Given that \(sin x=\frac{4}{5}\).
cos²x + sin²x = 1 gives \(cos\ x=\pm\sqrt{1-sin^2x}=\pm\sqrt{1-\frac{16}{25}}=\pm\frac{3}{5}\)
In the first quadrant, cos x is always positive. Thus \(cos\ x=\frac{3}{5}\)
Also, given that cos \(y=-\frac{12}{13}\) in the II quadrant. We have 
\(sin\ y=\pm\sqrt{1-cos^2y}=\pm\sqrt{1-\frac{144}{169}}=\pm\frac{5}{13}\)
In the second quadrant, sin y is always positive, Thus sin \(y=\frac{5}{13}\)
(i) sin(x - y) = sin x cos y - cos x sin y = \(\frac{4}{5}(\frac{-12}{13})-\frac{3}{5}(\frac{5}{13})=-\frac{63}{65}\)

To check whether compositions can be defined, let us find the domain and range of these functions.
Domain of f = {1, 2, 3}, Range of f = {2, 4}, Domain of g = {2, 3, 4} and Range of g = {1, 2}. Since the range of f is contained in the domain of g we can define g o f, so as to find the image of 1 under g o f, we first find the image of 1 under f and then its image under g. The image of 1 under f is 2 and its image under g is 1. So (g o f) (1) = g(f(1)) = g(2) = 1.
Similarly we find that (g o f) (2) = 1 and (g o f) (3) = 2. So g o f = {(1, 1), (2, 1), (3, 2)}.
Similarly f o g = {(2, 2), (3, 2), (4, 2)}.

Given equation of line is 3x + 4y - 6 = 0.
Any line perpendicular to 3x + 4y - 6 = 0 will be of the form 4x - 3y + k = 0 ....(1)
Given perpendicular distance is 4 units from (2,1) to line (1)
\(\therefore\)4 = \(\pm \frac { (4(2)-3(1)+k) }{ \sqrt { { 4 }^{ 2 }+{ \left( -3 \right) }^{ 2 } } } \)
\(\Rightarrow\) 4 = \(\pm \left( \frac { 8-3+k }{ \sqrt { 16+9 } } \right) \)
\(\Rightarrow\) 4 = \(\pm \left( \frac { 5+k }{ 5 } \right) \)
\(\therefore\) 20 = +(5 + k) or 20 = -(5+ k)
\(\Rightarrow\) k = 20 - 5  or k = -(20 + 5)
\(\Rightarrow\) k = 15 or k = -25
\(\therefore\) Required equation of the lines are 4x - 3y + 15 = 0 and 4x - 3y - 25 = 0.

Given x = a cos³ \(\theta\) , y = a sin³ \(\theta\)
\(\Rightarrow \quad \frac { x }{ a } ={ cos }^{ 3 }\theta \quad and\quad \frac { y }{ a } ={ sin }^{ 3 }\quad \theta \)
Taking power \(\left( \frac { 2 }{ 3 } \right) \)for both the equations, we get
\({ \left( \frac { x }{ a } \right) }^{ \frac { 2 }{ 3 } }={ \left( { cos }^{ 3 }\theta \right) }^{ \frac { 2 }{ 3 } }and\)
\({ \left( \frac { y }{ a } \right) }^{ \frac { 2 }{ 3 } }={ { (sin }^{ 3 }\theta })^{ \frac { 2 }{ 3 } }\)

\({ \left( \frac { x }{ a } \right) }^{ \frac { 2 }{ 3 } }={ cos }^{ 2 }\theta \quad and\quad { \left( \frac { y }{ a } \right) }^{ \frac { 2 }{ 3 } }={ sin }^{ 2 }\theta \)
We know that cos² \(\theta\) + sin² \(\theta\) = 1
\(\therefore { \left( \frac { x }{ a } \right) }^{ \frac { 2 }{ 3 } }+{ \left( \frac { y }{ a } \right) }^{ \frac { 2 }{ 3 } }=1\)
\(\Rightarrow \quad \frac { { x }^{ \frac { 2 }{ 3 } } }{ { x }^{ \frac { 2 }{ 3 } } } +\frac { { x }^{ \frac { 2 }{ 3 } } }{ { x }^{ \frac { 2 }{ 3 } } } =1\)
\(\Rightarrow \quad \frac { { x }^{ \frac { 2 }{ 3 } }+{ y }^{ \frac { 2 }{ 3 } } }{ { a }^{ \frac { 2 }{ 3 } } } =1\)
\(\Rightarrow \quad { x }^{ \frac { 2 }{ 3 } }+{ y }^{ \frac { 2 }{ 3 } }={ a }^{ \frac { 2 }{ 3 } }\)
\(\therefore \) The required point is (0, 12)

Given a²+ b² = 7ab
Adding 2ab both sides we get,
a²+b²+2ab = 7ab + 2ab
⇒ (a+b)² = 9ab
⇒ \(\frac { (a+b)^{ 2 } }{ 9 } \)  = ab
⇒  \(\left( \frac { a+b^{ 2 } }{ 9 } \right) ^{ 2 }\)  = ab
Taking square root,we get
\(\frac { a+b }{ 3 } \)  = ab
\(log\left( \frac { a+b }{ 3 } \right) =log(ab)^{ \frac { 1 }{ 2 } }\)
= \(\frac { 1 }{ 2 } \) log (ab)
log \(\left( \frac { a+b }{ 3 } \right) \) = \(\frac { 1 }{ 2 } \) [log a + log b]
Hence proved.

\(LHS=\frac { sinx+sin3x+sin5x+sin7x }{ cosx+cos3x+cos5x+cos7x } \)
\(=\frac { 2sin\left( \frac { x+3x }{ 2 } \right) cos\left( \frac { x-3x }{ 2 } \right) +2sin\left( \frac { 7x+5x }{ 2 } \right) cos\left( \frac { 7x-5x }{ 2 } \right) }{ 2cos\left( \frac { 3x+x }{ 2 } \right) cos\left( \frac { 3x-x }{ 2 } \right) +2cos\left( \frac { 7x+5x }{ 2 } \right) cos\left( \frac { 7x-5x }{ 2 } \right) } \)
\(=\frac { sin2xcos(x)+sin6xcosx }{ cos2xcox+cos6xcosx } \)
\(=\frac { cosx(sin2x+sin6x) }{ cosx(cos2x+cos6x) } =\frac { sin4x.cos2x }{ cos4x.cos2x } \)
= tan 4x = RHS

\(LHS=\frac { sin8x\quad cosx-sin6x\quad cos3x }{ cos2x\quad cosx-sin3x\quad sin4x } \)
\(=\frac { \frac { 1 }{ 2 } \left[ sin9x+sin(7x) \right] -\frac { 1 }{ 2 } \left[ sin9x+sin3x \right] }{ \frac { 1 }{ 2 } \left[ cos\quad 3x+cos\quad x) \right] -\frac { 1 }{ 2 } \left[ cos\quad (x)-cos7x) \right] } \quad \left[ \therefore sinAcosB=\frac { 1 }{ 2 } (sin(A+B)+sin(A-B)) \right] \)
\(=\frac { \frac { 1 }{ 2 } \left[ sin9x+sin7x-sin9x-sin3x \right] }{ \frac { 1 }{ 2 } \left[ cos3x+cosx-cosx+cos7x \right] } =\frac { sin7x-sin3x }{ cos3x+cos7x } \)
\(=\frac { 2cos\left( \frac { 7x+3x }{ 2 } \right) sin\left( \frac { 7x-3x }{ 2 } \right) }{ 2cos\left( \frac { 7x+3x }{ 2 } \right) .cos\left( \frac { 7x-3x }{ 2 } \right) } =\frac { 2cos5x.sin2x }{ 2cos5x.cos2x } \)
\(\left[ \because sinC-sinD=2cos\left( \frac { C+D }{ 2 } \right) sins\left( \frac { C-D }{ 2 } \right) andcosC+cosD=2cos\left( \frac { C+D }{ 2 } \right) coss\left( \frac { C-D }{ 2 } \right) \right] \)
= tan 2x = RHS

Let A₁ be the event that the items are produced by Machine-I, A₂ be the event that items are produced by Machine-II. Let B be the event of drawing a defective item. Now we are asked to find the conditional probability P (A₂/B). Since A₁, A₂ are mutually exclusive and exhaustive events, by Bayes’ theorem,

P(A₂/B) \(={P(A_2)P(B/A_2)\over P(A_1)P(B\ A_1)+P(A_2)P(B/A_2)}\)
We have, P(A₁) = 0.40,P(B/A₁) = 0.04 
P(A₂) = 0.60,P(B/A₂) = 0.05
P(A₂/B) \(={P(A_2)P(B/A_2)\over P(A_1)P(B\ A_1)+P(A_2)P(B/A_2)}\)
P(A₂/B) \(={(0.60)(0.05)\over(0.40)(0.04)+(0.60)(00.05)}={15\over23}\)

Let \(\frac{x+1}{(x+2)(x+3)}=\frac{A}{x+2}+\frac{B}{x+3}\)
Multiplying both sides by (x + 2)(x + 3)
x + 1 =  A(x + 3) + B(x + 2) ............(1)
Putting x = -2 in (1)
-2 + 1 = A(-2 + 3) + 0
-1 = A(1) ⇒ A = -1
Putting x = -3 in (1)
-3 +1 = 0 + B(-3 + 2)
-2 = B(-1) ⇒ B = 2
\(\therefore \frac{x+1}{(x+2)(x+3)}=\frac{-1}{x+2}+\frac{2}{x+3}\)
\(\therefore \int \frac{x+1}{(x+2)(x+3)} d x=\int\left[\frac{-1}{x+2}+\frac{2}{x+3}\right] d x\)
\( =-\int \frac{1}{x+2} d x+2 \int \frac{1}{x+3} d x\)
\(=-\log |x+2|+2 \log |x+3|+c \)
\(=2 \log |x+3|-\log |x+2|+c \)

y = e^{tan-1 x} ...........(1)
⇒ y'=\({ e }^{ tan^{ -1 }x }(\frac { 1 }{ 1+{ x }^{ 2 } }) \)
\(\left(1+x^2\right) y^{\prime}=e^{\tan ^{-1} x} \)
\(\left(1+x^2\right) y^{\prime}=y\) (using (1))
Again Diff w. r. to x.
(1 + x²) y" + y' (2x) = y'
(1 + x²) y" + y' (2x) - y' = 0
(1 + x²) y" + (2x - 1) y' = 0
Hence proved.

\(lim_{x\rightarrow 0}{sin \ x(1-cos \ x)\over x^3}\)\(=lim_{x\rightarrow 0}{sin \ x \times (2sin^2{x\over2})\over x^3}\)
\(=2[lim_{x\rightarrow0}{sin \ x\over x}]\times lim_{x\rightarrow0}{sin^2{x\over 2}\over x^2}=2(1)\times lim _{{x\over 2}\rightarrow 0}{sin^2{x\over2}\over {x^2\over 4}\times 4}\)
\(2\times [lim_{{x\over 2}\rightarrow 0}{sin^2{x\over 2}\over {x^2\over 4}}]\times {1\over4}=2\times 1\times {1\over4}={1\over2}\)
\(\therefore lim_{x\rightarrow 0}{sin \ x(1-cos \ x)\over x^3}={1\over2}\)

\(lim_{x\rightarrow0}{e^x-e^{-x}\over sin x}\)\(=lim_{x\rightarrow0}{e^x-1-e^{-x}+1\over {sin x\over x}\times x}\)
\(=lim_{x\rightarrow0}{{e^x-1\over x}-{(e^{-x}+1)\over x}\over {sin x\over x}}\)
\(={(lim_{x\rightarrow 0}{e^x-1\over x})-(lim_{x\rightarrow 0}{e^{-x}-1\over x})\over (lim_{x\rightarrow 0}{sin x\over x})}\)  \((\because lim_{x\rightarrow 0}{e^x-1\over x})=1\)
\({1-(-1)\over 1}={2\over1}=2\)

(i) f(-5^-)
For x < - 5, |x + 5| = - (x + 5)
Thus  f(-5^-) = \(lim_{x\rightarrow-5^- {-(x+5)\over (x+5)}}=-1\)
(ii) f(-5⁺)
For x > - 5, |x + 5| = (x + 5)
Thus  f(-5⁺) = \(lim_{x\rightarrow-5^+{(x+5)\over (x+5)}}=1\)
Note that f( -5^-)  ≠ f( -5⁺). Hence the limit does not exist.

Let the vertices of the triangle be A, B, C.
Then, given \(\overrightarrow { OA } =\hat { i } +2\hat { j } +3\hat { k } ,\overrightarrow { OB } =3\hat { i } -4\hat { j } +5\hat { k } \) and \(\overrightarrow { OC } =-2\hat { i } +3\hat { j } -7\hat { k } \)
\(\overrightarrow { AB } =\overrightarrow { OB } -\overrightarrow { OA } =(3\hat { i } -4\hat { j } +5\hat { k } )-(\hat { i } +2\hat { j } +3\hat { k } )=2\hat { i } -6\hat { j } +2\hat { k } \)
\(|\overrightarrow { AB } |=\sqrt { { 2 }^{ 2 }+{ (-6 })^{ 2 }+{ 2 }^{ 2 } } =\sqrt { 4+36+4 } =\sqrt { 44 } \)
\(\overrightarrow { BC } =\overrightarrow { OC } -\overrightarrow { OB } =(-2\hat { i } +3\hat { j } -7\hat { k } )-(3\hat { i } -4\hat { j } +5\hat { k } )=-5\hat { i } +7\hat { j } -12\hat { k } \)
\(|\overrightarrow { BC } |=\overrightarrow { OC } -\overrightarrow { OB } =(-2\hat { i } +3\hat { j } -7\hat { k } )-(3\hat { i } -4\hat { j } +5\hat { k } )=-5\hat { i } +7\hat { j } -12\hat { k } \)
\(|\overrightarrow { BC } |=\sqrt { { (-5) }^{ 2 }+{ 7 }^{ 2 }+{ (-12) }^{ 2 } } =\sqrt { 25+49+144 } =\sqrt { 218 } \)
\(\overrightarrow { CA } =\overrightarrow { OA } -\overrightarrow { OC } =(\hat { i } +2\hat { j } +3\hat { k } )-(-2\hat { i } +3\hat { j } -7\hat { k } )=3\hat { i } -\hat { j } +10\hat { k } \)
\(|\overrightarrow { CA } |=\sqrt { { 3 }^{ 2 }+{ (-1) }^{ 2 }+{ 1 }0^{ 2 } } =\sqrt { 9+1+100 } =\sqrt { 110 } \)
\(\therefore\) Perimeter of \(\Delta ABC,\)
\(|\overrightarrow { AB } |+|\overrightarrow { BC } |+|\overrightarrow { CA } |=\left( \sqrt { 44 } +\sqrt { 218 } +\sqrt { 110 } \right) \) units

Let the position vector of the vertices of the quadrilateral ABCD be \(\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\)and\(\overrightarrow{d}\) respectively.
\(\therefore \overrightarrow{OA}=\overrightarrow{a}, \overrightarrow{OB}=\overrightarrow{b},\overrightarrow{OC}=\overrightarrow{c}\) and \(\overrightarrow{OD}=\overrightarrow{d}.\)
Since E and F are the mid-points of AC and BD respectively, we have
\(\overrightarrow{OE}={\overrightarrow{a}+\overrightarrow{c}\over 2}\) and \(\overrightarrow{OF}={\overrightarrow{b}+\overrightarrow{d}\over 2}\)
To prove that \(\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{CB}+\overrightarrow{CD}=4\overrightarrow{EF}\)
\(LHS=\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{CB}+\overrightarrow{CD}\)
\(=\overrightarrow{OB}-\overrightarrow{OA}+\overrightarrow{OD}-\overrightarrow{OA}+\overrightarrow{OB}-\overrightarrow{OC}+\overrightarrow{OD}-\overrightarrow{OC}\)
\(=\overrightarrow{b}-\overrightarrow{a}+\overrightarrow{d}-\overrightarrow{a}+\overrightarrow{b}-\overrightarrow{c}+\overrightarrow{d}-\overrightarrow{c}\)

\(=-2\overrightarrow{a}+2\overrightarrow{b}-2\overrightarrow{c}+2\overrightarrow{d}\)
\(=2[(\overrightarrow{b}+\overrightarrow{d})-(\overrightarrow{a}+\overrightarrow{c})]\)
\(=2[2\overrightarrow{OF}+2\overrightarrow{OE})\) [From (1)]
\(=4[\overrightarrow{OF}-\overrightarrow{OE}]=4.\overrightarrow{EF}=RHS\)
Hence proved.

We have |B| = \(\begin{vmatrix} 2(a+b+c) & 2(a+b+c) &2(a+b+c) \\ c+a & a+b &b+c \\ a+b & b+c & c+a \end{vmatrix}\)\((R_1 \rightarrow R_1+R_2+R_3)\)
= 2\(\begin{vmatrix} a+b+c & a+b+c & a+b+c \\ c+a & a+b &b+c \\ a+b & b+c & c+a \end{vmatrix}\)
= 2\(\begin{vmatrix} a+b+c & a+b+c & a+b+c \\ -b & -c &-a \\ -c & -a & -b \end{vmatrix}\)\((R_2 \rightarrow R_2-R_1andR_3\rightarrow R_3-R_1)\)

= 2\(\begin{vmatrix}a &b &c \\ -b & -c & -a \\ -c & -a & -b \end{vmatrix}\)\((R_1 \rightarrow R_1+R_2+R_3)\)
= 2(-1)² \(\begin{vmatrix}a &b &c \\b &c & a \\ c & a & b \end{vmatrix}\)
= 2| A |.

\(Given A=\left[\begin{array}{ccc}1 & 2 & 2 \\ 2 & 1 & -2 \\ x & 2 & y\end{array}\right]\)
\(A^T=\left[\begin{array}{ccc} 1 & 2 & x \\ 2 & 1 & 2 \\ 2 & -2 & y \end{array}\right]\)
\(A A^T=9 I\)
\(\left[\begin{array}{ccc} 1 & 2 & 2 \\ 2 & 1 & -2 \\ x & 2 & y \end{array}\right]\left[\begin{array}{ccc} 1 & 2 & x \\ 2 & 1 & 2 \\ 2 & -2 & y \end{array}\right]=9\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\)
\(\left[\begin{array}{ccc} 1+4+4 & 2+2-4 & x+4+2 y \\ 2+2-4 & 4+1+4 & 2 x+2-2 y \\ x+4+2 y & 2 x+2-2 y & x^2+4+y^2 \end{array}\right]=\left[\begin{array}{lll} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{array}\right]\)
\(\left[\begin{array}{ccc} 9 & 0 & x+2 y+4 \\ 0 & 9 & 2 x-2 y+2 \\ x+4+2 y & 2 x+2-2 y & x^2+4+y^2 \end{array}\right]=\left[\begin{array}{lll} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{array}\right]\)
\(x+2 y+4=0 \Rightarrow x+2 y=-4 ....(1)\)
\(2 x-2 y+2=0 \Rightarrow 2 x-2 y=-2 ....(2)\)
\(\begin{aligned} &\text { Add (1) & (2) } \Rightarrow 3 x=-6 \\ &x=-2 \end{aligned}\)
\(\text { Put } x=-2 \text { in }(1) \Rightarrow-2+2 y=-4\)
\(-2+4=-2 y\)
\(2=-2 y\)
\(y=-1\)
\(\therefore x=-2\)

\(A=\left[\begin{array}{ccc} 1 & 3 & 5 \\ -6 & 8 & 3 \\ -4 & 6 & 5 \end{array}\right] \Rightarrow A^T=\left[\begin{array}{ccc} 1 & -6 & -4 \\ 3 & 8 & 6 \\ 5 & 3 & 5 \end{array}\right]\)
\( Let \ P=\frac{1}{2}\left(A+A^T\right)=\frac{1}{2}\left[\begin{array}{ccc}2 & -3 & 1 \\ -3 & 16 & 9 \\ 1 & 9 & 10\end{array}\right] \)
\(Now \ P^T=\frac{1}{2}\left[\begin{array}{ccc}2 & -3 & 1 \\ -3 & 16 & 9 \\ 1 & 9 & 10\end{array}\right]=P\)
\(\text { Thus, } P=\frac{1}{2}\left(A+A^T\right)\) is a symmetric matrix.
\(\text { Let } Q=\frac{1}{2}\left(A-A^T\right)\)
\(=\frac{1}{2}\left[\begin{array}{ccc} 0 & 9 & 9 \\ -9 & 0 & -3 \\ -9 & 3 & 0 \end{array}\right]\)
\(\text { Then } Q^T=\frac{1}{2}\left[\begin{array}{ccc} 0 & -9 & -9 \\ 9 & 0 & 3 \\ 9 & -3 & 0 \end{array}\right]=-Q\)
\(\text { Thus } Q=\frac{1}{2}\left(A-A^T\right)\) is a skew-symmetric matrix.
\(A=P+Q=\frac{1}{2}\left[\begin{array}{ccc} 2 & -3 & 1 \\ -3 & 16 & 9 \\ 1 & 9 & 10 \end{array}\right]+\frac{1}{2}\left[\begin{array}{ccc} 0 & 9 & 9 \\ -9 & 0 & -3 \\ -9 & 3 & 0 \end{array}\right]\)
Thus A is expressed as the sum of symmetric and skew-symmetric matrices.

RHS = tan (60° + \(\theta\)) tan (60° - \(\theta\))
Let 60° + \(\theta\) = \(\alpha\) and 60° - \(\theta\) = \(\beta\)
RHS = \(tan\ \alpha\ tan\ \beta=\frac{sin\alpha\ sin\beta}{cos\alpha\ cos\beta}\)
\(=\frac{2sin\alpha\ sin\beta}{2cos\alpha\ cos\beta}=\frac{cos(\alpha-\beta)-cos(\alpha+\beta)}{cos(\alpha+\beta)+cos(\alpha-\beta)}\)
\(=\frac{cos2\theta-cos120​​°}{cos120°+cos20\theta}=\frac{cos2\theta-(-\frac{1}{2})}{-\frac{1}{2}+cos2\theta}\)
[\(\because\)cos 120° = cos (90° + 30°) = - sin 30° = \(-\frac{1}{2}\)]
\(=\frac{2cos2\theta+1}{2cos2\theta-1}=LHS\)

S₁₀ = 52, S₁₅ = 77, S₂₀ =?
\({S}_{10}={10\over 2}(2a(10-1)d)=52\)
= 5 (2a + 9d) = 52
\(2a+9d={52\over 5}\)
\({S}_{15}={15\over 2}(a+(15-1)d)=77\)
\(={15\over 2}(2a+14d)=77\)
\(2a+14d={154\over 15}\)
\(2a+9d={52\over 2}\)
(2)-(1) \(\Rightarrow\)    \({2a+14d={154\over 15}\over 5d={154\over 15}-{52\over 2}={154-156\over 15}}\)
\(5d={-2\over 15}\Rightarrow d={-2 \over 75}\)
substituting in (1) ​​​​​​​\(\Rightarrow\) \(2a-{18\over 75}={52\over 5}\)
\(2a={52\over 5}+{18\over 75}={780+18\over 75}\)
\(2a={798\over 75}\Rightarrow a={399\over 75}={133 \over 25}\)
\({S}_{20}={20 \over 2}\left[ \left( {798 \over 75} \right)+19\left( {-2\over 75} \right) \right]\)
\(=10{(798-38)\over 75}=10\left( {760\over 75} \right)\)
\(={1520\over 15}={304\over 3}\)

The family of equations of straight lines through the point of intersection of the lines is of the form
\((a_1x+b_1y+c_1)+\lambda(a_2x+b_2y+c_2)=0\)
That is, \((3x+2y+5)+\lambda(3x-4y+6)=0\)
Since the required equation passes through the point (1, 1), the point satisfies the above equation.
Therefore {3 + 2(1) + 5} + λ {3(1) - 4(1) + 6} = 0 ⇒ λ= -2
Substituting λ = -2 in the above equation we get the required equation as 3x - 10y + 7 = 0 (verify the above problem by using two points form)

The sine formula is, \(({a\over sin A})=({b\over sin B})=({c\over sin C})=2R\)
Now, \({b-c\over a}cos {A\over2}={2Rsin B-2R sin C\over 2Rsin A}cos {A\over2}\)
\(={2sin({B-C\over2})cos({B+C\over 2})\over 2sin {A\over2}cos{A\over2}}cos{A\over 2}\)
\(={sin({B-C\over2})cos(90^o-{A\over2})\over sin {A\over2}}\)
\(={sin({B-C\over2}) sin {A\over2}\over sin {A\over2}}=sin ({B-C\over2})\)

Let, P(n) = 1² + 2² + 3² +...+k² = \(\frac { n(n+1)(2n+1) }{ 6 } \)
Substituting n = 1 in the statement we get, P(1) = \(\frac { 1(1+1)(2(1)+1) }{ 6 } \)  = 1.
Hence, P(1) is true. Let us assume that the statement is true for n = k. Then
P(k) = 1² + 2² + 3²+...+k² = \(\frac { k(k+1)(2K+1) }{ 6 } \)
We need to show that P(k + 1) is true. Consider
\(P(k+1) =\underbrace{1^2+2^2+3^3+...k^2}+(k+1)^2\)
= P(k) + (k+1)²
= \(\frac { k(k+1)(2K+1) }{ 6 } +(k+1)^2\)
= \(\frac { k(k+1)(2k+1)+6(k+1)^{ 2 } }{ 6 } =\frac { (k+1)(k(2k+1)+6(k+1) }{ 6 } \)
= \(\frac { (k+1)(2k^{ 2 }+7k+6) }{ 6 } =\frac { (k+1)[(k+2)(2k+3)] }{ 6 } \)
= \(\frac { (k+1)[(k+1)+1)(2(k+1)+1)] }{ 6 } \)
That is, P(k+1) = \(\frac { (k+1)((k+1)+1)(2(k+1)+1) }{ 6 } \)
This implies P(k + 1) is true. The validity of P(k + 1) follows from that of P(k). Therefore by the principle of mathematical induction,
1² +2² + 3²+...n² = \(\frac { n(n+1)(2n+1) }{ 6 } \), for all n ≥ 1.

\(\sqrt{3}tan^2\theta+(\sqrt{3}-1)tan\theta-1=0\)
\(\sqrt{3}tan^2\theta+\sqrt{3}tan\theta-tan\theta-1=0\)
\((\sqrt{3}tan\theta-1)(tan\theta+1)=0\)
Thus, either \(\sqrt{3}tan\theta-1=0(or)tan\theta+1=0\)

If\(\sqrt{3}tan\theta-1=0, \ then\)  \(tan\theta={1\over\sqrt{3}}=tan {\pi\over 6}\) \(\Rightarrow \theta =n\pi+{\pi\over6},n \in Z ....(i)\)

If \(tan\theta+1=0 \ then\)  \(tan\theta=-1=tan ({-\pi\over 4})\) \(\Rightarrow \theta =n\pi-{\pi\over4},n \in Z ....(ii)\)

From (i) and (ii) we have the general solution.

Given p = length of perpendicular from (0, 0) to \(x\sec { \theta } +y{ cosec }{ \theta }=a\)
\(\Rightarrow p=\left| \frac { 0\left( \sec { \theta } \right) +0\left( cosec{ \theta } \right) -a }{ \sqrt { \sec ^{ 2 }{ \theta } +{ cosec }^{ 2 }{ \theta } } } \right| \)
\(\Rightarrow p=\frac { a }{ \sqrt { \frac { 1 }{ \cos ^{ 2 }{ \theta } } +\frac { 1 }{ \sin ^{ 2 }{ \theta } } } } =\frac { a }{ \sqrt { \frac { \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } }{ \sin ^{ 2 }{ \theta } \cos { \theta } } } } \)
\(\Rightarrow p=\frac { a\sin { \theta } \cos { \theta } }{ \sqrt { \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } } } =\frac { a\sin { \theta } \cos { \theta } }{ 1 } \)
\(\Rightarrow p=a\sin { \theta } \cos { \theta } \)
Also, it is given that p¹ = length of perpendicular from (0, 0) to \(x\cos { \theta } -y\sin { \theta } =a\cos { 2\theta } \)
\(\Rightarrow { p }^{ 1 }=\left| \frac { 0\left( \cos { \theta } \right) -0\left( \sin { \theta } \right) -a\cos { \theta } }{ \sqrt { \cos ^{ 2 }{ \theta } +{ \left( -\sin { \theta } \right) }^{ 2 } } } \right| \)
\(=\frac { a\cos { 2\theta } }{ 1 } \)
\(\Rightarrow { p }^{ 1 }=a\cos { 2\theta } \)
LHS = 4p² + (p¹)²
= 4 (a cos \(\theta\) cos \(\theta\))² + (a coos 2\(\theta\))²
= a² (2 sin \(\theta\) cos \(\theta\))² + a² (cos 2\(\theta\))²
= a² ((sin 2\(\theta\))² + (cos 2\(\theta\))²) = a² (sin² 2\(\theta\) + cos² 2\(\theta\))
= a² (1) Hence Proved.

Given equation of pair of lines is 6x² + 5xy - py² + 7x + qy - 5 = 0.
6x²+ 5xy - py²+ 7x + qy - 5 = 0
Here a = 6, b = -p, 2h = 5, 2g = 7, 2f = q
h = \(\frac { 5 }{ 2 } \) g = \(\frac { 7 }{ 2 } \) f = \(\frac { q }{ 2 } ,\) c = -5
The condition to represent perpendicular lines is a + b = 0 \(\Rightarrow\) 6 - p = 0 \(\Rightarrow\) p = 6
The condition to represent pair of lines is abc + 2fgh - af²- bg²- ch² = 0
\(\Rightarrow \ 6-(-6)(-5)+\left( \frac { 7 }{ 2 } \right) \left( \frac { 5 }{ 2 } \right) -6\left( \frac { { q }^{ 2 } }{ 4 } \right) +6\left( \frac { 4q }{ 4 } \right) +5\left( \frac { 25 }{ 4 } \right) =0\)
Multiplying by 4 throughout we get,
720 + 35q - 6q²+294 + 125 = 0
\(\Rightarrow \quad 35q-6q^{ 2 }+1139=0\)
\(\Rightarrow \quad q=\frac { 35\pm \sqrt { { (-35) }^{ 2 }-4(6)(1139) } }{ 2\times 6 } \)
\(\Rightarrow \quad q=\frac { 35\pm \sqrt { 1225+27336 } }{ 12 } \)
\(\Rightarrow \quad q=\frac { 35\pm \sqrt { 28561 } }{ 12 } =\frac { 35\pm 169 }{ 12 } \)
\(\Rightarrow \quad q=\frac { 204 }{ 12 } or-\frac { 134 }{ 12 } \)
\(\Rightarrow \quad q=17\quad or\quad -\frac { 67 }{ 6 } \)
Hence P = 6, q = 17 or \(-\frac{67}{6}\)

Since \(\alpha ,\beta \) are the roots of the equation \(​​{ x }^{ 2 }-px+q=0\),we have
\(\alpha +\beta =-\frac { \left( -p \right) }{ 1 } =p\quad and\quad \alpha \beta =\frac { q }{ 1 } q\)
\(\therefore \log { (1+px+q{ x }^{ 2 }) } =\log { \left[ 1+(\alpha +\beta )x+\alpha \beta { x }^{ 2 } \right] } \)
\(=\log { \left[ (1+\alpha x)(1+\beta x) \right] } \)
\(=\log { (1+\alpha x)+log(1+\beta x) } \)
\(=\left( \alpha x-\frac { { \alpha }^{ 2 } }{ 2 } { x }^{ 2 }+\frac { { \alpha }^{ 3 } }{ 2 } { x }^{ 3 }+......+\infty \right) +\left( \beta x-\frac { { \beta }^{ 2 }{ x }^{ 2 } }{ 2 } +\frac { { \beta }^{ 3 }{ x }^{ 3 } }{ 3 } +.....\infty \right) \)
\(=(\alpha +\beta )x-\frac { { (\alpha }^{ 2 }+\beta ^{ 2 }) }{ 2 } { x }^{ 2 }+\frac { { (\alpha }^{ 3 }+\beta ^{ 3 }) }{ 3 } { x }^{ 3 }-...\infty \)
Hence proved. 

Given equation is 2x² - xy - 3y² - 6x + 19y - 20 = 0
Consider 2x² - xy - 3y² = (2x - 3y)(x + y) (By factorising)
\(\therefore\)  (2x² - xy - 3y²- 6x + 19y - 20) = (2x-3y+1)(x + y + m)
Equating the co-efficients of x and y both sides we get,

-6 = 2m + l   ......(1)
-      +      -
19 = -3m + l  ......(2)
__________________
 -25  = 5m
\(\Rightarrow\)  m = -5
Substituting m = -5 in (1) we get,
-6 = -10 + l \(\Rightarrow\) -6 + 10 = l
\(\Rightarrow\)  l = 4
Hence the separate equations are 2x - 3y + 4 = 0 ......(3)
and x + y - 5 = 0  .....(4)
To get the point of intersection solve (3) and (4)
   (3) \(\rightarrow\)           2x - 3y + 4 = 0
                           -       -    +
   (4) \(\rightarrow\) \(\times\)2\(\Rightarrow\) 2x + 2y - 10 = 0 
                   ______________
                         -5y+ 14 = 0
\(\Rightarrow\)  -5y = -14
\(\Rightarrow\)  \(y=\frac { -14 }{ -5 } \)
\(\Rightarrow\) \(y=\frac { 14 }{ 5 } \)
Substituting y = \(\frac { 14 }{ 5 } \) in (3) we get
\(2x-3\left( \frac { 14 }{ 5 } \right) +4=0\)
\(\Rightarrow \quad 2x-\frac { 42 }{ 5 } +4=0\)
\(2x=\frac { 42 }{ 5 } -4\)
\(\Rightarrow \quad 2x=\frac { 42-20 }{ 5 } =\frac { 22 }{ 5 } \)
\(\Rightarrow \quad x=\frac { 11 }{ 5 } \)
\(\therefore\)  The point of intersection is \(\left( \frac { 11 }{ 5 } ,\frac { 14 }{ 5 } \right) \)
Hence the given lines are intersecting lines.
Let \(\theta\) be the angle between the given lines.
Then tan \(\theta\) = \(\pm \frac { 2\sqrt { { h }^{ 2 }-ab } }{ a+b } \)
\(=\pm \frac { 2\sqrt { { \left( -\frac { 1 }{ 2 } \right) }^{ 2 } } -2(-3) }{ 2-3 } \begin{bmatrix} \because \quad a=2,\quad 2h=-1\quad b=-3 & \\ \quad \quad \quad \quad \quad \quad h=\frac { 1 }{ 2 } , & \\ & \end{bmatrix}\)
\(=\pm \frac { 2\sqrt { \frac { 1 }{ 4 } +6 } }{ -1 } =2\sqrt { \frac { 25 }{ 4 } } =2\left( \frac { 5 }{ 2 } \right) =5\)
\(\because \quad \theta ={ tan }^{ -1 }(5).\)

Given series is \(\sum_{n=1}^{\infty}{1\over 2^{n-1}}\left({1\over 9^{n-1}}+{1\over9^{2n-1}}\right)\)
\(S_\infty=1\left(1+{1\over9}\right)+{1\over2}\left( {1\over9}+{{1\over9^3}} \right)+{1\over 2^2}\left( {1\over9^2}+{1\over9^5} \right)+...\)
\(=\left( 1+{1\over9} \right)+{1\over2}\left( {1\over9}+{1\over 9^3} \right)+{1\over2^2}\left( {1\over2^2}+{1\over 9^5} \right)+..\)
Separating the first term and second term from each bracket, we get
\(=\left[ 1+{1\over 2}+{1\over 9}+{1\over 2^2}\left( {1\over 9^2}\right)+..\right]+\left[ {1\over9}+{1\over2}\left( {1\over 9^5} \right)+...\right]\)
\(=\left[1+{1\over2}\left(1\over 9\right)+{1\over 2^2}\left( {1\over 9^2} \right)+...\right]+{1\over9}\left[1+{1\over2}\left( {1\over 9^2}\right) +{1\over2^2} \left( {1\over 9^4} \right)+...\right]\)
Now consider \({1+{1\over2}}\left(1\over9\right)+{1\over 2^2}\left(1\over 9^2\right)+...\)
Here a = 1, \(r={1\over 18}\)
\(∴\ S_\infty={a\over 1-r}={1\over 1-{1\over 8}}={18\over 17}\)
In \(1+{1\over2}\left(1\over9^2\right)+{1\over 2^2}\left(1\over 9^4\right)+...\)
\(a=1,r={1\over 81\times2}={1\over162}\)
\(∴\ S_\infty =1, r={a\over 1-r}={1\over 1-{1\over 162}}={162\over 161}\)
Substituting these values in (1) we get
\(S_\infty={18\over 17}+{1\over 9}\times{162\over161}={18\over17}+{162\over 1449}={26082+2754\over 17\times1499}={28836\over 24633}\)
\({ S }_{ \infty }=1.170\)

Given series is \(\sqrt { 3 } +\sqrt { 75 } +\sqrt { 243 } +.... .\) and \(S_n =435\sqrt { 3 }\)
Given series is \(1(\sqrt3)+5(\sqrt3)+9(\sqrt3)+...\)
Here a = √3, d = 4√3
∴ The given series an arithmetic progression
\(∴\ S_n={n\over2}[2a+(n-1)d]\)
\(435\sqrt3={n\over2}[2\sqrt3 +(n -1)4\sqrt3]\)  [∵ given S_n = 435√3J]
\(435\sqrt3={n\over2}[2\sqrt3+4n\sqrt3-4\sqrt4]\)

\(⇒\ 435\sqrt3={n\over2}[4n\sqrt3-2\sqrt3]\)
\(⇒\ 435\sqrt3=2{\sqrt3.n\over2}[2n-1]\)
⇒ 435 = 2n²-n
⇒ 2n²- n - 435 = 0
⇒ (n = 15)(2n + 29) = 0
⇒ \(n-15\ or\ n={-29\over2}\) which is not possible
⇒ n = 15

Let S_n = 8 + 88 + 888 + 8888 + .... upto n terms
= 8 (1 + 11 + 111 + 1111 + ....) upto n terms
\(={8\over9}(9 + 99 + 999 + ...)\)
\({ S }_{ n }=\frac { 8 }{ 81 } \left[ \left( { 10 }^{ n }-1 \right) -9n \right] \) [multiplying and dividing by 9]
\(={8\over9}[10 -1) + (100 -1) + (1000 -1) + ...]\)
\(S_n={8\over 9}[(10^1 +10^2 +10^3 + ... +10^n)-(1+1+1+ ... +1n\ terms)]\)
In 10 + 10² + 10³ + ... + 10ⁿ, a = 10, r= 10, and it forms a G.P.
\(∴\ S_n={a(r^n-1)\over r-1}=10{(10^n-1)\over 10-1}={10\over 9}(10^n)-1\) and 1 + 1 + 1 ... + upto n terms = n
Substituting these values in (1) we get
\(S_n={8\over 9}\left[ 10(10^n-1)n\over 9\right]\)
\(S_n={8\over 81}[(10^n-1)-9n]\)

Let p(n) denote the statement
\(1^2+3^2+5^2+...+(2n-1)^2={n(2n-1)(2n+1)\over 3}\)

Step1: Putting n = 1
\(1^2={1(2-1)(2+1)\over3}={3\over 3}=1\)
∵ p(1) is true
Step 2: Let us assume that p(K) is true
i.e. \(1^2+3^2+5^2+..+(2K-1)^2={K(2K-1)\over3}\)
To prove thatp(K + 1) is true
\(∵\ 1^2+3^2 +5^2 + ..+ (2K-1)^2 + [2(K + 1)-1]^2={(K+1)(2(K+1)-1)(2(K+1)+1)\over 3}\)
i.e. to P.T. \(1^2 + 3^2 + 5^2 + ... + (2K -1)^2(2K + 1)^2 ={(K+1)(2K+1)(2K+3)\over 3}\)
LHS = 1²+ 3² + 5² + ... + (2K - 1)² + (2K + 1)²
\(={K(2K-1)(2K+1)\over3}+(2K+1)^2\)
\(=(2K+1)\left[{K(2K-1)\over 3}+2K+1\right]\)
\(=(2K+)\left[2K^2-K+6K+3\over3\right]={(2K+1)(2K^2+5K+3)\over3}\)
\(={(2K+1)(2K^2+5K+3)\over3}\)=RHS
∵ p(K+1)is true
∵ By mathematical induction, P(n) is true for all values of n.

(i) Either starts with L or ends with S.

1	4	3	2	1
L

Since the words starts with L, the remaining 4 boxes can be filled in 4 x 3 x 2 x 1 ways by the remaining letters 0, T, U, S.
∴ Number of words starting with L
= 1 \(\times\) 4 \(\times\) 3 \(\times\) 2 \(\times\) 1 = 24.

1	2	3	4	1
				S

Here also, the remaining 4 boxes can be filled in 4 \(\times\) 3 \(\times\) 2 \(\times\) 1 ways = 24.......(1)
Number of words ending with S = 24 ....(2)
Number of words starting with L and end with S are 3 \(\times\) 2 \(\times\) 1 = 6...(3)
∴  By fundamental principle of addition, number of words either starts with L nor ends with S = 24 + 24 - 6 = 48 - 6 = 42

	3	2	1
F				S

(ii) Neither starts with L nor ends with S.
Total number of words formed by the letters of the word LOTUS is 5 \(\times\) 4 \(\times\) 3 \(\times\) 2\(\times\) 1 = 120.
Now, number of words neither starts with L nor end with S.
= (Total number of words) - (Number of words starts with either L nor ends with S)
= 120 - 42
= 78.

\({{6x^2-x+1}\over{x^3+x^2+x+1}}={6x^2-x+1\over x^2(x+1)+1(x+1)}={6x^2-x+1\over (x^2+1)(x+1)}\)
⇒ \({6x^2-x+1\over (x^2+1)}={Ax+B\over x^2+1}+{C\over x+1}\)
⇒ 6x²-x+1 = (Ax+B)(x+1)+C(x²+1)
Putting x=-1 in (1) we get
6 + 1 + 1 = C(1 + 1) ⇒ 8 = 2C ⇒ C = 4
EEquating the Co-efficient of x² in (1) we get,
6 = A + C ⇒ 6 = A + 4 ⇒ A = 6 - 4
⇒ A = 2
Putting x=0 in (1) we get
1 = B + C ⇒​​​​​​​ 1 = B + 4 ⇒​​​​​​​ 1 - 4 = B
⇒​​​​​​​ B = -3
\(\therefore\ {{6x^2-x+1}\over{x^3+x^2+x+1}}={{2x-3}\over{x^2+1}}+{{4}\over{x+1}}\)

\({1\over x^4-1}={1\over (x^2+1)(x^2-1)}={1\over (x^2+1)(x+1)(x-1)}\)
\({1\over x^4-1}={Ax+B\over x^2+1}+{C\over x+1}+{D\over x-1}\)
\(⇒ {1\over x^2-1}={(Ax+B)(x+1)(x-1)+C(x^2+1)(x-1)+D(x^2+1)(x+1)\over (x^2+1)(x^2+1)}\)
⇒ 1= (Ax + B)(x + 1)(x - 1) + C(x² + 1) (x - 1) + D(x² + 1) (x + 1)
Putting x=1 in (1) we get
1 = D (2) (2) ⇒  \(D={1\over 4}\)
Putting x = -1 in we get
1 = C(2)(-2) ⇒  \(C=-{1\over 4}\)
Equating the coefficient of x³ we get
0 = A + C + D ⇒ A = - C - D
⇒ \(A={1\over 4}-{1\over 4}=0\)
⇒ A = 0
Putting x = 0 in (1) we get
1 = -B - C +D
⇒\(1=-B+{1\over 4}+{1\over 4}\)
⇒​​​​​​​ \(B=-1+{1\over 2}⇒B=-{1\over 2}\)
\(∴\ \ {1\over x^4-1}={0x-{1\over2}\over x^2+1}+{-{1\over 4}\over x+1}+{{1\over 4}\over x-1}\)
\(\Rightarrow\)  \({{1}\over{x^4-1}}={{-{{1}\over{2}}}\over{x^2+1}}-{{{{1}\over{4}}}\over{x+1}}+{{{{1}\over{4}}}\over{x-1}}=-{{1}\over{2(x^2+1)}}-{{1}\over{}4(x+1)}+{{1}\over{4(x-1)}}\)

LHS = a(cos B + cos C)
= \(a\left( \frac { { c }^{ 2 }+{ a }^{ 2 }-{ b }^{ 2 } }{ 2ac } +\frac { { a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 } }{ 2ab } \right) \)
\(\frac { 1 }{ 2 } \left[ \frac { { c }^{ 2 }+{ a }^{ 2 }-{ b }^{ 2 } }{ c } +\frac { { a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 } }{ b } \right] \)
= \(\frac { 1 }{ 2bc } \left[ { bc }^{ 2 }+{ a }^{ 2 }b-{ b }^{ 3 }+{ a }^{ 2 }c+{ b }^{ 2 }c-{ c }^{ 3 } \right] \)
= \(\frac { 1 }{ 2bc } \left[ -\left( { b }^{ 3 }+{ c }^{ 3 } \right) +\left( { a }^{ 2 }b+{ a }^{ 2 }c \right) +{ bc }^{ 2 }+{ b }^{ 2 }c \right] \)
= \(\frac { b+c }{ 2bc } \left[ -{ b }^{ 2 }+bc-{ c }^{ 2 }+{ a }^{ 2 }+bc \right] \)
= \(\frac { b+c }{ 2bc } \left[ { a }^{ 2 }-{ b }^{ 2 }-{ c }^{ 2 }+2bc \right] \)
= \(\left( \frac { b+c }{ 2bc } \right) \left[ \left( a+b-c \right) +\left( a-b+c \right) \right] \)
= \(\left( \frac { b+c }{ 2bc } \right) \left( a+b+c-2c \right) \left( a+b+c-2b \right) \)
= \(\left( \frac { b+c }{ 2bc } \right) \left[ \left( 2s-2c \right) \left( 2s-2b \right) \right] \)
= \(\frac { b+c\left( 4 \right) }{ 2bc } \left[ \left( s-c \right) \left( s-b \right) \right] =2\left( b+c \right) \left[ \frac { \left( s-b \right) \left( s-c \right) }{ bc } \right] \)
= \(2\left( b+c \right) { sin }^{ 2 }\frac { A }{ 2 } \)
= RHS
Hence proved.

Let x be the smaller of two positive odd integers so that other one is x + 2
Given x > 10,.....(1)    and  x + 2 > 10.
\(\Rightarrow\) x > 10-2
\(\Rightarrow\) x > 8.....(2)
And (x) + (x + 2) < 40 .....(3)
From (1) and (2) we get, x > 10.....(4)
From (3) we get
2x+2 <40
\(\Rightarrow\) 2x < 40 -2
\(\Rightarrow\) 2x < 38
\(\Rightarrow\) x < \(\frac { 38 }{ 2 } \)
\(\Rightarrow\) x = 19....(5)
From (4) and (5) we get,
10 < x < 19.
Since x is an odd natural number, x can take the values 11, 13, 15, 17.
Hence the required possible consecutive pairs will be (11, 13), (13, 15), (15, 17)

LHS = sin 2A + sin 2B + sin 2C
= 2sin \(\left( \frac { 2A+2B }{ 2 } \right) \)cos\(\left( \frac { 2A-2B }{ 2 } \right) \) + 2sin C cos C.
= 2sin(A+B) cos(A-B) + 2sin C cos C
= 2sin\(\left( \frac { \pi }{ 2 } -C \right) \)cos(A-B) + 2sin C cos C
= 2cos C cos(A-B) + 2sin C cos C
= 2 cos C\(\left[ cos(A-B)+sin\left( \frac { \pi }{ 2 } (A+B) \right) \right] \)
= 2cos C 2cos A cos B
= 4cos A cos B cos C = RHS

Given A = {a, b, c, d}, B = {a, c, e}, C = {a, e}
B∩C = {a, c}
A∩(B∩C) = {a}
A ∩ B = {a, e}
(A ∩ B) ∩ C = {a}
From (1) and (2), it is clear that A ∩ (B ∩ C) = (A ∩ B) ∩ C.

Given \(\sqrt{x+5}+\sqrt{x+21}=\sqrt{6x+40}\)
Squaring both sides we get
\((\sqrt{x+5}+\sqrt {x+21})^2=(\sqrt{6x+40})^2\)
⇒ \(z+5+z+21+2\sqrt{(x+5)(x+21)}=6x+40\)
⇒ \( 2x+26+2\sqrt{(x+5)(x+21)}=6x+40\)
⇒ \(2\sqrt{(x+5)(x+21)}=6x+40-2x-26\)
⇒ \(2\sqrt{(x+5)(x+21)}=4x+14\)
 \(\sqrt{(x+5)(x+21)}=2x+7\)
Squaring again we get
(x + 5)(x + 21) = (2x + 7)²
⇒ x² + 21x + 5x + 105 = 4x²+ 49 + 28x
⇒ x² + 26x + 105 = 4x² +49 + 28x
⇒ 3x² + 2x- 56 = 0
\(x = {-2 \pm \sqrt{4-4(3)(-56)} \over 6}\)
\(x = {-2 \pm \sqrt{4+672)} \over 6}\)
\(x={-2\pm26\over 6}⇒x=4,{-14\over 3}\)
⇒ When x = 4
Case (i) :
\(\sqrt{4+5}+\sqrt{4+21}=\sqrt{6(4)}+40\)
\(\sqrt9+\sqrt{25}=\sqrt{64}\)
3 + 5 = 8
8 = 8 which is true  ⇒ x = 4 is a root
Case (ii) : When x = \(-14\over 3\)
\(\sqrt{{1-\over3}+5}+\sqrt{{-14\over 3}+21}=\sqrt{+6\left(-14\over 3\right)+40}\)
\(\sqrt{1\over 3}+\sqrt{49\over 3}=\sqrt{12}\) which is not true
\(\therefore\) x  \(={{-14}\over{3}}\) is not a root.

(i) \(y=-x^{ \left( \frac { 1 }{ 3 } \right) }\)

\(Let \ y=-x^{ ^{ \frac { 1 }{ 3 } } }\)
\(Then \ y=-x^{ ^{ \frac { 1 }{ 3 } } }\) is the reflection of the graph of \(y=x^{ ^{ \frac { 1 }{ 3 } } }\)  about the x-axis.
(ii) \(y=x^{1\over 3}+1\)

Let \(y=x^{ ^{ \frac { 1 }{ 3 } } }\)
Then \(y=x^{ ^{ \frac { 1 }{ 3 } } }+1\) is the x graph of \(y=x^{ ^{ \frac { 1 }{ 3 } } }\) shifts to the upward for one unit
(iii) \(y=x^{1\over 3}-1\)

Let \(y=x^{1\over 3}\)
Then \(y=x^{1\over 3}\)-1 is the graph of \(x^{ ^{ \frac { 1 }{ 3 } } }\) shifts to the downward for one unit.
(iv)  \(y=(x+1)^{1\over 3}\)

x	0	1	7	-9
y	1	1	2	-2

\(y=(x+1)^{1\over 3}\) causes the graph of \({x}^{\frac{1}{3}}\), shifts to the left for one unit.

Given f(x) = 3x - 4
Let y = 3x - 4 ⇒ y + 4 = 3x
\(⇒ x={y+4\over 3}\)
Let g(y) = \(y+4\over 3\)
Now gof(n) = g(f(n)) = g(3\(\times\) -4) = \({3x-4+4\over 3}={3x\over 3}=x\)
and fog(y) = f(g(y)) = \(f\left(y+4\over 4\right)=3\left(y+4\over 3\right)-4=y+4-4=y\)
Thus, gof(x) = I_x and fog (y) = I_y
This implies that f and g are bijections and inverses to each other
Hence f is bijection and \(f^{-1} (x)={y+4\over 3}\)
Replacing y by x, we get f^-1 (x) = \(\frac { x+4 }{ 3 } \)

Hence, the graph of y = f^-1(x) is the reflection of the graph of f in y = x

Given f(x) = 1.23x where x represents the number of American dollars.
and g(y) = 50.50y where y represents the number of Singapore dollars.

To convert American dollars to Indian rupees, we have to find out go f(x)
∴ go f(x) = g(f(x))
= g(1.23x)
= 50.50[1.23x]
= 62.115x
∴ The function for exchange rate of American dollars in terms of Indian rupee is g o f (x) = 62.115x.

Given P(A) = 0.5, P(B) = 0.8
⇒ P(B/A) = 0.8
We kmow P(B/A) = \(\frac{P(A\cap B)}{P(A)}\)
⇒ 0.8 = \(\frac{P(A\cap B)}{0.5}\)
⇒ \(P(A\cap B)=(0.8)(0.5)=0.4\)
(i) Now P(A./B) = \(\frac{P(A\cap B)}{P(B)}=\frac{0.4}{0.8} =\frac{1}{2}=0.5\)
(ii) \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\)
= 0.5 + 0.8 - 0.4 = 1.3 - 0.4
= 0.9

\( \int\left(1-x^2\right)^{-\frac{1}{2}} d x =\int \frac{1}{\left(1-x^2\right)^{1 / 2}} d x \)
\(=\int \frac{1}{\sqrt{1-x^2}} d x=\sin ^{-1} x+c\)

\(y={e^{3x}\over 1+e^x}\)
\(\frac { dy }{ dx } =\frac { (1+{ e }^{ x }).\frac { d }{ dx } ({ e }^{ 3x })-{ e }^{ 3x }.\frac { d }{ dx } (1+{ e }^{ x }) }{ (1+{ e }^{ x })^{ 2 } } =\frac { (1+{ e }^{ x }).{ e }^{ 3x }.3-{ e }^{ 3x }(0+{ e }^{ x }) }{ (1+{ e }^{ x })^{ 2 } } \)
=\(\frac { 3{ e }^{ 3x }+3{ e }^{ 4x }-{ e }^{ 4x } }{ (1+{ e }^{ x })^{ 2 } } =\frac { 3e^{ 3x }+2e^{ 4x } }{ (1+{ e }^{ x })2 } \) .

\(lim_{x\rightarrow0}{\sqrt{1-x}-1\over x^2}\)
Multiplying and dividing by \((\sqrt{1-x}+1)\)we get,
\(lim_{x\rightarrow0}{\sqrt{1-x}-1\over x^2}\times {\sqrt{1-x}+1\over \sqrt{1-x}+1}\)\(=lim_{x\rightarrow0}{({1+x})-1\over x^2[\sqrt{1-x}+1]}\)
\(=lim_{x\rightarrow0}{{-x}\over x^2[\sqrt{1-x}+1]}= lim_{x\rightarrow0}{{-1}\over x[\sqrt{1-x}+1]} =-\infty\)
Since f(x) -\(\infty\) as x \(\rightarrow\) 0, the limit of the given function does not exist.

(i) If f(n) = f(m), then n + 2 = m + 2 and hence m = n. Thus f is one-to-one. As 1 has no pre-image, this function is not onto.
(ii) As above, this function is one-to-one. If m is in the co-domain, then m − 2 is in the domain and f(m − 2) = (m − 2) + 2 = m; thus m has a pre-image and hence this function is onto.

cos 35^o - cos 75^o = \(2\sin { \left( \frac { 35+75 }{ 2 } \right) } .\sin { \left( \frac { 75-35 }{ 2 } \right) } \)
= 2 sin 55^o sin 20^o

Let l, m, n ∈ p.
Reflexivity: We cannot say l is perpendicular to l itself.
∴ l R l \(\Rightarrow \) R  is not reflexive.

Symmetry:  lRm ≠ mRl
I is perpendicular to m ⇒ m is perpendicular to l
∴  R is symmetric
Transitive: lRm and mRn ≠ lRn.
l is perpendicular to m and m is perpendicular to n.
⇒ l is perpendicular to n.
∴ R is not transitive.
⇒ R is only symmetric.