y = sin 5 + log₁₀ x + 2 sec x
\(\therefore \cfrac { dy }{ dx } =0+\left( \cfrac { 1 }{ x } \right) { log }_{ 10 }e+2\left[ secxtanx \right] =\cfrac { { log }_{ 10 }e }{ x } +2secxtanx\) 

\(\lim _{ x\rightarrow 0 }{ \frac { \sqrt { 1+x } +\sqrt { 1-x } }{ 1+x } } =\frac { \sqrt { 1+0 } +\sqrt { 1-0 } }{ 1+0 } =\frac { \sqrt { 1 } +\sqrt { 1 } }{ 1 } =\frac { 2 }{ 1 } =2\)

Given \(P(\bar { A } )=0.6\)
\(\Rightarrow P(A)=1-P(\bar { A } )=1-0.6=0.4\)
P(B) = 0.7 and P(B/A) = 0.4
We know that P(B/A) = \(\frac { P(A\cap B) }{ P(A) } \)
\(\Rightarrow 0.4=\frac { P(A\cap B) }{ 0.4 } \)
\(\Rightarrow P(A\cap B)=0.16\)
Now, P(A/B) = \(\frac { P(A\cap B) }{ P(B) } =\frac { 0.16 }{ 0.7 } =0.2286\)
Also, P(A\(\cup \) B) = P(A) + P(B) - P(A\(\\ \cap \)B)
= 0.4 + 0.7 - 0.16 = 1.1 - 0.16 = 0.94

Given \(\begin{bmatrix} x-y & 2z+w \\ 2x-y & 2x+w \end{bmatrix}=\begin{bmatrix} 5 & 3 \\ 12 & 15 \end{bmatrix}\)
Equating the corresponding entries on both sides, we get

Substituting x=7 in (1) we get,
7-y = 5  \(\Rightarrow\)y=2
2z + w = 3
2x + w =15 ....(3)
2(7) + w =15   \(\Rightarrow\)4+w=15...(4)
\(\Rightarrow\) w=1
Substituting w=1 in (3) we get,
2z + 1 =3   \(\Rightarrow\)2z= 2 \(\Rightarrow\)z = 1
\(\therefore\)x=7, y=2, z=1 and w=1

\(\int \frac{1}{\cos ^2 x} d x=\int \sec ^2 x d x\)
= tan x + c

\(y=x^{\cos x}\)
Take log on both sides.
\(\log y =\log \left(x^{\cos x}\right) \)
\( \log y =\cos x \cdot \log x \)
\( \frac{1}{y} \frac{d y}{d x} =\cos x \frac{d}{d x}(\log x)+\log x \frac{d}{d x}(\cos x) \)
\( \frac{1}{y} \frac{d y}{d x} =\cos x \cdot \frac{1}{x}+\log x(-\sin x) \)
\( \frac{d y}{d x} =y\left[\frac{\cos x}{x}-\log x(\sin x)\right] \)
\(=x^{\cos x}\left[\frac{\cos x}{x}-\log x(\sin x)\right] \)

For 3-digit even numbers, the unit's place can be occupied by one of the 3 digits 2, 4 or 6. The remaining 5 digits can be arranged int he remaining 2 places in ₅P² ways.
\(\therefore\) By the multiplication rule, the required number of 3-digit even number is 3 \(\times\) ⁵P₂ = 3 \(\times\) 5 \(\times\) 4 = 60.

Given series is 3 - 6 + 9 - 12+ ...
= 3 (1) + 6 (- 1) + 9 (-1)² + 1² (- 1)³+ . . .
This is an arithmetic geometric (AG) series with correspondingA.P 3, 6, 9, 12 ... and G.P 1, -1, (-1)²,(-1)³.
\(\therefore\). n^th term of the given A. G. series is
= (n^th term of 3, 6, 9, ... ) (n^th term of 1, - 1, (-1)², ... )
= [3 + (n - 1)3] [1 (-1)^n-1]      [\(\because\) For AP, a = 3, d = 3 for GP = a = 1, r = -1]
= (3 + 3n - 3) (-1)^n-1
= 3n (-1)^n-l.

Given L₁ = 124.942, C₁ = 20 \(\Rightarrow\) (124.942,20)
and L₂ = 125.134, C₂ = 110 \(\Rightarrow\) (125.134,110)
Using two point form,
\(\frac{L-L_1}{L_2-L_1}=\frac{C-C_1}{C_2-C_1}\Rightarrow\frac{L-124.942}{125.134-124.942}\)
\(=\frac{C-20}{110-20}\)
\(\Rightarrow\frac{L-124.942}{0.192}=\frac{C-20}{90}\)
\(\Rightarrow L-124.942=\frac{0.192}{90}(C-20)\)
\(\Rightarrow\) L-124.942 = 0.0021 (C - 20)
\(\Rightarrow\) L-124.942 = 0.0021 (C - 20)
\(\Rightarrow\) L-124.942 = 0.0021C-.042
\(\Rightarrow\) L = 0.0021C - 0.042 + 124.942
\(\Rightarrow\) L = 0.0021C + 124.9

Number of ways of buying a toy car from 3 types of car = 3
Number of ways of buying a toy train from 2 types of train = 2.
∴ By fundamental principle of multiplication, number of ways of buying a toy car and a toy train = 3 \(\times\) 2 = 6.

\(\left( \left[ (256)^{ \frac { -1 }{ 2 } } \right] ^{ \frac { -1 }{ 4 } } \right) ^{ 3 }\)  =  \((256)^{ \frac { -1 }{ 2 } \times \frac { -1 }{ 4 } \times 3 }\)   \([\because \frac { { a }^{ m } }{ { a }^{ n } } ={ a }^{ m-n }]\)
=  \((256)^{ \frac { 3 }{ 8 } }=({ 2 }^{ 8 })^{ \frac { 3 }{ 8 } }={ 2 }^{ 8\times \frac { 3 }{ 8 } }={ 2 }^{ 3 }=8\)

Let A, Band C be the three possible events of selections. The number of combinations are shown below:

n(S) = Selecting 4 from 10 students = \(^{ 10 }{ C }_{ 4 }=\cfrac { 10\times 9\times 8\times 7 }{ 4\times 3\times 2\times 1 } =210\) 
\(n\left( A\cup B\cup \right) C=n(A)+n(B)+n(C)\)
= ⁴C₂C⁴C₂+⁴C1X⁶C1+⁴C₀X⁶C₄
\(^{ 4 }{ C }_{ 2 }=\cfrac { 4\times 3 }{ 2\times 1 } =6;^{ 4 }{ C }_{ 1 }=4;^{ 4 }{ C }_{ 0 }=1\) 
\(^{ 6 }{ C }_{ 2 }=\cfrac { 6\times 5 }{ 2\times 1 } =15;^{ 6 }{ C }_{ 3 }=\cfrac { 6\times 5\times 4 }{ 3\times 2\times 1 } =20.^{ 6 }{ C }_{ 4 }=^{ 6 }{ C }_{ 2 }=15\) 
\(n\left( A\cup B\cup C \right) =(6)(15)+(4)(20)+(1)(15)=90+80+15=185\) 
\(\therefore P(A\cup B\cup C)=\cfrac { 185 }{ 210 } =\cfrac { 37 }{ 42 } \)

Distance between two parallel lines a₁x + b₁y + c₁ = 0 and a₁x + b₁y + c₂ = 0 is
D = \(\frac{|c_1-c_2|}{\sqrt{a_1^2+b_1^2}}\)
Given lines can be written as 3x + 4y - 12 = 0 and 3x + 4y+\(\frac{1}{2}\)=0
Here a₁ = 3, b₁ = 4, c₁ = \(\frac{1}{2}\)
D = \(\frac{|c_1-c_2|}{\sqrt{a_2^1+b_2^1}}=\left|\frac{-12-\frac{1}{2}}{\sqrt{3^2+4^2}}\right|=\frac{25}{2\times5}\) = 2.5 units

Since polynomial function and a constant function are continuous, the given function is continuous for all x < 2 and for all x>2.
At x=2, 
\(\lim _{ x\rightarrow { 2 }^{ - } }{ f\left( x \right) } =\lim _{ x\rightarrow { 2 }^{ - } }{ k{ x }^{ 2 } } =k{ (2) }^{ 2 }=4k\)
\(\lim _{ x\rightarrow { 2 }^{ + } }{ f\left( x \right) } =\lim _{ x\rightarrow { 2 }^{ + } }{ 3 } =3\)
\(Also,\quad f(2)=k{ (2) }^{ 2 }=4k\)
\(Since\quad f\left( x \right) is\quad continuous\quad at\quad x=2,\)
\(\lim _{ x\rightarrow { 2 }^{ - } }{ f\left( x \right) } =\lim _{ x\rightarrow { 2 }^{ + } }{ f\left( x \right) } =f(2)\)
\(\Rightarrow 4k=3\)
\(k=\frac { 3 }{ 4 } \)

Given \(\left| \begin{matrix} 1 & a & b \\ 1 & c & a \\ 1 & b & c \end{matrix} \right| \)=0
Applying R₂-R₂-R₁ and R₃-R₃-R₁ we get,
\(\left| \begin{matrix} 1 & a & b \\ 1 & c-a & a-b \\ 1 & b-a & c-b \end{matrix} \right| =0\)
Expanding along C₁ we get,
\(1\left| \begin{matrix} c-a & a-b \\ b-a & c-b \end{matrix} \right| =0\)
\(\Rightarrow\) (c - a)(c - b)(b - a)(a - b) = 0
\(\Rightarrow\) (c²- cb - ac + ab)-(ab - b²- a²+ ab) = 0

\(\Rightarrow\) a² + b² + c² - ab - bc - ca = 0
Multiplying both sides by 2, we get,
\(\Rightarrow\) 2a² + 2b² + 2c² - 2ab - 2bc - 2ca = 0
\(\Rightarrow\) (a - b)²+(b - a)²+ (c - a)² = 0
\(\Rightarrow\) a- b = 0, b - c = 0, c - a = 0
\(\Rightarrow\) a = b = c 
ABC is equilateral
A = B = C =\(\frac{\pi}{3}\)
sin²A + sin²B + sin²C =3 sin²\(\frac{\pi}{3}\) = 3\({ \left( sin\frac { \pi }{ 3 } \right) }^{ 2 }\)
= 3 \(\times\)\({ \left( \frac { \sqrt { 3 } }{ 3 } \right) }^{ 2 }=\frac { 3\times 3 }{ 4 } =\frac { 9 }{ 4 } \)

\(\int \frac{x+2}{\sqrt{x^2-1}} d x=\int \frac{x}{\sqrt{x^2-1}} d x+\int \frac{2}{\sqrt{x^2-1}} d x \)
\(=\frac{1}{2} \int \frac{1}{\sqrt{x^2-1}} \times 2 x d x+2 \int \frac{1}{\sqrt{x^2-1}} d x \)
\(=\frac{1}{2}\left(2 \sqrt{x^2-1}\right)+2 \log \left|x+\sqrt{x^2-1}\right|+c\)
\(=\sqrt{x^2-1}+2 \log \left|x+\sqrt{x^2-1}\right|+c \)

Given that the position vector of the \(\triangle\)ABC is \(\vec{a}\), \(\vec{b}\)and \(\vec{c}\)
\(\therefore \vec { OA } =\vec { a } ,\vec { OB } =\vec { b } \ and \ \vec { OC } =\vec { c } \)
\(\vec { AB } =\vec { OB } -\vec { OA } =\vec { b } -\vec { a } \)
\(\vec { AC } =\vec { OC } -\vec { OA } =\vec { c } -\vec { a } \)
\(\therefore \vec { AB } \times \vec { AC } =(\vec { b } -\vec { a } )\times (\vec { c } -\vec { a } )=\vec { b } \times \vec { c } -\vec { b } \times \vec { a } -\vec { a } \times \vec { c } +\vec { a } \times \vec { a } \)
= \(\vec { b } \times \vec { c } +\vec { a } \times \vec { b } +\vec { c } \times \vec { a } +\vec { 0 } \)
\(\left| \vec { AB } \times \vec { AC } \right| =\left| \vec { a } \times \vec { b } +\vec { b } \times \vec { c } +\vec { c } \times \vec { a } \right| \)
\(\therefore\) Area of \(\triangle\)ABC = \(\frac { 1 }{ 2 } \left| \vec { a } \times \vec { b } +\vec { b } \times \vec { c } +\vec { c } \times \vec { a } \right| \)
Condition for the points A, B, C to be collinear is area of \(\triangle\)ABC = 0
\(\Rightarrow \frac { 1 }{ 2 } \left| \vec { a } \times \vec { b } +\vec { b } \times \vec { c } +\vec { c } \times \vec { a } \right| =0\)
\(\Rightarrow \left| \vec { a } \times \vec { b } +\vec { b } \times \vec { c } +\vec { c } \times \vec { a } \right| =0\) which is the required condition.

The given inequality is x+y≤9
Draw the graph of the line x+y =9

Table of values satisfying the equation
x+y = 9

Putting (0, 0) in the given inequation, we have 0+0≤9⇒0≤9, is towards is origin
∴ Half plane of x+y≤9 is away from origin
Also, the given inequality is x-y < 0
Draw the graph of the line x - y = 0.
Table of values satisfying the equation
 x-y =0

Putting (0, 3) in the given inequation, we have 0-3-0<0⇒-3<0 which is true.
∴  Half plane of x-y < 0 containing the points (0,3).

\(\frac { (2n)! }{ n! } =\frac { 1.2.3.4...(2n-2).(2n-1)2n }{ n! } \)
= \(\frac { (1.3.5...(2n-1))(2.4.6...(2n-2).2n) }{ n! } \) (Grouping the odd and even numbers separately)
= \(\frac { (1.3.5...(2n-1))\times 2^{ n }\times (1.2.3...(n-1).n) }{ n! } \) (taking out the 2's)
= \(\frac { (1.3.5...(2n-1)\times 2^{ n }\times n! }{ n! } \)
= 2ⁿ(1.3.5...(2n - 1))

Let p(n) be the statement 41n - 1411 is a multiple of 27.
Step 1: p(1): 41¹- 14¹ is a multiple of 27.
⇒ p(1): 27 is a multiple of27.
⇒ p(1) is true.
Step 2: Let p(m) be true.
Then 41^m - 14^m is a multiple of 27.
⇒ 41^m -14^m=⋋.27
Step 3: To prove that p(m + 1) is true.
i.e. to prove that 41^m+1- 14^m+1 is a multiple of 27
Consider 41^m+1- 14^m+1
41^m+1 - 41 x 14^m + 41 x 14^m - 14^m+1    [Adding and subtracting 41 x 14^m]
= 41^m . 41 - 41.14^m + 41 x 14^m - 14^m. 14
= 41(41^m -14^m) + 14^m (41 - 14)
= 41 (⋋.27) + 14^m(27)
= 27 (41⋋ + 14^m) which is a multiple of27
⇒ p(m + I) is true. ,
Hence, by the principle of mathematical induction,p(n) is true for all n E N.

We have (1 +x)ⁿ = nCo(1)ⁿ + nC₁(1)^n-1 . x¹ + nC₂(1)^n-2 x² + ...+ nC_nxⁿ;
⇒ (1 +x)ⁿ = C₀ + C₁x + C₂x₂+ ...+ Cn_nxⁿ  ...(i)
Similarly, (x + 1)ⁿ = C₀xⁿ + C₁x^n-1 + C₂x^n-1 + ...+ Cn
Multiplying equations (1) and (2) we get
(1 + x)ⁿ (x + 1)ⁿ = [Co + C₁x + C₂x² + ...+C_nxⁿ]  [C₀xⁿ + C₁x^n-1 + C₂x^n-2+...+ C_n]
(1 +x)²ⁿ = [C₀ + C₁x + C₂x²+ ... +C_nxⁿ][C₀xⁿ + C₁x^n-1 + C₂x^n-2 +...+ C_n]     ...(3)
This relation is true for all values of n. Let us equate the co-efficients of xⁿ both sides​​​​​​​
General term in (1 + x)²ⁿ is T_r+1 = 2nC_r(1)^2n-rx^r​​​​​​​
Putting r = n we get​​​​​​​
T_n+1= 2_nC_n.xⁿ
⇒ Co-efficient of xⁿ of LHS of(3) is 2_nC_n
Now, co-efficient of xⁿ in RHS of (3) is​​​​​​​
\(C_0^2+C_1^2+C_2^2+...+C_n^2\)
\(∴\ C_0^2+C_1^2+C_2^2+...+C_n^2=2_nC_n\)
\(⇒\ C_0^2+C_1^2+C_2^2+...+C_n^2={(2n)!\over n!n!}\left[ ∵ nC_r={n!\over r!(n-r)!}\right]\)
\(={(2n)!\over (n!)^2}\)
Hence \({ C }_{ 0 }^{ 2 }+{ C }_{ 1 }^{ 2 }+{ C }_{ 2 }^{ 2 }+...C_{ n }^{ 2 }=\frac { (2n)! }{ (n)! } \)

LHS = \(\sqrt { \frac { 1-x }{ 1+x } } \)
\(\sqrt { \frac { 1-x }{ 1+x } } =\sqrt{(1-x)(1-x)\over(1+x)(1-x)}\)
\(={1-x\over\sqrt{1-x^2}}\)
\(=(1-x)(1-x^2)^{-1\over2}\)
\(=(1-x)\left[ 1+{-1\over2}(-x^2)-{\left(-1\over2\right)\left({-1\over2}-1\right)\over2.1}(-x^2)+... \right]\)∵ x is very small, [x² is also very small]
\(=(1-x)\left[ 1+{x^2\over2}+{\left(1\over2\right)\left(3\over2\right)\over1.2}(x^4)+... \right]\)
\(=(1-x)\left[1+{x^2\over2}+{3\over8}x^4+...\right]=1-x+{x^2\over2}-{x^2\over2}+{3\over8}x^4+...\)
\(\left( 1-x+\frac { { x }^{ 2 } }{ 2 } \right) \) approximately
RHS.
Hence proved.

Let y = 3x -5.
\(\Rightarrow y+5=3x\Rightarrow \frac { y+5 }{ 3 } =x\)
Let g(y) = \(\frac { y+5 }{ 3 } \)
\(gof(x)=g(f(x))=g(3x-5)=\frac { 3x-5+5 }{ 3 } =\frac { 3x }{ 3 } =y\)
Also f o g(y) = f(g(y)) = \(f\left( \frac { y+5 }{ 3 } \right) =3\left( \frac { y+5 }{ 3 } \right) -5=y+5-5=y\)
Thus g o f = Ix and fog = Iy.
This implies that f and g are bijections and inverses to each other.
Hence f is a bijection and f^-1(y) = \(\frac { y+5 }{ 3 } \)
Replacing y by x we get, f^-1(x) = \(\frac { x+5 }{ 3 } \)

2a + 3b = 30
R = {(3,8), (6,6), (9,4). (12,2)}
Not reflexive, Not Symmetric, transitive, hence not an equivalence relation.

LHS = \(log2+16log{16\over 15}+12log{25\over 24}+7log{81\over 80}\)
\(=log2+log\left(16\over 15\right)^{16}+log\left(25\over 24\right)^{12}+log \left(81\over 80\right)^7\)
\(=log2\times{(2^4)^{16}\over (3\times5)^{16}}\times{(5^2)^{12}\over (2^2\times3)^{12}}\times{(3^4)^7\over 2^{28}\times5^7}\)
\(=log2^1\times{2^{64}\over 3^{16}}\times{5^{24}\over 2^{36}\times3^{12}}\times{3^{28}\over 2^{28}\times5^7}\)
\(=log{2^{1+64}.5^{24}.3^{28}\over 3^{16+12}.5^{16+7}.2^{36+28}}\)  \(\left[∵\ {a^m\over a^n}=a^{m-n} \right]\)

= log 2^65-64 x 5^24-23 = log 2¹ \(\times\) 5¹ = log₁₀¹⁰ = 1 = RHS

a sec θ - c tan θ = b...(1)
b sec θ + d tan θ = c...(2)

tan θ = \(\frac{ac-b^2}{bc+ad}\)

secθ = \(\frac{bd+c^2}{ad+bc}\)
We know sec²θ - tan²θ = 1
⇒ \((\frac{bd+c^2}{ad+bc})^2-(\frac{ac-b^2}{bc+ad})^2=1\)
⇒ \(\frac{(bd+c^2)^2}{(ad+bc)^2}-\frac{(ac-b^2)^2}{(bc+ad)^2}=1\)
⇒ \(\frac{(bd+c^2)^2-(ac-b^2)^2}{(ad+bc)^2}=1\)
⇒ (bd + c²)²- (ac - b²)² = (ad + bc)²
⇒ (c²+bd)² = (ad + bc)² + (ac - b²)²
Thus θ is eliminated.

Assistant    Statistics
\(\begin{array}{ll} A=2 & S=3 \\ S=3 & T=3 \\ T=2 & A=1 \\ I=1 & I=2 \\ N=1 & C=1 \end{array}\)
\(\mathrm{n}(\mathrm{S})=9 \mathrm{C}_{1} \times 10 \mathrm{C}_{1}=90\)
One letter in I word and another letter (same) in II word the combinations are
\(\begin{array}{ll} \mathrm{AA} & \mathrm{A} \\ \mathrm{SSS} & \mathrm{SSS} \\ \mathrm{TT} & \mathrm{TTT} \\ \mathrm{I} & \mathrm{II} \end{array}\)
No. of doublets are
AA, AA, SS, SS, SS, SS, SS, SS, S$ SS, SS, TT TT,
TT, TT, TT, TT, II, II
n(A) = 19
\(P(A) =\frac{n(A)}{n(S)} \)
\(=\frac{19}{90} \)

\(\int \frac{1}{x \sqrt{(\log x)^{2}-5}} d x =\int \frac{1}{\sqrt{(\log x)^{2}-5}} \times \frac{1}{x} d x \)
\(=\int \frac{1}{\sqrt{u^{2}-5}} d u, \quad u=\log x \)
\(=\log \left|u+\sqrt{u^{2}-5}\right|+c \)
\(=\log \left|\log x+\sqrt{(\log x)^{2}-5}\right|+c \)

\(f^{\prime}\left(3^{-}\right)=\lim _{x \rightarrow 3^{-}} \frac{f(x)-f(3)}{x-3}=\lim _{x \rightarrow 3^{-}} \frac{x+2-5}{x-3}\)
\(=\lim _{x \rightarrow 3^{-}} \frac{x-3}{x-3}=1\)
\(f^{\prime}\left(3^{+}\right)=\lim _{x \rightarrow 3^{+}} \frac{f(x)-f(3)}{x-3}=\lim _{x \rightarrow 3^{+}} \frac{8-x-5}{x-3}\)
\(=\lim _{x \rightarrow 3^{+}} \frac{3-x}{x-3} \)
\(=\lim _{x \rightarrow 3^{+}} \frac{-(x-3)}{(x-3)}=-1 \)
\(\text { by }(1) \text { and }(2), f^{\prime}\left(3^{-}\right) \neq f^{\prime}\left(3^{+}\right)\)
\(\text { It is not differentiable }\)
\(\therefore f^{\prime}(x) \text { does not exist }\)

\(f(x)=\frac{|2 x-3|}{2 x-3} \text { is not defined at } x= \frac{3}{2}\)
\(\therefore f(x) \text { is not continuous at } x=\frac{3}{2}\)

\(\text { Area of triangle }=\frac{1}{2}\left|\begin{array}{lll} \frac{x_{1}}{a} & \frac{y_{1}}{a} & 1 \\ \frac{x_{2}}{b} & \frac{y_{2}}{b} & 1 \\ \frac{x^{3}}{c} & \frac{y_{3}}{c} & 1 \end{array}\right|\)
\(=\frac{1}{2 a b c}\left|\begin{array}{lll} x_{1} & y_{1} & a \\ x_{2} & y_{2} & b \\ x_{3} & y_{3} & c \end{array}\right| \text { Multiply }R_{1}, R_{2} \& R_{3} \text { by } a, b, c\text { respectively }\)
\(=\frac{1}{2 a b c} \frac{1}{(2)}\left|\begin{array}{ccc} 2 a & x_{1} & y_{1} \\ 2 b & x_{2} & y_{2} \\ 2 c & x_{3} & y_{3} \end{array}\right|=\frac{1}{4 a b c}\left(\frac{a b c}{2}\right)\)
\(=\frac{1}{8}\)

\(\frac{{ }^{2 n} C_{3}}{{ }^{n} C_{3}} =\frac{11}{1} \)
\(\frac{(2 n)(2 n-1)(2 n-2)}{n(n-1)(n-2)} =\frac{11}{1} \)
\(\frac{2 n(2 n-1) 2(n-1)}{n(n-1)(n-2)} =11 \)
\(4(2 n-1) =11(n-2) \)
\(8 n-4 =11 n-22 \)
\(18 =3 n \)
\(n = 6\)

\(\text { Let } \log _{\sqrt{2}} 512=x\)
\(\text { Then }(\sqrt{2})^{x}=2^{9}\)
\(\Rightarrow 2^{\frac{x}{2}}=2^{9} \Rightarrow x / 2=9 \Rightarrow x=18\)

\(I=\int { \cfrac { 1 }{ 7-(4x+1)^{ 2 } } dx=\int { \cfrac { dx }{ \left( \sqrt { 7 } \right) ^{ 2 }-(4x+1)^{ 2 } } } } \) 
= \(\cfrac { 1 }{ 2\sqrt { 7 } } =\cfrac { log\frac { \sqrt { 7 } +\left( 4x+1 \right) }{ \sqrt { 7 } -\left( 4x-1 \right) } }{ 4 } =\cfrac { 1 }{ 8\sqrt { 7 } } log\left( \cfrac { \sqrt { 7 } +\left( 4x+1 \right) }{ \sqrt { 7 } -\left( 4x+1 \right) } \right) +c\)

(i) If A is a subset of B, then
A\(\cap \)B = A
\(\Rightarrow\) n(A\(\cap \)B) = n(A)
\(\Rightarrow\) P(A\(\cap \)B) = P(A)
\(\therefore P(B/A)=\frac { P(A\cap B) }{ P(A) } =\frac { P(A) }{ P(A) } =1\)
(ii) If \(A\cap B=\phi \) then n(A\(\cap \)B) = 0 \(\Rightarrow\) P(A\(\cap \)B) = 0
\(\therefore P(B/A)=\frac { P(A\cap B) }{ P(A) } =\frac { 0 }{ P(A) } =0\)

\(\lim _{ x\rightarrow 1 }{ \frac { (2x-3)\sqrt { x } -1 }{ { 2x }^{ 2 }+x-3 } } =\lim _{ x\rightarrow 1 }{ \frac { (2x-3)(\sqrt { x } -1)(\sqrt { x } +1) }{ (x-1){ (2x }+3)(\sqrt { x } +1) } } =\lim _{ x\rightarrow 1 }{ \frac { (2x-3)(x-1) }{ (x-1){ (2x }+3)(\sqrt { x } +1) } } \)
\(=\lim _{ x\rightarrow 1 }{ \frac { 2x-3 }{ { (2x }+3)(\sqrt { x } +1) } } =\frac { -1 }{ (2+3)(1+1) } \)
\(=\frac { -1 }{ 5(2) } =\frac { -1 }{ 10 } \)

Let \(\overrightarrow{a}\) and \(\overrightarrow{b}\) be the unit vectors and\(\theta\) is the angle between \(\overrightarrow{a}\) and \(\overrightarrow{b}\).
Consider \(|\overrightarrow{a}-\overrightarrow{b}|^2=|\overrightarrow{a}|^2+|\overrightarrow{b}|^2-2(\overrightarrow{a}.\overrightarrow{b})\)      \([\because |\overrightarrow{a}|=1;|\overrightarrow{b}|=1]\)
\(=1+1-2|\overrightarrow{a}||\overrightarrow{b}|cos \theta =2-2cos \theta\)
\(=2(1-cos \theta)=2.2sin^2{\theta \over2}=4sin^2{\theta \over2}\)
\(|\overrightarrow{a}-\overrightarrow{b}|=2sin{\theta \over2}\)
\(sin{\theta \over2}={1\over 2}|\overrightarrow{a}-\overrightarrow{b}|\)

\(\frac { 1 }{ 2+\sqrt { 3 } } \)=\(\frac { 1 }{ 2+\sqrt { 3 } } \times \frac { 2-\sqrt { 3 } }{ 2-\sqrt { 3 } } =\frac { 2-\sqrt { 3 } }{ 4-3 } =2-\sqrt { 3 } \)
\(\frac { 3 }{ 4-\sqrt { 5 } } =\frac { 3 }{ 4-\sqrt { 5 } } \times \frac { 4+\sqrt { 5 } }{ 4+\sqrt { 5 } } =\frac { 12+3\sqrt { 5 } }{ 16-5 } =\frac { 12+3\sqrt { 5 } }{ 11 } \)
\(\frac { 6 }{ 7-\sqrt { 8 } } =\frac { 6 }{ 7-\sqrt { 8 } } \times \frac { 7+\sqrt { 8 } }{ 7+\sqrt { 8 } } =\frac { 42+6\sqrt { 8 } }{ 49-8 } =\frac { 42+6\sqrt { 8 } }{ 41 } \)
∴ \(\frac { 1 }{ 2-\sqrt { 3 } } +\frac { 3 }{ 4-\sqrt { 5 } } +\frac { 6 }{ 7-\sqrt { 8 } } =2-\sqrt { 3 } +\frac { 12+3\sqrt { 5 } }{ 11 } +\frac { 42+6\sqrt { 8 } }{ 41 } \)

Let the coordinates of the points 0, A and B are (0, 0), (a, 0) and (0, b) respectively.
Observed that the points A and B are moving points.

Let (h, k) be a centroid of ΔOAB
Centroid of ΔOAB is \((\frac{0+a+0}{3},\frac{0+0+b}{3})=(h,k)\)
\(\frac{a}{3}\) = h ⇒ a = 3h, \(\frac{b}{3}\) = k ⇒ b = 3k
From right ΔOAB, OA² + OB² = AB²
(3h)² + (3k)² = (6)² ⇒ h² + k² = 4
Locus of (h, k) is a circle, x² + y² = 4

\(\frac { (2n)! }{ n! } =\frac { (2n)(2n-1)(2n-2)...4.3.2.1 }{ n! } \)
\(=\frac { (2n)(2n-2)(2n-4)..4.2(2n-1)(2n-3)....3.1 }{ n!\quad n! } \)   [Separating odd and even terms]
\(=\frac { { 2 }^{ n }.(2n-1)(2n-3)...3.1 }{ n! } \)

Given that ㄥA, ㄥB, ㄥC are in A.P.
ஃ 2ㄥB = ㄥA + ㄥC
3ㄥB = 180°
ㄥB = \(\frac{180}{3}\) = 60°
Now \(\frac { b }{ c } =\frac { sinB }{ sinC } \)
\(\Rightarrow \frac { \sqrt { 3 } }{ \sqrt { 2 } } =\frac { sin60° }{ sinC } \)
\(\Rightarrow \frac { \sqrt { 3 } }{ \sqrt { 2 } } =\frac { \frac { \sqrt { 3 } }{ \sqrt { 2 } } }{ sinC } \)
\(\Rightarrow sinC=\frac { \sqrt { 3 } }{ 2\sqrt { 3 } } \times \sqrt { 2 } =\frac { \sqrt { 2 } }{ 2 } =\frac { \sqrt { 2 } }{ \sqrt { 2 } \times \sqrt { 2 } } =\frac { 1 }{ \sqrt { 2 } } \)
⇒ ㄥC = 45°
ㄥA = 180°-(ㄥB+ㄥC) = 180 - (60 + 45) = 75°
ㄥA = 75°, ㄥB = 60° and ㄥC = 45°