Let y = x³+8x²+5x-2
Differentiating w.r.t. 'x' we get
\({dy\over dx}=3x^2+16x+5\)
\({dy\over dx}=0⇒3x^2 + 16x + 5=0\)
⇒ (x+5)(3x+1)=0
⇒ x = - 5, -1/3

The possible intervals are (-∞, - 5), (-5, -1/3) and (-1/3, ∞)

Hence the given function is increasing in the intervals (-∞, - 5), (-1/3, ∞) and decreasing in (-5, -1/3).

Given a = Rs.500,i = 5% =0.05,n =10
A=\(\cfrac { a }{ i } \left[ 1-\left( 1+i \right) ^{ -n } \right] \)
=\(\cfrac { 500 }{ 0.05 } \left( 1.05 \right) \left[ \left( 1.05 \right) ^{ 10 }-1 \right] \)
=10,500 [1.629-1]
=10,500(0.629)
= Rs.6604.50
\(\therefore\) At the end of 10 years. I will receive Rs.6604.5
(1.05)¹⁰ = 10 log(1.05)
= 10(0.0212)
= 0.2120
Antilog of 0.2120 is 1.629

We haveP(not A or not B) = 1/4
\(P(\overline { A } U\overline { B } )=\frac { 1 }{ 4 } \) \(\Rightarrow P(\overline { A\cap B } )=\frac { 1 }{ 4 } \)
\(\Rightarrow 1-P(A\cap B)=\frac { 1 }{ 4 } \)   \(\Rightarrow P(A\cap B)=1-\frac { 1 }{ 4 } =\frac { 3 }{ 4 } \)
\(\therefore P(A\cap B)=\frac { 3 }{ 4 } \)
Now, P(A). P(B) = \(\frac { 1 }{ 2 } \times \frac { 7 }{ 12 } =\frac { 7 }{ 24 } \)
\(\therefore P(A\cap B)\neq P(A).P(B)\)
So, A and B are not independent events.

The regression co-efficient are given by
b_xy= \(r.\frac { { \sigma }_{ x } }{ { \sigma }_{ y } } \)and b_yx=\(r.\frac { { \sigma }_{ y } }{ { \sigma }_{ x } } \)
where r is the co-efficient of correlation.

\(r=\sqrt { { b }_{ xy }.{ b }_{ yx } } \)

r(x,y) =\(\frac { N\sum { XY-(\sum { X)(\sum { Y) } } } }{ \sqrt { N.{ \sum { X } }^{ 2 }-({ \sum { X) } }^{ 2 } } .\sqrt { N.{ \sum { Y } }^{ 2 }-{ (\sum { Y) } }^{ 2 } } } \)
=\(\frac { 9(597)-45(108) }{ \sqrt { 9(285)-{ (45) }^{ 2 } } .\sqrt { 9(1356)-({ 108) }^{ 2 } } } \)
=0.95
\(\therefore\)X and Y are highly positively correlated.

LHS\(=\left\{1+\cot x-\sec\left(\frac{\pi}{2}+x\right)\right\}\left\{1+\cot x+\sec\left(\frac{\pi}{2}+x\right)\right\}\)
=(1 + cot x + cosec x) (1 + cot x - cosec x)
(1 + cot x)² - (cosec²x) [\(\therefore\)(a + b) (a - b) = a² - b²]
1 + cot²x + 2 cot x - cosec²x
= cosec²x + 2 cot x - cosec²x [\(\therefore\) 1 + cot²x = cosec²x]
= 2 cot x = RHS. Hence proved.

Given f(x) = x² -x² -sin x + 5
\(\underset { x\rightarrow \pi }{ lim } f(x)=\underset { x\rightarrow \pi }{ lim } ({ x }^{ 2 }-sinx+5)\)
\(={ \pi }^{ 2 }-sin\quad \pi +5={ \pi }^{ 2 }-0+5+5=5{ \pi }^{ 2 }\)  \([\therefore sin \ \pi=0]\)
Also, f(\(\pi\)) = \({ \pi }^{ 2 }-sin\quad \pi +5=\pi \quad -0+5=\pi +5\)
ஃ  \(\underset { x\rightarrow \pi }{ lim } f(x)=f(x)\)
ஃ  f(x) is continuous at x = \(\pi\)

Let p(x₁,y₁) be the point on the locus and A(-1, 0) B(0, 2) are the fixed points
Given pA = 3 pB
\(\Rightarrow\)  pA² = 9 pB²
\(\Rightarrow ({ x }_{ 1 }+1)^{ 2 }+({ y }_{ 1 }-0)^{ 2 }=9[({ x }_{ 1 }-0)^{ 2 }+({ y }_{ 1 }-2)^{ 2 }]\)
\(\Rightarrow { x }_{ 1 }^{ 2 }+2{ x }_{ 1 }+1+{ y }_{ 1 }^{ 2 }=9[{ x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 }-4{ y }_{ 1 }+4]\)
\(\Rightarrow { x }_{ 1 }^{ 2 }+2{ x }_{ 1 }+1+{ y }_{ 1 }^{ 2 }=9{ x }_{ 1 }^{ 2 }+9{ y }_{ 1 }^{ 2 }-36{ y }_{ 1 }+36\)
\(\Rightarrow 8{ x }_{ 1 }^{ 2 }+8{ y }_{ 1 }^{ 2 }-{ 2x }_{ 1 }-36{ y }_{ 1 }+35=0\)
\(\therefore Locus\quad of\quad (x_{ 1 },{ y }_{ 1 })\quad is\) 8x² + 8y² - 2x - 36y + 35 = 0

In "AGAIN" the letters in ascending order are A, A, G, I, N
∴  The first word is AAGIN
Number of words beginning with A A
= No of ways of arranging G, 1, N = 3! = 6
The next word begin with AG and it is AGAIN
∴  No. of words before AGAIN = 6.
∴  Rank of word AGAIN = 7

Given A \(=\begin{bmatrix} 2&3\\-1&4 \end{bmatrix}\)
\(|A|=\begin{bmatrix} 2&3\\-1&4 \end{bmatrix}=8+3=11\)
Now, A₁₁ = 4, A₁₂ = - (-1) = 1, A₂₁  = 3, A₂₂ = 2
\(\therefore\ adj\ A={\begin{bmatrix} 4&1\\-3&2 \end{bmatrix}}^{T}=\begin{bmatrix} 4&-3\\1&2 \end{bmatrix} \)
\(\therefore\) A (adj A)\(=\begin{bmatrix} 2 & 3 \\ -1 & 4 \end{bmatrix} { }\begin{bmatrix} 4 & -3 \\ 1 & 2 \end{bmatrix}\)
\(=\begin{bmatrix} 8+3&-6+6\\-1+4&3+8 \end{bmatrix}=\begin{bmatrix} 11&0\\0&11 \end{bmatrix}=11\begin{bmatrix} 1&0\\0&1 \end{bmatrix}=|A|I_2\)   ...(1)
Also ( adj A ) A = \(\begin{bmatrix} 4&-3\\1&2 \end{bmatrix}\begin{bmatrix} 2&3\\-1&4 \end{bmatrix}\)
\(=\begin{bmatrix} 8+3&12-12\\2-2&3+8 \end{bmatrix}=\begin{bmatrix} 11&0\\0&11 \end{bmatrix}=11\begin{bmatrix}1&0\\0&1 \end{bmatrix}=|A|I_2\)   ....(2)
From (1) and (2), A( adj A) = (adj A) A = |A|.I₂