\({ x }^{ 2 }-3x+10={ x }^{ 2 }-2\left( \cfrac { 3 }{ 2 } \right) x+10+\cfrac { 9 }{ 4 } -\cfrac { 9 }{ 4 } \) 
= \(\left( x-\cfrac { 3 }{ 2 } \right) ^{ 2 }+\cfrac { 31 }{ 4 } =\left[ \left( x-\cfrac { 3 }{ 2 } \right) ^{ 2 }+\cfrac { \sqrt { 31 } }{ 2 } \right] ^{ 2 }\) 
\(\therefore \int { \sqrt { { x }^{ 2 }-3x+10 } dx } =\int { \sqrt { \left( x-\cfrac { 3 }{ 2 } \right) ^{ 2 }+\left( \cfrac { \sqrt { 31 } }{ 2 } \right) ^{ 2 } } dx } \) 
= \(\cfrac { 1 }{ 2 } \left\{ x-\cfrac { 3 }{ 2 } \right\} \sqrt { \left( x-\cfrac { 3 }{ 2 } \right) ^{ 2 }+\left( \cfrac { \sqrt { 31 } }{ 2 } \right) ^{ 2 } } +\cfrac { 31 }{ 4 } log\left[ \left( x-\cfrac { 3 }{ 2 } \right) +\sqrt { \left( x-\cfrac { 3 }{ 2 } \right) ^{ 2 }+\left( \cfrac { \sqrt { 31 } }{ 2 } \right) ^{ 2 } } \right] \) 
= \(\cfrac { 2x-3 }{ 4 } \sqrt { { x }^{ 2 }-3x+10 } +\cfrac { 31 }{ 4 } log\left[ \left( x-\cfrac { 3 }{ 2 } \right) +\sqrt { { x }^{ 2 }-3x+10 } \right] +c\)

Let the given vectors be \(\vec { a } =3\vec { i } -5\vec { j } +8\vec { k } \) and \(\vec { b } =-2\vec { i } -2\vec { k } \) 
Now \(\vec { a } +\vec { b } =\left( 3\hat { i } -5\hat { j } +8\hat { k } \right) +\left( -2\hat { j } -2\hat { k } \right) \) 
= \(\hat { i } (3)+\hat { j } (-5-2)+\hat { k } (8-2)\) 
= \(3\hat { i } -7\hat { j } +6\hat { k } \) 
\(\left| \vec { a } +\vec { b } \right| =\sqrt { 9+49+36 } =\sqrt { 94 } units\) 
The unit vectors parallel \(\vec { a } +\vec { b } \) are \(\pm \cfrac { \vec { a } +\vec { b } \quad }{ \left| \vec { a } +\vec { b } \right| } =\pm \cfrac { 3\hat { i } -7\hat { j } +6\hat { k } }{ \sqrt { 94 } } \) 

\(LHS=\left| \begin{matrix} 1 & a & { a }^{ 3 } \\ 1 & b & { b }^{ 3 } \\ 1 & c & { c }^{ 3 } \end{matrix} \right| \begin{matrix} { C }_{ 1 }\rightarrow { C }_{ 1 }-{ C }_{ 2 } \\ { C }_{ 2 }\rightarrow { C }_{ 2 }-{ C }_{ 3 } \end{matrix}\) 
= \(\left| \begin{matrix} 0 & a-b & { a }^{ 3 }-{ b }^{ 3 } \\ 1 & b-c & { b }^{ 3 }-{ c }^{ 3 } \\ 1 & c & { c }^{ 3 } \end{matrix} \right| \) Expanding along C₁
= \((0)-(0)+1\left| \begin{matrix} a-b & { a }^{ 3 }-{ b }^{ 3 } \\ b-c & { b }^{ 3 }-{ c }^{ 3 } \end{matrix} \right| \) 
= (a - b) (b³ - c³) - (b - c) (a³ - b³)
= (a - b) (b - c) (b² + bc + 2) - (b - c) (a - b) (a² + ab + b²)
= (a - b) (b - c) [(b² + bc + c²) - (a² + ab + b²)]
= (a - b) (b - c) [b² + bc + 2 - a² - ab - b²]
= (a - b) (b - c) [(bc - ab) + (c^₂ - a²)]
= (a - b) (b - c) [(b (c - a) + (c + a) (c - a)]
= (a - b)(b - c)[b + c + a] (c - a)
= (a - b) (b - c) (c - a) (a + b + c) = RHS

f'(x) = 4x + 3
f'(0) = 3
f'(-1) = -1
\(\therefore\) f'(0) + 3f(-1) = 3 + 3(-1) = 3 - 3 = 0

Let I = \(\int { \frac { \left( { a }^{ x }+{ b }^{ x } \right) ^{ 2 } }{ { a }^{ x }{ b }^{ x } } } \) dx = \(\int { \frac { { a }^{ 2x }+{ b }^{ 2x }+{ 2a }^{ x }{ b }^{ x } }{ { a }^{ x }{ b }^{ x } } } dx\)
= \(\int { \frac { { a }^{ x } }{ { b }^{ x } } dx } +\int { \frac { { a }^{ x } }{ { b }^{ x } } dx+2 } \int { dx } =\int { \left( \frac { a }{ b } \right) ^{ x } } dx+\left( \frac { a }{ b } \right) ^{ x }dx+2\int { dx } \)
= \(\frac { \left( \frac { a }{ b } \right) ^{ x } }{ log\left( \frac { a }{ b } \right) } +\frac { \left( \frac { a }{ b } \right) ^{ x } }{ log_{ e }\left( \frac { a }{ b } \right) } +2x+c\)

\(\lim _{ x\rightarrow a }{ \frac { { (x+2) }^{ \frac { 3 }{ 2 } }-{ (a+2) }^{ \frac { 3 }{ 2 } } }{ x-a } } =\lim _{ x+2\rightarrow a+2 }{ \frac { { (x+2) }^{ \frac { 3 }{ 2 } }-{ (a+2) }^{ \frac { 3 }{ 2 } } }{ x-a } } \left[ \because \quad \lim _{ x\rightarrow a }{ \frac { { x }^{ n }-{ a }^{ n } }{ n-a } =n.{ a }^{ n-1 } } \right] \)
\( =\frac { 3 }{ 2 } (a+2{ ) }^{ \frac { 3 }{ 2 } -1 }=\frac { 3 }{ 2 } (a+2{ ) }^{ \frac { 1 }{ 2 } }\)

n(S) = 52
There are 4 kings and 4 queens in a pack of cards
\(\therefore\) n(E) = 8
There are 4 queens and 4 aces in a pack of cards
\(\therefore\) n(F) = 8
\(\therefore P(E)=\frac { 8 }{ 52 } =\frac { 2 }{ 13 } \) and \(P(F)=\frac { 8 }{ 52 } =\frac { 2 }{ 13 } \) and \(P(E\cap F)=\frac { 4 }{ 52 } =\frac { 1 }{ 13 } \)
\(P(E/F)=\frac { P(E\cap F) }{ P(F) } =\frac { \frac { 1 }{ 13 } }{ \frac { 2 }{ 13 } } =\frac { 1 }{ 13 } \times \frac { 13 }{ 2 } =\frac { 1 }{ 2 } \)

Let 'a' be the first term and 'd' the common difference of A.P.
Then, a₁ = a, a₂ = a + (2 -1) d = a + d
a₃ = a + (3-1) d = a + 2d, a₄ = a + (4-1) d = a + 3d
We have to show that a₂.a₃ - a₁.a₄ = 2 (a₂ - a₁)2
LHS = a₂a₃ -a,a₄ = (a + d)( a+ 2d) - a(a + 3d)
= a₂ + 3ad + 2d² - a² - 3ad = 2d²
RHS = 2 ( a₂ - a₁ )2 = 2 ( a + d - a)² = 2d²
Since LHS = RHS. Hence proved.

Let h_n be the harmonic progression and let \(a_n={1\over h_n}.\) 
Then a₅ = 19 and a₉ = 35.
As a_n's from an arithmetic progression, we have a + 4d = 19 and a + 8d = 35.
Solving these two equations, we get a = 3 and d = 4. 
Thus a₁₂ = a + 11d = 47.
Thus the 12^th term of the harmonic progression is \({1\over 47}.\)

LHS = b cos A + c cos A + c cos B + a cos B + a cos C + b cos C
= b cos C + c cos B + c cos A + a cos C + b cos A + a cos B
= a + b + c [by projection formula]

log₃5 log₂₅27 = log₃5 log₂₅3³
= log₃5 x 3 log₂₅3 (by exponent rule)
= 3 log₂₅5 =\(\frac{3}{log25}=\frac{3}{2log_55}=\frac{3}{2}\)

In the word RAMANUJAN there are 4 vowels (A, A, U, A) in that 3 A's, 1 U and 5 consonants (R, M, N, J, N) in that two N's and rest are distinct. The 4 vowels (A, A, A, U) can be arranged themselves in \(\frac { 4! }{ 3! } \) = 4 ways.  The 5 consonants (R, M, N, J, N) can be arranged themselves in \(\frac { 5! }{ 2! } \) = 60 ways. Therefore the number of required arrangements are \(\frac { 4! }{ 3! } \times \frac { 5! }{ 2! } \) = 4 \(\times\) 60 = 240.

LHS =¹⁵C₃ + 2 \(\times\) ¹⁵C₄ +¹⁵C₄ +¹⁵C₅
= (¹⁵C₃ +¹⁵C₄) + (¹⁵C₄ + ¹⁵C₅)
=(¹⁵C₃ +¹⁵C₄) + (¹⁵C₄ + ¹⁵C₅)       [∴ ⁿC_r-1 + ⁿC_r = n +¹C_r]
= ¹⁶C₄ + (¹⁵C₄ +¹⁵C₅)
=¹⁶C₄ + ¹⁶C₅
= ¹⁷C₅ = RHS.

Let f(x) = \(\frac { 1 }{ 1-2sinx } \)
When the denominator is 0,
1-2 sin x = 0
\(\Rightarrow\) 1 = 2 sin x
\(\Rightarrow sin\quad x=\frac { 1 }{ 2 } \)
\(\Rightarrow sin\quad x=sin\frac { \pi }{ 6 } \)
\(\Rightarrow x=n\pi +{ (-1) }^{ n }\frac { \pi }{ 6 } n\in Z\)   \(\left[ \because sin\quad x=sin\alpha \Rightarrow x=n\pi +{ (-1) }^{ n }\alpha \quad n\in Z \right] \)
Domain of f(x) is R - \(\left( n\pi +{ (-1) }^{ n }\frac { \pi }{ 6 } \right) ,n\in Z\)

(i)Repetition of digits is allowed

The unit place can be filled in 4 ways.
Since repetition is allowed, the tens place and hundreds place can also be filled in 4 ways each.
∴ Total number one-digit numbers = 4 x 4 x 4 = 64
(ii) Repetition of digits is not allowed

The unit place can be filled in 4 ways.
Since repetition of digits is not allowed, the tens place can be filled in 3 ways.
Hundreds place can be filled in 2 ways .
∴ Total number of 3-digit numbers without repetition = 4 \(\times\) 3 \(\times\) 2 = 24

Given diameter of the circle is 40 cm
r = 20cm

Let  AB = 20 cm be a chord of the circle
Since OA = OB = AB = 20 cm, ΔAOB is equilateral
ஃ θ = ㄥAOB = 60°= 60 \(\times\) \(\frac{\pi}{180}=\frac{\pi}{3}\) radians
Let I be the length of the minor arc of the chord AB.
Then θ = \(\frac{l}{r}\) ⇒ l = rθ ⇒ l = 20(\(\frac{\pi}{3}\))
l = \(\frac{20\pi}{3}\) cm = 20 \(\times\) \(\frac{22}{7}\times\frac{1}{3}\) = 20.95 cm (app)

395⁰
395⁰ = 360⁰ + 35⁰
\(\Rightarrow \) 395 - 35⁰ = 360⁰
∴ Coterminal angle For 395⁰ is 35⁰ 

Suppose the loan in cleared in n months. Clearly the amount forms an. A.P. with a = 20 and d = 15
∴ Sum of the amounts = 3250
S_n = 3250

\(⇒\ {n\over2}[2a + (n -1)d]=3250\)
\(⇒\ {n\over2}[40+(n-1)15]=3250\)
⇒ n(40 + 15n - 15) = 6500
⇒ n (15n + 25) 6500
⇒ 15n² + 25n = 6500
⇒ 15n² + 25n = 6500
⇒ 3n² + 5n - 1300 = 0
⇒ (n - 20) (3n + 65) = 0
⇒ n = 20 or \(n={-65\over 3}\) which is not possible
∴ n = 20
Thus, the amount is cleared in 20 months.

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.

(i) This is not at all a function because f(x) is not defined for x = 0.
(ii) This function is one-to-one (verify) but not onto because 0 has no pre-image.