\( \sqrt{x^2+y^2} =\tan ^{-1} \frac{y}{x} \)
\(\left(x^2+y^2\right)^{1 / 2} =\tan ^{-1} \frac{y}{x}\)
\( \frac{1}{2 \sqrt{x^2+y^2}}\left(2 x+2 y \frac{d y}{d x}\right)=\frac{1}{1+\left(\frac{y}{x}\right)^2}\left(\frac{x \frac{a y}{d x}-y}{x^2}\right) \)
\(\frac{2\left(x+y \frac{d y}{d x}\right)}{2 \sqrt{x^2+y^2}}=\frac{1}{\left(\frac{x^2+y^2}{x^2}\right)}\left(\frac{x \frac{d y}{d x}-y}{x^2}\right)\)
\( x+y \frac{d y}{d x}=\frac{\sqrt{x^2+y^2}}{\left(x^2+y^2\right)}\left(x \frac{d y}{d x}-y\right) \)
\(x+y \frac{d y}{d x}=\frac{1}{\sqrt{x^2+y^2}}\left(x \frac{d y}{d x}-y\right) \)
\(x+y \frac{d y}{d x}=\frac{x}{\sqrt{x^2+y^2}} \frac{d y}{d x}-\frac{y}{\sqrt{x^2+y^2}}\)
\( \frac{x}{\sqrt{x^2+y^2}} \frac{d y}{d x}-y \frac{d y}{d x} =x+\frac{y}{\sqrt{x^2+y^2}}\)
\(\frac{d y}{d x}\left(\frac{x}{\sqrt{x^2+y^2}}-y\right) =\frac{x \sqrt{x^2+y^2}+y}{\sqrt{x^2+y^2}} \)
\(\frac{d y}{d x}\left(\frac{x-y \sqrt{x^2+y^2}}{\sqrt{x^2+y^2}}\right) =\frac{x \sqrt{x^2+y^2}+y}{\sqrt{x^2+y^2}} \)
\(\frac{d y}{d x} =\frac{x \sqrt{x^2+y^2}+y}{x-y \sqrt{x^2+y^2}}\)

The order of A is 3 \(\times\) 3 and the order of B is 3 \(\times\) 2. Therefore the order of AB is 3 \(\times\) 2.
A and B are conformable for the product AB. Call C = AB. Then,
c₁₁ = (first row of A) (first column of B)
\(\Rightarrow c_{11}=[1\ -1 \ 2 ]\begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}=1+1+2=4,\)since c₁₁ is an element.
Similarly c₁₂ = 0, c₂₁ = 0, C₂₂ = 13, c₃₁ = 7, c₃₂ = 5.
Therefore, AB = C = [c_ij] =\(\begin{bmatrix} 4 &0 \\ 0 & 13 \\ 7 & 5 \end{bmatrix}\)
The product BA does not exist, because the number of columns in B is not equal to the number of rows in A.

\(\frac { b+c }{ a } sin\frac { A }{ 2 } =\frac { 2RsinB+2RsinC }{ 2RsinA } sin\frac { A }{ 2 } \)
\(=\frac { sinB+sinC }{ sinA } sin\frac { A }{ 2 } \)
\(=\frac { 2sin\frac { B+C }{ 2 } cos\frac { B-C }{ 2 } }{ 2sin\frac { A }{ 2 } cos\frac { A }{ 2 } } sin\frac { A }{ 2 } =\frac { sin\frac { B+C }{ 2 } cos\frac { B-C }{ 2 } }{ cos\frac { A }{ 2 } } \)
\(=\frac { sin\left( \frac { 180-A }{ 2 } \right) cos\frac { B-C }{ 2 } }{ cos\frac { A }{ 2 } } =\frac { sin\left( 90-\frac { A }{ 2 } \right) cos\frac { B-C }{ 2 } }{ cos\frac { A }{ 2 } } \)
\(=cos\frac { B-C }{ 2 } \ \because sin\left( 90-\frac { A }{ 2 } \right) =cos\frac { A }{ 2 } \) 

If the line 2x - y = 5 makes an angle \(\theta\) with x - axis. Then, tan \(\theta\) = 2. Let P (a, a) be a point on the line 2x - y = 5. Then, 2 a - a = 5 \(\Rightarrow\) a = 5

So, the coordinates of Pare (5, 5). If the line 2x - y - 5 = 0 is rotated about point P through 45° in anti-clockwise direction, then the line in its new position makes angle 8 + 45° with x -axis. Let m be the slope of the line in its new position. Then,
 \(m'=tan(\theta+45^o)=\frac{\tan\theta+\tan45^o}{1-\tan\theta\tan45^o}=\frac{2+1}{1-2\times 1}=-3\)
Thus, the line in its new position passes through P (5, 5) and has slope m' = -3
So, its equationy -5 = m' (x - 5) or, y -5 = -3 (x - 5) or, 3x + y - 20 = 0.

\((x+2)^{-2\over3}=(2+x)^{-2\over3}\)
\(=2^{-2\over3}\left(1+{x\over2}\right)\left[∵(1+x)^{-p\over q}=1-\left(p\over q\right)x+{\left(p\over q\right)\left({p\over q}-1\right)\over2!}x^2-{\left(p\over q\right)\left({p\over q}-1\right)\left({p\over q}-2\right)\over3!}x^3+... \right]\)
\(={1\over 2^{2\over3}} \left( 1-{2\over3}\left(x\over2\right)+{\left(-{2\over3} \right)\left(-{2\over3}-1 \right)\over2!} \left( x\over2 \right)^2+{\left(-{2\over3} \right)\left(-{2\over3}-1\right)\left(-{2\over3}-2 \right)\over3!\left({x\over2} \right)^3}+{\left(-{2\over3} \right)\left(-{2\over3} -1 \right)\left(-{2\over3}-2 \right)\left(-{2\over3}-3 \right)\over 4!}\left({x\over2} \right)^4+{\left( - {2\over3} \right)\left( - {2\over3}-1 \right)\left( -{2\over3}-2 \right)\left(- {2\over3}-3 \right)\left( - {2\over3} -4 \right)\over5!}\left( {x\over2} \right)^5+.... \right)\)\(={1\over 2^{2\over3}}\left[ 1-{x\over3}+{\left( -{2\over3}\right)\left( -{5\over3}\right)\over2 }{x^2\over4}+{\left( -{2\over3}\right)\left( -{5\over3}\right)\left( -{8\over3}\right)\over6}{x^3\over8}+{\left( -{2\over3}\right)\left( -{5\over3}\right)\left( -{8\over3}\right)\left( -{11\over3}\right)\over24}{x^4\over16}+... \right]\)
\(={ 2 }^{ \frac { -2 }{ 3 } }\left[ 1-\frac { x }{ 3 } +\frac { { 5x }^{ 2 } }{ 36 } -\frac { 5 }{ 81 } .{ x }^{ 3 }+\frac { 55 }{ 1944 } { x }^{ 4 }-.. \right] \)

Given R is the set of all real numbers. Then R x R is the set of all ordered pairs (x, y) where x, y ∈ R.
R \(\times\)R = {(x,y): x, y ∈ R}
Clearly R \(\times\)R is the set of all points in xy-plane. Now R \(\times\)R \(\times\)R = {(x, y, z) : x,y, x ∈ R}.
∴ R \(\times\)R \(\times\)R represents the set of all points in space.

\(LHS=cos({ 30 }-A)cos({ 30 }+A)cos({ 45 }-A)cos(45+A)\)
\(={ cos }^{ 2 }30-{ sin }^{ 2 }A+{ cos }^{ 2 }45-{ sin }^{ 2 }A \quad \left[ \because cos(A+B)cos(A-B=cos^{ 2 }A-{ sin }^{ 2 }B \right] \)
\(={ \left( \frac { \sqrt { 3 } }{ 2 } \right) }^{ 2 }-2{ sin }^{ 2 }A+{ \left( \frac { 1 }{ \sqrt { 2 } } \right) }^{ 2 }\)
\(=\frac { 3 }{ 4 } +\frac { 1 }{ 2 } -2{ sin }^{ 2 }A=\frac { 3 }{ 4 } -(1-cos2A)+\frac { 1 }{ 2 } \)
\(=\frac { 3 }{ 4 } -1+cos2A+\frac { 1 }{ 2 } \quad \left[ \because 2{ sin }^{ 2 }A=1-cos2A \right] \)
\(=\frac { 3-4+2 }{ 4 } +cos2A=\frac { 1 }{ 4 } +cos2A=RHS.\)

Given \(\left( { x }^{ +\frac { 1 }{ 2 } }+{ x }^{ -\frac { 1 }{ 2 } } \right) ^{ 2 }=\frac { 9 }{ 2 } \)
⇒ \(x+\frac { 1 }{ 2 } +{ 2x }^{ \frac { 1 }{ 2 } }.\frac { 1 }{ { x }^{ -\frac { 1 }{ 2 } } } =\frac { 9 }{ 2 } \)
⇒  \(x+\frac { 1 }{ x } +2=\frac { 9 }{ 2 } \)
⇒ \(x+\frac { 1 }{ x } =\frac { 9 }{ 2 } -2=\frac { 9-4 }{ 2 } =\frac { 5 }{ 2 } \)
Consider \(\left( { x }^{ \frac { 1 }{ 2 } }-{ x }^{ -\frac { 1 }{ 2 } } \right) ^{ 2 }\)
= \(x+\frac { 1 }{ x } -2x^{ \frac { 1 }{ 2 } }.\frac { 1 }{ { x }^{ \frac { 1 }{ 2 } } } \)
= \(x+\frac { 1 }{ x } -2\)    [From(1)]
=   \(\frac { 5 }{ 2 } -2=\frac { 5-4 }{ 2 } =\frac { 1 }{ 2 } \)       [using 1]
∴  \(x^{ \frac { 1 }{ 2 } }-x^{ -\frac { 1 }{ 2 } }=\pm \frac { 1 }{ \sqrt { 2 } } \)
⇒ \(x^{ \frac { 1 }{ 2 } }-x^{ -\frac { 1 }{ 2 } }=\frac { 1 }{ \sqrt { 2 } } \)   since x > 1

Since \(\pi,x<{{3\pi}\over{2}},\) x lies in the III quadrant only cot x and tan x are positive
Also \(\pi,x<{{3\pi}\over{2}},\) y is also lies in the III quadrant
Only cot y and tan y are positive
 
\(sinx=-\frac { 3 }{ 5 } \quad siny=-\frac { 24 }{ 25 } \)
\(cosx=-\frac { 4 }{ 5 } \quad cosy=-\frac { 7 }{ 25 } \)
ஃ cos(x - y) = cosx cosy + sinx siny
\(=\left( -\frac { 4 }{ 5 } \right) \left( -\frac { 7 }{ 25 } \right) +\left( -\frac { 3 }{ 5 } \right) \left( -\frac { 24 }{ 25 } \right) \)
\(=\frac { 28 }{ 125 } +\frac { 72 }{ 125 } =\frac { 100 }{ 125 } =\frac { 4 }{ 5 } \)

\(\text { The right hand derivative of } f(x) \text { at } x=2 \text { is }\)
\(f^{\prime}\left(2^{+}\right) =\lim _{x \rightarrow 2^{+}} \frac{f(x)-f(2)}{x-2} \)
\(=\lim _{x \rightarrow 2^{+}} \frac{(3 x+4)-(3(2)+4)}{x-2} \)
\(=\lim _{x \rightarrow 2^{+}} \frac{3 x+4-6-4}{x-2} \)
\(=\lim _{x \rightarrow 2^{+}} \frac{3(x-2)}{(x-2)}=3 \)

\(y =\cos \left(\sin x^{2}\right) \)
\(\frac{d y}{d x} =-\sin \left(\sin x^{2}\right) \cos \left(x^{2}\right)(2 x) \)
\(\text { At } x =\sqrt{\pi / 2}, \frac{d y}{d x}=-\sin \sin \left(\frac{\pi}{z}\right) \cos (\pi / 2) 2(\sqrt{\pi / 2}) \)
\(=(\sin 1)(0) 2\left(\frac{\sqrt{\pi}}{2}\right)=0 \quad[\because \cos \pi / 2=0] \)

\(\lim _{\theta \rightarrow 0} \frac{\sin \sqrt{\theta}}{\sqrt{\sin \theta}} =\lim _{\theta \rightarrow 0}\left(\frac{\sin \sqrt{\theta}}{\sqrt{\theta}} \cdot \frac{\sqrt{\theta}}{\sqrt{\sin \theta}}\right) \)
\(=\lim _{\theta \rightarrow 0} \frac{\sin \sqrt{\theta}}{\sqrt{\theta}} \cdot\left(\lim _{\theta \rightarrow 0} \frac{\sin \sqrt{\theta}}{\theta}\right)^{\frac{1}{2}}=1.1=1 \)

\(\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1 \)
\(\cos ^{2} 60^{\circ}+\cos ^{2} 45^{\circ}+\cos ^{2} \gamma=1 \)
\(\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{\sqrt{2}}\right)^{2}+\cos ^{2} \gamma=1 \Rightarrow \cos ^{2} \gamma=1-\frac{1}{4}-\frac{1}{2} \)
\(\cos ^{2} \gamma=\frac{4-1-2}{4}=\frac{1}{4} \Rightarrow \cos \gamma=\frac{1}{2} \Rightarrow \gamma=60^{\circ} \)
\(\text { Angle between } \overrightarrow{O P} \text { and } z \text { axis is } 60^{\circ}\)

\(A+B =\left[\begin{array}{ll} 1+a & 0 \\ 2+b & -2 \end{array}\right] \)
\((A+B)^{2} =\left[\begin{array}{ll} 1+a & 0 \\ 2+b & -2 \end{array}\right]\left[\begin{array}{ll} 1+a & 0 \\ 2+b & -2 \end{array}\right] \)
\(=\left[\begin{array}{ll} (1+a)^{2} & 0 \\ (2+b)(1+a)-2(2+b) & 4 \end{array}\right] \)
\(A^{2} =\left[\begin{array}{ll} 1 & -1 \\ 2 & -1 \end{array}\right]\left[\begin{array}{ll} 1 & -1 \\ 2 & -1 \end{array}\right]=\left[\begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array}\right] \)
\(B^{2} =\left[\begin{array}{cc} a & 1 \\ b & -1 \end{array}\right]\left[\begin{array}{cc} a & 1 \\ b & -1 \end{array}\right]=\left[\begin{array}{ll} a^{2}+b & a-1 \\ a b-b & b+1 \end{array}\right] \)
\(A^{2}+B^{2} =\left[\begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array}\right]+\left[\begin{array}{cc} a^{2}+b & a-1 \\ a b-b & b+1 \end{array}\right] \)
\(=\left[\begin{array}{cc} a^{2}+b-1 & a-1 \\ a b-b & b \end{array}\right]\)
\((A+B)^{2} =A^{2}+B^{2} \)
\(\therefore a-1 =0 \)
\(a =1 \)
\(b =4\)

\((1,2) \text { lies on } 3 x^{2}+3 y^{2}-8 x-12 y+17=0\)
\(\text { Because, } 3(1)^{2}+3(2)^{2}-8(1)-12(2)+17\)
\(=3+12-8-24+17=0\)

4 rings can be worn in,3 fingers in 3⁴ ways

\((38)^{5} =(39-1)^{15} \)
\(=(39)^{15}-{ }^{15} C_{1}(39)^{14}+\ldots+{ }^{15} C_{14}(39)-1 \)
The remainder will be 12 because all the terms except the last is divisible by 39 and so by 13, -1 remains

\(\cos 2 \theta \cos 2 \phi+\sin ^{2}(\theta-\phi)-\sin ^{2}(\theta+\phi) \)
\(=\cos 2 \theta \cos 2 \phi+\frac{1}{2}-\frac{1}{2} \cos (2 \theta-2 \phi)-\frac{1}{2}+\frac{1}{2} \cos (2 \theta+2 \phi) \)
\(=\cos 2 \theta \cos 2 \phi+\frac{1}{2}[\cos (2 \theta+2 \phi)-\cos (2 \theta-2 \phi)] \)
\(=\cos 2 \theta \cos 2 \phi-\sin 2 \theta \cos 2 \phi \)
\(=\cos (2 \theta+2 \phi) \)
\(=\cos 2(\theta+\phi) \)

\(\text { If } 4 \in X \text { then }(4,4) \notin \mathrm{R}\)
R is not reflexive
Symmetric can be easily checked

Given y = (cos^-1x)²
Differentiating with respect to 'x' we get
y' = 2.cos^-1x\(\left(\frac{-1}{\sqrt{1-x^2}}\right)\)
\(\sqrt{1-x^2} y_1=-\left(2 \cos ^{-1} x\right)\)
Squaring on both sides
\( \left(1-x^2\right) y_1^2=4\left(\cos ^{-1} x\right)^2 \)
\(\left(1-x^2\right) \dot{y}_1^2=4 y \) (using(1))
Differentiate W. R. T x
\( \left(1-x^2\right)\left(2 y_1 y_2\right)+y_1^2(-2 x)=4 y_1 \)
\(\left(1-x^2\right) 2 y_1 y_2-2 x y_1^2-4 y_1=0\)
Divide it by 2y₁
\(\left(1-x^2\right) y_2-x y_1-2=0\)
when x = 0
\( (1-0) y_2-0 y_1-2=0 \)
\(y_2-2=0 \)
\(y_2=2 \)

\(lim_{x\rightarrow a}{\sqrt{x-b}-\sqrt{a-b}\over x^2-a^2}(a>b)\)
Multiplying and dividing by \((\sqrt{x-b}+\sqrt{a-b})\)we get,
\(=lim_{x\rightarrow a}{\sqrt{x-b}-\sqrt{a-b}\over x^2-a^2}{\sqrt{x-b}+\sqrt{a-b}\over\sqrt{x-b}+\sqrt{a-b}}\)
\(=lim_{x\rightarrow a}{({x-b})-({a-b})\over (x^2-a^2)[\sqrt{x-b}+\sqrt{a-b}]}\)

\(={1\over (a+a)[\sqrt{a-b}+\sqrt{a-b}]}={1\over2a[2\sqrt{a-b}]}\)
\(={1\over4a[\sqrt{a-b}]}\)

Let O be the origin and let \(\overrightarrow{OA}\), \(​​\overrightarrow{OB}\), and\(​​\overrightarrow{OC}\) be the vectors 2\(\hat{i}\) + 3\(\hat{j}\) − 5\(\hat{k}\), 3\(\hat{i}\) + \(\hat{j}\) − 2\(\hat{k}\) and, 6\(\hat{i}\) − 5\(\hat{j}\) + 7\(\hat{k}\) respectively. Then
\(\overrightarrow{AB}=\hat{i}-2\hat{j}+3\hat{k} \ and \ \overrightarrow{AC}=4\hat{i}-8\hat{j}+12\hat{k}\).
Thus \(\overrightarrow{AC}=4\overrightarrow{AB}\) and hence \(​​\overrightarrow{AB}\) and\(​​\overrightarrow{AC}\) are parallel. They have a common point namely A. Thus, the three points are collinear.
Alternative method
Let O be the point of reference.
Let \(\overrightarrow {OA} = 2\hat i+3\hat j-5\hat k, \) \(\overrightarrow {OB} = 3 \hat j+\hat j-2\hat k\ and\ \overrightarrow {OC} = 6\hat i-5\hat j+7\hat k \)
\(\overrightarrow {AB} = \hat i- 2\hat j+3\hat k; \overrightarrow {BC} = 3\hat i-6\hat j+9\hat k; \overrightarrow {CA} = -4\hat i+8\hat j-12 \hat k\\ |\overrightarrow {AB}| = \sqrt 14; |\overrightarrow {BC}|= \sqrt 126 = 3 \sqrt 14; |\overrightarrow {CA}|= \sqrt 224 = 4 \sqrt 4\)
Thus, AC = AB + BC.
Hence A, B, C are lying on the same line. That is, they are collinear.

Minors : M₁₁ = \(\begin{vmatrix} -5 & 6 \\ 5 & 2 \end{vmatrix}\) = -10 - 30 = -40
M₁₂ = \(\begin{vmatrix}4 & 6 \\ -3 & 2 \end{vmatrix}\) = 8 + 18 = 26
M₁₃ = \(\begin{vmatrix}4 & -5 \\ -3 & 5 \end{vmatrix}\) = 20 - 15 = 5
M₂₁ = \(\begin{vmatrix}3 & -2 \\ 5 & 2 \end{vmatrix}\) = 6 + 10 = 16
M₂₂ = \(\begin{vmatrix}1 & -2 \\ -3 & 2 \end{vmatrix}\) = 2 - 6 = -4
M₂₃ = \(\begin{vmatrix}1 & 3 \\ -3 & 5 \end{vmatrix}\) = 5 + 9 = 14
M₃₁ = \(\begin{vmatrix}3 & -2 \\ -5 & 6 \end{vmatrix}\) = 18 - 10 = 8
M₃₂ = \(\begin{vmatrix}1 & -2 \\ 4 & 6 \end{vmatrix}\) = 6 + 8 = 14
M₃₃ = \(\begin{vmatrix}1 & 3 \\ 4 & -5 \end{vmatrix}\) = -5 - 12 = -17
Cofactors:
A₁₁ = (−1)¹⁺¹(−40) = −40
A₁₂ = (−1)¹⁺² (+26) = −26
A₁₃ = (−1)¹⁺³ (5) = 5
A₂₁ = (−1)²⁺¹(16) = −16
A₂₂ = (−1)²⁺² (−4) = −4
A₂₃ = (−1)²⁺³ (14) = −14
A₃₁ = (−1)³⁺¹(8) = 8
A₃₂ = (−1)³⁺² (14) = −14
A₃₃ = (−1)³⁺³ (−17) = −17
Expanding along R₁ yields
|A| = a₁₁ A₁₁ + a₁₂ A₁₂ + a₁₃ A₁₃ .
|A| = 1(−40) + (3)(−26) + (−2)(5) = −128 ....(1)
Expanding along C₁ yields
|A| = a₁₁ A₁₁ + a₂₁ A₂₁ + a₃₁ A₃₁
= 1(−40) + 4(−16) + −3(8) = −128....(2)
From (1) and (2), we have
|A| obtained by expanding along R₁ is equal to expanding along C₁.

The given equation is x cos \(\alpha\) + y sin \(\alpha\) = p or \(\frac{x}{p/\cos\alpha}+\frac{y}{p/sin\alpha}=1\) ........(i)
This cuts the coordinate axes at \(A(p/\cos\alpha,0)\) and \(B(0,p/\sin\alpha)\). Let P (h, k)be the mid point of the intercept AB. Then,
\(h=\frac{p/\cos\alpha+0}{2},k=\frac{0+p/\sin\alpha}{2}\)
\(\Rightarrow h=\frac{p}{2\cos\alpha},k=\frac{p}{2\sin\alpha}\)
\(\Rightarrow \cos\alpha=\frac{p}{2h},\sin\alpha=\frac{p}{2k}\)
Here, u is a variable. to find the locus of P (h, k), we have to eliminate c.
From (i), we obtain
\(\cos^2\alpha+\sin^2\alpha=\frac{p^2}{4h^2}+\frac{p^2}{4k^2}\Rightarrow 1=\frac{p^2}{4h^2}+\frac{p^2}{4k^2}\Rightarrow \frac{4}{p^2}=\frac{1}{h^2}+\frac{1}{k^2}\)
Hence, the locus of (h, k) is \(\frac{1}{x^2}+\frac{1}{y^2}=\frac{4}{p^2}\)

By unrolling the hollow cylinder and flattening it into a rectangle, and with a single reflection allows us to determine the ant's path, as shown the figure. Let the baseline x-axis in cm and the vertical line through A (initial position of the ant) be the y-axis. Let H be the position of honey drop and E be the entry point of ant inside the vessel. From the given information we have
Let A(x₁, y₁) and H(x₂, y₂) be (0, 4) and (12, 13) respectively.
(i) The shortest distance between A and H is
\(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{12^2+9^2}=15\)
the ant would need to crawl 15 cm to get the honey.
(ii) The equation of the path AH is \(\frac{y-4}{13-4}=\frac{x-0}{12-0}\)
y = 0.75x + 4
(iii) At the entry point E, y = 10 ⇒ x = 8
E = (8, 10)

Let p(n): be \(\left( 1+\frac { 1 }{ 1 } \right) \left( 1+\frac { 1 }{ 2 } \right) \left( 1+\frac { 1 }{ 3 } \right) ...\left( 1+\frac { 1 }{ n } \right) =\left( n+1 \right) \)
Step 1: p(1): \(( 1+\frac { 1 }{ 1 }) =\left( 1+1 \right) \)
\(\Rightarrow\) p(1):2 = 2
\(\Rightarrow\) p(1) is true.
Step 2: Let p(m) be true
\(\left( 1+\frac { 1 }{ 1 } \right) \left( 1+\frac { 1 }{ 2 } \right) \left( 1+\frac { 1 }{ 3 } \right) ...\left( 1+\frac { 1 }{ m } \right) =\left( m+1 \right) \)
Step 3: To prove that p(m + 1) is true
i.e. to prove that \(\left( 1+\frac { 1 }{ 1 } \right) \left( 1+\frac { 1 }{ 2 } \right) ...\left( 1+\frac { 1 }{ m } \right)\left( 1+\frac { 1 }{ m+1 } \right) =\left( m+2 \right) \)
LHS = \(\left( 1+\frac { 1 }{ 1 } \right) \left( 1+\frac { 1 }{ 2 } \right) ...\left( 1+\frac { 1 }{ m } \right)\left( 1+\frac { 1 }{ m+1 } \right) \)
= (m + 1) \(\left( 1+\frac { 1 }{ m+1 } \right)\) = m + 1 + 1
= (m + 2) = RHS
p(m + 1) is true.
By the principle of mathematical induction, p(n) is true for all \(n\in N\)

(i) Let the x - axis represent the weight and the y-axis represent the length

(ii) Consider the point \(\begin{pmatrix} { x }_{ 1 } & { y }_{ 1 } \\ 2 & 3 \end{pmatrix}\begin{pmatrix} { x }_{ 2 } & { y }_{ 2 } \\ 4 & 4 \end{pmatrix}\)
Equation of the straight line using two point form is
\(\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \)
\(\Rightarrow \quad \frac { y-3 }{ 4-3 } =\frac { x-2 }{ 4-2 } \)
\(\Rightarrow \quad \frac { y-3 }{ 1 } =\frac { x-2 }{ 2 } \)
\(\Rightarrow\) 2y - 6 = x-2
 \(\Rightarrow\) x - 2y + 4 = 0
(iii) To find the actual length of the spring,
Put x = 0
\(\Rightarrow\) 0 - 2y + 4 = 0
\(\Rightarrow\) -2y = -4
\(\Rightarrow\) y = 2 cm
(iv) Put y = 9 in (1) we get
\(\Rightarrow\) x - 18 + 4 = 0
\(\Rightarrow\) x - 14 = 0
\(\Rightarrow\) x = 14 kg
\(\therefore\) 14 Kg must be added
(v) Put x = 6 in (1) we get,
\(\Rightarrow\)  6 - 2y + 4 = 0
\(\Rightarrow\) 10 - 2y = 0
\(\Rightarrow\) 10 = 2y 
\(\Rightarrow\) y = 5
\(\therefore\) Strength of the spring = 5 cm

Let R(h, k) be the locus of the mid-point of PQ where, P is (2, -7) and Q is a point on (2x² + 9y² = 18)
Given equation is 2x² + 9y² = 18
Dividing by 18 we get,
\(\Rightarrow \quad \frac { { x }^{ 2 } }{ 9 } +\frac { { y }^{ 2 } }{ 2 } =1\)
\(\Rightarrow \quad \frac { { x }^{ 2 } }{ { 3 }^{ 2 } } +\frac { { y }^{ 2 } }{ { \left( \sqrt { 2 } \right) }^{ 2 } } =1\)
\(\Rightarrow \quad a=3\quad and\quad b=\sqrt { 2 } \)
Any point on the ellipse is (a cos \(\theta\), b sin \(\theta\))
\(\therefore\) Q is (3 cos \(\theta\), \(\sqrt { 2 } \) sin \(\theta\)
Since R is the mid-point of PQ, we get, (h,k) = \(\left( \frac { 2+3cos\theta }{ 2 } ,\frac { -7+\sqrt { 2 } sin\theta }{ 2 } \right) \)
\(\Rightarrow \quad h=\frac { 2+3\quad cos\quad \theta }{ 2 } \)
\(\Rightarrow \quad 2h=2+3cos\theta \)
\(\Rightarrow \quad 2h-2=3cos\theta \)
\(\Rightarrow \quad \frac { 2h-2 }{ 3 } =cos\quad \theta \)
\(k=\frac { -7+\sqrt { 2 } sin\theta }{ 2 } \)
\(\Rightarrow \quad 2k=-7+\sqrt { 2 } sin\theta \)
\(\Rightarrow \quad 2k+7\quad =\quad \sqrt { 2 } sin\theta \)
\(\Rightarrow \quad \frac { 2k+7 }{ \sqrt { 2 } } =sin\quad \theta \)
Squaring and adding we get,
\({ \left( \frac { 2h-2 }{ 3 } \right) }^{ 2 }+{ \left( \frac { 2k+7 }{ \sqrt { 2 } } \right) }^{ 2 }={ cos }^{ 2 }\theta +{ sin }^{ 2 }\theta \)
\(\Rightarrow \quad \frac { { 4h }^{ 2 }+4-8h }{ 9 } +\frac { { 4k }^{ 2 }+49+28k }{ 2 } =1\quad [\because { cos }^{ 2 }\theta +{ sin }^{ 2 }\theta =1]\)
\(\Rightarrow \quad 2({ 4h }^{ 2 }+4-8h)=9({ 4k }^{ 2 }+49+28k)=18\)
\(\Rightarrow \quad { 8h }^{ 2 }+8-16h+36{ k }^{ 2 }+441+252k-18=0\)
\(\Rightarrow \quad { 8h }^{ 2 }+36{ k }^{ 2 }-16h+252k+431=0\)
\(\therefore\)   Locus of (h,k) is
8x²+36y²-16x+252y+431=0

Let = 6 + 66 + 666 + ... upto n terms
= 6 (I + 11 + 111+ ....) upto n terms
\(={6\over9}(9+99+999+ ...)\) upto n terms
\(={63\over 6}[(10 -1) + (10^2-1) + (10^3 -1) + ...]\) upto n terms
\(={6\over 9}[(10+ 10^2 + 10^3+ ...) - (1+ 1+1...)]\) upto n terms​​​​​​​
\(={6\over9}\left[{ 10(10^n-1)\over 10-1}n\right]\)[In a G.P with a = 10 r = 10, \(S_n={(r^n-1)\over r-1}\)]
\(={6\over9}\left[{ 10(10^n-1)\over 10-1}n\right]={6\over9}\left[ 10(10^n-1)-9n\over9\right]\)
\({ S }_{ n }=\frac { 6 }{ 81 } \left[ 10\left( { 10 }^{ n }-1 \right) -9n \right] \)

Given equations are x² - ax + b = 0 .....(1)
and x² - ex + f = 0 .....(2)
Let \(\alpha\) be the common root.
Let \(\alpha ,\beta \) be the roots of x² - ax + b = 0
Then \(\alpha +\beta =a\) and \(\alpha \beta =b\) ....(3)
Let \(\alpha\), \(\alpha\) be the roots of x² - ex +f = 0 [\(\because\) the roots are equal]
\(\therefore\) \(\alpha\)+\(\alpha\) = e,   \(\alpha \times \alpha =f\)
\(\Rightarrow\)  \(2\alpha =e\)     and   \({ \alpha }^{ 2 }=f\) .....(4)
Now LHS = \(ae=\left( \alpha +\beta \right) 2\alpha \)
LHS = \(ae=2{ \alpha }^{ 2 }+2\alpha \beta \)
= 2(f) + 2b    [From (4)]
= 2 (b + f) = RHS. Hence proved.

Given \(x={{\sqrt{3}-\sqrt{2}}\over{\sqrt{3}+\sqrt{2}}}\)
Multiplying by the conjugate of the denominator we get,
\(x={{(\sqrt{3}-\sqrt{2})(\sqrt{3}-\sqrt{2})}\over{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}}={{3-2\sqrt{6}+2}\over{({\sqrt{3}})^{2}-{(\sqrt{2})}^{2}}}\)
\(={{5-2\sqrt{6}}\over{3-2}}=5-2\sqrt{6}\)
Similarly, \(y={{\sqrt{3}+\sqrt{2}}\over{\sqrt{3}-\sqrt{2}}}\times{{\sqrt{3}+\sqrt{2}}\over{\sqrt{3}+\sqrt{2}}}\)
y \(={{3+2+2\sqrt{6}}\over{{(\sqrt{3})}^{2}-{(\sqrt{2})}^{2}}}={{5+2\sqrt{6}}\over{3-2}}=5+2\sqrt{6}\)
Now, x+y\(=5-2\sqrt{6}+5+2\sqrt{6}=10\)
and xy \(= (5-2\sqrt{6})(5+2\sqrt{6})=5^2-(2\sqrt{6})^2=25-4(6)=25-24=1. \)
We know that
x² + xy + y² = (x + y)² - xy = 10² - 1 = 100 - 1 = 99

f(-4) = +4 + 4    [\(\therefore\) f(x) = -x + 4 when x = -4]
=8
f(1) = 1-1² [\(\therefore\) f(x) = x-x² when x = 1]
f(1) = 0
f(-2) = (-2)²-(-2) [\(\therefore\) f(x) = x²-x when x = -2]
= 4+2 = 6
f(7) = 0   [\(\therefore\) f(x) = 0 when x = 7]
f(0 = 0²-0  [\(\therefore\) f(x) = x² - x when x = 0]
=0
\(\therefore\) f(-4) = 8, f(1) = 0, f(-2) = 6, f(7) = 0 and f(0) = 1

Take u = x³-1 so that
y = u¹⁰⁰
and \({dy \over dx}={dy\over du}\times {du\over dx}\)
= 100u^100-1 \(\times\) (3x²-0)
= 100(x³-1)⁹⁹ \(\times\) 3 x²
= 300 x²(x³-1)⁹⁹.

\(lim_{x\rightarrow3}(4-x)\)

At x = 3, the value of the curve on y-axis is 1.
\(\therefore lim_{x\rightarrow3}(4-x)=1\)

Let \( f(x)=\frac{\cos x-1}{x} \)
\(\therefore \lim _{x \rightarrow 0} \frac{\cos x-1}{x}=0\)

Let \(l={2\over \sqrt{2}},m={1\over2}\) and \(n={1\over2}\)
\(\therefore l^2+m^2+n^2=({1\over \sqrt{2}})^2+({1\over2})^2+({1\over2})^2\)\(={1\over2}+{1\over4}+{1\over 4}={2+1+1\over4}={4\over4}=1\)
Hence, the given ratios are direction cosines of some vector.

By the definition of sum of matrices, we have
\(A+B+C=\left[\begin{array}{cc} \sin ^2 \theta+\cos ^2 \theta+0 & 1+0-1 \\ \cot ^2 \theta-\operatorname{cosec}^2 \theta-1 & 0+1+0 \end{array}\right]=\left[\begin{array}{cc} 1 & 0 \\ -2 & 1 \end{array}\right]\)

= \(\frac { 12\times 11\times 10\times 9! }{ 9!\quad 3\times 2 } =\frac { 12\times 11\times 10 }{ 6 } \)
= 2 \(\times\) 11 \(\times\) 10 = 220