45^o)
⇒ y = \(-\frac{1}{2}\) ⇒ x = \(\frac14\)
∴ Point as \(\left( \frac { 1 }{ 4 } ,-\frac { 1 }{ 2 } \right) \)

As \(\frac{dy}{dt}\) = 2x.\(\frac{dx}{dt}\)
⇒ 6.\(\frac{dx}{dt}\) = 2x.\(\frac{dx}{dt}\) ⇒ x = 3
From curve, y = 9. Point is (3, 9)