\(\frac { 3 }{ 2 } \)

\(\frac { 1 }{ loga } -\frac { 1 }{ logb } \)

\(\frac { { \vartheta }^{ 2 }u }{ \vartheta x\vartheta y } ={ e }^{ xy }\left[ 3xy+1+{ x }^{ 2 }{ y }^{ 2 } \right] \)

Rolle’s theorem is not valid.

Given \(\frac{dy}{dx}=1+e^{x-y}\) ...(1)
putting x - y = z ⇒ 1 - \(\frac{dy}{dx}=\frac{dz}{dx}\)
\(\Rightarrow \frac { dy }{ dx } =1-\frac { dz }{ dx } \)
∴ (1) becomes,
\(1-\frac { dz }{ dx } =1+{ e }^{ z }\)
\(\Rightarrow -\frac { dz }{ dx } ={ e }^{ z }\)
\(\Rightarrow -\frac { dz }{ { e }^{ z } } =dx\)
\(-\int { { e }^{ -z }dz } =\int { dx } \)
\(\Rightarrow \frac { { e }^{ -z } }{ -1 } =x+c\)
\(\Rightarrow { e }^{ y-x }=x+c\)

∴p v (q Λ r) is a contingency

Let I = \(\int _{ 1 }^{ 2 }{ \frac { 3x }{ { 9x }^{ 2 }-1 } dx } \) ⇒ IA³| = \(\left| I \right| \)
Put t = 9x² - 1 ⇒ dt = 18x dx
\(\frac{d t}{6}=3 x d x\)

∴ \(\int _{ 8 }^{ 35 }{ \frac { dt }{ 6t } } \)
= \(\frac { 1 }{ 6 } { \left[ log \ t \right] }_{ 8 }^{ 35 }\)
= \(\frac { 1 }{ 6 } [log35-log8]\)
= \(\frac { 1 }{ 6 } \left[ log\left( \frac { 35 }{ 8 } \right) \right] \)

Let y = f(x) = \(\sqrt x\)
Let x_o = 25, dx = 25.2 - 25 = 0.2
y = \(\sqrt x\)
dy = \(\frac{1}{2\sqrt{x}}\) dx
dy = \(\frac{1}{2\sqrt{x}}\) (0.2) = 0.02
∴\(\sqrt{25.2}\) = f(x₀) + f'(x₀) dx
= \(\sqrt{25}\) + 0.02
= 5 + 0.02 = 5.02

Y = x² is continuous on [-2, 2] and differentiable on [-2, 2]
f(a) = f(-2) = (-2)² = 4
f(b) = f(2) = 2² = 4
∴ f(a) = f(b)
Since the tangent is parallel to X - axis, f'(c) = 0
⇒ 2c = 0
⇒ c = 0
∴ When c = 0, y = 0
∴ AE (0, 0) the tangent is parallel to X- axis.

(cos θ + i sin θ )² = cos 2θ + isin 2θ
[By De moivre's theorem]
⇒ cos 2θ + isin 2θ = x + iy
Equating the real and imaginary parts we get,
x = cos 2θ, y = sin 2θ
∴ x² + y² = cos²2θ + sin²2θ = 1
Hence proved

Since the foci are one the x-axis, the equation of the hyperbola is \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
Given conjugate axis = \(\frac34\) (transverse axis)
⇒ 2b = \(\frac34\) (2a) ⇒ b = \(\frac{3a}4\) ⇒ b² = \(\frac { { 9a }^{ 2 } }{ 16 } \)
∴ e = \(\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1-\frac { 9{ a }^{ 2 } }{ { 16a }^{ 2 } } } =\sqrt { 1+\frac { 9 }{ 16 } } =\sqrt { \frac { 25 }{ 16 } } =\frac { 5 }{ 4 } \)
∴ e = \(\frac { 5 }{ 4 } \)

Given circle is x² + y² - 8x - 9y + 12 = 0
Length of the tangent = \(\sqrt { { 2 }^{ 2 }+({ -3) }^{ 2 }-8(2)-9(-3)+12 } \)
= \(\sqrt { 4+9-16+27+12 } \)
= \(\sqrt { 36 } \) = 6 unit

Let \({ cos }^{ -1 }\left( \frac { 3 }{ 5 } \right) =\theta \Rightarrow \frac { 3 }{ 5 } =cos\theta \)
\(\therefore sin\theta =\sqrt { 1-{ cos }^{ 2 }\theta } =\sqrt { 1-\frac { 9 }{ 25 } } =\sqrt { \frac { 25-9 }{ 25 } } \)
= \(\sqrt { \frac { 16 }{ 25 } } =\frac { 4 }{ 5 } \)
\(\therefore sin\left( { cos }^{ -1 }\left( \frac { 3 }{ 5 } \right) \right) =\frac { 4 }{ 5 } \)

A =\(\left[ \begin{matrix} 4 \\ 7 \end{matrix}\begin{matrix} 5 \\ -3 \end{matrix}\begin{matrix} -6 \\ 0 \end{matrix}\begin{matrix} 1 \\ 8 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 4 \\ -1 \end{matrix}\begin{matrix} 5 \\ -3 \end{matrix}\begin{matrix} -6 \\ 12 \end{matrix}\begin{matrix} 1 \\ 6 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} -1 \\ 4 \end{matrix}\begin{matrix} -3 \\ 5 \end{matrix}\begin{matrix} 12 \\ -6 \end{matrix}\begin{matrix} 6 \\ 1 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }+(-1){ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 4 \end{matrix}\begin{matrix} 13 \\ 5 \end{matrix}\begin{matrix} -12 \\ -6 \end{matrix}\begin{matrix} 6 \\ 1 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-4{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \end{matrix}\begin{matrix} 13 \\ -47 \end{matrix}\begin{matrix} -12 \\ 42 \end{matrix}\begin{matrix} -6 \\ 25 \end{matrix} \right] \)
The equivalent row-echelon matrix hats two non zero rows.
∴ \(\rho\) (A) = 2

Let ∝, β are the roots of the equation
Sum of the roots \(\alpha +\beta =\frac { -b }{ a } \)
 \(=\frac { [-(a-2)] }{ 1 } =a-2\)
Product of the roots \(=\alpha \beta =\frac { c }{ a } \)
\(=\frac { -(a+1) }{ 1 } =-(a+1)\)
we have \({ \alpha }^{ 2 }{ \beta }^{ 2 }=({ \alpha +\beta ) }^{ 2 }-2\alpha \beta \)
\(={ (a-2) }^{ 2 }+2(a+1)\)
\(={ a }^{ 2 }-4a+4+2a+2\)
\(=(a-1{ ) }^{ 2 }+5\)
Thus \(\\ { \alpha }^{ 2 }+{ \beta }^{ 2 }\) is least if a = 1

The planes \(\overset { \rightarrow }{ r } .\overset { \rightarrow }{ { n }_{ 1 } } ={ d }_{ 1 }\) and \(\overset { \rightarrow }{ r } .\overset { \rightarrow }{ { n }_{ 2 } } ={ d }_{ 2 }\) are perpendicular if \(\overset { \rightarrow }{ { n }_{ 1 } } .\overset { \rightarrow }{ { n }_{ 2 } } =0\)
Here \(\overset { \rightarrow }{ { n }_{ 1 } } =\overset { \wedge }{ i } +2\overset { \wedge }{ j } +3\overset { \wedge }{ k } \) and \(\overset { \rightarrow }{ { n }_{ 2 } } =\lambda \overset { \wedge }{ i } +2\overset { \wedge }{ j } -7\overset { \wedge }{ k } \)
\(\therefore \overset { \rightarrow }{ { n }_{ 1 } } .\overset { \rightarrow }{ { n }_{ 2 } } =\left( \overset { \wedge }{ i } +2\overset { \wedge }{ j } +3\overset { \wedge }{ k } \right) .\left( \lambda \overset { \wedge }{ i } +2\overset { \wedge }{ j } -7\overset { \wedge }{ k } \right) =0\)
⇒ λ + 4 - 21 = 0
⇒ λ - 17 = 0
⇒ λ = 17