\(n(S)=n(A \times A) \quad=16\)
\(\text {To find } n(A)\)
\(\text {Let } a=4, b=1 \text { then } a^{2}-4 b \text { is }\)
A = 1,b = 1 then 16 - 4 > 0
b = 2. then 16 - 8 > 0
b=3 then 16-L2>0
b = 4 then 16 - 16 = 0
Let Q = 3,b = 1 thg,l, 9-4 > 0
b = 2 then 9 - 8 > 0
Let a = 2,b = l then 4 - 4 > 0
n(A).= 7,
\(P(A)=\frac{n(A)}{n(S)}=\frac{7}{16}\)

\(\text { Let } \sqrt{x}=t\)
\(\text { Then } x=t^{2}\)
\(\therefore d x =2 t d t \)
\(\therefore \int e^{\sqrt{x}} d x =\int e^{t} \times 2 t d t \)
\(=2 \int t e^{t} d t \)
Applying Bernoulli's formula
\(=2\left[t\left(e^{t}\right)-1\left(e^{t}\right)\right]+c \)
\(=2(t-1) e^{t}+c \)
\(\text { Putting } t=\sqrt{x}\)
\(=2(\sqrt{x}-1) e^{\sqrt{x}}+c\)
 

\(\text { Given } f(x)=x^{2}-3 x\)
\(f^{\prime}(x) =2 x-3 \)
\(f(x) =f^{\prime}(x) \)
\(x^{2}-3 x =2 x-3 \)
\(x^{2}-3 x-2 x+3 =0 \)
\(x^{2}-5 x+3 =0 \)
\(\text { It has irrational roots. }\)

\(\text { WKT }\left\lfloor k^{-}\right\rfloor=k-1 \text { and }\left\lfloor k^{+}\right\rfloor=k \text { where } k \text { is an integer }\)
\(\therefore \lim _{x \rightarrow k^{-}} x-\lfloor x\rfloor=k-(k-1)=1\)

\(\text { Given } \text {matrix } A=\left[\begin{array}{cc} \alpha & \beta \\ \gamma & -\alpha \end{array}\right] \text { is unit matrix }\)
\(|A|=1 \)
\(-\alpha^{2}-\beta \gamma=1 \)
\(1+\alpha^{2}+\beta \gamma=0 \)

\(\sqrt{2}+\sqrt{8}+\sqrt{18}+\sqrt{32} \ldots . . =\sqrt{2}+2 \sqrt{2}+3 \sqrt{2}+4 \sqrt{2} . \)
\(=\sqrt{2}[1+2+3+\ldots .] \)
\(S_{n} =\frac{\sqrt{2}[n(n+1)]}{2} \)
\(=\frac{n(n+1)}{\sqrt{2}} \)

Number of ways of selecting 8 people from 12 in ¹²C₈ ways.
Let A and B both attend the party
Out of 10 remaining people 8 can attend  in ¹⁰C₆ ways.
.'. Number of ways in which two of them do not attend together = ¹²C₈ - ¹⁰C₆

\(x^{2}-\mathrm{k} x+\mathrm{c}=0\)
a and b are the roots
\(\therefore a+b=k, a b=c\)
To find
\(\sqrt{(a-b)^{2}+0^{2}}=a-b=\sqrt{(a+b)^{2}-4 a b}=\sqrt{k^{2}-4 c}\)
 

\(\frac{\tan 140^{\circ}-\tan 130^{\circ}}{1+\tan 140^{\circ} \tan 130^{\circ}} =\tan \left(140^{\circ}-130^{\circ}\right) \)
\(=\tan 10^{\circ} \ldots \ldots(1) \)
\(\tan 40^{\circ} =\lambda \)
\(\tan 80^{\circ}=\frac{2 \tan 40^{\circ}}{1-\tan ^{2} 40^{\circ}} =\frac{2 \lambda}{1-\lambda^{2}} \)
\(\tan \left(90^{\circ}-10^{\circ}\right) =\frac{2 \lambda}{1-\lambda^{2}} \)
\(\cot 10^{\circ} =\frac{2 \lambda}{1-\lambda^{2}} \)
\(\Rightarrow \tan 10^{\circ} =\frac{1-\lambda^{2}}{2 \lambda} \)
\(\frac{\tan 140^{\circ}-\tan 130^{\circ}}{1+\tan 140^{\circ} \tan 130^{\circ}} \)
\(=\frac{1-\lambda^{2}}{2 \lambda}\)

It is onto not one-one

\(\text { Since } \sin 30^{\circ}=\frac{1}{2}\)
\(\sin 150^{\circ}=\frac{1}{2}\)

Let I = \(\int { \frac { { x }^{ 3 }dx }{ { x }^{ 4 }+{ 3x }^{ 2 }+2 } } \) = \(\int { \frac { { x }^{ 2 }.xdx }{ \left( { x }^{ 2 } \right) ^{ 2 }+{ 3x }^{ 2 }+2 } } \)
Let t = x² \(\Rightarrow\) dt = 2x dx \(\Rightarrow\) \(\frac { dt }{ 2 } \) = x dx
\(\therefore\) I = \(\int { \frac { t.\frac { dt }{ 2 } }{ { t }^{ 2 }+3t+2 } =\frac { 1 }{ 2 } \int { \frac { tdt }{ { t }^{ 2 }+3t+2 } } }\)
Now t = A \(\frac { d }{ dx } \) (t² + 3t + 2) + B \(\Rightarrow\) t = A (2t  + 3)  + B
Equating the x - term constant terms we get,
1 = 2A \(\Rightarrow\) A = \(\frac { 1 }{ 2 } \) 0 = 3A + B \(\Rightarrow\) 0 = \(\frac { 3 }{ 2 } \) + B\(\Rightarrow\) B = \(-\frac { 3 }{ 2 } \);    \(\therefore\) t = \(\frac { 1 }{ 2 } \)(2t + 3) -\(\frac { 3 }{ 2 } \)
I = \(\int { \frac { tdt }{ { t }^{ 2 }+3t+2 } } =\frac { 1 }{ 2 } \left[ \int { \frac { 2t+3 }{ 2\left( { t }^{ 2 }+3t+2 \right) } dt-\frac { 3 }{ 2 } \int { \frac { dt }{ { t }^{ 2 }+3t+2 } } } \right] \)
\(\Rightarrow\) I = \(\frac { 1 }{ 4 } \left[ log\left| { t }^{ 2 }+3t+2 \right| -3\int { \frac { dt }{ { t }^{ 2 }+3t+\frac { 9 }{ 4 } -\frac { 9 }{ 4 } +2 } } \right] =\frac { 1 }{ 4 } \left[ log\left| { t }^{ 2 }+3t+2 \right| -3\int { \frac { dt }{ \left( t+\frac { 3 }{ 2 } \right) -\left( \frac { 1 }{ 2 } \right) ^{ 2 } } } \right] \)
Now I₁ = \(\int { \frac { dt }{ { t }^{ 2 }+3t+2 } } =\int { \frac { dt }{ { t }^{ 2 }+3t+\left( \frac { 3 }{ 2 } \right) ^{ 2 }-\left( \frac { 3 }{ 2 } \right) ^{ 2 }+2 } } \) = \(3\frac { 1 }{ 4 } \left[ log\left| { t }^{ 2 }+3t+2 \right| -3log\left| \frac { t+\frac { 3 }{ 2 } -\frac { 1 }{ 2 } }{ t+\frac { 3 }{ 2 } +\frac { 1 }{ 2 } } \right| \right] \)
I = \(\frac { 1 }{ 4 } log\left| { t }^{ 2 }+3t+2 \right| -3log\left| \frac { t+1 }{ t+2 } \right| +c\)
\(\therefore\) I = \(\frac { 1 }{ 4 } log\left| { x }^{ 4 }+3{ x }^{ 2 }+2 \right| -3log\left| \frac { { x }^{ 2 }+1 }{ { x }^{ 2 }+1 } \right| +c\) 

Given P(Simultaneous occurrence of atleast one of A and B) =p
\(\Rightarrow P(AUB)\) = p and P (occurrence of exactly one of A and B) = q
\(\Rightarrow P(AUB)-P(A\cap B)=q\)
\(\therefore p-P(A\cap B)=q\)
\(\Rightarrow P(A\cap B)=p-q\)
\(\Rightarrow 1-P(\overline { A\cap B) } =p-q\)   \(\left[ \because P(A\cap B)+P(\overline { A\cap B } )=1 \right] \)
\(\Rightarrow 1-P(\bar { A } \cup \bar { B } )=p-q\)
\(\Rightarrow P(\bar { A } \cup \bar { B } )=p-q\)
\(\Rightarrow P(\bar { A } )+P(\bar { B } )-P(\bar { A } \cap \bar { B } )=1-p+q\)
\(\Rightarrow P(\bar { A } )+(P\bar { B } )=1-p+q+p(\bar { A } \cap \bar { B } )=1-p+q+P(\overline { A\cup B } )\)
= 1 - p + q +1 - p     \(\left[ \because P\overline { A\cup B } =1-P(AUB)=1-p \right] \)
= 2 - p+q
\(\therefore P(\bar { A } )+P(\bar { B } )\) = 2 - 2 p+q   Hence proved.

\(f\left( x \right) =\begin{cases} x{ e }^{ -\left( \frac { 1 }{ \left| x \right| } +\frac { 1 }{ x } \right) }=x{ e }^{ -\frac { 2 }{ x } },\quad x>0 \\ \quad \quad \ 0,\quad \quad \quad \quad \quad \quad \quad x=0\quad \\ x{ e }^{ -\left( -\frac { 1 }{ x } +\frac { 1 }{ x } \right) }=x,\quad \quad \quad \quad x<0 \end{cases}\)
\(\therefore f\left( x \right) =\begin{cases} x{ e }^{ -\frac { 2 }{ x } }\quad x>0 \\ 0,\quad \quad \quad x=0 \\ x,\quad \quad x<0 \end{cases}\)
\( \therefore f^{ ' }\left( { 0 }^{ - } \right) =\lim _{ x\rightarrow { 0 }^{ + } }{ \frac { f\left( x \right) -f\left( 1 \right) }{ x-0 } } =\lim _{ x\rightarrow { 0 }^{ + } }{ \frac { x-0 }{ x-0 } } =\lim _{ x\rightarrow { 0 }^{ - } }{ \frac { x }{ x } } =1\quad \left[ \because f\left( x \right) =x\quad for\quad x<0\quad and\quad f\left( 0 \right) =0 \right] \)
\(\therefore f^{ ' }\left( { 0 }^{ + } \right) =\lim _{ x\rightarrow { 0 }^{ + } }{ \frac { f\left( x \right) -f\left( 0 \right) }{ x-0 } } =\lim _{ x\rightarrow { 0 }^{ + } }{ \frac { x{ e }^{ -\frac { 2 }{ x } }-0 }{ x } } =\lim _{ x\rightarrow { 0 }^{ + } }{ { e }^{ -\frac { 2 }{ x } } } =0\quad \left[ \because f\left( x \right) =x{ e }^{ -2x }\quad for\quad x>0\quad and\quad f\left( 0 \right) =0 \right] \)
\(\therefore f^{ ' }\left( { 0 }^{ - } \right) \neq f^{ ' }\left( { 0 }^{ + } \right) \)
\(\therefore f\left( x \right) \)is not differentiate at x = 0.

Let A₁, A₂ and A₃ be the event that the units of PVC pipes produced by plants X, Y, Z respectively.
Let B be the event of selected item is defective.
Then \(P\left(A_1\right)=\frac{2000}{10,000}=0.2, P\left(B / A_1\right)=0.03\)
\(P\left(A_2\right)=\frac{3000}{10,000}=0.3, P\left(B / A_2\right)=0.04\)
\(P\left(A_3\right)=\frac{5000}{10,000}=0.5, P\left(B / A_3\right)=0.02\)
(i) P (the selected pipe is defective) = P(B)
\(=P\left(A_1\right) \cdot P\left(B / A_1\right)+P\left(A_2\right) \cdot P\left(B / A_2\right)+P\left(A_3\right) \cdot P\left(B / A_3\right)\)
\(=0.2(0.03)+0.3(0.04)+0.5(0.02)\)
\(=0.006+0.012+0.010\)
\(=0.028=\frac{28}{1000}=\frac{7}{250}\)
(i) By Bayes theorem
\(P\left(A_2 / B\right)=\frac{P\left(A_2\right) \cdot P\left(B / A_2\right)}{P\left(A_1\right) P\left(B / A_1\right)+P\left(A_2\right) P\left(B / A_2\right)} +P\left(A_3\right) P\left(B / A_3\right)\)
\(=\frac{0.3(0.04)}{0.2(0.03)+0.3(0.04)+0.5(0.02)}\)
\(=\frac{0.012}{0.028}=\frac{12}{28}=\frac{3}{7}\)

i) Given: f(0) is undefined
\(\lim _{x \rightarrow 0} f(x)=4, f(2)=6, \lim _{x \rightarrow 2} f(x)=3\)
To sketch the graph of y = f(x)
(i) Consider the points (0, 4) and (2, 3)
(ii) Draw a curve straight. line passing through (0, 4) and (2, 3)
(iii) Remove (0, 4), (2, 3) and plot the point (2, 6)

Given \(\overrightarrow{a}=\lambda \hat{i}+\hat{j}+4\hat{k}\) and \(\overrightarrow{b}=2\hat{i}+6\hat{j}+3\hat{k}\)
\(|\overrightarrow{b}|=\sqrt{2^2+6^2+3^2}=\sqrt{4+36+9}=\sqrt{49}=7\)
\(\overrightarrow{a}.\overrightarrow{b}=(\lambda \hat{i}+\hat{j}+4\hat{k}).(2\hat{i}+6\hat{j}+3\hat{k})=2\lambda+6+12=2\lambda +18\)
Also projection of \(\overrightarrow{a}\) on \(\overrightarrow{b}\) is \(\overrightarrow{a}.\overrightarrow{b}\over |\overrightarrow {b}|\)
\(4={2\lambda +18\over 7}\)
\(28=2\lambda +18\)
\(28-18=2\lambda \)
\(10=2\lambda \)
\(\lambda ={10\over2}=5\)
\(\therefore \lambda =5\)

\(\left|\begin{array}{ccc} 1 & \log _x y & \log _x z \\ \log _y x & 1 & \log _y z \\ \log _z x & \log _z y & 1 \end{array}\right|\)
\(=\left|\begin{array}{lll} \log _x x & \log _x y & \log _x z \\ \log _y x & \log _y y & \log _y z \\ \log _z x & \log _z y & \log _z z \end{array}\right|\)
\(\left.=\left|\begin{array}{lll} \frac{\log _a x}{\log _a x} & \frac{\log _a y}{\log _a x} & \frac{\log _a z}{\log _a x} \\ \frac{\log _a x}{\log _a y} & \frac{\log _a y}{\log _a y} & \frac{\log _a z}{\log _a y} \\ \frac{\log _a x}{\log _a z} & \frac{\log _a y}{\log _a z} & \frac{\log _a z}{\log _a z} \end{array}\right| \therefore \log _a b=\frac{\log _{\mathrm{a}} b}{\log _{\mathrm{a}} \mathrm{a}}\right)\)
\(=\frac{1}{\log _a x \log _a y \log _a z}\left|\begin{array}{lll} \log _a x & \log _a y & \log _a z \\ \log _a x & \log _a y & \log _a z \\ \log _a x & \log _a y & \log _a z \end{array}\right|\)
\(=\frac{1}{\log _a x \log _a y \log _a z}(0)\)
= 0

Let p(n) be the statement that 3n² > (n + 1)² with n\(\ge\) 2. Therefore the first stage is n = 2.
Now, P(2) = 3 x 2² = 12 and 3² = 9. As 12> 9 we get P(2) is true.
We assume that p(n) is true for n = k.
Now,
P(k+ 1) 3(k+ 1)² = 3k² + 6k+ 3
P(k) + 6k+ 3
> (k + 1)² + 6k + 3
k² + 8k+4
k² + 4k+4 + 4k
(k+ 2)² + 4k
> (k+ 2)² since k> 0.
This is the statement P(k + 1). The validity of P(k + 1) follows from that of P(k).
Therefore by the principle of mathematical induction, for all n\(\ge\) 2, 3n² > (n + 1)².

\(\frac { { x }^{ 2 }-2x-9 }{ (x+1)({ x }^{ 2 }+x+6) } =\frac { A }{ (x+1) } +\frac { Bx+C }{ ({ x }^{ 2 }+x+6) } \)
= \(\frac { A({ x }^{ 2 }+x+6)+(Bx+C)(x+1) }{ (x+1)({ x }^{ 2 }+x+6) } \)
Equating numerator on b/s
x²-2x-9 = A(x²+x+6)+B(Bx+C)(x+1)
put x=-1
1+2-9 = A(1-1+6)+0
-6 = 6A ⇒ A=-1
Equating co-eff of x²
1 = A+B
a =1+B ⇒B =2
Equating co-eff of x
-2 = A+B+C
-2 =-1+2+C
-2 =1+C ⇒ C=-2-1 =-3
\(\frac { { x }^{ 2 }-2x-9 }{ (x+1)({ x }^{ 2 }+x+6) } =\frac { -1 }{ x+1 } +\frac { 2x-3 }{ { x }^{ 2 }+x+6 } \)

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 \(\frac{1}{x^2}+\frac{1}{y^2}=\frac{4}{p^2}\)

Let ABC be a triangle with constant perimeter 2s. Thus s is constant.
We know that \(\triangle =\sqrt{s(s-a)(s-b)(s-c)}\)
Observe  that \(\triangle\) is maximum, when (s - a) (s - b) (s - c) is maximum.
Now, (s-a)(s-b)(s-c)\(\le\)\(({(s-a)+(s-b)+(s-c)\over 3})^3\) = \({s^3\over 27}\) [G.M\(\le\) A.M]
Thus, we get (s - a) (s - b) (s - c) \(\le\) \({s^3\over 27}\)
Equality occurs when s - a = s - b = s - c. That is, when a = b = c maximum of (s - a) (s - b) (s - c) is \({s^3\over 27}\)
Thus, for a fixed perimeter 2s, the area of a triangle is maximum when a = b = c.
Hence, for a fixed perimeter, the equilateral triangle has the maximum area and the maximum area is given by \(\triangle =\sqrt {s(s)^3\over 27}={s^2\over3\sqrt{3}}sq.units\)

sin 9\(\theta\) = sin \(\theta\) \(\Rightarrow\)sin 9\(\theta\) - sin \(\theta\) = 0
2cos 5\(\theta\) sin 4\(\theta\) = 0
Either cos 5\(\theta\) = 0 (or) sin 4θ = 0

Thus, the general solution of the given equation is \(\theta=(2n+1){\pi\over10},\theta={n\pi\over 4},n\in Z.\)

Given lines are y = mx - 3
\(\Rightarrow\)   mx - y = 3 ......(1)
and      x-y = 6.....(2)
Solving (1) and (2), we get
(1) \(\Rightarrow\) mx-y = 3
           -    +    -
(2) \(\Rightarrow\)  x-y = 6
______________
       x(m-1)  = -3      \(\Rightarrow \quad x=\frac { -3 }{ m-1 } =\frac { 3 }{ 1-m } \)
Substituting x = \(\frac { 3 }{ 1-m } \) in (2) we get,
\(\frac { 3 }{ 1-m } -y=6\quad \Rightarrow \quad \frac { 3 }{ 1-m } -6=y\)
\(y=\frac { 3-6+6m }{ 1-m } \Rightarrow \quad y=\frac { 6m-3 }{ 1-m } \)
The point of intersection of the given line is \( \left( \frac { 3 }{ 1-m } ,\frac { 6m-3 }{ 1-m } \right) \)
It is given that the slope m and the x-Co-ordinates are integers.
\(\therefore \ \frac { 3 }{ 1-m } \) is an integer    \(\Rightarrow\)  (1-m) is a divisor of 3.
\(\Rightarrow\) 1-m = \(\pm 1,1-m=\pm 3\) where m is an integer
\(\Rightarrow\) 1-m = 1 or 1 - m = -1 \(\Rightarrow\)  m = 0 or m = -2
\(\Rightarrow\) 1-m = \(\pm 3\ or\ 1-m=-3\ \Rightarrow \ m=-2\ or\ m=4\)
\(\therefore\) The equations are y =0 x-3 (\(\because\) y = mx-3)
y = -2x-3 and y = 4x-3
\(\Rightarrow\) y = -3, y = -2x-3 and y = 4x-3
\(\Rightarrow\)  y+3 = 0 or 2x + y +3 = 0, or 4x - y - 3 = 0

We have P = Principal = Rs. 500, R = rate of interest = 10%
Amount at the end of one year = \(\left(P+{R\over100}\right)\)
\(=P\left(1+{R\over100}\right)\)
Amount at the end of second year
\(=P\left(1+{R\over100}\right)+P\left(1+{R\over100}\right)\left(R\over 100\right)\)
 \(=P\left(1+{R\over100}\right)\left(1+{R\over100}\right)=P\left(1+{R\over 100}\right)^2\) 
and so on.
Clearly amount at the end of various year form a G.P. with first term and common ratio \(1+{R\over100}\)
∴ Amount at the end of 10thyear
= 11^th term of the G.P
\(=P\left(1+{R\over100}\right)^{10}\)
 \(=500\left(1+{10\over100}\right)^{10}=500\left(11\over10\right)^{10}\)
 Amount at the end of the 10^th  year = 500 (1.1)¹⁰ = 1296.87 of the 10 ^th year.

The relation is defined by aRb if a + b \(\le\) 6 for all a, b  \(\in \)N.
a+b \(\le\)6 \(\Rightarrow\) a \(\le\) 6 - b

\(\therefore\) The list of ordered pairs are (5,1) (4, 2) (3, 3) (2, 4) and (1, 5).
Symmetric : (5, 1) \(\in \) R \(\Rightarrow\) (1, 5) \(\in \) R
(4, 2) \(\in \) R \(\Rightarrow\) (2, 4) \(\in \) R
\(\therefore\) R is symmetric

f(x) = |x| − 1 causes the graph of the functionf(x) = |x| shifts to the downward for one unit. f(x) = |x| + 1 causes the graph of the function f(x) = |x| shifts to the upward for one unit.

\(f^{ ' }\left( 1^{ - } \right) =\lim _{ x\rightarrow 1^{ - } }{ \frac { f\left( x \right) -f\left( 1 \right) }{ x-1 } } =\lim _{ x\rightarrow 1^{ - } }{ \frac { { x }^{ 2 }-1 }{ x-1 } } =\lim _{ x\rightarrow 1^{ - } }{ \frac { (x+1)(x-1) }{ x-1 } } =\lim _{ x\rightarrow 1^{ - } }{ (x+1) } =1+1=2\quad ...(1)\)
\(f^{ ' }\left( 1^{ + } \right) =\lim _{ x\rightarrow 1^{ + } }{ \frac { f\left( x \right) -f\left( 1 \right) }{ x-1 } } =\lim _{ x\rightarrow 1^{ + } }{ \frac { { x }^{ 2 }-1 }{ x-1 } } =\lim _{ x\rightarrow 1^{ + } }{ \frac { (x+1)(x-1) }{ x-1 } } =\lim _{ x\rightarrow 1^{ + } }{ x+1 } =1+1=2 ...(2)\)
From (1)and (2), \(f^{ ' }\left( 1^{ - } \right) =f^{ ' }\left( 1^{ + } \right) \)
\(\therefore f\left( x \right) \)is differentiable at x = 1 and \(f^{ ' }\left( 1 \right) =2\)

\(=\int x \sin 3 x d x \)
\(=x\left(\frac{-\cos 3 x}{3}\right)-1\left(\frac{-\sin 3 x}{3 \times 3}\right)+c \)
\(=-\frac{x}{3} \cos 3 x+\frac{1}{9} \sin 3 x+c\)

Let \(I=\int \frac{1}{x \log x \cdot \log (\log x)} d x\)
\(=\int \frac{\frac{1}{x \log x}}{\log (\log x)} d x \)
\(t =\log (\log x) \)
\(d t =\frac{1}{\log x} \cdot \frac{1}{x} d x \)
\(I =\int \frac{d t}{t}=\log |t|+c \)
\(I =\log |\log (\log x)|+c\)

\(\int{x^3+2\over x-1}dx=\int {x^3-1+3\over x-1}dx=\int ({x^3-1\over x-1}+{3\over x-1})dx\)
\(=\int\left[\frac{(x-1)\left(x^2+x+1\right)}{x-1}+\frac{3}{x-1}\right] d x\)
\(\int (x^2+x+1+{3\over x-1})dx\)
\(={x^3\over 3}+{x\over 2}+x+3log|(x-1)|+c.\)

tan(x + y) + tan(x - y) = x
Differentiate w.r. to x
\(\sec ^2(x+y)\left[1+\frac{d y}{d x}\right]+\sec ^2(x-y)\left[1-\frac{d y}{d x}\right]=1\)
\( \sec ^2(x+y)+\sec ^2(x+y) \frac{d y}{d x}+\sec ^2(x-y) -\sec ^2(x-y) \frac{d y}{d x}=1\)
\( \frac{d y}{d x}\left[\sec ^2(x+y)-\sec ^2(x-y)\right] =1-\sec ^2(x+y)-\sec ^2(x-y) \)
\(\frac{d y}{d x}=\frac{1-\sec ^2(x+y)-\sec ^2(x-y)}{\sec ^2(x+y)-\sec ^2(x-y)}\)

\(\text { (i) Civen } A=\left[\begin{array}{ccc} 0 & 1 & -2 \\ -1 & 0 & x^3 \\ 2 & -3 & 0 \end{array}\right] \text { is skew-symmetric. }\)
\(A^T=-A\)
\(aA^T=\left(\begin{array}{ccc} 0 & -1 & 2 \\ 1 & 0 & -3 \\ -2 & x^3 & 0 \end{array}\right)\)
\(A^T=-A\)
\(\left[\begin{array}{ccc} 0 & -1 & 2 \\ 1 & 0 & -3 \\ -2 & x^3 & 0 \end{array}\right]=\left[\begin{array}{ccc} 0 & -1 & 2 \\ 1 & 0 & -x^3 \\ -2 & 3 & 0 \end{array}\right]\)
\(x^3=3\)
\(\therefore x=3^{1 / 3}\)

\(\sqrt { 98 } =\sqrt { 49\times 2 } =7\sqrt { 2 } ;\sqrt { 18 } =\sqrt { 9\times 2 } 3\sqrt { 2 } \)
\(\sqrt { 50 } =\sqrt { 25\times 2 } =5\sqrt { 2 } ;\sqrt { 75 } =\sqrt { 25\times 3 } 5\sqrt { 3 } \)
\(\sqrt { 27 } =\sqrt { 9\times 3 } 3\sqrt { 3 } \)
∴ \(\sqrt { 98 } +\sqrt { 50 } -\sqrt { 18 } +\sqrt { 75 } -\sqrt { 27 } =7\sqrt { 2 } +5\sqrt { 2 } -3\sqrt { 2 } +5\sqrt { 3 } -3\sqrt { 3 } \)
= \(9\sqrt { 2 } +2\sqrt { 3 } \)

Let a₁ be the first term and d, the common difference. Here n = 21.
∴ The three middle terms are a₁₀, a_11, a₁₂
Now, a₁₀ + a₁₁ + a₁₂ = 129 [Given]
∴ \((a_1+9d)+(a_1+10d)+(a_1+11d)=129\)
⇒ \(3a_1+30d=129 \Rightarrow a_1+10d=43\)   --- (i)
The last three terms are a₁₉, a₂₀, a₂₁
∴ a₁₉ + a₂₀ + a₂₁  [Given]
∴ \((a_1+8d)+(a_1+19d)+(a_1+20d)=237\)
(i.e)\(3a_1+57d=237, \) 
∴ a₁ + 19d = 79   -- (ii)
Subtracting (i) from (ii), we get 9d = 36, ∴ d = 4
∴ From(i), a₁ + 40 = 43, ∴ a₁ = 3 
Hence, the series is 3, 7, 11, 15 ....

We know that for |x|I< 1
(1+x)n = \(1+nx+\frac { n(n-1) }{ 2! } { x }^{ 2 }+\frac { n(n-1)(n-2) }{ 3! } { x }^{ 3 }+...\)
\(\sqrt [ 3 ]{ 65 } \) =65^1/3
=64+1)^1/3
=\(\left( 1+\frac { 1 }{ 64 } \right) ^{ 1/3 }\Rightarrow \left( 1+\frac { 1 }{ 64 } \right) ^{ 1/3 }\)
=\(4\left( 1+\frac { 1 }{ 3 } \times \frac { 1 }{ 64 } +\frac { \frac { 1 }{ 3 } \left( \frac { 1 }{ 3 } -1 \right) }{ 2! } \times \left( \frac { 1 }{ 64 } \right) ^{ 2 }+... \right) \)
=\(4+\frac { 1 }{ 48 } -4\times \frac { 1 }{ 9 } \times \frac { 1 }{ 64 } \times \frac { 1 }{ 64 } +...\)
=\(4+\frac { 1 }{ 48 } -\frac { 4 }{ 36864 } +...\)
=\(4+\frac { 1 }{ 48 } -\frac { 4 }{ 9216 } +...\)
= 4+0.2    (since \(\frac { 1 }{ 9216 } \) +....is very small)
\(\sqrt [ 3 ]{ 65 } \)=4.02 (approximately)

Let r₁ and r₂ be the radii of the two given circles and l be the length of the arc,
\(\theta_1=30^0=\frac{\pi}{6}\ radians\)
\(\theta_2=80^0=\frac{4\pi}{9}\ radians\)
Given that \(l=r_1\theta_1=r_2\theta_2\)
Thus, \(\frac{\pi}{6}r_1=\frac{4\pi}{9}r_2\)
\(\frac{r_1}{r_2}=\frac{8}{3}\) Which implies r₁ : r₂ = 8 : 3

(i) two units from the X-axis
(h, k) be any point on the required path

Any line parallel to x - axis will be of the form y = c.
By the given condition, the distance from x - axis is 2 units ⇒ c = 2.
ஃ Locus of the point P is y = 2.
(ii) Any line parallel to y-axis will be of the formy = c.

By the given condition, the distance from y-axis is 3 units ⇒ c = 2
ஃ Locus of the point P is x = 3.

Let m be the event the selected student passes in mathematics and E be the event that the selected student passes in English.
\(\therefore P(M)=\frac { 2 }{ 3 } \) and P(M\(\cap \)E)=\(\frac { 1 }{ 3 } \)
\(\therefore P(E/M)=\frac { P(M\cap E) }{ P(M) } =\frac { \frac { 1 }{ 3 } }{ \frac { 2 }{ 3 } } =\frac { 1 }{ 3 } \times \frac { 3 }{ 2 } =\frac { 1 }{ 2 } \)

\(Given,\quad \lim _{ x\rightarrow 2 }{ \frac { { x }^{ n }-{ 2 }^{ n } }{ x-2 } } =80\)
\( \Rightarrow n.{ 2 }^{ n-1 }=80\)
By trial method, put n 5
\(\Rightarrow 5({ 2 }^{ 5-1 })=80\)
\(\Rightarrow 5({ 2 }^{ 4 })=80\)
\( \Rightarrow 5(16)=80\)
\( \Rightarrow 80=80\)
\(\therefore n=5\)

\(lim_{x\rightarrow 0}{1-cosx\over x^2}=lim_{x\rightarrow 0}{2sin^2{x \over 2}\over x^2}[\because cos 2x=1-2sin^2x]\)
\(=2.lim_{x\rightarrow0}{sin^2({x\over 2})\over {x^2\over 4}\times 4}={2\over4}[lim{{x\over 2}\rightarrow 0}{sin({x\over2})\over ({x\over2})}]^2\)
\(={2\over4}\times 1^2={2\over4}={1\over2}\)  \([\because lim_{\theta \rightarrow o}{sin \theta \over \theta}=1]\)
\(\therefore lim_{x\rightarrow 0}{1-cosx\over x^2}={1\over2}\)

We start with the assumption f(x) = f(y). Then
\({x \over x^2-1}={y \over y^2-1}\)
\(\Rightarrow x(y^2-1)=(x^2-1)\)
\(\Rightarrow xy^2-x-yx^2+y-0\Rightarrow (y-x)(xy+1)=0\)
This implies that x =y or xy = - 1. So if we select two numbers x and y so that xy = -1, then f(x) = f(y).  \(f(x)=f(y) \cdot\left(2,-\frac{1}{2}\right),\left(7,-\frac{1}{7}\right),\left(-2, \frac{1}{2}\right)\)are some among the infinitely many possible pairs. Thus \(f(2)=f\left(\frac{-1}{2}\right)=\frac{2}{3}\). That is, f(x) = f(y) does not imply x = y. Hence it is not one-to-one.

Given series is  1² + 3² + 5² +...
Let Tn be the nth term
T_n = (nth term of 1, 3, 5,...)²
= [1+(n-1)2]² = (1 + 2n - 2)² = (2n-1)²
= 4n² + 1 - 4n
∴ Sum of n terms =  \(\sum { 4{ n }^{ 2 } } -4n+1=4\sum { n^{ 2 } } -4\sum { n } +n\)
= \(4\frac { (n)(n+1)(2n+1) }{ 6 } -\frac { 4n(n+1)+n }{ 2 } \)
= \(\frac { n }{ 2 } \) [2(n + 1)(n + 1) - 6(n + 1) + 3]
= \(\frac { n(4{ n }^{ 2 }-1) }{ 3 } \)

Given p(n) :"7 divides 2³ⁿ- 1"
\(\Rightarrow\) 2³ⁿ- 1 = 7k    [where k is a constant]
\(\Rightarrow\) 2³ⁿ = 7k+1 ...(1)
Now, p(n+1) is 7 divides 2³⁽ⁿ⁺¹⁾-1
2³⁽ⁿ⁺¹⁾-1 = 7k₁
2³ⁿ.2³-1 = 7k₁
8(2³ⁿ) = 7k₁+1 which is the required statement.