\(\underset { x\rightarrow a }{ lim } \cfrac { { x }^{ \frac { 3 }{ 5 } }-{ a }^{ \frac { 3 }{ 5 } } }{ { x }^{ \frac { 1 }{ 5 } }-{ a }^{ \frac { 1 }{ 5 } } } =\underset { x\rightarrow a }{ lim } \cfrac { { x }^{ \frac { 3 }{ 5 } }-{ a }^{ \frac { 3 }{ 5 } } }{ \frac { { x }^{ \frac { 4 }{ 5 } }-{ a }^{ \frac { 1 }{ 5 } } }{ x-a } } \) 
= \(\cfrac { \frac { 3 }{ 5 } \left( a \right) ^{ -\frac { 2 }{ 5 } } }{ \frac { 1 }{ 5 } \left( a \right) ^{ -\frac { 4 }{ 5 } } } =3{ a }^{ \frac { -2 }{ 5 } +\frac { 4 }{ 5 } }=3a^{ \frac { 2 }{ 5 } }\)

Let \({ \tan }^{ -1 }\left( \cfrac { 1 }{ 8 } \right) =\theta \)
Then \(\tan\theta =\cfrac { 1 }{ 8 } \) 
\(\therefore \tan\left[ \cfrac { \pi }{ 4 } -{ \tan }^{ -1 }\left( \cfrac { 1 }{ 8 } \right) \right] =\tan\left( \cfrac { \pi }{ 4 } -\theta \right) \) 
= \(\cfrac { \tan\frac { \pi }{ 4 } -\tan\theta }{ 1 + \tan\frac { \pi }{ 4 } \tan \theta } \) 
= \(\cfrac { 1-\frac { 1 }{ 8 } }{ 1+\frac { 1 }{ 8 } } =\cfrac { 7 }{ 9 } \)

\(\left( n+2 \right) { C }_{ n }=45\)
\(\left( n+2 \right) { C }_{ n+2-n }=45\)
\( \left( n+2 \right) { C }_{ 2 }=45\)
\(\cfrac { \left( n+2 \right) \left( n+1 \right) }{ 2 } =45\)
\({ n }^{ 2 }+3n-88=0\)
\(\left( n+11 \right) \left( n-8 \right) =0\)
\( n=-11,8\)
\(n=-11\) is not possible
\(\therefore n=8\)

The given equation is \({ x }^{ 2 }+4xy+{ y }^{ 2 }=0\)
Here a = 1, b = 1 and h = 2
If \(\theta \) is the angle between the given straight lines, the
\(\theta ={ tan }^{ -1 }\left[ \left| \cfrac { 2\sqrt { { h }^{ 2 }-ab } }{ a+b } \right| \right] \)
\(={ \tan }^{ -1 }\left[ \left| \cfrac { 2+\sqrt { 4-1 } }{ 2 } \right| \right]\)
\(= { \tan }^{ -1 }\left( \sqrt { 3 } \right) \)
\(\theta =\cfrac { \pi }{ 3 } \)

Given the lines are concurrent
Therefore \(\left| \begin{matrix} { a }_{ 1 } & { b }_{ 1 } & { c }_{ 1 } \\ { a }_{ 2 } & { b }_{ 2 } & { c }_{ 2 } \\ { a }_{ 3 } & { b }_{ 3 } & { c }_{ 3 } \end{matrix} \right| =0\) 
\(\left| \begin{matrix} 3 & -5 & -11 \\ 5 & 3 & -7 \\ 1 & k & 0 \end{matrix} \right| =0\)
\(1(35+33)-k(-21+55)=0\)
\( \Rightarrow 34k=68\therefore k=2\)

\(|A|=\left| \begin{matrix} -2 & 6 \\ 3 & -9 \end{matrix} \right| =0\)
Since A is a singular matrix, A ^-1 does not exist.

(i) S = {1, 2, 3, 4, 5, 6}
n(S) = 6
(i) Let A be the event of getting a prime number
A = {2, 3, 5}
n(A) = 3
\(P(A)=\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \)
(ii) Let B be the event that the number is greater than or equal to 3.
B = {3, 4, 5, 6}    
n(B) = 4
\(P(B) =\frac { 4 }{ 6 } =\frac { 2 }{ 3 } \)

Let us define the following events.
E₁: X speaks truth
E₂: X tells a lie
E: X reports a six
From the data given in the problem, we have
P(E₁) = \(\frac{4}{5}\); P(E₂) = \(\frac{1}{5}\);
P(E/E₁) = \(\frac{1}{6}\); P(E/E₂) = \(\frac{5}{6}\)
The required probability that actually there was six (by Bayes theorem) is 
\(P(E_1/E)=\frac{P(E_1)P(E/E_2)}{P(E_1)P(E/E_1)+P(E_2)P(E/E_2)}\) = \(\frac{\frac{4}{5}\times \frac{1}{6}}{(\frac{4}{5}\times \frac{1}{6})+(\frac{1}{5}\times \frac{5}{6})}=\frac{4}{9}\)

Let Investment in each stock be 120 x 135
At 7% stock
Investment = 120, Income = 7
Investment = 120 x 135
Income = \(\cfrac { 7\times120\times135 }{ 120 }=Rs.945\)
At 8% stock
Investment = 135, Income = 8
Investment = 120 x 135
Income =\(\cfrac { 8 \times 120 \times135 }{ 135 }=Rs.960\)
\(\therefore\) 8% of Rs.100 shares at Rs.135 is a better investment .

Market value of one share = 25 - 7 = 18
Market value of 125 shares = 125 x 18
= Rs. 2250

Face value = Rs. 100
Market value = Rs. (100 – 17 + 1) = Rs. 84
\(\therefore\) percentage of his income = \(\frac{(12\times 100)}{84}\)
= \(\frac{100}{7}=14\frac{2}{7}\)
\(\therefore\) % of Income = \(14\frac{2}{7}\%\)

P = 4LK-L²+K², L>0, k>0
\({\partial P\over \partial L}=4K-2L\)
\({\partial P\over \partial K}=4LK+2K\)
\(L{\partial P\over \partial L}+K{\partial P\over \partial K}=4KL-2L^2+4KL+2K^2\)
= 8KL-2L²+ 2K²  = 2(4LK-L²+ K²)
= 2p = RHS
Hence proved.

Using the precedence relationships and following the rules of network construction, the required network is shown in following figure.

At equilibrium, demand = Supply
\(\Rightarrow\)  100 - 2p = 3p - 50
\(\Rightarrow\) 150 = 5p
\(\Rightarrow\) p = 30
Equilibrium price is P_E = 30
x = 100 - 2p
= 100 - 60 = 40
x = 40
\(\therefore\) Equilibrium quantity is x_E = 40 units.

Given \(\frac{d y}{d x}=2 x\)
\(y=x^3+10 x^2-48 x+8\)
\(\frac{d y}{d x}=3 x^2+20 x-48\)
\(2 x = 3x^2+20 x-48\)
\(3 x^2+18 x-48=0 \)
Dividing by 3 we get,
 x² + 6x - 16 = 0
\(\Rightarrow\)(x + 8)(x - 2) = 0
\(\Rightarrow\) x = -8,2

x = 2p² + 5
\({dx\over dp}=4p\)
Elasticity of supply \(\eta_s={p\over x}.{dx\over dp}\)
\(\eta_s={p\over 2p^2+5}(4p)={4p^2\over 2p^2+5}\)
when p = 3, \(\eta_s={36\over 18+5}\)\(={36\over23}\)

5x² + 5y² +4x - 8y - 16 = 0
[Divide by 5]
x² + y² +  \(\frac { 4 }{ 5 } x-\frac { 8 }{ 5 } y-\frac { 16 }{ 5 } =0\)
Here 2g  = \(\frac { 4 }{ 5 } \) \(\Rightarrow\)  \(g=+\frac { 2 }{ 5 } \)
2f = \(-\frac { 8 }{ 5 } \)     \(\Rightarrow\) \(f=-\frac { 4 }{ 5 } \)
and c = \(-\frac { 16 }{ 5 } \)
Center of the circle is (-g, -f) \(\Rightarrow \) \(\left( -\frac { 2 }{ 5 } ,\frac { 4 }{ 5 } \right) \)
Radius of the circle is \(\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } \)
\(\Rightarrow\) r = \(\sqrt { \frac { 4 }{ 25 } +\frac { 16 }{ 25 } +\frac { 16 }{ 5 } } =\sqrt { \frac { 20 }{ 25 }+ { \frac { 16 }{ 5 }} } \)
\(\Rightarrow\) \(r=\sqrt { \frac { 20 }{ 5 } } \) = \(\sqrt { 4 } \) = 2 units

f(x) = \(\frac { x+1 }{ x-1 } \)
\(\text { LHS }=f(f(x)) =f\left(\frac{x+1}{x-1}\right)=\frac{\frac{x+1}{x-1}+1}{\frac{x+1}{x-1}-1} \)
\(=\frac{x+1+x-1}{x+1-x+1}=\frac{2 x}{2}=x=R H S \)

 

\(\left(x+\frac{1}{y}\right)^7 =7 C_0 x^7+7 C_1 x^6\left(\frac{1}{y}\right)+7 C_2 x^5\left(\frac{1}{y}\right)^2 +7 C_4 x^3\left(\frac{1}{y}\right)^4+7 C_5 x^2\left(\frac{1}{y}\right)^5 +7 C_6 x\left(\frac{1}{y}\right)^6+7 C_7\left(\frac{1}{y}\right)^7\)
\(=x^7+\frac{7 x^6}{y}+\frac{21 x^5}{y^2}+\frac{35 x^4}{y^3} +\frac{35 x^3}{y^4}+\frac{21 x^2}{y^5}+\frac{7 x}{y^6}+\frac{1}{y^7}\)

n{S) = 1000
Let A be the event that the student selected is from XI std and B be the event that the selected student is a girl.
n(B) = 450
\(P(B)=\frac { 450 }{ 1000 } \)
n(A\(\cap \)B) = \(\frac { 20 }{ 100 } \times 450=90\)
P(A\(\cap \)B) \(=\frac { 90 }{ 1000 } \)
\(P(A/B)=\frac { P(A\cap B) }{ P(B) } =\frac { \frac { 90 }{ 1000 } }{ \frac {450 }{ 1000} } =\frac { 1 }{ 5 } =0.20\)

u = xy+sin(xy)
\(\frac { \partial u }{ \partial x } \) = y+ycos(xy)
\(\frac { \partial u }{ \partial y } \) = x+x cos(xy)
\(\frac { \partial ^{ 2 }u }{ \partial y\partial y } =\frac { \partial }{ \partial x } \left( \frac { \partial u }{ \partial y } \right) \)
= 1 + x (−sin(xy) .y) + cos(xy)
= 1 – xy sin(xy) + cos(xy).... (1)
\(\frac { \partial ^{ 2 }u }{ \partial x\partial y } =\frac { \partial }{ \partial y } \) (y + cos(xy))
1+ cos(xy)+y (−sin(xy)x )
1−xy sin(xy)+ cos(xy) ... (2)
From (1) and (2), we get
\(\frac { \partial ^{ 2 }u }{ \partial x\partial y } =\frac { { \partial }^{ 2 }u }{ \partial y\partial x } \quad \)

LHS =  \(\begin{vmatrix} {1\over a}&bc&b+c\\{1\over b}&ca&c+a\\{1\over c}&ab&a+b \end{vmatrix}\)
Multiplying R₁ by a, R₂ by b, and R₃ by c respectively and dividing the determinant by abc we get.
\(LHS={1\over abc}\begin{vmatrix} 1&abc&ab+ac\\1&abc&bc+ab\\{1}&abc&ac+bc \end{vmatrix}\)
\(={1\over abc}(abc)\begin{vmatrix} 1&1&ab+ac\\1&1&bc+ab\\{1}&1&ac+bc \end{vmatrix}\) = 0  \([\because C_1\equiv C_2]\) = RHS.
Hence Proved.

\(\frac { 1 }{ { x }^{ 2 }-1 } =\frac { 1 }{ (x+1)(x-1) } =\frac { A }{ x+1 } +\frac { B }{ x-1 } \)
⇒ \(\frac{1}{(x+1)(x-1)}=\frac{A(x-1)+B(x+1)}{(x+1)(x-1)}\)
⇒ 1 = A (x - 1) + B (x + 1)       ...(1)
If x = -1 in (1) we get
1 = A(-2) + 0 ⇒ A = \(\frac { -1 }{ 2 } \)
If x = 1 in (1) we get,
1 = 0 + B(1 + 1) ⇒ 1 = 2B ⇒ B = \(\frac { 1 }{ 2 } \)
\(\frac{1}{x^2-1}=\frac{-1}{2(x+1)}+\frac{1}{2(x-1)}\).

y = A sin x + B cos x
y₁ = A cos x – B sin x
y₂ = –A sin x –B cos x
y₂ = –y
y₂ + y = 0

Differentiating y = x² sin x with respect to x.
\(\cfrac { dy }{ dx } ={ x }^{ 2 }\cfrac { d }{ dx } \left( \sin x \right) +\sin x\cfrac { d }{ dx } \left( { x }^{ 2 } \right) \)
\(={ x }^{ 2 }\cos x+2x \sin x\)
\(=x\left( x \cos x+2\sin x \right) \)

f(x) = 2^x (Given)
\(\therefore \) f(x + y) = 2^x+y
= 2^x.2^y
= f(x).f(y)

The supply price relation is given by
\({ x }^{ 2 }=5p-15\) 
= \(5(p-3)\) 
\(\Rightarrow { X }^{ 2 }aP\)  where X = x and \(P=p-3\) 
\(\\ \therefore \) the supply curve is a parabola whose vertex is \(\left( X=0,\ P=0 \right) \) 
i.e., The supply curve is a parabola whose vertex is (0, 3)

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

nP _r= 360 = 36 \(\times\) 10
= 3 \(\times\) 3 \(\times\) 4 \(\times\) 5 \(\times\) 2
= 6 \(\times\) 5 \(\times\) 4 \(\times\) 3 = 6P₄
Therefore n = 6 and r = 4

\(\left| \begin{matrix} x & x+1 \\ x-1 & x \end{matrix} \right| \) = x² - (x – 1)(x + 1)
= x² – (x² – 1)
= x² – x² + 1 = 1

Let A, B be the events of getting a black ball in the first and second draw
Probability of drawing a black ball in the first attempt is
P(A) = \(\frac{3}{5+3}=\frac{3}{8}\)
Probability of drawing the second black ball given that the first ball drawn is black
P(B/A) = \(\frac{2}{5+2}=\frac{2}{7}\)
\(\therefore\) The probability that both balls drawn are black is given by
\(P\left( A\cap B \right) =P(A)P(B/A)=\frac { 3 }{ 8 } \times \frac { 2 }{ 7 } =\frac { 3 }{ 28 } \)

We know that the sample space is S = {1,2,3,4,5,6}
Now, A = {3,6} ; B = { 2,4,6} then (A∩B) = {6}
P(A) = \(\frac { 2 }{ 6 } =\frac { 1 }{ 3 } \)
P(B) = \(\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \) and P(A∩B) = \(\frac{1}{6}\)
Clearly P(A∩B) = P(A) P(B)
Hence A and B are independent events.

S = {(1, 1) (1, 2) (1,3), ..... (6, 6)}
(2, 1), (2, 2) ...... (2, 6)
(6, 1), (6, 2)  ..... (6, 6)}
n(S) = 36
Let B be the event that sum of numbers appearing is 6
B = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}
\(P(B)=\frac{5}{36}\)
Let A be the event that 4 has appeared atleast once
A = {(2, 4), (4, 2)}
\(A\cap B\)  = {(2, 4),(4, 2)}
\(P(A\cap B)=\frac{2}{36}\)
\(P(A/B)=\frac { P(A\cap B) }{ P(B) } =\frac { 2/36 }{ 5/36 } =\frac { 2 }{ 5 } \)

S {(G₁, G₂), (G₁, B₂), (B₁, G₂), (B₁, B₂)}
n(S) = 4
Let A be the event that both are girls
A= {(G₁ G₂)}
n(A) = 1
\(P(A)=\frac{1}{4}\)
Let B to the event that atleast one of the children is a girl.
B = {(G₁ G₂), (G₁ B₁), (G₂ B₂)}
\(P(B)=\frac{3}{4}\)
\( A\cap B=\{ ({ G }_{ 1 }{ G }_{ 2 })\}, P(A\cap B)=\frac{1}{4}\)
\(P(A / B)=\frac { P(A\cap B) }{ P(B) } =\frac{\frac{1}{4}}{\frac{3}{4}}=\frac { 1 }{ 3 } \)

\(r =\frac{N \Sigma X Y-(\Sigma X)(\Sigma Y)}{\sqrt{N \Sigma X^2-(\Sigma X)^2} \sqrt{N \Sigma Y^2-(\Sigma Y)^2}} \)
\(=\frac{10(-115)-(50)(-30)}{\sqrt{10(290)-(50)^2} \sqrt{10(300)-(-30)^2}} \)
\(=\frac{-1150+1500}{\sqrt{400 \times 2100}}=\frac{350}{916.52}=0.382\)

Let x be the number of shares.
Face value of ‘x’ shares = Rs. 25 x
Now \(\frac{10}{100}\times 25x=Rs. 2,000\)
\(\Rightarrow x=\frac{2000\times 100}{25\times 10}=800\)
Hence the number of shares = 800

a = 50, i = \(\frac{5}{100}\) = .05
A = \(\cfrac { a }{ i } =\cfrac { 50 }{ 0.05 } \) Rs.1000

Here n = 11 observations are arranged into ascending order
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22
D₂ = size of 2 \(\left( \frac { n+1 }{ 10 } \right) \)^th value
D₆ = size of 6  \(\left( \frac { n+1 }{ 10 } \right) \)^th value
D₂ = size of 2.4^th value ≈ size of 2^nd value = 4
D₆ = size of 7.2^th value ≈ size of 7^th value = 14

Given Σxy = 120, Σx² = 90, Σy² = 640
Then r = \(\frac { \Sigma xy }{ \sqrt { \Sigma { x }^{ 2 }\Sigma { y }^{ 2 } } } =\frac { 120 }{ \sqrt { 90(640) } } =\frac { 120 }{ \sqrt { 57600 } } =\frac { 120 }{ 240 } \) = 0.5

We know that correlation coefficient
r= \(\frac { N\Sigma XY-(\Sigma X)(\Sigma Y) }{ \sqrt { N\Sigma X^{ 2 }-(\Sigma X)^{ 2 }\sqrt { N\Sigma { Y }^{ 2 }-(\Sigma Y)^{ 2 } } } } \)
= \(\frac { 9(597)-(45\times 108) }{ \sqrt { 9(285)-(45)^{ 2 }\times \sqrt { 9(1356)-(108^{ 2 }) } } } \)
r = +0.95

Using the immediate precedence relationships and following the rules of network construction the required network is shown in the following figure

Using the immediate precedence relationships and following the rules of network construction, the required network is shown in following figure.

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

\(\tan 15^o=\tan(45^o-30^o)\)
\(=\frac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ} \tan 30^{\circ}}=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}} \)
\(=\frac{\sqrt{3}-1}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}=\frac{(\sqrt{3}-1)^2}{(\sqrt{3})^2-1^2} \)
\(=\frac{3-2 \sqrt{3}+1}{3-1}=\frac{4-2 \sqrt{3}}{2}=\frac{2(2-\sqrt{3})}{2} =2-\sqrt{3}\)

Number of words \(=7 C_3 \times 4 C_2 \times 5 !\)
\(=\frac{7 \times 6 \times 5}{3 \times 2 \times 1} \times \frac{4 \times 3}{2 \times 1} \times 120\)
\(=25200\)
Since out of 7 consonants and 4 vowels 3 consonants and 2 vowels can be chosen in \(7 C_3 \times 4 C_2 \text { ways }\)
Thus there are \(7 C_3 \times 4 C_2\) each containing 3 consonants and 2 vowels. Since each group contains 5 letters which can be arranged among themselves in 5! Ways.

\(\tan \left(-855^{\circ}\right)=-\tan 855^{\circ}\)
\(=-\tan \left(2 \times 360^{\circ}+135^{\circ}\right)\)
\(=-\tan 135^{\circ}\)
\(=-\tan \left(180^{\circ}-45^{\circ}\right)(\text { II quad })\)
\(=\tan 45^{\circ}=1\)

\(\mathrm{RHS}=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\)
Let \(x=\tan \theta \Rightarrow \theta=\tan ^{-1} x\)
\(=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)=\sin ^{-1}(\sin 2 \theta)\)
\(=2 \theta=2 \tan ^{-1} x=\text { LHS }\)
Hence proved.

Let A = \(\begin{vmatrix}5&20\\ 0&-1 \end{vmatrix}\)
Minor of 5 = M₁₁ = -1
Minor of 20 = M₁₂ = 0
Minor of 0 = M₂₁ = 20
Minor of -1 = M₂₂ = 5
Co-factor of 5 = A₁₁ = -1
Co-factor of 20 = A₁₂ = 0
Co-factor of 0 = A₂₁ = -20
Co-factor of -1 = A₂₂ = 5

B = \(\begin{bmatrix} 0.6 & 0.9 \\ 0.20 & 0.80 \end{bmatrix}\) 
I - B = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}-\begin{bmatrix} 0.6 & 0.9 \\ 0.20 & 0.80 \end{bmatrix}=\begin{bmatrix} 0.4 & -0.9 \\ -0.20 & 0.20 \end{bmatrix}\)
|I - B| =\(\begin{bmatrix} 0.4 & -0.9 \\ -0.20 & 0.20 \end{bmatrix}\)
= 0.08 - 0.18 = - 0.1 < 0
Since |I - B| is negative, Hawkins - Simon conditions are not satisfied.
Therefore the given system is not viable.