We have, \(f(x)=a x+\frac{b}{x}(a>0, b>0, x>0)\)
\(\therefore \quad f^{\prime}(x)=a-\frac{b}{x^2}\) and \(f^{\prime \prime}(x)=\frac{2 b}{x^3}\)
For maxima and minima of f(x),
\(f^{\prime}(x) =0 \)
\(\Rightarrow  a-\frac{b}{x^2} =0\)
\(\Rightarrow  x^2  =\frac{b}{a} \)
\(\Rightarrow  x =\sqrt{\frac{b}{a}} \quad[\because x>0(\text { given })]\)
Again, \(f^{\prime \prime}\left(\sqrt{\frac{b}{a}}\right)=\frac{2 b}{\left(\frac{b}{a}\right)^{3 / 2}}=\frac{2 a^{3 / 2}}{\sqrt{b}}>0\) \([\because a>0, b>0]\)
So, f(x) has least value at \(x=\sqrt{\frac{b}{a}}\)
\(\therefore f_{\min }(x) =f\left(\sqrt{\frac{b}{a}}\right)=a \sqrt{\frac{b}{a}}+\frac{b}{\sqrt{\frac{b}{a}}} \)
\( =\sqrt{a b}+\sqrt{a b}=2 \sqrt{a b}\)

Given, $f(x)=x^3+x, \forall x \in R$
On differentiating w.r.t. x, we get
\( f^{\prime}(x)=3 x^2+1 \)
\(\because x^2>0, \forall x \in R\)
\(\therefore f^{\prime}(x)\) can never be zero or undefined, \(\forall x \in R\).
\(\Rightarrow \quad f^{\prime}(x)>0\)
Hence, no critical point exists.

Given, \(f(x)=\frac{1}{4 x^2+2 x+1}\)
On differentiating w.r.t. x, we get
\(f^{\prime}(x)=\frac{0-1(8 x+2)}{\left(4 x^2+2 x+1\right)^2}\)
For maxima or minima, we put
\( f^{\prime}(x) =0\)
\(\Rightarrow 8 x+2 =0\)
\(\Rightarrow x  =-\frac{1}{4}\)
Again, differentiating Eq. (i) w.r.t. x, we get
\( -8\left(4 x^2+2 x+1\right)^2+(8 x+2) \times 2\)
\(f^{\prime \prime}(x)=\frac{\left(4 x^2+2 x+1\right)(8 x+2)}{\left(4 x^2+2 x+1\right)^4}\)
At \(x=\frac{-1}{4}, f^{\prime \prime}\left(\frac{-1}{4}\right)<0\)
\(\Rightarrow f(x)\) is maximum at \(x=\frac{-1}{4}\)
Therefore Maximum value of f(x) is
\(f\left(\frac{-1}{4}\right)=\frac{1}{4\left(\frac{-1}{4}\right)^2+2\left(\frac{-1}{4}\right)+1}=\frac{1}{\frac{1}{4}+\frac{1}{2}}=\frac{4}{3}\)

We have, \(C(x)=0.005 x^3-0.02 x^2+30 x+5000\) Clearly, the marginal cost, MC \((x)=\frac{d}{d x} C(x)\)
\(=\frac{d}{d x}\left(0.005 x^3-0.02 x^2+30 x+5000\right) \)
\(=0.005 \times 3 x^2-0.02 \times 2 x+30+0 \)
\(=0015 x^2-0.04 x+30\)
Now, marginal cost when 3 units are produced
\(=\mathrm{MC}(3)=0.015(9)-0.04(3)+30\)
=0.135-0.12+30=30.015

Given, f(x) = xe^x
On differentiating w.r.t. x, we get
\(f^{\prime}(x)=x e^x+e^x=e^x(x+1)\)
For f(x) to be increasing, we have
Hence, the required interval, where f(x) increases is [-1, \(\infty\)).
\( f^{\prime}(x)=e^x(x+1) \geq 0\)
\( \Rightarrow \quad x \geq-1 \text { as } e^x>0, \forall x \in R\)

Let r be the radius of the circle, A be the area and C be the circumference at any time t.
Then, A = \(\pi\)r² and C = 2 \(\pi\)r
It is given that \(\frac{d A}{d t}=2 \mathrm{~cm}^2 / \mathrm{s}\)
\(\begin{aligned} \Rightarrow & \frac{d}{d t}\left(\pi r^2\right) =2 \end{aligned}\)
\(\begin{aligned} \Rightarrow & 2 \pi r \cdot \frac{d r}{d t} =2 \end{aligned}\)
\(\begin{aligned} \Rightarrow & \frac{d r}{d t}=\frac{1}{\pi r} \end{aligned}\)
Now,  C = 2 \(\pi\)r
\(\begin{aligned} \Rightarrow \quad \frac{d C}{d t} & =2 \pi \cdot \frac{d r}{d t} \end{aligned}\)
\(\begin{aligned} =2 \pi \times \frac{1}{\pi r} \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \frac{d C}{d t} & =\frac{2}{r} \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad\left(\frac{d C}{d t}\right)_{r=5} & =\frac{2}{5} \mathrm{~cm} / \mathrm{s} \end{aligned}\)

Equation of tangent is \(8 x+3 \sqrt{5} y=36\) meet at X-axis at \(\left(\frac{9}{2}, 0\right)\) and equation of normal is \(24 y-9 \sqrt{5} x=14 \sqrt{5}\) meet X-axis at \(\left(-\frac{14}{9}, 0\right)\)

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.

1.8πcm/s

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, \(m_1=m_2\)
\(\therefore 2x-a=2\)
\(\Rightarrow 2(-1)-a=2\)
\(\Rightarrow -2-2=a\)
\(\Rightarrow a=-4\)

Let a and b be the sides of right angled triangle and c be the hypotenuse.

From \(\triangle A B C\), we have
\(c^2=a^2+b^2\)
Area of \(\triangle A B C,(A)=\frac{1}{2} a \cdot b=\frac{1}{2} a \sqrt{c^2-a^2}\)
\(\left[\because b=\sqrt{c^2-a^2}\right]\)
On differentiating both sides w.r.t. a, we get
\(\frac{d A}{d a}  =\frac{1}{2} \cdot 1 \cdot \sqrt{c^2-a^2}+\frac{1}{2} \cdot a \cdot \frac{1}{2} \cdot \frac{(-2 a)}{\sqrt{c^2-a^2}} =\frac{1}{2}\left(\sqrt{c^2-a^2}-\frac{a^2}{\sqrt{c^2-a^2}}\right)\)
For maxima or minima, put \(\frac{d A}{d a}=0\)
\( \Rightarrow \frac{1}{2}\left(\sqrt{c^2-a^2}-\frac{a^2}{\sqrt{c^2-a^2}}\right)=0 \Rightarrow c^2-a^2-a^2=0 \)
\( \Rightarrow \quad c^2=2 a^2 \Rightarrow a=\frac{c}{\sqrt{2}}\)
\(\text { Now, } \quad \frac{d^2 A}{d a^2}=\frac{1}{2}\left[\frac{-a}{\sqrt{c^2-a^2}}-\frac{a^3}{\left(c^2-a^2\right)^{3 / 2}}\right]\)
\(=-\frac{1}{2} a\left[\frac{c^2-a^2+a^2}{\left(c^2-a^2\right)^{3 / 2}}\right] \)
\(=-\frac{1}{2} \frac{c^2 a}{\left(c^2-a^2\right)^{3 / 2}}<0\)
\(\therefore\) Area of \(\triangle A B C\) is maximum and
\(b=\sqrt{c^2-a^2}=\sqrt{2 a^2-a^2}=a\)
Hence, the triangle is isosceles.
Hence proved.

Let we have a sphere of radius R and a cylinder is inscribed in it. EB = r be the radius of the cylinder and BC = h be the height of cylinder.
In right angled \(\Delta A B C\) we have
\((A C)^{2}=(A B)^{2}+(B C)^{2}\) [using Pythagoras theorem]
\(\Rightarrow(2 R)^{2}=(2 r)^{2}+h^{2} \Rightarrow 4 R^{2}=4 r^{2}+h^{2} \)
\(\Rightarrow \cdot r^{2}=\frac{1}{4}\left(4 R^{2}-h^{2}\right) \) 
volume of cylinder \(V=\pi r^{-2} h\) 
\(\Rightarrow V=\pi\left[\frac{1}{4}\left(4 R^{2}-h^{2}\right)\right] h\) 
\(\Rightarrow V=\frac{\pi}{4}\left(4 R^{2} h-h^{3}\right)\)   ...(ii)
On differentiating V w.r.t. II, we get
\(\frac{d V}{d h}=\frac{\pi}{4}\left(4 R^{2}-3 h^{2}\right)\) 
For maxima or minima put \(\frac{d V}{d h}=0\) 
\(\Rightarrow \frac{\pi}{4}\left(4 R^{2}-3 h^{2}\right)=0 \Rightarrow 4 R^{2}=3 h^{2}\) 
\(\Rightarrow h=\frac{2 R}{\sqrt{3}} \ [\because \text { height cannot be negative }] \) 
On differentiating both sides of Eq. (iii) w.r.t. h, we get
\( \frac{d^{2} V}{d h^{2}} =-\frac{3}{2} \pi h \)
\(\text { At } h=\frac{2 R}{\sqrt{3}}, \frac{d^{2} V}{d h^{2}} =-\frac{3}{2} \pi \times \frac{2 R}{\sqrt{3}}=-\sqrt{3} \pi R<0[\because R>0] \) 
\(\therefore\) Volume of cylinder is maximum at \(h=\frac{2 R}{\sqrt{3}}\) 
From Eq. (i), we get \(r^{2}=\frac{1}{4}\left(4 R^{2}-\frac{4 R^{2}}{3}\right)=\frac{2 R^{2}}{3}\) 
\(\therefore\) Volume of largest cylinder inscribed in a sphere,
\( V =\pi\left(\frac{2 R^{2}}{3}\right) \times \frac{2 R}{\sqrt{3}} \)
\(=\frac{1}{\sqrt{3}} \times\left(\frac{4}{3} \pi R^{3}\right) \)
\(=\frac{1}{\sqrt{3}} \times \text { Volume of sphere } \) 
Hence, height of the cylinder of maximum volume is \(\frac{2 R}{\sqrt{3}}\) 
and volume of largest cylinder inscribed in sphere is \(\frac{1}{\sqrt{3}}\)  times volume of sphere. 

Let wire is cut at x m and made into a rectangle,
whose breadth is a and length 2a.
For rectangle,
\(\therefore 2(a+2 a)=x \Rightarrow 6 a=x \)
\(\Rightarrow a=\frac{x}{6} \mathrm{~m} \)
\(\therefore \text { Area of rectangle }=a(2 a)=2 a^{2}=2\left(\frac{x}{6}\right)^{2}=\frac{x^{2}}{18} \)
For square,
\(4 \times \text { side }=(34-x) \Rightarrow \text { side }=\frac{34-x}{4} \mathrm{~m} \)
\(\therefore \text { Area of square }=\frac{1}{16}(34-x)^{2}\)
\(\text { Combined area }(A)=\frac{x^{2}}{18}+\frac{1}{16}(34-x)^{2} \)
\(\frac{d A}{d x} =\frac{2 x}{18}+\frac{2}{16}(34-x) \cdot(-1) \)
\(=\frac{x}{9}-\left(\frac{34-x}{8}\right)=\frac{8 x-306+9 x}{72} \)
\(\frac{d A}{d x} =\frac{17 x-306}{72}\)
\(\text { For minimum area, } \frac{d A}{d x}=0\)
\(\Rightarrow 17 x-306=0 \)
\(\Rightarrow x=\frac{306}{17}=18 \)
\(\frac{d^{2} A}{d x^{2}} =\frac{17}{72} \)
\(\left.\frac{d^{2} A}{d x^{2}}\right|_{x=18} =\frac{17}{72}>0\)
For x = 18, A is minimum. Hence, wire must be cut at 18 m to made into a rectangle and remaining 16 m piece to made into a square for a minimum combined area.

p=4

Given, \(\frac{d h}{d t}=100 \mathrm{~km} / \mathrm{h}, \text { then find } \frac{d A}{d t}\) 
Height of the plane after 3 rnin \(100 \times \frac{3}{60}=5 \mathrm{~km}\) 
\(\frac{200 \pi r^{3}}{(r+5)^{2}}\)

Let radius of cone be x and its height be h.
\(\therefore\) OD = (h - r)

Volume of cone (V)
\(=\frac { 1 }{ 3 } \pi { x }^{ 2 }h\) ...(i)
In \(\Delta OCD,\quad { x }^{ 2 }+({ h-r) }^{ 2 }={ r }^{ 2 }or\quad { x }^{ 2 }={ r }^{ 2 }-{ (h-r) }^{ 2 }\)
\(\therefore V=\frac { 1 }{ 3 } \pi h\{ { r }^{ 2 }-(h-r{ ) }^{ 2 }\} \)
\(=\frac { 1 }{ 3 } \pi (-{ h }^{ 3 }+{ 2h }^{ 2 }r)\)
\(\Rightarrow \frac { dV }{ dh } =\frac { \pi }{ 3 } (-3{ h }^{ 2 }+4hr)\)
\(\therefore \quad \frac { dV }{ dh } =0\Rightarrow h=\frac { 4r }{ 3 } \)
\(\frac { { d }^{ 2 }V }{ { dh }^{ 2 } } =\frac { \pi }{ 3 } (-6h+4r)\)
\(=\frac { \pi }{ 3 } \left( -6\left( \frac { 4r }{ 3 } \right) +4r \right) \)
\(=-\frac { 4\pi r }{ 3 } <0\)
\(\therefore \ at\quad h=\frac { 4r }{ 3 } \), Volume is maximum
Maximum volume
\(=\frac { 1 }{ 3 } \pi .\left\{ -{ \left( \frac { 4r }{ 3 } \right) }^{ 3 }+2{ \left( \frac { 4r }{ 3 } \right) }^{ 2 }r \right\} \)
\(=\frac { 8 }{ 27 } .\left( \frac { 4 }{ 3 } \pi { r }^{ 3 } \right) \)
\(=\frac { 8 }{ 27 } \) (volume of sphere)

We have,\(y^{2}=12 x \Rightarrow 2 y \frac{d y}{d x}=12\) 
\(\Rightarrow \quad \frac{d y}{d x}=\frac{6}{y}\) 
Let x + y = K be normal to y² = 12x at point P(x₁, y₁) then
\(\begin{array}{l} \left(\frac{-1}{d y / d x}\right)_{\text {at } p}=(\text { Slope of the line } x+y=K) \\ \Rightarrow \quad-\frac{y_{1}}{6}=-1 \Rightarrow y_{1}=6 \end{array}\) 
Since, (x₁ y_I) lies on l = 12x, therefore
\( y_{1}^{2}=12 x_{1} \Rightarrow 12 x_{1}=36 \\ \Rightarrow x_{1}=3 \) 
Also P( x₁ y₁) lies on x + y = K, therefore 
\(x_{1}+y_{1}=K \Rightarrow K=9\) 

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 \(\left( \frac { 1 }{ 4 } ,-\frac { 1 }{ 2 } \right) \)

(i) (d) : In order to make least expensive water tank, Nitin need to minimize its cost.
(ii) (d) : Let 1ft be the length and h ft be the height of the tank. Since breadth is equal to 5 ft. (Given)
\(\therefore\) Two sides will be 5h sq. feet and two sides will be
1h sq. feet. So, the total area of the sides is (10 h + 2 1h)ft²
Cost of the sides is Rs.10 per sq. foot. So, the cost to build the sides is (10h + 21h) x 10 = Rs.(100h + 20lh)
Also, cost of base = (5l) x 20 = Rs. 100 l
\(\therefore\) Total cost of the tank in Rs. is given by
c = 100 h + 20 I h + 100 l
Since, volume of tank = 80ft³
\(\therefore \quad 5 l h=80 \mathrm{ft}^{3} \quad \therefore l=\frac{80}{5 h}=\frac{16}{h}\) 
\(\therefore \quad c(h)=100 h+20\left(\frac{16}{h}\right) h+100\left(\frac{16}{h}\right)\) 
\(=100 h+320+\frac{1600}{h}\) 
(iii) (b) : Since, all side lengths must be positive
\(\therefore \quad h>0\) and \(\frac{16}{h}>0\) 
Since, \(\frac{16}{h}>0, \text { whenever } h>0\) 
\(\therefore \text { Range of } h \text { is }(0, \infty)\) 
(iv) (a) : To minimize cost, \(\frac{d c}{d h}=0\) 
\(\Rightarrow \quad 100-\frac{1600}{h^{2}}=0\) 
\(\Rightarrow 100 h^{2}=1600 \Rightarrow h^{2}=16 \Rightarrow h=\pm 4\) 
\(\Rightarrow h=4\)  [\(\therefore\) height can not be negative]
(v) (c) : Cost of least expensive tank is given by
\(c(4)=400+320+\frac{1600}{4}\) 
= 720 + 400 = Rs. 1120

(i) (d) : Given, r cm is the radius and h cm is the height of required cylindrical can
Given that, volume = 3 l= 3000 cm³ \(\left(\because 1 l=1000 \mathrm{~cm}^{3}\right)\) 
\(\Rightarrow \pi r^{2} h=3000 \Rightarrow h=\frac{3000}{\pi r^{2}}\) 
Now, the surface area, as a function of r is given by
\(S(r)=2 \pi r^{2}+2 \pi r h=2 \pi r^{2}+2 \pi r\left(\frac{3000}{\pi r^{2}}\right)\) 
\(=2 \pi r^{2}+\frac{6000}{r}\) 
(ii) (c) : Now, \(S(r)=2 \pi r^{2}+\frac{6000}{r}\) 
\(\Rightarrow S^{\prime}(r)=4 \pi r-\frac{6000}{r^{2}}\) 
To find criti£al points, put S'(r) = 0
\(\Rightarrow \frac{4 \pi r^{3}-6000}{r^{2}}=0\) 
\(\Rightarrow r^{3}=\frac{6000}{4 \pi} \Rightarrow r=\left(\frac{1500}{\pi}\right)^{1 / 3}\) 
Also, \(\left.S^{\prime \prime}(r)\right|_{r=} \sqrt[3]{\frac{1500}{\pi}}=4 \pi+\frac{12000 \times \pi}{1500}\) 
\(=4 \pi+8 \pi=12 \pi>0\) 
Thus, the critical point is the point of minima.
(iii) (b) : The cost of material for the tin can is minimized when \(r=\sqrt[3]{\frac{1500}{\pi}} \mathrm{cm}\) and the height is  \(\frac{3000}{\pi\left(\sqrt[3]{\frac{1500}{\pi}}\right)^{2}}=2 \sqrt[3]{\frac{1500}{\pi}} \mathrm{cm} .\) .
(iv) (a) : We have,minimum surface area = \(\frac{2 \pi r^{3}+6000}{r}\) .
\(=\frac{2 \pi \cdot \frac{1500}{\pi}+6000}{\sqrt[3]{\frac{1500}{\pi}}}=\frac{9000}{7.8}=1153.84 \mathrm{~cm}^{2}\) 
Cost of 1 m² material = Rs.100
\(\therefore \ \text { Cost of } 1 \mathrm{~cm}^{2} \text { material }=Rs. \frac{1}{100}\) 
\(\therefore \ \text { Minimum cost }=Rs. \frac{1153.84}{100}=Rs. 11.538\) 
(v) (c) : To minimize the cost we need to minimize the total surface area.

(i) (c) : If P is the rent price per apartment and N is the number of rented apartment, the profit is given by
NP - 500 N = N(P- 500)
[\(\therefore\)Rs. 500/month is the maintenance charges for each occupied unit]
(ii) (c) : Now, if x be the number of non-rented apartments, then N= 50 - x and P = 10000 + 250 x Thus, profit = N(P- 500) = (50 - x ) (10000 + 250x - 500 
= (50 - x) (9500 + 250 x) = 250(50 - x) (38 + x)
(iii) (b) : Clearly, if P = 10500, then 
\(10500=10000+250 x \Rightarrow x=2 \Rightarrow N=48\) 
(iv) (a) : Also, if P = 11000, then
\(11000=10000+250 x \Rightarrow x=4\) and so profit
(v) (b) : We have, P(x) = 250(50 - x) (38 + x)
Now, P'(x) = 250[50 - x - (38 + x)] = 250[12 - 2x]
For maxima/minima, put P'(x) = 0
\(\Rightarrow 12-2 x=0 \Rightarrow x=6\) 
Thus, price per apartment is, P = 10000 + 1500 = 11500
Hence, the rent that maximizes the profit is Rs. 11500.

(i) (a) : Let x be the charges per bike per day and n be the number of bikes rented per day.
R(x) = n x x = (2000 - lOx) x = -10x² + 2000x
(ii) (b) : We have, R(x) = 2000x - 10x²
\(\Rightarrow R^{\prime}(x)=2000-20 x\) 
For R(x) to be maximum or minimum, R'(x) = 0
\(\Rightarrow 2000-20 x=0 \Rightarrow x=100\) 
Also,\(R^{\prime \prime}(x)=-20<0\) 
Thus, R(x) is maximum at x = 100
(iii) (c) : If company charge ~ 200 or more, they will not rent any bike. Therefore, revenue collected by him will be zero.
(iv) (c) : If x = 105, number of bikes rented per day is given by
n = 2000 - 10 x 105 = 950
(v) (d) : At x = 100, R(x) is maximum
\(\therefore\) Maximum revenue = R(100)
= -10(100)² + 2000(100) = Rs. 1,00,000

(i) (a) : Let x be the number of extra days after 1^st July.
\(\therefore\) Price = Rs.(300 - 3Xx) = Rs.(300 - 3x)
Quantity = 80 quintals + x(1 quintal per day)
= (80 + x) quintals
(ii) (b) : R(x) = Quantity x Price
= (80 + x) (300 - 3x) = 24000 - 240x + 300x -3x²
= 24000 + 60x - 3x²
(iii) (a) : We have, R(x) = 24000 + 60x - 3x²
\(\begin{equation} \Rightarrow R^{\prime}(x)=60-6 x \Rightarrow R^{\prime \prime}(x)=-6 \end{equation}\)
For R(x) to be maximum, R'(x) = 0 and R"(x) < 0
\(\begin{equation} \Rightarrow 60-6 x=0 \Rightarrow x=10 \end{equation}\)
(iv) (a) : Shyams father will attain maximum revenue after 10 days.
So, he should harvest the onions after 10 days of 1^st July i.e., on 11^th July.
(v) (c) : Maximum revenue is collected by Shyams father when x = 10
\(\therefore\) Maximum revenue = R(10)
= 24000 + 60(10) - 3(10)2 = 24000 + 600 - 300 = 24300

(c) Assertion is correct, Reason is incorrect

(a) Assertion is correct, Reason is correct; Reason is a correct explanation for assertion.

(b) Assertion is correct, Reason is correct; Reason is not a correct explanation for Assertion

(a) Assertion is correct, Reason is correct; Reason is a correct explanation for assertion.

(a) Assertion is correct, Reason is correct; Reason is a correct explanation for assertion.