Given lines 3x - 4y - 13 = 0...(1)
8x - 11y = 33  ...(2)
2x - 3y - 7 = 0 ...(3)
Conditon for concurrent lines is
\(\left| \begin{matrix} { a }_{ 1 } & { b }_{ 1 } & { c }_{ 1 } \\ { a }_{ 2 } & { b }_{ 2 } & { c }_{ 2 } \\ { a }_{ 3 } & { b }_{ 3 } & { c }_{ 3 } \end{matrix} \right| =0\) 
i.e.,\(\left| \begin{matrix} 3 & -4 & -13 \\ 8 & -11 & -33 \\ 2 & -3 & -7 \end{matrix} \right| =3\left( 77-99 \right) +4\left( 56+44 \right) -13\left( -24+22 \right) \) 
= -66 + 40 + 26 = 0
\(\Rightarrow \) Given lines are concurrent. To get the point of concurrency solve the equations (1) and (3)

When y = 5 from (2) 8x = 88
x = 11
Point of concurrency is (11, 5)

Given e^y(x+1)=1    ....(1)
Differentiating with respect to 'x' we get,
\({ e }^{ y }(1)+(x+1){ e }^{ y }\frac { dy }{ dx } =0\)   [product rule]
\(\Rightarrow { e }^{ y }+(1)\frac { dy }{ dx } =0\quad [using\quad (1)]\)
\(\Rightarrow \frac { dy }{ dx } =-{ e }^{ y }\)...(2)
Differentiating again with respect to 'x' we get,
\(\frac { d }{ dx } \left( \frac { dy }{ dx } \right) =\frac { d }{ dx } \left( -{ e }^{ y } \right) \)
\(\Rightarrow \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =-{ e }^{ y }.\frac { dy }{ dx } \)
\(=\left( \frac { dy }{ dx } \right) \left( \frac { dy }{ dx } \right) \) [using (2)]
\(={ \left( \frac { dy }{ dx } \right) }^{ 2 }\)
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } ={ \left( \frac { dy }{ dx } \right) }^{ 2 }\)
Hence proved.

The circle touches the co-ordinate axes and the line x = a is shown in the diagram.
\(\therefore \ centre\ is\ \left( \frac { a }{ 2 } ,\frac { a }{ 2 } \right) and\ r=\frac { a }{ 2 } \)
\(\therefore\) There may be two such circles, one lying

above X-axis and other below X-axis.
Equation of the circle lying above the X-axis is
\({ \left( x-\frac { a }{ 2 } \right) }^{ 2 }+{ \left( y+\frac { a }{ 2 } \right) }^{ 2 }={ \left( \frac { a }{ 2 } \right) }^{ 2 }\)

Equation of the circle lying below the X-axis is \({ \left( x-\frac { a }{ 2 } \right) }^{ 2 }+{ \left( y+\frac { a }{ 2 } \right) }^{ 2 }={ \left( \frac { a }{ 2 } \right) }^{ 2 }\)

y= cos³x
\(\frac{d y}{d x}=-3 \cos ^2 x \sin x=\frac{-3}{2} \cos x(2 \sin x \cos x)\)
\(=\frac{-3}{2} \cos x \sin 2 x\)

Let y = \(\frac { { e }^{ x } }{ 1+x } \)
\(\frac{d y}{d x}=\frac{(1+x) e^x-e^x(1)}{(1+x)^2}=\frac{e^x+x e^x-e^x}{(1+x)^2}=\frac{x e^x}{(1+x)^2}\)

Let y = (ax² + bx + c)ⁿ
dy/dx = n(ax² + bx + c)^n-1(2ax + b)

Given equations are 2x + 5y = 1; 3x + 2y = 7
This system of equations can be written in matrix form as \(\begin{pmatrix}2 & 5 \\3 & 2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} 1\\7 \end{pmatrix}\Rightarrow Ax=B\)
where A = \(\begin{pmatrix} 2 & 5 \\3 & 2 \end{pmatrix},X=\begin{pmatrix} x\\y \end{pmatrix}\) and B = \(\begin{pmatrix} 1 \\ 7 \end{pmatrix}\)
\(\therefore\)  X = A^-1 B.
\(|A|=\begin{vmatrix} 2 &5 \\3 & 2 \end{vmatrix}=4-15=-11\)
A₁₁ = 2, A₁₂ = -3, A₂₁ = -5, A₂₂ = 2
\(\therefore\)   adj A = \(\begin{bmatrix} 2 & -3 \\ -5 & 2 \end{bmatrix}^{T}=\begin{bmatrix} 2 & -5 \\-3 & 2 \end{bmatrix}\)
\(\therefore\)  \({A}^{-1}={{1}\over{|A|}}\) . adj A = \({{-1}\over{11}}\begin{bmatrix} 2 & -5\\ -3& 2 \end{bmatrix}\)
X = A^{- 1} B \(={{-1}\over{11}}\begin{bmatrix} 2 & -5\\ -3&2 \end{bmatrix}\begin{bmatrix} 1\\7 \end{bmatrix}\)
\(=\frac { 1 }{ 11 } \left[ \begin{matrix} +2-35 \\ -3+14 \end{matrix} \right] ={{-1}\over{11}}\begin{bmatrix} -33\\11 \end{bmatrix}=\begin{bmatrix} 3\\-1 \end{bmatrix}\)
\(\therefore\)  x = 3.and y = -1.

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

\(\frac{d y}{d x}=1, \frac{d z}{d x}=\frac{-1}{x^2} \quad \frac{d y}{d z}=-x^2\)

\(f(x+1) =(x+1)^2-(x+1)+1 \)
\(=x^2+2 x+1-x-1+1 \)
\(=x^2+x+1 \)

\(p =\frac{\tan 50^{\circ}}{\sec 50^{\circ}}=\frac{\sin 50^{\circ} / \cos 50^{\circ}}{1 / \cos 50^{\circ}} =\sin 50^{\circ}\)

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

\(h^2-a b=0\)
\(4-2 a=0 \Rightarrow a=2\)

(parallel line)

The ninth digit cannot be filled with zero
Ninth place can be filled in 9 ways
Eighth place can be filled in 9 ways
Seventh place can be filled in 8 ways and so on No. of ways
= 9 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2
= 9 x 9!

The number of parallelograms = 4C₂ x 3C₂

\(n P_r =720(\mathrm{nCr}) \)
\(\frac{n !}{(n-r) !} =720\left(\frac{n !}{r !(n-r) !}\right) \)
\(r ! =720 \)
\(r ! =6 ! \)
r = 6

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

(Since \(\mathrm{X}=\mathrm{A}^{-1} \mathrm{~B}\) is the correct answer)

\(\left|\begin{array}{lll} 0 & 1 & 0 \\ x & 2 & x \\ 1 & 3 & x \end{array}\right|=0 \Rightarrow-1\left[x^2-x\right]=0\)
\(\Rightarrow x(x-1)=0 \Rightarrow x=0,1\)

Let f(x) = \({ x }^{ \frac { 2 }{ 3 } }\)
f(x+h) = (x+h)\(^{ \frac { 2 }{ 3 } }\)
\(\frac { d }{ dx } (f(x))=\underset { h\rightarrow 0 }{ lim } \frac { f(x+h)-f(x) }{ h } \)
= \(\underset { h\rightarrow 0 }{ lim } \frac { (x+h)^{ \frac { 2 }{ 3 } }-x^{ \frac { 2 }{ 3 } } }{ x+h-x } \) [adding and subtracting x in the denominator]
= \(\frac { 2 }{ 3 } .x^{ \frac { 2 }{ 3 } -1 }\)
\(\left[ \therefore \underset { x\rightarrow a }{ Lt } \frac { { x }^{ n }-{ a }^{ n } }{ x-a } =n-a^{ n-1 } \right] =\frac { 2 }{ 3 } { x }^{ \frac { -1 }{ 3 } }\)
\(\therefore \frac { d }{ dx } \left( x^{ \frac { 2 }{ 3 } } \right) -\frac { 2 }{ 3 } .x^{ \frac { -1 }{ 3 } }\)

\(\cos\left(\frac{5\pi}{12}\right)=\cos\left(\frac{\pi}{4}+\frac{\pi}{6}\right)\)
\(=\cos\frac{\pi}{4}\cos\frac{\pi}{6}-\sin\frac{\pi}{4}\sin\frac{\pi}{6}[\therefore \cos(A+B)=\cos A\cos B-\sin A\sin B]\)
\(=\frac{1}{\sqrt2}\times\frac{\sqrt3}{2}-\frac{1}{\sqrt2}\times\frac{1}{2}=\frac{\sqrt3-1}{2\sqrt2}\)

Given pair of lines is
3x²+10xy+8y²+14x+22y+15=0
2h=10
Here a=3, h=5, b=8,
Let \(\theta\) be the angle between the pair of lines
Then \(tan\quad \theta =\frac { \pm 2\sqrt { { h }^{ 2 }-ab } }{ a+b } =\frac { \pm 2\sqrt { 25-3(8) } }{ 3+8 } =\frac { \pm 2\sqrt { 1 } }{ 11 } =\frac { \pm 2 }{ 11 } \)
\(\therefore \quad tan\quad \theta =\frac { 2 }{ 11 } \Rightarrow \theta ={ tan }^{ -1 }\left( \frac { 2 }{ 11 } \right) \)

Given f(x) = 5x - \(\left| x \right| \)
\(\therefore f(x)=\begin{cases} 5x-x\quad if\quad x\ge 0 \\ 5x-(-x)\quad if\quad x>0 \end{cases}=\begin{cases} 4x\quad ifx\ge 0 \\ 6x\quad if\quad x<0 \end{cases}\)
\(L\left[ f\left( x \right) \right] _{ x=0 }=\underset { x\rightarrow 0 }{ lim }f\left( x \right) =\underset { h\rightarrow 0- }{ lim } \quad f(o-h)\)
\(=\underset { h\rightarrow 0 }{ lim } f(-h)=\underset { h\rightarrow 0 }{ lim } 6(-h)\quad \quad \left[ \therefore f(x)=6x\quad ifx\ge 0 \right] \)
= 0
\(R\left[ f\left( x \right) \right] _{ x=0 }=\underset { x\rightarrow 0 }{ lim } lif\left( x \right) =\underset { h\rightarrow 0 }{ lim } \quad f(o+h)\)
\(\underset { h\rightarrow 0 }{ lim } f(h)=\underset { h\rightarrow 0 }{ lim } 4(h)\quad \left[ \therefore f(x)=4xi\quad fx\ge 0 \right] \)
=0
∴ \(L\left[ f\left( x \right) \right] _{ x=0 }=R\left[ f\left( x \right) \right] _{ x=0 }\)
∴ f(x) is continuous at x = 0

Let |A| = \(\begin{vmatrix} 6 &5 &12 \\ 2 & 4 &4 \\2 & 1 & 4 \end{vmatrix}\)
Taking 2 common from C₁ and 4 common from C₃, we get,
\(|A|=2\times4\begin{vmatrix} 3 & 5&3 \\ 1 & 4 & 1\\1 &1 &1\end{vmatrix}=8\times 0\ [\because C_1\equiv C_3]=0\)

The technology matrix is
\(B=\begin{bmatrix} 0.50 & 0.25 \\ 0.40 & 0.67 \end{bmatrix}\)
I - B =\(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}-\begin{bmatrix} 0.50 & 0.25 \\ 0.40 & 0.67 \end{bmatrix}=\begin{bmatrix} 0.50 & -0.25 \\ -0.40 & 0.33 \end{bmatrix}\)
\(|\mathrm{I}-\mathrm{B}|\) = 0.165 - 0.1 = 0.065 > 0
Since the diagonal elements of (I - B) are positive and |I - B| is positive, Hawkins - Simon conditions are satisfied.
Therefore the given system is viable.

Let \(y=\sqrt { \cfrac { \left( x-3 \right) \left( { x }^{ 2 }+4 \right) }{ { 3x }^{ 2 }+4x+5 } } \) 
= \(\left| \cfrac { \left( x-3 \right) \left( { x }^{ 2 }+4 \right) }{ { 3x }^{ 2 }+4x+5 } \right| ^{ \frac { 1 }{ 2 } }\) 
Taking logarithm on both sides,
\(logy=\cfrac { 1 }{ 2 } \left[ log\left( x-3 \right) +log\left( { x }^{ 2 }+4 \right) -log\left( { 3x }^{ 2 }+4x+5 \right) \right] \)\({[\because \log a b} =\log a+\log b \text { and } \log \frac{a}{b} =\log a-\log b]\) 
Differentiating with respect to x
\(\cfrac { 1 }{ y } .\cfrac { dy }{ dx } =\cfrac { 1 }{ 2 } \left[ \cfrac { 1 }{ x-3 } +\cfrac { 2x }{ { x }^{ 2 }+4 } -\cfrac { 6x+4 }{ { 3x }^{ 2 }+4x+5 } \right] \)
\(\cfrac { dy }{ dx } =\cfrac { 1 }{ 2 } \sqrt { \cfrac { \left( x-3 \right) \left( { x }^{ 2 }+4 \right) }{ { 3x }^{ 2 }+4x+5 } } \)\( \left[ \cfrac { 1 }{ x-3 } +\cfrac { 2x }{ { x }^{ 2 }+4 } -\cfrac { 6x+4 }{ { 3x }^{ 2 }+4x+5 } \right] \)

cosec A + sec A = cosec B + sec B
cosec A - cosec B  = sec B - sec A
\(\frac{1}{\sin A}-\frac{1}{\sin B}=\frac{1}{\cos B}-\frac{1}{\cos A}\)
\(\frac{\sin B-\sin A}{\sin A \sin B}=\frac{\cos A-\cos B}{\cos B \cos A}\)
\(\frac{\sin B-\sin A}{\cos A-\cos B}=\frac{\sin A \sin B}{\cos A \cdot \cos B}\)
\(\frac{2 \sin \frac{B-A}{2} \cos \frac{B+A}{2}}{-2 \sin \frac{A-B}{2} \sin \frac{A+B}{2}}=\tan A \tan B\)
\(\cot \left(\frac{A+B}{2}\right)=\tan A \tan B\)

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

Let \(y=\frac { sinx+cosx }{ sinx-cosx } \)
Differentiating with respect to 'x' we get,
\(\frac { dy }{ dx } =\frac { (sinx-cosx).\frac { d }{ dx } (sinx+cosx)-(sinx+cosx).\frac { d }{ dx } (sinx-cosx) }{ { (sinx-cosx) }^{ 2 } } \)
\(=\frac { (sinx-cosx)(cosx-sinx)-(sinx+cosx)(cosx+sinx) }{ { (sinx-cosx) }^{ 2 } } \)
\(=\frac { (sinxcosx-{ sin }^{ 2 }x-cos^{ 2 }x+sinx.cosx)-({ sin }^{ 2 }x+cos^{ 2 }x+2sinxcosx) }{ { (sinx-cosx) }^{ 2 } } \)
\(=\frac { 2sinxcox-1-1-2sinxcox }{ { (sinx-cosx) }^{ 2 } } \)
\(=\frac { -2 }{ { (sinx-cosx) }^{ 2 } } \left[ \therefore sin^{ 2 }x+cos^{ 2 }x=1 \right] \)

Step-1 : Let P (n) denote the statement a - b is divisible by a - b for all \(n\in N\)
∴ a¹ - b¹ = a - b is divisible by a - b
Put n = 1
P(1) : a - b is divisible by a - b
∴ P(1) is true.
Step-2:
Let us assume that P(k) is true.
∴ a^k - b^k is divisible by a - b
⇒ a^k - b^k = C(a - b) where C is a constant     ...(1)
Step-3:
To prove that P(k+1) is true
P(k + 1) = a^k+1 - b^k+1
= a^k+1 - a^kb + a^kb - b^k+1  [Adding and subtracting ak b]
= a^k (a - b) + b (a^k - b^k)
= a^k (a - b) + b.c. (a - b) [using (1)]
= (a - b) (a^k + bc) which is divisible by a - b.
∴ P (k + 1) is true whenever P(k) is true
∴ P(n) is true for all \(n\in N\).

\(|A|=\left| \begin{matrix}1 & tan\quad x \\ -tan\quad x & 1 \end{matrix} \right| =1+{ tan }^{ 2 }x={ sec }^{ 2 }x\neq 0\)
\(\Rightarrow\) A^-1 exists
Let C_ij be the cofactor of a_ij  in A
C₁₁ = (-1)¹⁺¹ M₁₁ = (-1)²(1) = 1
C₁₂ = (-1)¹⁺² (-tan x) = tan x
C₂₁ = (-1)²⁺²(1) = 1
\(\therefore \quad adj\quad A={ \left[ \begin{matrix} 1 & tan\quad x \\ -tan\quad x & 1 \end{matrix} \right] }^{ T }=\left[ \begin{matrix} 1 & -tan\quad x \\ tan\quad x & 1 \end{matrix} \right] \)
\({ A }^{ -1 }=\frac { 1 }{ |A| } adjA=\frac { 1 }{ 1+{ tan }^{ 2 }x } \left[ \begin{matrix} 1 & -tan\quad x \\ tan\quad x & 1 \end{matrix} \right] =\left[ \begin{matrix} \frac { 1 }{ 1+{ tan }^{ 2 }x } & \frac { -tan\quad x }{ 1+{ tan }^{ 2 }x } \\ \frac { tan\quad x }{ 1+{ tan }^{ 2 }x } & \frac { 1 }{ 1+{ tan }^{ 2 }x } \end{matrix} \right] \)
\(\therefore \quad { A }^{ T }{ A }^{ -1 }=\left[ \begin{matrix} 1 & -tan\quad x \\ tan\quad x & 1 \end{matrix} \right] \left[ \begin{matrix} \frac { 1 }{ 1+{ tan }^{ 2 }x } & \frac { -tan\quad x }{ 1+{ tan }^{ 2 }x } \\ \frac { tan\quad x }{ 1+{ tan }^{ 2 }x } & \frac { 1 }{ 1+{ tan }^{ 2 }x } \end{matrix} \right] =\left[ \begin{matrix} \frac { 1 }{ 1+{ tan }^{ 2 }x } \frac { -tan\quad x }{ 1+{ tan }^{ 2 }x } & \frac { -tan\quad x }{ 1+{ tan }^{ 2 }x } \frac { -tan\quad x }{ 1+{ tan }^{ 2 }x } \\ \frac { tan\quad x }{ 1+{ tan }^{ 2 }x } +\frac { tan\quad x }{ 1+{ tan }^{ 2 }x } & \frac { -{ tan }^{ 2 }x }{ 1+{ tan }^{ 2 }x } \frac { 1 }{ 1+{ tan }^{ 2 }x } \end{matrix} \right] \)
\(=\left[ \begin{matrix} \frac { 1-tan\quad x }{ 1+{ tan }^{ 2 }x } & \frac { -2tan\quad x }{ 1+{ tan }^{ 2 }x } \\ \frac { 2tan\quad x }{ 1+{ tan }^{ 2 }x } & \frac { 1-{ tan }^{ 2 }x }{ 1+{ tan }^{ 2 }x } \end{matrix} \right] =\left[ \begin{matrix} cos\quad 2x & -sin2x \\ sin\quad 2x & cos\quad 2x \end{matrix} \right] \)  (Using multiple angle formula)

Let A \(=\begin{vmatrix} 1 & a&a^2&-bc \\1 &b&{b}^{2}&-ca\\1&c&c^2&-ab \end{vmatrix}\)
\(=\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|+\left|\begin{array}{ccc} 1 & a & -b c \\ 1 & b & -c a \\ 1 & c & -a b \end{array}\right|\)
\(A=\begin{vmatrix} 1 & a&{a}^{2} \\ 1 &b&b^2\\1&c&c^2 \end{vmatrix}-\begin{vmatrix} 1 & a&bc \\1 &b&ca\\1&c&ab \end{vmatrix}\)
\(=\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|-\frac{1}{a b c}\left|\begin{array}{ccc} a & a^2 & a b c \\ b & b^2 & a b c \\ c & c^2 & a b c \end{array}\right|\)
(Multiplying R₁, R₂ and R₃ of II det by a, b, c respectively)
\(=\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|-\frac{a b c}{a b c}\left|\begin{array}{lll} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{array}\right|\)
\(\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|-\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|=0\)