Here \({ r }^{ 2 }=25\Rightarrow r=5\) 
Parametric equations are \(x=rcos\theta ,y=rsin\theta \) 
\(\Rightarrow x=5cos\theta ,y=5sin\theta ,0\le \theta \le 2\pi \)

\(\cos 20^{\circ}+\cos 100^{\circ}+\cos 140^{\circ}\)
\(\cos 20^{\circ}+\left(\cos 100^{\circ}+\cos 140^{\circ}\right) =\cos 20^{\circ}+2 \cos \left(\frac{100^{\circ}+140^{\circ}}{2}\right) \cos \left(\frac{100^{\circ}-140^{\circ}}{2}\right) \)
\(=\cos 20^{\circ}+2 \cos 120^{\circ} \cos \left(-20^{\circ}\right)\)
\(=\cos 20^{\circ}+2 \cos \left(180^{\circ}-60\right) \cos 20^{\circ}\)
\(=\cos 20^{\circ}-2 \cos 60^{\circ} \cos 20^{\circ}\)
\(=\cos 20^{\circ}-2\left(\frac{1}{2}\right) \cos 20^{\circ}\)
\(=\cos 20^{\circ}-\cos 20^{\circ}=0\)

Given f(x) = 5x - 3
\(L\left[ f\left( x \right) \right] _{ x=3 }\) =\(\underset { x\rightarrow 3 }{ lim } f(x)=\underset { h\rightarrow 0 }{ lim } f(3-h)\)
= \(\underset { h\rightarrow 0 }{ lim } 5(3-h)-3=\underset { h\rightarrow 0 }{ lim } (15-5h-3)\)
= \(\underset { h\rightarrow 0 }{ lim } \) (12-5h) = 12-0=12
\(R\left[ f\left( x \right) \right] _{ x=3 }=\underset { x\rightarrow 3^{ + } }{ lim } f(x)=\underset { h\rightarrow 0 }{ lim } f(3+h)\)
= \(\underset { h\rightarrow 0 }{ lim } 5(3+h)-3=\underset { h\rightarrow 0 }{ lim } 15+5h-3\)
= \(\underset { h\rightarrow 0 }{ lim } 12+5h=12-0=12\)
\(L\left[ f\left( x \right) \right] _{ x=3 }=R\left[ f\left( x \right) \right] _{ x=3 }\) 
ஃ f(x) is continous at x =3

Equation of the parabola is x²+y=6x-14
⇒ x²-6x=-y-14
⇒ x²-6x+9=-y-14+9 (Adding 9 both sides)
⇒ (x-3)²=-y-5
⇒ (x-3)²=-1(y+5)
⇒ x²=-y where X=x-3, Y=y+5

Number of ways of forming necklace
\(\frac { (10-1)! }{ 2! } =\frac { 9! }{ 2 } =181440\)

Number of ways for first passenger to occupy seat = 6. ....(1)
Number of ways for the second passenger to occupy seat = 5
Number of ways for the third passenger to occupy seat = 4.
∴By fundamental principle of counting, the total number of ways for all the three to occupy seats
= 6 x 5 x 4 = 120.

Let A = \(\begin{vmatrix} x &a &x+a \\ y & b &y+b \\z & c & z+c \end{vmatrix}\)
Applying the elementary transformation, \(C_1\rightarrow C_1+C_2\) we get,
\(A=\begin{vmatrix} x+a&a&a+x\\y+b&b&y+b\\z+c&c&z+c\end{vmatrix}=0[C_1\equiv C_3]\)
\(\therefore\) |A| = 0.

Given = \(\begin{bmatrix} 1 & 2 \\ 4 & 2\end{bmatrix},\) then 2A = \(\begin{bmatrix} 2 & 4 \\ 8 & 4 \end{bmatrix}\)
\(\therefore\)   |2A| = \(\begin{vmatrix}2 & 4 \\8 & 4 \end{vmatrix}\) = 8 - 32 = -24   ...(1)
Also, |A| = \(\begin{vmatrix} 1&2 \\4 & 2 \end{vmatrix}\) = 2 - 8 =  -6
\(\therefore\)  4|A| = 4(-6) = -24   ....(2)
From (1) and (2), |2A| = 4.|A|

\(\frac { 3x+7 }{ { x }^{ 2 }-3x+2 } =\frac { 3x+7 }{ (x-1)(x-2) } =\frac { A }{ x-2 } +\frac { B }{ x-1 } \)
\(\frac{3 x+7}{(x-1)(x-2)}=\frac{A(x-2)+B(x-1)}{(x-1)(x-2)}\)
\(3 x+7=A(x-2)+B(x-1)\)  ...(1)
If x = 1
\(3+7=\mathrm{A}(1-2) \Rightarrow \mathrm{A}=-10\)
If x = 2
\(6+7=\mathrm{B}(2-1) \Rightarrow \mathrm{B}=13\) 
\(\therefore\)\(\frac { 3x+7 }{ { x }^{ 2 }-3x+2 } \)=\(\frac { -10 }{ x-1 } + \frac { 13 }{ x-2 } \)

GM = \(\sqrt{8 \times 18} = \sqrt{144} = 12\)

If F.V = 100, Investment = 90
FV 10,000,
Investment \(= \frac{ 90 \times 10,000}{100} = 9000\)

Brokerage = 100 x 400 x \(\frac{1}{100}=400\)

Investment = 200 x 100 x \(\frac{8}{100}\)= Rs. 1600

\(\frac{\partial u}{\partial x} =8 x+4 y+4 \)
\(\frac{\partial^2 u}{\partial y \partial x} =4 \)

R = px = 20x- 3x²
M.R = dR/dx = 20 - 6x

1 - 2sin²45^o = cos 2(45^o)
= cos 90^o = 0

\(\sin15^o = \sin(45^o - 30^o) = \sin45^o \cos 30^o - \cos 45^o \sin 30^o\)
\(= \frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}}\)

\(a+b=0 \Rightarrow k-2=0\)

Since if n = 1 then (1) (2) (3) (4) = 24 is divisible by = 24

\(|A|=\cos ^2 \theta+\sin ^2 \theta=1\)
\(|2 A|=2^2(1)=4\)

\(\text {Since }|A|=4-4=0\)

\(\text {Co-factor of }-7=\left|\begin{array}{cc} 2 & -3 \\ 6 & 0 \end{array}\right|=0+18=18\)

Given R = 8x, \(C = 150000 + 60\left(x^2\over 900\right)\)
Profit = Revenue - Cost
\(P=8x- 150000- 60\left(x^2\over900\right)\)
Differentiating w.r.t. 'x' we get,
\({dP\over dx}=8-{120x\over 900}=8-{2x\over 15}\)
Condition for maximum is \(dp\over dx\)and \({d^2P\over dx^2}<0\)
\({dP\over dx}=0⇒8={2x\over 15}⇒{8\times15\over2}=x\)
⇒x=60
Also, \({d^2P\over dx^2}={-2\over 15}<0\)
∴ Profit is maximum when x = 60.
So, to get the maximum profit, we should sell 60 units.

Total Number of bolts produced = 1000 + 2000 + 3000 = 6000
\(P({ A }_{ 1 })=\frac { 1000 }{ 6000 } =\frac { 1 }{ 6 } \)
\(P({ A }_{ 2 })=\frac { 2000 }{ 6000 } =\frac { 1 }{ 3 } \)
\(P({ A }_{ 3 })=\frac { 3000 }{ 6000 } =\frac { 1 }{ 2 } \)
Let B be the event of selecting defective bolts.
\(\therefore \) P(B/A₁) = 1% = \(\frac { 1 }{ 100 } \) = 0.01
P(B/A₂) = 1.5% = 0.015
and (P(B/A₃) = 2% = 0.02
\(\therefore P(A_{ 1 }/B)=\frac { P({ A }_{ 1 }).P\left( B/{ A }_{ 1 } \right) }{ P({ A }_{ 1 }).P(B/{ A }_{ 1 })+P({ A }_{ 2 }).P\left( B/{ A }_{ 2 } \right) +P({ A }_{ 3 }).P\left( B/{ A }_{ 3 } \right) } \)
\(=\frac { \frac { 1 }{ 6 } \times 0.01 }{ \frac { 1 }{ 6 } \times 0.01+\frac { 1 }{ 3 } \times 0.015+\frac { 1 }{ 2 } \times 0.02 } =\frac { 1 }{ 600 } \times 60=\frac { 1 }{ 10 } \)
\(\therefore P(A_{ 1 }/B)=0.1\)

Given A = Rs.3000,r =\(\cfrac { 8 }{ 100 } \times \cfrac { 1 }{ 4 } \) = 0.02,n = 3 x 4 =12
A=\(\cfrac { a }{ i } \left[ \left( 1+i \right) ^{ n }-1 \right] \)
3000=\(\cfrac { a }{ 0.02 } \left[ \left( 1.02 \right) ^{ 12 }-1 \right] \)
3000 x 0.02 = a[1.2690-1]
60 = a[0.2690]
a=\(\cfrac { 60 }{ 0.2690 } =223.04\)
a = Rs.223 (app)
(1.02)¹² = 12 log (1.02)
= 12 (0.0086)
= 0.1032
Antilog of 0.1032 is 1.2690

Let the events E₁, E₂, E₃ and A be defined as
E₁ = I urn is chosen
E₂ = II urn is chosen
E₃ = III urn is chosen
A - Ball drawn is white in color
\(\therefore\) P(E₁) = P(E₂) = P(E₃) = \(\frac{1}{3}\)
P(A/E₁) \(=\frac{7}{17},\)
P(A/E₂ ) \(=\frac{5}{17},\)
P(A/E₃ )  \(=\frac{17}{17},\)
(i)  \(P(E_{ 1 }/A)\)
\(=\frac { P({ E }_{ 1 }).P(A/{ E }_{ 1 }) }{ P({ E }_{ 1 }).P(A/{ E }_{ 1 })+P({ E }_{ 2 }).P(A/{ E }_{ 2 })+P({ E }_{ 3 }).P(A/{ E }_{ 3 }) } \)
(By Baye's theorem)
\(=\frac { \frac { 1 }{ 3 }. \frac { 7 }{ 17 } }{ \frac { 1 }{ 3 }. \frac { 7 }{ 17 } +\frac { 1 }{ 3 } . \frac { 5 }{ 17 } +\frac { 1 }{ 3 } . \frac { 17 }{ 17 } } \)
\(=\frac {\frac {7}{51}}{\frac {7+5+17}{51}}=\frac{7}{29}\)  
(ii) \(P(E_{ 2 }/A)\) \(=\frac {\frac {1}{3}.\frac{5}{17}}{\frac {29}{51}}=\frac{5}{29}\)
(iii)  \(P(E_{ 3 }/A)\) \(=\frac {\frac {1}{3}.\frac{17}{17}}{\frac {29}{51}}=\frac{17}{29}\)

\(\tan \frac{A}{2}=\pm \sqrt{\frac{1-\cos A}{1+\cos A}}\)
Put \(A=\frac{\pi}{4}\)
\(\tan \frac{\pi}{8}=\sqrt{\frac{1-\cos \pi / 4}{1+\cos \pi / 4}}\left[\because \tan \frac{\pi}{8} \text { is }+v e\right]\)
\(=\sqrt{\frac{1-1 / \sqrt{2}}{1+1 / \sqrt{2}}}\)
\(=\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1} \times \frac{\sqrt{2}-1}{\sqrt{2}-1}}\)
\(=\sqrt{\frac{(\sqrt{2}-1)^2}{2-1}}=\sqrt{2}-1\)

Taking vertex of the parabola as reflector at origin, x-axis along the axis of the parabola, equation of the parabola is y² = 4ax
Given depth = 5 cm, diameter = 20 cm
\(\therefore\) (5, 10) lies on the parabola
\(\therefore\) 10² = 4a(5) ⇒ 100 = 20a ⇒ a = \(\frac{100}{20}\) = 5

\(\therefore\) Focus is (a, 0) = (5, 0) which is the mid-point of the given diameter

Given xy + y² = tan x + y
Differentiating with respect to 'x' we get,
\(x.\frac { dy }{ dx } +y\left( 1 \right) +2y\frac { dy }{ dx } =\sec ^{ 2 }{ x } +\frac { dy }{ dx } \)
\(\Rightarrow x.\frac { dy }{ dx } +y+2y\frac { dy }{ dx } =\sec ^{ 2 }{ x } +\frac { dy }{ dx } \)
\(\Rightarrow \frac { dy }{ dx } \left( x+2y-1 \right) =\sec ^{ 2 }{ x-y } \Rightarrow \frac { dy }{ dx } =\frac { \sec ^{ 2 }{ x-y } }{ x+2y-1 } \)

In \({\left( ax+{{1}\over{x}} \right)}^{n},n=n,x=ax,a={{1}\over{x}}\)
Compare \(\left(ax+{{1}\over{x}} \right)^n\) with (x+a)ⁿ
\(\therefore\)  General term is t_r+1 = nC_rX^n-r a^r
\(\Rightarrow\)    t_r+1 = nCr (ax)^n-r \({\left( {{1}\over{x}} \right)}^{r}=nC_r{a}^{n-r}{x}^{n-2r}\)
To find the fourth term, put r = 3
\({t}_{4}=nC_3{a}^{n-3}{x}^{n-2(3)}={{5}\over{2}}\)
\(\Rightarrow\)    \(nC_3{a}^{n-3}{x}^{n-6}={{5}\over{2}}\)
Equating the powers of x both sides, we get
n - 6 = 0  \(\Rightarrow\)  n = 6
Putting n = 6 in (1) we get,
\(6{C}_{x}{a}^{6-3}{x}^{0}={{5}\over{2}}\)
\(\Rightarrow\)    \(6C_3a^3={{5}\over{2}}\)    \([\because\ n{C}_{r}=n{C}_{n-r}]\)
\(\Rightarrow\)                        
\(\Rightarrow\)   20 a³ = \({{5}\over{2}}\Rightarrow{a}^{3}={{5}\over{2\times20}}\)
\(\Rightarrow\)  \({a}^{3}={{1}\over{8}}={\left({{1}\over{2}} \right)}^{3}\)  
\(\Rightarrow\)   \(a={{1}\over{2}}\)
 \(\therefore\) n = 6 and \(a={{1}\over{2}}\)

Equation of parabola \(y^2=-4 a x\)
It passes through (-2, -3)
\(9=-4 a(-2) \Rightarrow 4 a=\frac{9}{2}\)
Required equation \(y^2=\frac{-9 x}{2}\)

Give A = \(\left| \begin{matrix} 1 & a & b \\ 1 & c & a \\ 1 & b & c \end{matrix} \right| =0\)
Applying R₂ \(\rightarrow\) R₂-R₁ and R₃ \(\rightarrow\) R₃ - R₁ 
we get A = \(\left| \begin{matrix} 1 & a & b \\ 0 & c-a & a-b \\ 0 & b-a & c-b \end{matrix} \right| =0\)
Expanding along C₁ we get
\(1\left| \begin{matrix} c-a & a-b \\ b-a & c-b \end{matrix} \right| =0\)
\(\Rightarrow\) (c - a) (c - b) - (b - a) (a - b) = 0
\(\Rightarrow\) c² - bc - ac + ab - (ab - b² - a² + ab) = 0
\(\Rightarrow\) a² + b² + c² - ab - bc - ca = 0
Multiplying both sides by 2 we get, 2a² + 2b² + 2c² - 2ab - 2bc - 2ca = 0
\(\Rightarrow\) (a - b)² + (b - c)² + (c - a)² = 0
\(\Rightarrow\) a = b = 0, b - c = 0,c - a = 0
\(\Rightarrow\) a = b = c
\(\Rightarrow\) \(\Delta\) ABC is equilateral.
\(\therefore A=B=C=\frac { \pi }{ 3 } \)
\(\therefore { sin }^{ 2 }A+{ sin }^{ 2 }B+{ sin }^{ 2 }C=3{ sin }^{ 2 }\frac { \pi }{ 3 } =3{ \left( sin\frac { \pi }{ 3 } \right) }^{ 2 }=3{ \left( \frac { \sqrt { 3 } }{ 2 } \right) }^{ 2 }=3\times \frac { 3 }{ 4 } =\frac { 9 }{ 4 } \)

Given \({ A }^{ -1 }=\left[ \begin{matrix} 1 & 0 & 3 \\ 2 & 1 & -1 \\ 1 & -1 & 1 \end{matrix} \right] \)
\(\left|A^{-1}\right|=1[1-1]-0+3[-2-1]\)
\(=-9 \neq 0\)
\(\therefore\left(\mathrm{A}^{-1}\right)^{-1} \text { exists }\)
\(A_{11}=\text {Co-factor of } 1=1-1=0 \)
\(A_{12}=\text {Co-factor of } 0=-(2+1)=-3 \)
\(A_{13}=\text {Co-factor of } 3=(-2-1)=-3\)
\(A_{21}=\text {Co-factor of } 2=-(0+3)=-3\)
\(\mathrm{A}_{22}= \text {Co-factor of }1=1-3=-2\)
\(A_{23}= \text {Co-factor of }-1=-(-1-0)=1\)
\(A_{31}=\text {Co-factor of }1=0-3=-3\)
\(A_{32}= \text {Co-factor of } -1=-1(-1-6)=7\)
\(\mathrm{A}_{33}= \text {Co-factor of }1=1-0=1\)
\(\text {Co-factor matrix }=\left(\begin{array}{ccc} 0 & -3 & -3 \\ -3 & -2 & 1 \\ -3 & 7 & 1 \end{array}\right)\)
\(\text {Since }\left(A^{-1}\right)^{-1}=A\)
\(\mathrm{A}=\frac{1}{\left|A^{-1}\right|} \text { adj } A^{-1}\)
\(=\frac{1}{-9}\left(\begin{array}{ccc} 0 & -3 & -3 \\ -3 & -2 & 7 \\ -3 & 1 & 1 \end{array}\right)\)
\(A=\frac{1}{9}\left(\begin{array}{ccc} 0 & 3 & 3 \\ 3 & 2 & -7 \\ 3 & -1 & -1 \end{array}\right)\)

\({ \left( x+\frac { 1 }{ x } \right) }^{ 2n }\)
t_{r+ 1} = nC_r x^n-r a^r
\(\therefore\) t_r+1 = 2nC_r x^2n-r \({\left( {{1}\over{x}} \right)}^{r}\) = 2nC_r x^2n-r \({\left( {{1}\over{x}} \right)}^{r}\) = 2nC_r x^2n-2r
To find the term independent of x, equate the power of x as 0
 2n = 2r \(\Rightarrow\) r = n
\(\therefore\) t_r+1 = 2nC_n.x^2n-2n = 2nC_n
\(={{2n!}\over{n!n!}}={{(2n)(2n-1)(2n-2)(2n-3)......4,3,2,1}\over{n!n!}}\)
\(={{[(2n)(2n-2)(2n-4)...4.2][(2n-1)(2n-3)......3.1]}\over{n!n!}}\)
\(={{2^n(n)(n-1)(n-2).2.1.[(2n-1)(2n-3)....3.1]}\over{n!n!}}\) 
\(=\frac{2^n . n !(2 n-1) \ldots 5.3 .1}{n ! n !}\)
\(=\frac{1.3 .5 \ldots(2 n-1) 2^n}{n !}\) 
Hence proved.

\(\sin 9 A+\sin 7 A= 2 \sin \left(\frac{9 A+7 A}{2}\right) \cos \left(\frac{9 A-7 A}{2}\right)\)
\(=2 \sin \left(\frac{16 A}{2}\right) \cos \left(\frac{2 A}{2}\right)\)
\(=2 \sin 8 A \cos A\)

A_ij = (–1)^i+jM_ij
A₁₁ = (–1)¹⁺¹M₁₁ = \(\left| \begin{matrix} 1 & 0 \\ 1 & 0 \end{matrix} \right| \) = 0
A₁₂ = (–1)¹⁺²M₁₂ = \(-\left| \begin{matrix} 0 & 0 \\ -4 & 0 \end{matrix} \right| \) = 0
A₁₃ = (–1)¹⁺³M₁₃ = \(\left| \begin{matrix} 0 & 1 \\ -4 & 1 \end{matrix} \right| \)= 0 - (-4) = 4
A₂₁ = (–1)²⁺¹M₂₁ = \(-\left| \begin{matrix} -2 & -3 \\ 1 & 0 \end{matrix} \right| \)= -(0 - (-3)) = -3
A₂₂ = (–1)²⁺²M₂₂ = \(\left| \begin{matrix} 1 & -3 \\ -4 & 0 \end{matrix} \right| \) = 0 - 12 = -12
A₂₃ = (–1)²⁺³M₂₃ = \(-\left| \begin{matrix} 1 & -2 \\ -4 & 1 \end{matrix} \right| \) = -(1-8) = 7
A₃₁ = (–1)³⁺¹M₃₁ = \(\left| \begin{matrix} -2 & -3 \\ 1 & 0 \end{matrix} \right| \) = 0 - (-3) = 3
A₃₂ = (–1)³⁺²M₃₂ = \(-\left| \begin{matrix} 1 & -3 \\ 0 & 0 \end{matrix} \right| \) = 0 - 0 = 0
A₃₃ = (–1)³⁺³M₃₃ = \(\left| \begin{matrix} 1 & -2 \\ 0 & 1 \end{matrix} \right| \) = 1 - 0 = 1
\(\left[ { A }_{ ij } \right] =\left[ \begin{matrix} 0 & 0 & 4 \\ -3 & -12 & 7 \\ 3 & 0 & 1 \end{matrix} \right] \)
Adj A = [A_ij]^T
\(=\left[ \begin{matrix} 0 & -3 & 3 \\ 0 & -12 & 0 \\ 4 & 7 & 1 \end{matrix} \right] \)

Let y = x³+8x²+5x-2
Differentiating w.r.t. 'x' we get
\({dy\over dx}=3x^2+16x+5\)
\({dy\over dx}=0⇒3x^2 + 16x + 5=0\)
⇒ (x+5)(3x+1)=0
⇒ x = - 5, -1/3

The possible intervals are (-∞, - 5), (-5, -1/3) and (-1/3, ∞)

Intervals	Sign of \(dy\over dx\)	Nature of Function
(-∞, - 5) say x = - 6	3(-6)2 + 16(-6) + 5 = 17 (Positive)	Increasing Function
(-5, -1/3) say x = -1	3(-1)2 + 16(-1) + 5 = - 8 (Negative)	Decreasing Function
(-1/3, ∞) say x = 0	3(0)2 + 16(0) + 5 = 5 (Positive)	Increasing Function

Hence the given function is increasing in the intervals (-∞, - 5), (-1/3, ∞) and decreasing in (-5, -1/3).

Cost of one share = 100-4 = 96
Stock = 9300
Number of shares purchased = \(\frac { \text {Stock }}{ FV } \)
=\(\frac { 9300 }{ 100 } \) = 93
\(\therefore\)  Purchase price of 93 shares = 93 x 96 = Rs.8928

S = {1, 2, 3,.....10}
n(S) = 10
Let B be the event of drawing a card greater than 4
B = {5, 6,.... 10}
\(P(B) =\frac { 6 }{ 10 } \)
Let A be the event of getting an even number
A = {2, 4, 6, 8, 10}
\(P(A) =\frac { 5 }{ 10 } \)
\(A\cap B\) = {6, 8, 10}  
\(P(A\cap B)=\frac { 3 }{ 10 } \)
\(P(A/B)=\frac {P(A\cap B)}{P(B)}=\frac{\frac{3}{10}}{\frac{6}{10}}=\frac{3}{6}=\frac{1}{2}\)

Let the number of shares be x.
Dividend = No.of shares x F.V x Rate percentage.
1620 =  \(x \times 20 \times \frac { 9 }{ 100 } \)
x = \(\frac { 1620\times 100 }{ 20\times 9 } \) = 900 shares

p = 200 - \(\frac { x }{ 100 } \) 
R = px \(=200x-{x^2\over100}\)
C = 40x + 120
(i) Profit function = R - C
\(=200x-{x^2\over100}-40x-120\)
\(={x^2\over100}-160x-120\)
(ii) Averange Profit \(=\mathrm{A} \cdot \mathrm{P}=\frac{p(x)}{x}=\frac{R(x)-c(x)}{x}\)
\(=\frac{200 x \frac{-x^2}{100}+40 x+1}{x}=160-\frac{x}{100}-\frac{120}{x}\)
At x = 10 units
\(\text {A.P } =160-\frac{1}{10}-\frac{120}{10} \)
\(=160-12.1= 147.90\)
(iii) Marginal profit \(=\mathrm{M} \cdot \mathrm{P}=\frac{d p}{d x}=\frac{d}{d x} p(x)\)
\(=160-\frac{2 x}{100} \)
\(=160-\frac{x}{50}\)
At x = 10 units
\(\text {M.P }=160-\frac{1}{5}=160-0.2= 159.80\)
(iv) Marginal average profit = M.A.P \( =\frac{d(A P)}{d x}\)
\(=-\frac{1}{100}+\frac{120}{x^2}\)
At x = 10 units
\(\text {M.A.P }=-\frac{1}{100}+\frac{120}{100}= 1.19\)

\(p={(a-bx)^{1\over 2}}\)
Differentiating with respect to the price ‘p’ ,
we get \(1={1\over2}(a-bx)^{1\over2}(-b).{dx\over dp}\)
\(∴\ {dx\over dp}={2(a-bx)^{1\over2}\over-b}\)
Elasticity of demand: \(η_d=-{p\over x}.{dx\over dp}\)
\(=-{(a-bx)^{1\over2}\over x}.{2(a-bx)^{1\over2}\over -b}\)\(={2(a-bx)\over bx}\)
When \(η_d=1,\ {2(a-bx)\over bx}=1\)
2(a - bx) = bx ⇒ output \( x={2a\over 3b}\)units

Let y = sin x. sin 2x. sin 3x
Taking logarithms on both sides we get,
log y = log (sin x. sin 2x. sin 3x)
= log (sin x) + log (sin 2x) + log (sin 3x)    [\(\therefore\) log ab = log a + log b]
Differentiating with respect to 'x' we get,
\(\frac { 1 }{ y } \frac { dy }{ dx } =\frac { 1 }{ sin\quad x } .\frac { d }{ dx } (sin\quad x)+\frac { 1 }{ sin\quad 2x } .\frac { d }{ dx } (sin\quad 2x)+\frac { 1 }{ sin\quad 3x } .\frac { d }{ dx } (sin\quad 3x)\)
\(\frac { 1 }{ y } \frac { dy }{ dx } =\frac { cos\quad x }{ sin\quad x } +\frac { 2cos2x }{ sin\quad 2x } =3.\frac { cos\quad 3x }{ sin\quad 3x } \)
= cot x + 2 cot 2x + 3 cot 3x
\(\Rightarrow \frac { dy }{ dx } =y[cotx+2cot2x+3cot3x]\)
\(\Rightarrow \frac { dy }{ dx } \)= sin x sin 2x sin 3x [cot x + 2 cot 2x + 3 cot 3x]

The given lines are
2x - 5y + 3 = 0
5x - 9y + \(\lambda \) = 0
x - 2y + 1 = 0
The condition for the lines to be concurrent is
\(\left| \begin{matrix} 2 & -5 & 3 \\ 6 & -9 & \lambda \\ 1 & -2 & 1 \end{matrix} \right| =0\quad \)
Expanding along R₁ , we get
\(2\left| \begin{matrix} -9 & \lambda \\ -2 & 1 \end{matrix} \right| +5\left| \begin{matrix} 5 & \lambda \\ 1 & 1 \end{matrix} \right| +3\left| \begin{matrix} 5 & -9 \\ 1 & -2 \end{matrix} \right| =0\)
\(\Rightarrow 2(-9+2\lambda )+5(5-\lambda )+3(-10+9)=0\)
\(\Rightarrow -18+4\lambda +25-5\lambda -30+27=0\)
\(\Rightarrow -\lambda +4=0\Rightarrow \lambda =+4\)

y = 2 sin x + 3cos x
y₁ = 2 cos x - 3 sin x
y₂ = 2 (-sin x) -3 cos x
y₂ = -2 sin x -3 cos x = -y
y₂ + y = 0.

\({{(2x+1)!}\over{(x+2)!}}.{{(x-1)!}\over{(2x-1)!}}={{3}\over{5}}\)
\(\Rightarrow\)  \({{(2x+1)(2x)(2x-1)!}\over{(x+2)(x+1)(x-1)!}}.{{(x-1)!}\over{(2x-1)!}}={{3}\over{5}}\)
\(\Rightarrow\)  \({{(2x+1)(2)}\over{(x+2)(x+1)}}={{3}\over{5}}\)
\(\Rightarrow\)   10 (2x+1) = 3 (x+2) (x+1)
\(\Rightarrow\)   20x+10=3 (x²+3x+2)
\(\Rightarrow\)   20x+10+3x²-9x-6=0
\(\Rightarrow\)   -3x²+11x+4 = 0
\(\Rightarrow\)   -3x²-11x-4=0
\(\Rightarrow\)   (x-4)(3x+1)=0
\(\Rightarrow\)   x = 4 or x \(={{-1}\over{3}}\)
Since \(x={{-1}\over{3}}\) is not possible, x = 4.

Let A = \(\left| \begin{matrix} x & sin\theta & cos\theta \\ -sin\theta & -x & 1 \\ cos\theta & 1 & x \end{matrix} \right| \)
Expanding along R₁ we get
|A| = x\(\left| \begin{matrix} -x & 1 \\ 1 & x \end{matrix} \right| -sin\theta \left| \begin{matrix} -sin\theta & 1 \\ cos\theta & x \end{matrix} \right| +cos​​\theta \begin{vmatrix} -sin\theta & -x \\ cos\theta & 1 \end{vmatrix}\)
= x(-x² - 1) - sin \(\theta\) (-x sin \(\theta\) - cos \(\theta\)) + cos \(\theta\) (-sin \(\theta\) + x cos \(\theta\))
\(=-x^{ 3 }-x+xsin^{ 2 }\theta +sin\theta cos\theta +xcos^{ 2 }\theta \)
= -x³ - x + x(sin²\(\theta\) + cos²\(\theta\))
= -x³ - x + x(1)    [\(\because\) sin²\(\theta\) + cos²\(\theta\) ] = -1
= -x³ which is independent of \(\theta\)