33x – 150

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

concave downward everywhere

\({ e }^{ 2x }=1+\frac { 2x }{ 1! } +\frac { \left( 2x \right) ^{ 2 } }{ 2! } +\frac { \left( 3x \right) ^{ 3 } }{ 3! } +...\)

\(\theta =\frac { \pi }{ 2 } ,\frac { 2\pi }{ 2 } \varepsilon \left( 0,2\pi \right) \)

x = 4

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 \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ log(tan \ x)dx } \)  ....(1)
Applying the property \(\int _{ 0 }^{ a }{ f(x) } dx=\int _{ 0 }^{ a }{ f(a-x)dx } \)
we get
I = \(\int _{ 0 }^{ \pi /2 }{ log(tan(\frac { \pi }{ 2 } -x))dx } \)
= \(\int _{ 0 }^{ \pi /2 }{ log(cot \ x) } dx\) ......(2)
\((1)+(2)\longrightarrow 2I\int _{ 0 }^{ \pi /2 }{ log(tan \ x)+log(cot \ x)dx } \)
\( =\int_{0}^{\pi / 2} \log \tan x \cdot \cot x d x \)
\(\int _{ 0 }^{ \pi /2 }{ log1dx } =0\)
⇒ I = 0 Hence proved

Given u(x, y) = x² + 3xy + y²
u(x_o, y_o) = u(2,1)
= 2² + 3(2)(1) + 1²
= 4 + 6 + 1 = 11
\(\frac { \partial u }{ \partial x } \) = 2x+ 3y
\({ \left( \frac { \partial u }{ \partial x } \right) }_{ (2,1) }\)= 2 + 3 = 5
\(\frac { \partial u }{ \partial y } \) = 3x+ 2y
\({ \left( \frac { \partial u }{ \partial y } \right) }_{ (2,1) }\) = 6 + 2 = 8
Linear approximation 
L(x,y) = U(x_o, y_o) + \({ \left( \frac { \partial u }{ \partial x } \right) }_{ ({ x }_{ 0 },{ y }_{ 0 }) }\) (x - x_o) + \({ \left( \frac { \partial u }{ \partial y} \right) }_{ ({ x }_{ 0 }{ ,y }_{ 0 }) }\)(y - y_o)
L (x,y) = 11 + 5 (x - 2) + 8 (y - 1)
= 11 + 5x - 10 + 8y - 8
L(x,y) = 5x + 8y - 7

Given f(x, y) = 2x³ - 11x²y + 3y³
f(tx, ty) = 2t³ x3 - 11 t² x²ty + 3t³y³
= t³(2x³ - 11x²y + 3y³)
= t³. f(x,y)
∴ f (x, y) is a homogeneous function of degree 3.
∴ By Euler's theorem,
\(x\frac { \partial f }{ \partial x } +y\frac { \partial f }{ \partial y } =3f\)

Let r be the radius and A be the area of the plate
Given \(\frac { \triangle r }{ r } \times 100=2\)
when r = 10
\(\frac { \triangle r }{ r } \times 100=2\)
⇒ \(\triangle r=\frac { 2r }{ 100 } =\frac { 2\times 10 }{ 100 } =\frac { 2 }{ 10 } \)
\(\therefore dr=\frac { 2 }{ 10 } \)
A= ㅠr²
dA = 2πr(dr) = 2π(10)\(\left( \frac { 2 }{ 10 } \right) \)
= 4πcm²

f(x) -|x+3| =  ∀ \(x\in R\).
Now, |x+3| ≥ 0  ∀ \(x\in R\).
⇒ f(x) ≥ 0  ∀ \(x\in R\).
So, the minimum value of f(x) is 0
Also, f(x) = |x + 3| does not have the maximum value.

Z = \(\left( \frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } \right) ^{ 107 }+\left( \frac { \sqrt { 3 } }{ 2 } -\frac { i }{ 2 } \right) ^{ 107 }\)
\(\bar { z } =\left( \overline { \frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } } \right) ^{ 107 }+\left( \overline { \frac { \sqrt { 3 } }{ 2 } -\frac { i }{ 2 } } \right) ^{ 107 }\)
= \(\left( \frac { \sqrt { 3 } }{ 2 } -\frac { i }{ 2 } \right) ^{ 107 }+\left( \frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } \right) ^{ 107 }\) = z
Since z = \(\bar { z } \), Im(z) = 0

(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

Let z₁ = a+ib and z₂ = c+id
Given |z₁| = Iz₂|
⇒ \(\sqrt { { a }^{ 2 }+{ b }^{ 2 } } =\sqrt { { c }^{ 2 }+{ d }^{ 2 } } \)
Squaring both sides we get, a² + b² = c² + d²
This cannot imply that a = c and b = d
∴ z₁ and z₂ need not be equal

Let P(x, y) be the movable point.
By focal property of ellipse, PA + PB = 2a
∴ 2a = 10 ⇒ a = 5
Since focus is (4, 0), ae = 4 ⇒ 5e = 4 ⇒ e = \(\frac45\)
Also b² = a²(1 - e²) = 25\(\left( 1-\frac { 16 }{ 25 } \right) =25\left( \frac { 9 }{ 25 } \right) \) = 9
Equation of ellipse is \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
\(\frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 9 } =1\)

Let AOB be the vertical section of the reflector and m is the mid-point of AB. Let the equation of the parabola be y² = 4ax A(6, 12) lies on (1)

∴ 12² = 4a(6) ⇒ a = 6
∴ Focus is (a, 0) = (b, 0)
Hence focus coincides with m, the mid-point of AB.

Let \({ cos }^{ -1 }\left( \frac { 4 }{ 5 } \right) =\theta \Rightarrow cos\theta =\frac { 4 }{ 5 } \)
\(\therefore sin\left( \frac { 1 }{ 2 } { cos }^{ -1 }\left( \frac { 4 }{ 5 } \right) \right) =sin\left( \frac { 1 }{ 2 } \theta \right) \)
= \(sin\frac { \theta }{ 2 } =\sqrt { \frac { 1-cos\theta }{ 2 } } \)
= \(\sqrt { \frac { 1-\frac { 4 }{ 5 } }{ 2 } } =\sqrt { \frac { 5-4 }{ 5(2) } } =\sqrt { \frac { 1 }{ 10 } } \)

L.H.S = \({ tan }^{ -1 }\left( \frac { 1 }{ 7 } \right) +{ tan }^{ -1 }\left( \frac { 1 }{ 13 } \right) \)
= \({ tan }^{ -1 }\left( \cfrac { \frac { 1 }{ 7 } +\frac { 1 }{ 3 } }{ 1-\left( \frac { 1 }{ 7 } \right) \left( \frac { 1 }{ 13 } \right) } \right) ={ tan }^{ -1 }\left( \cfrac { \frac { 13+7 }{ 91 } }{ \frac { 91-1 }{ 91 } } \right) \)

\(=\tan ^{-1}\left(\frac{\frac{20}{91}}{\frac {90}{91}}\right)=\tan ^{-1}\left(\frac{20}{\not 91} \times \frac{\not 91}{90}\right) \)
\(=\tan ^{-1}\left(\frac{\not 20^{2}}{\not 90^{9}}\right)=\tan ^{-1}\left(\frac{2}{9}\right)=\mathrm{RHS} \)
Hence proved.

Given
\({ cot }^{ -1 }\left( \frac { 1 }{ 7 } \right) =0\Rightarrow \theta =\frac { 1 }{ 7 } \)
\(\Rightarrow tan\theta =7\)
\(\Rightarrow sec\theta =\sqrt { 1+{ tan }^{ 2 }\theta } =\sqrt { 1+{ 7 }^{ 2 } } =\sqrt { 50 } =5\sqrt { 2 } \)
\(\Rightarrow cos\theta ={ \frac { 1 }{ 5\sqrt { 2 } } }\)

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

The parametric form of vector equation of a line passing through a point \(\left( \overset { \rightarrow }{ a } \right) \) and parallel to \(\overset { \rightarrow }{ b } \) is
\({ \overset { \rightarrow }{ r } }=\overset { \rightarrow }{ a } +t\overset { \rightarrow }{ b } \), t ∈ R
⇒ Here \(\overset { \rightarrow }{ a } =2\overset { \wedge }{ i } -\overset { \wedge }{ j } +3\overset { \wedge }{ k } \) and \(\overset { \rightarrow }{ b } =2\overset { \wedge }{ i } +\overset { \wedge }{ j } -2\overset { \wedge }{ k } \)
\(\therefore { \overset { \rightarrow }{ r } }=\left( 2\overset { \wedge }{ i } -\overset { \wedge }{ j } +3\overset { \wedge }{ k } \right) +t\left( 2\overset { \wedge }{ i } +\overset { \wedge }{ j } -2\overset { \wedge }{ k } \right) \), t ∈ R which is the required equation of a line.

\(\overset { \rightarrow }{ F } =\frac { 6\left( \overset { \wedge }{ 2i } -2\overset { \wedge }{ j } +\overset { \wedge }{ k } \right) }{ \sqrt { 4+4+1 } } =\frac { 6 }{ 3 } \left( \overset { \wedge }{ 2i } -2\overset { \wedge }{ j } +\overset { \wedge }{ k } \right) =\overset { \wedge }{ 4i } -4\overset { \wedge }{ j } +2\overset { \wedge }{ k } \)
\(\overset { \rightarrow }{ d } \) = (5, 3, 7) - (1, 2, 3) = (4, 1, 4) =\(\overset { \wedge }{ 4i } +\overset { \wedge }{ j } +4\overset { \wedge }{ k } \)
∴ Work done (w)
= \(\overset { \rightarrow }{ F } .\overset { \rightarrow }{ d } =\left( \overset { \wedge }{ 4i } -4\overset { \wedge }{ j } +2\overset { \wedge }{ k } \right) .\left( \overset { \wedge }{ 4i } +\overset { \wedge }{ j } +4\overset { \wedge }{ k } \right) \)
= 16 - 4 + 8 = 20 units

Sum of the roots = sin ∝ + cos ∝ = \(\frac{-b}{a}\)
Product of the roots = sin ∝ cos ∝ = \(\frac{c}{a}\)
Now 1 = cos²∝ + sin² ∝
= (sin ∝ +cos ∝)² - 2 sin ∝ cos ∝
\(1=\frac { { b }^{ 2 } }{ { a }^{ 2 } } -\frac { 2c }{ a } \Rightarrow 1=\frac { { b }^{ 2 }-2ac }{ { a }^{ 2 } } \)
⇒ a² = b² - 2ac ⇒ a² + 2ac = b²
Adding c² both sides, a² +2ac+c² = b²+c²
⇒ (a+c)² = b² + c²