Let E₁ = missing card is a heart card
E₂ = missing card is a spade card
E₃ = missing card is a club card
E₄ = missing card is a diamond card
and A = drawing 2 heart cards from the remaining cards.
\(\therefore\) P(E₁) = P(E₂) = P(E₃) = P(E₄)
\(=\frac { 13 }{ 52 } =\frac { 1 }{ 4 } \)
P(A/E₁) = P(two heart cards given that one heart card is missing)
\(=\frac { 12{ C }_{ 2 } }{ 51{ C }_{ 2 } } =\frac { \frac { 12\times 11 }{ 2\times 1 } }{ \frac { 51\times 50 }{ 2\times 1 } } =\frac { 66 }{ 1275 } \)
P(A/E₂) = P(2 heart cards given that one spade card is missing)
\(=\frac { 13{ C }_{ 2 } }{ 51{ C }_{ 2 } } =\frac { \frac { 13\times 12 }{ 2 } }{ \frac { 51\times 50 }{ 2\times 1 } } =\frac { 78 }{ 1275 } \)
Similarly, P(A/E₃) = \(\frac { 13{ C }_{ 2 } }{ 51{ C }_{ 2 } } =\frac { 78 }{ 1275 } \)
and P(A/E₄) = \(\frac { 13{ C }_{ 2 } }{ 51{ C }_{ 2 } } =\frac { 78 }{ 1275 } \)
\(\therefore\) By Baye's theorem,
\(P({ E }_{ 1 }/A)=\frac { P({ E }_{ 1 }).P\left( A/{ E }_{ 1 } \right) }{ P({ E }_{ 1 }).P\left( A/{ E }_{ 1 } \right) +P({ E }_{ 2 }).P\left( A/{ E }_{ 2 } \right) +P({ E }_{ 3 }).P(A/{ E }_{ 3 }) } \)
\(=\frac { \frac { 1 }{ 4 } \times \frac { 66 }{ 1275 } }{ \frac { 1 }{ 4 } \times \frac { 66 }{ 1275 } +\frac { 1 }{ 4 } \times \frac { 78 }{ 1275 } +\frac { 1 }{ 4 } \times \frac { 78 }{ 1275 } +\frac { 1 }{ 4 } \times \frac { 78 }{ 1275 } } \)
\(=\frac { \frac { 1 }{ 4 } \times \frac { 66 }{ 1275 } }{ \frac { 1 }{ 4 } \left[ \frac { 66 }{ 1275 } +\frac { 78 }{ 1275 } +\frac { 78 }{ 1275 } +\frac { 78 }{ 1275 } \right] } \)
\(=\frac { 66 }{ 1275 } \times \frac { 1275 }{ 66+78+78+78 } \)
\(=\frac { 66 }{ 300 } =\frac { 11 }{ 50 } \)

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).

Since the decision variables are non-negative, the solution lies in the I quadrant of the plane.
Consider the equations 
\({ x }_{ 1 }-{ x }_{ 2 }=-1\)

\({ -x }_{ 1 }+{ x }_{ 2 }=0\)

The feasible region is not common. Thus, there is no maximum value of Z.

Given a = Rs.1000,i = 7% = 0.07.n = 5
A = \(\cfrac { a }{ i } \left[ \left( 1+i \right) ^{ n }-1 \right] \)
=\(\cfrac { 1000 }{ 0.07 } \) x [(1.07)⁵-1]
=14,285.71[1.4028-1]
= (14,285.71)[0.4028]
= Rs.5754.28
(1.07)⁵ = 5 log(1.07)
= 5(0.0294)
= 0.1470
Antilog of 0.1470 is 1.4028

n=12
Rank correlation co-efficient
\(\rho =1-\frac { 6\sum { { d }^{ 2 } } }{ N({ N }^{ 2 }-1) } \)
=1-\(\frac { 6\times 416 }{ 12({ 12 }^{ 2 }-1) } =1-\frac { 16 }{ 11 } =\frac { -5 }{ 11 } \)
\(\rho\)=-0.45

\(\tan{{\pi}\over{8}}=\sqrt{2}-1.\)

LHS\(=\tan\left(\frac{\pi}{3}+x\right)\tan\left(\frac{\pi}{3}-x\right)\)
\(=\frac{2\sin\left(\frac{\pi}{3}+x\right).\sin\left(\frac{\pi}{3}-x\right)}{2\cos\left(\frac{\pi}{3}+x\right)\cos\left(\frac{\pi}{3}-x\right)}\)
\([\because2\sin A\sin B=\cos(A-B)-\cos(A+B)\ and\ \ 2\cos A\cos B=\cos(A+B)+\cos(A-B)]\)
\(=\frac{\cos\left(\frac{\pi}{3}+x-\frac{\pi}{3}+x\right)-\cos\left(\frac{\pi}{3}+x+\frac{\pi}{3}-x\right)}{\cos\left(\frac{\pi}{3}+x+\frac{\pi}{3}-x\right)+\cos\left(\frac{\pi}{3}+x-\frac{\pi}{3}+x\right)}\)
\(=\frac{\cos2x-\cos\frac{2\pi}{3}}{\cos\frac{2\pi}{3}+\cos2x}=\frac{\cos2x+\frac{1}{2}}{-\frac12+\cos2x}\)
\(=\frac{2\cos2x+1}{2\cos2x-1}\)
=RHS
Hence proved.

Given x = 15(t - sint)
Differentiating with respect to 't' we get,
\(\frac{dy}{dx}\) = 15(1 - cos t) Also y = 18(1 - cos t)
\(\frac{dy}{dx}\) = 18(sin t)
Now  \(\frac { dy }{ dx } =\frac { \frac { dy }{ dt } }{ \frac { dx }{ dt } } =\frac { 18\sin { t } }{ 15\left( 1-\cos { t } \right) } =\frac { 6\sin { t } }{ 5\left( 1-\cos { t } \right) } \)
\(=\frac { 6\times 2\sin { \frac { t }{ 2 } } \cos { \frac { t }{ 2 } } }{ 5\times 2\sin ^{ 2 }{ \frac { t }{ 2 } } } \)      \(\left[ \because sin2A=2sinAcosA\ \ and\ 1-cosA=2{ sin }^{ 2 }\frac { A }{ 2 } \right] \quad \quad \)
\(=\frac { 6 }{ 5 } \cot { \left( \frac { t }{ 2 } \right) } \)

Let P(x₁, y₁) be any point on the locus and let A(0, 2) B(0, -2) be the fixed points.
By the given condition, PA + PB =6
\(\Rightarrow \sqrt { { \left( { x }_{ 1 }-0 \right) }^{ 2 }+{ \left( { y }_{ 1 }-2 \right) }^{ 2 } } +\sqrt { { ({ x }_{ 1 }-0 })^{ 2 }+{ ({ y }_{ 1 }+2) }^{ 2 } } =6\)
\(\Rightarrow \sqrt { { x }_{ 1 }^{ 2 }+{ ({ y }_{ 1 }-2) }^{ 2 } } =6-\sqrt { { x }_{ 1 }^{ 2 }+{ ({ y }_{ 1 }+2) }^{ 2 } } \)
Squaring both sides we get,
\({ x }_{ 1 }^{ 2 }+{ ({ y }_{ 1 }-2) }^{ 2 }=36-12\sqrt { { x }_{ 1 }^{ 2 }+({ y }_{ 1 }+2)^{ 2 } } +{ x }_{ 1 }^{ 2 }+{ ({ y }_{ 1 }+2) }^{ 2 }\)
\(\Rightarrow \quad { x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 }+4-4{ y }_{ 1 }=36-12\sqrt { { x }_{ 1 }^{ 2 }+{ ({ y }_{ 1 }+2) }^{ 2 } } +{ x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 }+4+4{ y }_{ 1 }\)
\(\Rightarrow \quad { x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 }+4-4{ y }_{ 1 }-36-{ x }_{ 1 }^{ 2 }-{ y }_{ 1 }^{ 2 }-4-4{ y }_{ 1 }=-12\sqrt { { x }_{ 1 }^{ 2 }+{ ({ y }_{ 1 }+2) }^{ 2 } } \)
= -8y₁ - 36 = -12 \(\sqrt { { x }_{ 1 }^{ 2 }+{ ({ y }_{ 1 }+2) }^{ 2 } } \)
= 2y₁ + 9 = 3\(\sqrt { { x }_{ 1 }^{ 2 }+{ ({ y }_{ 1 }+2) }^{ 2 } } \)
Squaring again we get,
(2y₁+9)² = 9[\({ x }_{ 1 }^{ 2 }+({ y }_{ 1 }+2)^{ 2 }]\)
\(\Rightarrow 4{ y }_{ 1 }^{ 2 }+81+36{ y }_{ 1 }=9[{ x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 }+4+4{ y }_{ 1 }]\)
\(\Rightarrow 4{ y }_{ 1 }^{ 2 }+81+36{ y }_{ 1 }-9{ x }_{ 1 }^{ 2 }-9{ y }_{ 1 }^{ 2 }-36-36{ y }_{ 1 }=0\)
\(\Rightarrow -9{ x }_{ 1 }^{ 2 }-5{ y }_{ 1 }^{ 2 }+45=0\)
\(\Rightarrow 9{ x }_{ 1 }^{ 2 }+5{ y }_{ 1 }^{ 2 }=45\)
\(\therefore \)  Locus of (x₁ , y₁) is 9x² + 5y² = 45

Given A =\(\begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}\)
\(\therefore\)A² = A.A =\(\begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}\)\(\begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}\)=\(\begin{bmatrix} 9+2 & 6+2 \\ 3+1 & 2+1 \end{bmatrix}=\begin{bmatrix} 11 & 8 \\ 4 & 3 \end{bmatrix}\)
Given A² + aA + bI = 0
⇒ \(\begin{bmatrix} 11 & 8 \\ 4 & 3 \end{bmatrix}+a\begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}+b\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)= 0
⇒ \(\begin{bmatrix} 11+3a+b & 8+2a+0 \\ 4+a+0 & 3+a+b \end{bmatrix}=\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)
Equating the like terms we get,
4 + a = 0 ⇒ a = -4
3 + a + b = 0 ⇒ 3 - 4 + b = 0
-1 + b = 0 ⇒ b = 1