(i) Let y = x^x
Taking logarithm on both sides logy = x log x
Differentiating with respect to x,
\(\cfrac { 1 }{ y } .\cfrac { dy }{ dx } =x.\cfrac { 1 }{ x } +logx.1\)
\(\cfrac { dy }{ dx } =y\left[ 1+logx \right] \)
\( \therefore \cfrac { dy }{ dx } ={ x }^{ x }\left[ 1+logx \right] \)
 (ii) Let y = (logx)^cosx
Taking logarithm on both sides log y = cos x log(logx)
Differentiating with respect to x,
\(\cfrac { 1 }{ y } .\cfrac { dy }{ dx } =cosx\cfrac { 1 }{ logx } .\cfrac { 1 }{ x } +\left[ log\left( logx \right) \right] \left( -sinx \right) \)
\(\cfrac { dy }{ dx } =y\left[ \cfrac { cosx }{ xlogx } -sinxlog\left( logx \right) \right] \)
\(=\left( logx \right) ^{ cosx }\left[ \cfrac { cosx }{ xlogx } -sinxlog\left( logx \right) \right] \)

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

The given equations are \({ x }^{ 2 }+{ y }^{ 2 }-64=0\) and \(3x+4y-k=0\) 
The condition for the tangency is \(c^2=a^2\left(1+m^2\right)\) 
Here \({ a }^{ 2 }=64,m=\cfrac { -3 }{ 4 } \quad c=\cfrac { k }{ 4 } \) 
\({ c }^{ 2 }={ a }^{ 2 }\left( 1+m \right) ^{ 2 }\Rightarrow \cfrac { { k }^{ 2 } }{ 16 } =64\left( 1+\cfrac { 9 }{ 16 } \right) \)
\( { k }^{ 2 }=64\times 25\)
\(k=\pm 40\)

\(A=\left[ \begin{matrix} 2 & 4 \\ -3 & 2 \end{matrix} \right] \)
\(|A|=\left[ \begin{matrix} 2 & 4 \\ -3 & 2 \end{matrix} \right] \)
=16 ≠ 0
Since A is a nonsingular matrix, A ^-1 exists
Now  adj \(A=\left[ \begin{matrix} 2 & -4 \\3 & 2 \end{matrix} \right] \)
\({ A }^{ -1 }=\frac { 1 }{ \left| A \right| } adjA\)
\(=\frac { 1 }{ 16 } \left[ \begin{matrix} 2 & -4 \\ 3 & 2 \end{matrix} \right] \)

Given P= 5(L)^0.7(K)^0.3
Marginal Productivity of Labour (L) is
\({∂P\over ∂L}=5(0.7)(L)^{0.7-1} (K)^{0.3 }= 3.5(L)^{-o·3} (K)^{0.3} = 3.5\left(K\over L\right)^{0.3}\)
When L = 10 and K = 3,
\(\frac { \partial P }{ \partial L } =3.5\left( \frac { 3 }{ 10 } \right) ^{ 0.3 }=3.5(0.3)^{ 0.3 }=3.5\times 0.6968\)
= 2.438 = 2.44
\({∂P\over ∂L}=5(L)^{0.7} (0.3)(K)^{0.3-1 }= 1.5(L)^{0.7}(K)^-{0.7} = 1.5\left(L\over K\right)^{0.7}\)
When L= 10 and K=3,
\({∂P\over ∂L}=1.5\left(10\over3\right)^{0.7}=1.5(3.33)^{0.7} = 1.5(2.2322) = 3.481 = 3.48\)

cov(x, y)=\(\frac { n\sum { xy } -(\sum { x } )(\sum { y } ) }{ { n }^{ 2 } } =\frac { 10(-115)-50(-30) }{ { 10 }^{ 2 } } \)
=\(\frac { -1150+1500 }{ 100 } =\frac { 350 }{ 100 } \)=3.5
Again, co-efficient correlation
r(x,y)=\(\frac { n\sum { xy } -(\sum { x } )(\sum { y } ) }{ \sqrt { n{ \sum { x } }^{ 2 }-{ (\sum { x } ) }^{ 2 } } \sqrt { n{ \sum { y } }^{ 2 }-{ (\sum { y } ) }^{ 2 } } } \)
=\(\frac { 10(-115)-50(-30) }{ \sqrt { 10(290)-{ (50) }^{ 2 } } .\sqrt { 10(300)-{ (-30) }^{ 2 } } } \)
=\(\frac { -1150+1500 }{ \sqrt { 2900-2500 } .\sqrt { 3000-900 } } =\frac { 350 }{ 20\times 10\sqrt { 21 } } \)=0.3819
∴ r(x,y)=0.3819

Size x	No. of items f	Mid Value x	f/x
50-60	12	55	0.2182
60-70	15	65	0.2308
70-80	22	75	0.2933
80-90	18	85	0.2118
90-100	10	95	0.1053
	N=77		\(\sum { f/x=1.0594 } \)

\(\therefore\) Harmonic mean = \(\frac { N }{ \sum { \left( \frac { f }{ x } \right) } } =\frac { 77 }{ 1.0594 } =72.683\)
\(\therefore\) HM = 72.683

Let E₁ = missing card is a heart card
E₂ = missing card is a spade card
E₃ = missing card is a club card
E₄ = missing card is a diamond card
and A = drawing 2 heart cards from the remaining cards.
\(\therefore\) P(E₁) = P(E₂) = P(E₃) = P(E₄)
\(=\frac { 13 }{ 52 } =\frac { 1 }{ 4 } \)
P(A/E₁) = P(two heart cards given that one heart card is missing)
\(=\frac { 12{ C }_{ 2 } }{ 51{ C }_{ 2 } } =\frac { \frac { 12\times 11 }{ 2\times 1 } }{ \frac { 51\times 50 }{ 2\times 1 } } =\frac { 66 }{ 1275 } \)
P(A/E₂) = P(2 heart cards given that one spade card is missing)
\(=\frac { 13{ C }_{ 2 } }{ 51{ C }_{ 2 } } =\frac { \frac { 13\times 12 }{ 2 } }{ \frac { 51\times 50 }{ 2\times 1 } } =\frac { 78 }{ 1275 } \)
Similarly, P(A/E₃) = \(\frac { 13{ C }_{ 2 } }{ 51{ C }_{ 2 } } =\frac { 78 }{ 1275 } \)
and P(A/E₄) = \(\frac { 13{ C }_{ 2 } }{ 51{ C }_{ 2 } } =\frac { 78 }{ 1275 } \)
\(\therefore\) By Baye's theorem,
\(P({ E }_{ 1 }/A)=\frac { P({ E }_{ 1 }).P\left( A/{ E }_{ 1 } \right) }{ P({ E }_{ 1 }).P\left( A/{ E }_{ 1 } \right) +P({ E }_{ 2 }).P\left( A/{ E }_{ 2 } \right) +P({ E }_{ 3 }).P(A/{ E }_{ 3 }) } \)
\(=\frac { \frac { 1 }{ 4 } \times \frac { 66 }{ 1275 } }{ \frac { 1 }{ 4 } \times \frac { 66 }{ 1275 } +\frac { 1 }{ 4 } \times \frac { 78 }{ 1275 } +\frac { 1 }{ 4 } \times \frac { 78 }{ 1275 } +\frac { 1 }{ 4 } \times \frac { 78 }{ 1275 } } \)
\(=\frac { \frac { 1 }{ 4 } \times \frac { 66 }{ 1275 } }{ \frac { 1 }{ 4 } \left[ \frac { 66 }{ 1275 } +\frac { 78 }{ 1275 } +\frac { 78 }{ 1275 } +\frac { 78 }{ 1275 } \right] } \)
\(=\frac { 66 }{ 1275 } \times \frac { 1275 }{ 66+78+78+78 } \)
\(=\frac { 66 }{ 300 } =\frac { 11 }{ 50 } \)

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).

Since the decision variables are non-negative, the solution lies in the I quadrant of the plane.
Consider the equations 
\({ x }_{ 1 }-{ x }_{ 2 }=-1\)

\({ x }_{ 1 }\)	0	1
\({ x }_{ 2 }\)	1	2

\({ -x }_{ 1 }+{ x }_{ 2 }=0\)

\({ x }_{ 1 }\)	2	1
\({ x }_{ 2 }\)	2	1

 

The feasible region is not common. Thus, there is no maximum value of Z.

Given FV =Rs.100
Market value = 100-11+1 = Rs.90
Income=\(​​=\frac { \text {Investment }}{\text { MV } }\times {\text { Divicend Rate}}=\frac { 5400 }{ 90 } \times 9=Rs.540\)
\(\therefore\) Percentage of his income \(=\frac { \text {Income }}{\text { vestment } } \times 100=\frac { 540 }{ 5400 } \times 100=10\%\)

Given a = Rs.500,i = 5% = 0.05,n =10
A =\(\cfrac { a }{ i } \left( 1+i \right) \left[ \left( 1+i \right) ^{ n }-1 \right] \)
\(=\frac { 500 }{ 0.05 } (1.05)\left[ (1.05)^{ 10 }-1 \right] \)
\(=10,500[1.629-1]\)
\(=10,500(0.629)\)
\(\therefore\) = Rs. 6604.50
 At the end of 10 years, I will receive  Rs. 6604.50.
\((1.05)^{ 10 }=10log(1.05)\)
\(=0.2120\)
Antilog of 0.2120 is 1.629

The regression co-efficient are given by
b_xy= \(r.\frac { { \sigma }_{ x } }{ { \sigma }_{ y } } \)and b_yx=\(r.\frac { { \sigma }_{ y } }{ { \sigma }_{ x } } \)
where r is the co-efficient of correlation.

\(r=\sqrt { { b }_{ xy }.{ b }_{ yx } } \)

Let the investment in each case be Rs. (16× 24)
Step 1: Income on 12% shares:
Income from 12 % Rs. 20 shares at Rs. 16
= \(\frac{12}{16}\times (16\times 24)\) = Rs. 288
Step 2: Income on 15% shares:
Income from 15% Rs. 20 Shares at Rs. 24
= \(\frac{15}{24}\times (16\times 24)\) = Rs. 240
Hence, the first investment is better.

Here P = 30,000 ; a = 675 ; k = 2, i = ?
P = \({\frac{a}{{i}/k}} \)
30,000 = \(\frac{675}{\frac{i}{2}}\)
= \(\frac{1350}{i}\)
⇒ i = \(\frac{1350}{30,000}\) = \(\frac{135}{3000}\) = 0.045
Rate of interest (i) = 0.045 × 100% = 4.5%

X	\(\frac{1}{X}\)
1	1.0000
0.5	2.0000
10	0.1000
45	0.0222
175	0.0057
0.01	100.0000
4.0	0.2500
11.2	0.0893
	\(\sum { \left( \frac { 1 }{ X } \right) } \) = 103.4672

n = 8
\(HM=\frac { n }{ \sum { \left( \frac { 1 }{ X } \right) } } =\frac { 8 }{ 103.467 } =0.077\)

Let us consider X for price P and Y for stock S. Then the mean and SD for P is considered as \(\bar { X } \) = 100 and σ_x = 8 respectively and the mean and SD of S is considered as \(\bar { Y } \) = 103 and σ_y = 4. The correlation coefficient between the series is r(X, Y) = 0.4
Let the regression line X on Y be
\(X-\bar { X } =r\frac { { \sigma }_{ x } }{ { \sigma }_{ y } } (Y-\bar { Y } )\)
X-100 = (0.4)\(\frac{8}{4}\)(Y-103)
X–100 = 0.8(Y–103 )
X–0.8Y–17.6 = 0 (or)  X = 0.8Y+17.6
The regression line Y on X be \(Y-\bar { Y } =r\frac { { \sigma }_{ y } }{ { \sigma }_{ x } } (X-\bar { X } )\)
Y-103 = (0.4)\(\frac{4}{8}\)(X-100)
Y–103 = 0.2 (X–100 )
Y–103 = 0.2 X–20
Y = 0.2 X + 83 (or) 0.2 X–Y + 83 = 0

Marks are arranged in ascending order
12 15 20 28 30 40 50
n = number of observations = 7
Q₁ = Size of \(\left( \frac { n+1 }{ 4 } \right) ^{th}\)  value
= Size of \(\left( \frac { 7+1 }{ 4 } \right) ^{th}\) value
= Size of 2^nd value = 15
Q₃ = Size of \(\left( \frac { 3(n+1) }{ 4 } \right) ^{th}\) value
= Size of \(\\ \left( \frac { 3\times 8 }{ 4 } \right) ^{th}\) value
= Size of 6^th value = 40
D₆ = Size of \(\left( \frac { 6(n+1) }{ 10 } \right) \)^th value
= Size of \(\\ \left( \frac { 6\times 8 }{ 10 } \right) ^{th}\) value
= Size of 4.8^th value
= Size of 5^th value = 30
P₅₀ = Size of \(\left( \frac {50(n+1) }{ 100 } \right) ^{th}\) value
= Size of 4^th value = 28 
Hence Q₁ = 15, Q₃ = 40, D₆ = 30 and P₅₀ = 28

x	y	x2	y2	xy
5	1	25	1	5
10	6	100	36	60
5	2	25	4	10
11	8	121	64	88
12	5	144	25	60
4	1	16	1	4
3	4	9	16	12
2	6	4	36	12
7	5	49	25	35
1	2	1	4	2
\(\sum\)x = 60	\(\sum\)y = 40	\(\Sigma X^2=\) 494	\(\Sigma Y^2=\) 212	\(\Sigma XY=\) 288

Correlation Co-efficient r =\(\frac { N\sum { xy-(\sum { x)(\sum { y) } } } }{ \sqrt { N\sum { { x }^{ 2 }-({ \sum { x) } }^{ 2 }\times \sqrt { N{ \sum { y } }^{ 2 }-\left( { \sum { y } }^{ 2 } \right) } } } } \)
= \(\frac { 10(288)-(60)(40) }{ \sqrt { 10(494)-{ (60) }^{ 2 }\sqrt { 10(211)-{ (40) }^{ 2 } } } } \)
= \(\frac { 2880-2400 }{ \sqrt { 1340 } .\sqrt { 520 } } \)
= \(\frac { 480 }{ (36.61)(22.80) } =\frac { 480 }{ 834.71 } \)
r = 0.575

i) variables:
Let x₁ and x₂ denote the product A and B respectively.
ii) Objective function:
Profit on x₁ of the product A = 3x₁
Profit on x₂ of the product B = 4x₂
Total profit = 3x₁ + 4x₂
Let Z = 3x₁ + 4x₂, which is the objective function. Since the total profit is to be maximized, we have to maximize Z = 3x₁ + 4x₂.
(iii) Constraints:
We make the following table from the given data

Title	Requirement for A	Requirement for B	(min) Available time
(min) M1	1 min	2 min	7 x 60 + 30 = 450
(min) M2	1 min	1 min	10 x 60 = 600

 x₁ + x₂ \(\le \) 450 [\(\because\) Machine M₁ is available for not more than 7 hrs 30 minutes]
 2x₁ + x₂ \(\le \) 600 [\(\because\) Machine M₂ is available for 10 hrs during any working day]
(iv) Non-negative restrictions:
Since the product A and product B are non-negative, we have x₁, x₂\(\ge \) 0.
Thus, we have the following linear programming model.
Maximize \(Z=3{ x }_{ 1 }+{ 4x }_{ 2 }\)
subject to the constraints 
\(x_{ 1 }+{ x }_{ 2 }\le 450\)
\(2x_{ 1 }+{ x }_{ 2 }\le 600\)
\(x_1,x_2\ge0\)

\(A C =2 x-11+\frac{50}{x} \)
\(\frac{d(A C)}{d x} =2-\frac{50}{x^2} \)
\(\frac{d(A C)}{d x} =0 \Rightarrow 2-\frac{50}{x^2}=0 \Rightarrow 2=\frac{50}{x^2} \)
\(x^2 =25 \Rightarrow x=\pm 5\)
The real line is divided into 2 intervals namely  ( \(-\infty \), -5) (-5, 5) and  (5, \(\infty \))

Interval	Sign of \(\frac { d\left( AC \right) }{ dx } \)	Nature of function
(-5, 5)	-ve	Decreasing function
(5, \(\infty \))	+ ve	Increasing function
( \(-\infty \), -5)	-ve	Decreasing function

Average cost is increasing at x > 5

(i) Variables: Let x₁, x₂ and x₃ be the number of units of products P₁, P₂ and P₃ to be produced.
(ii) Objective function: Profit on x₁ units of the product P₁ = 20 x₁
Profit on x₂ units of the product P₂ = 25 x₂
Profit on x₃ units of the product P₃ = 15 x₃
Total profit = 20 x₁ + 25 x₂ + 15 x₃
Since the total profit is to be maximized, we have to maximize Z = 20 x₁ + 25 x₂ + 15 x₃
Constraints: 6x₁ + 3x₂ + 12x₃ ≤ 200
2x₁ + 5x₂ + 4x₃ ≤ 350
x₁ + 2x₂ + x₃ ≤ 100
Non-negative restrictions: Since the number of units of the products A, B and C cannot be negative, we have x₁, x₂, x₃ ≥ 0
Thus, we have the following linear programming model.
Maximize Z = 20 x₁ + 25 x₂ + 15 x₃
Subject to 6 x₁ + 3 x₂ + 12 x₃ ≤ 200
2x₁ + 5x₂ + 4x₃ ≤ 350
x₁ + 2x₂ + x₃ ≤ 100
x₁, x₂, x₃ ≥ 0

p = 40–x ⇒ x = 40–p
\({dx\over dp}=-1\)
Elasticity of demand: \(η_d=-{p\over x}.{dx\over dp}\)\(={40-x\over x}\)
Given that η_d = 1
\(∴\ {40-x\over x}=1\) ⇒ x = 20
∴ output x = 20 units.

\(\sin A=\frac{12}{13}\)
\(\sin 3 A=3 \sin A-4 \sin ^3 A\)
\(=3\left(\frac{12}{13}\right)-4\left(\frac{12}{13}\right)^3\)
\(=\frac{36}{13}-\frac{6912}{2197}=\frac{-828}{2197}\)

LHS\(=\left\{1+\cot x-\sec\left(\frac{\pi}{2}+x\right)\right\}\left\{1+\cot x+\sec\left(\frac{\pi}{2}+x\right)\right\}\)
=(1 + cot x + cosec x) (1 + cot x - cosec x)
(1 + cot x)² - (cosec²x) [\(\therefore\)(a + b) (a - b) = a² - b²]
1 + cot²x + 2 cot x - cosec²x
= cosec²x + 2 cot x - cosec²x [\(\therefore\) 1 + cot²x = cosec²x]
= 2 cot x = RHS. Hence proved.

We know that \(\cos^2x+\sin^2x=1\)
\(\Rightarrow\cos x=\pm\sqrt{1-\sin^2x}\)
In the III quadrant, cos x is negative
\(\therefore\cos x=-\sqrt{1-\sin^2x}=-\sqrt{1-\frac{24}{25}}=-\frac15\left[\because\sin^2x=\frac{4(6)}{25}=\frac{24}{25}\right]\)
In the III quadrant, tan x is positive
\(\therefore\tan x=\frac{\sin x}{\cos x}=\frac{-2\sqrt6}{5}\times\frac{-5}{1}=2\sqrt{6}\)
\(cosec x=\frac{1}{\sin x}=\frac{-5}{2\sqrt6}\)
\(\sec x=\frac{1}{\cos x}=-5\) and
\(\cot x=\frac{1}{\tan x}=\frac{1}{2\sqrt6}\)

\(y=\sin ^{ -1 }{ \left( \sqrt { \cos { x } } \right) } \)
Differentiating with respect to 'x' we have
\(\frac { dy }{ dx } =\frac { d }{ dx } .\sin ^{ -1 }{ { \left( \cos { x } \right) }^{ \frac { 1 }{ 2 } } } =\frac { 1 }{ \sqrt { 1-{ \left( \sqrt { \cos { x } } \right) }^{ 2 } } } .\frac { d }{ dx } { \left( \cos { x } \right) }^{ \frac { 1 }{ 2 } }\)
\(=\frac { 1 }{ \sqrt { 1-\cos { x } } } .\frac { 1 }{ 2\sqrt { 1-\cos { x } } } .\frac { d }{ dx } \left( \cos { x } \right) \)
\(=\frac { 1 }{ 2\sqrt { \cos { x } } .\sqrt { 1-\cos { x } } } \left( -\sin { x } \right) =\frac { -\sin { x } }{ 2\sqrt { \cos { x } } .\sqrt { 1-\cos { x } } } \)

Given lines are (b - c) x + (c - a) y + (a - b) = 0
(c - a) x + (a - b) y + (b - c) = 0
(a - b) x + (b - c) y + (c - a) = 0.
The condition for the given lines to be concurrent is
\(\left| \begin{matrix} b-c & c-a & a-b \\ c-a & a-b & b-c \\ a-b & b-c & c-a \end{matrix} \right| =0\)
Applying the elementary gains formation C₁⟶C₁+C₂+C₃
We get \(\left| \begin{matrix} 0 & c-a & a-b \\ 0 & a-b & b-c \\ 0 & b-c & c-a \end{matrix} \right| =0\)
Expanding along C₁ we get
\(\left| \begin{matrix} 0 & c-a & a-b \\ 0 & a-b & b-c \\ 0 & b-c & c-a \end{matrix} \right| =0\)
Hence the given lines are Concurrent

let p(x₁,y₁) be any point on the locus ​\(\therefore\)​  \(\sqrt { { \left( { x }_{ 1 }-4 \right) }^{ 2 }+{ \left( { y }_{ 1 }-0 \right) }^{ 2 } } =\frac { 1 }{ 2 } \left| \frac { { x }_{ 1 }-16 }{ \sqrt { { 1 }^{ 2 }+{ 0 }^{ 2 } } } \right| \)
\(\Rightarrow\)\(\sqrt { { \left( { x }_{ 1 }-4 \right) }^{ 2 }+{ y }_{ 1 }^{ 2 } } =\frac { 1 }{ 2 } \left| \frac { { x }_{ 1 }-16 }{ \sqrt { { 1 }^{ 2 }+{ 0 }^{ 2 } } } \right| \)
Squaring  both sides, (x₁ - 4)² + y₁² = \(\frac{1}{4}\)(x₁ - 16)²
⇒ 4[x₁² + 16 - 8 x₁ + y₁²] = x₁² - 32x₁ + 256
⇒ 3x₁² + 4y₁² - 192 = 0
Locus of (x₁, y₁) is 3x² + 4y² = 192

Given \(y=\sqrt { x } +\frac { 1 }{ \sqrt { x } } .\Rightarrow y=\frac { x+1 }{ \sqrt { x } } \)  ......(1)
Differentiating with respect to 'x' we get,
\(\frac { dy }{ dx } =\frac { \sqrt { x } .\frac { d }{ dx } (x+1)-(x+1).\frac { d }{ dx } (\sqrt { x } ) }{ { (\sqrt { x } })^{ 2 } } \)
\(=\frac { \sqrt { x } (1)-(x+1).\frac { 1 }{ 2\sqrt { x } } }{ x } =\frac { \frac { 2x-(x+1) }{ 2\sqrt { x } } }{ x } \)
\(\frac { dy }{ dx } =\frac { 2x-x-1 }{ 2x.\sqrt { x } } =\frac { x-1 }{ 2x\sqrt { x } } \)...(2)
LHS=\(2x\left( \frac { dy }{ dx } \right) +y\)
\(=2x\left( \frac { x-1 }{ 2x\sqrt { x } } \right) +\frac { x+1 }{ \sqrt { x } }\)
 [From (1) and (2)]
\(=\frac { x-1 }{ \sqrt { x } } +\frac { x+1 }{ \sqrt { x } } =\frac { x-1+x+1 }{ \sqrt { x } } =\frac { 2x }{ \sqrt { x } } =\frac { 2\sqrt { x } .\sqrt { x } }{ \sqrt { x } } =2\sqrt { x } \) = RHS.
Hence proved

P(n): "n² + n is even"
Given that P(k) is true.
\(\Rightarrow \)   k² + k is even
\(\Rightarrow \)  k² + k = 2\(\lambda \) for some \(\lambda \)  ....(1)
To prove that P (k + 1) is true.
P (k + 1) : (k + 1)²+ (k + 1) is even
Consider (k + 1)² + (k + 1)
= k²+2k+ 1 +k+ 1
= k²+2k+ 1 +k+ 1
= (k²+ k) + (2 k + 2)
= 2\(\lambda \) + 2 (k + 1)   [from (1)]
= even + even
= Sum of two even numbers is always an even number .
∴  P (k + 1) is true.

No. of ways of selecting 4 questions out of 6 questions (Part - I) = 6C₄ = 6C₂ =\(\frac { 6\times 5 }{ 1\times 2 } =15\)
No. of ways of selecting 4 questions out of 7 questions (Part - II) = 7C₄ = 7C₃ =\(\frac { 7\times 6\times 5 }{ 3\times 2\times 1 } =35\)
No. of ways of selecting 4 questions out of 8 questions (Part - III) = 8 C₄ =\(\frac { 8\times 7\times 6\times 5 }{ 4\times 3\times 2\times 1 } =70\)
∴ Total number of choices = 15 x 35 x 70 = 36750

Given equation of the parabola is
\({ y }^{ 2 }-8y-8x+24=0\)
\(\Rightarrow\) \({ y }^{ 2 }-8y=8x-24=0\)
\(\Rightarrow\)\({ y }^{ 2 }-8y+16=8x-24+16\)   [Adding 16 both sides]
\(\Rightarrow { \left( y-4 \right) }^{ 2 }=8x-8\)
\(\Rightarrow { \left( y-4 \right) }^{ 2 }=8(x-1)\)
X = x - 1 and Y = y - 4, 4a = 8 \(\Rightarrow\) a = 2
\(\Rightarrow\)  \(y^{ 2 }=8X\) 
 y² = 8X with y² = 4(2)X

		Referred to (x, y)	Referred to (x,y) x = x + 1, y = y + 4a
(i)	Vertex	(0, 0)	(1, 4)
(ii)	Focus	(a, 0),(2, 0)	(3, 4)
(iii)	Axis	Y = 0	y = 4
(iv)	Directrix	X = -a; X = -2	x = -1; x + 1 = 0
(v)	Length of latus rectum	4a	8 units

\(\sec \left(\frac{3 \pi}{2}-\theta\right) \sec \left(\theta-\frac{5 \pi}{2}\right) +\tan \left(\frac{5 \pi}{2}+\theta\right) \tan \left(\theta-\frac{5 \pi}{2}\right)=-1 \)
\(\operatorname{LHS}= \sec \left(\frac{3 \pi}{2}-\theta\right) \sec \left(\theta-\frac{5 \pi}{2}\right) +\tan \left(\frac{5 \pi}{2}+\theta\right) \tan \left(\theta-\frac{5 \pi}{2}\right) \)
\(= \sec \left(270^{\circ}-\theta\right) \sec \left(450^{\circ}-\theta\right) +\tan \left(450^{\circ}+\theta\right)\left(-\tan \left(450^{\circ}-\theta\right)\right) \)
\(=-\operatorname{cosec} \theta \sec \left(360^{\circ}+90^{\circ}-\theta\right) -\tan \left(360^{\circ}+90^{\circ}+\theta\right) \tan \left(360^{\circ}+90^{\circ}-\theta\right) \)
\(=-\operatorname{cosec} \theta \sec \left(90^{\circ}-\theta\right) -\tan \left(90^{\circ}+\theta\right) \tan \left(90^{\circ}-\theta\right) \)
\(=-\operatorname{cosec} \theta \operatorname{cosec} \theta+\cot \theta \cot \theta \)
\(=-\left(\operatorname{cosec}{ }^2 \theta-\cot { }^2 \theta\right)=-1 = \text { RHS } \)
Hence proved.

Let A = \(\left| \begin{matrix} 2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86 \end{matrix} \right| \)
Applying C₁➝C₁+9C₂ we get
A = \(\left| \begin{matrix} 2+63 & 7 & 65 \\ 3+72 & 7 & 65 \\ 5+81 & 9 & 86 \end{matrix} \right| =\left| \begin{matrix} 65 & 7 & 65 \\ 75 & 8 & 75 \\ 86 & 9 & 86 \end{matrix} \right| =0[ \because {C}_{1}\equiv{C}_{3}]\)

LHS = \(\begin{vmatrix}x+a &b&c \\a &x+b&c\\a&b&x+c \end{vmatrix}\)
Applying C₁ \(\rightarrow\) C₁ + C₂ + C₃ we get,
LHS = \(\begin{vmatrix}x+a+b+c &b&c \\x+a+b+c &x+b&c\\x+a+b+c&b&x+c \end{vmatrix}\)
Taking (x + a + b + c) common from C₁ we get,
= (x + a + b + c) \(\begin{vmatrix}1 &b&c \\1 &x+b&c\\1&b&x+c \end{vmatrix}\)
Applying R₂ \(\rightarrow\) R₂ - R₁ and R₃ \(\rightarrow\) R₁ we get,
= \((x+a+b+c)\begin{vmatrix}1 &b&c \\0&x&0\\0&0&x\end{vmatrix}\)
Expanding along C₁ we get,
\(=(x+a+b+c)\begin{bmatrix} 1\begin{bmatrix} x & 0 \\ 0 & x\end{bmatrix} +0+0\end{bmatrix}\)
= (x + a + b + C)(x²) = RHS
Hence proved.

\(\frac{5 x+7}{(x-1)(x+3)}=\frac{A}{x-1}+\frac{B}{x+3}\)
\(\frac{5 x+7}{(x-1)(x+3)}=\frac{A(x+3)+B(x-1)}{(x-1)(x+3)}\)
\(\text {If } x=1 \Rightarrow 12=A(4)\)
\(\dot{A}=\frac{12}{4}=3\)
\(\text {If } x=-3\)
\(-15+7=B(-3-1)\)
\(-8=-4 B \Rightarrow B=\frac{8}{4}=2\)
\(\frac{5 x+7}{(x-1)(x+3)}=\frac{3}{x-1}+\frac{2}{x+3}\)

(i) Number of one digit number using 5, 6, 7, 8 and 9 is 5
(ii) Number of two digit number using 5, 6, 7, 8 and 9 is 5 \(\times\) 4 = 20
(iii) Number of three digit number using 5, 6, 7, 8 and 9 is 5 \(\times\) 4 \(\times\) 3 = 60
(iv) By addition principle of counting total number of numbers = 5 + 20 + 60 = 85

\(\left|\begin{array}{ccc} 3 & -2 & 4 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{array}\right|=3[0-2]+2[6-1]+4[4-0]\)
\(=-6+10+16=20\)