c(x-y)2=|x|e-y/x,x≠0

3πa²

\(\frac { { 4a }^{ 2 } }{ 3 } sq.units\)

\(\frac { 9\sqrt { 3 } { a }^{ 2 } }{ 45 } \)

\( \frac { \partial w }{ \partial u } =\frac { 2u }{ \sqrt { 1-\left( { u }^{ 2 }-{ v }^{ 2 } \right) } } ;\frac { \partial w }{ \partial v } =\frac { -2v }{ \sqrt { 1-\left( { u }^{ 2 }-{ v }^{ 2 } \right) } } \)

local min value = \(\frac { 27 }{ 256 } \)
No local maximum

(−∞,−2)decreasing
(−2,−1)increasing
(−1,∞)decreasing
Local maximum value = 6
Local minimum value =5

100 m / s, t = 4 sec, 200 m / s, −100 m / s

Given \(\frac { dy }{ dx } \) =(3x+2y+1)²
Let z =3x+2y+1
\(\frac { dz }{ dx } =3+2\frac { dy }{ dx } \)
⇒ \(\frac { dz }{ dx } -3=2\frac { dy }{ dx } \)
⇒ \(\frac { dy }{ dx } =\frac { 1 }{ 2 } \left( \frac { dz }{ dx } -3 \right) \)
(1) becomes,
\(\frac { 1 }{ 2 } \left( \frac { dz }{ dx } -3 \right) \) =z²
⇒ \(\frac { dz }{ dx } \)-3 =2z²
⇒ \(\frac { dz }{ dx } \)=2x²+3
⇒ \(\frac { dz }{ 2{ z }^{ 2 }+3 } \) =dx
⇒ \(\frac { 1 }{ 2 } \int { \frac { dz }{ { z }^{ 2 }+\frac { 3 }{ 2 } } } =\int { dx } \)
⇒ \(\frac { 1 }{ 2 } \int { \frac { dz }{ { z }^{ 2 }+\left( \sqrt { \frac { 3 }{ 2 } } \right) ^{ 2 } } } \) =x+c
⇒ \(\frac { 1 }{ 2 } .\frac { 1 }{ \frac { \sqrt { 3 } }{ \sqrt { 2 } } } tan^{ -1 }\left( \frac { 1 }{ \frac { \sqrt { 3 } }{ \sqrt { 2 } } } \right) \) =x+c
⇒ \(\frac { 1 }{ \sqrt { 6 } } tan^{ -1 }\left( \frac { \sqrt { 2 } }{ \sqrt { 3 } } \right) \)= x+c
⇒ \(tan^{ -1 }\left( \frac { \sqrt { 2 } }{ \sqrt { 3 } } \right) =\sqrt { 6 } \) x+c
⇒ \(\frac { \sqrt { 2 } }{ \sqrt { 3 } } .z=tan(\sqrt { 6 } x+c)\)
⇒ \(\frac { \sqrt { 2 } }{ \sqrt { 3 } } (3x+2y+1)=tan(\sqrt { 6 } x+c)\).

Since the entries in the last column are F, (p ∧ ~p) ∧ (~q ∧ P) is a contradiction

Given sides are y = 2x + 1.....(1)
y = 3x + 1...(2)
x = 4...(3)
Solving (1) & (2), x = 0, y = 1
Solving (2) & (3), x = 4, y = 13
Solving (1) & (3), x = 4, y = 9

∴ Required area \(\int _{ 0 }^{ 4 }{ (3x+1) } dx-\int _{ 0 }^{ 4 }{ (2x+1) } dx\)
\({ =\left( \frac { { 3x }^{ 2 } }{ 2 } +x \right) }_{ 0 }^{ 4 }-{ \left( \frac { { 2x }^{ 2 } }{ 2 } +x \right) }_{ 0 }^{ 4 }\)
\(=\left( \frac { 48 }{ 2 } +4 \right) -(16+4)=28-20\)
Area = 8 sq. units

Let f(x) = cos x, x = 60° dx = 1°
f(x_o) = cos 60° = \(\frac12\) = 0.5
f'(x) = - sinx dx
f'(x_o) = - sin x_o dx
f'(60) = - sin 60° (1°)
= - (0.86603) (0.01745)
= - 0.0154
∴ f(x) = f(x_o) +f(x_o) dx 
f(61) = 0.5 - 0.0154
∴ tan 46° = f(x_o) +f(x_o) dx 
cos 61° = 0.4849

Given curves are 4x = y² ....(1)
⇒ 4xy = k....(2)
Substituting (1) in (2) we get
y³ = k  ⇒ y = \(k^{ \frac { 1 }{ 3 } }\)
∴ (1) becomes, 4x = \(\left( { k }^{ \frac { 1 }{ 3 } } \right) ^{ 2 }=k^{ \frac { 2 }{ 3 } }\)
⇒ x = \(\frac { { k }^{ \frac { 2 }{ 3 } } }{ 4 } \)
∴ The point of intersection of the given curves is \(\left( \frac { k^{ \frac { 2 }{ 3 } } }{ 4 } ,{ k }^{ \frac { 1 }{ 3 } } \right) \)
Differentiating 4x = y²
4 = 2y\(\frac { dy }{ dx } \)
⇒ \(\frac { dy }{ dx } =\frac { 2 }{ y } \)
m₁ = \(\frac { 2 }{ { k }^{ \frac { 1 }{ 3 } } } \)
Differentiating 4xy = k,
4x\(\frac { dy }{ dx } \)+4y = 0
⇒  \(\frac { dy }{ dx } =\frac { -y }{ x } \)
∴ m₂ = \(\frac { { -k }^{ \frac { 1 }{ 3 } } }{ { k }^{ \frac { 2 }{ 3 } } } \)(4)
Since the given curves cut at right angles, m₁m₂ = -1.
∴ \(\left( \frac { 2 }{ { k }^{ \frac { 1 }{ 3 } } } \right) \left( \frac { -4.k^{ \frac { 1 }{ 3 } } }{ k^{ \frac { 2 }{ 3 } } } \right) \) = 1
⇒ \(\frac { 8 }{ { k }^{ \frac { 2 }{ 3 } } } \) = 1 ⇒ \({ k }^{ \frac { 2 }{ 3 } }\) = 8
⇒ \(\left( { k }^{ \frac { 2 }{ 3 } } \right) ^{ 3 }\) = 8³
⇒ k² = 512.

arg(1+i) + arg(1-i) = arg[(1+i) (1-i)]
LHS = arg (1+i) + arg(1-i)
1+i = \(\sqrt { 2 } \left( \frac { 1 }{ \sqrt { 2 } } +\frac { i }{ \sqrt { 2 } } \right) \)
= \(\sqrt { 2 } \left( cos\frac { \pi }{ 4 } +isin\frac { \pi }{ 4 } \right) \)
∴ arg (1+i) = π/4
-1+i =\(\sqrt { 2 } \left( \frac { -1 }{ \sqrt { 2 } } +\frac { i }{ \sqrt { 2 } } \right) \)
= \(\sqrt { 2 } \left( cos3\frac { \pi }{ 4 } +isin3\frac { \pi }{ 4 } \right) \)
∴ (-1+i) = 3\(\frac { \pi }{ 4 } \)
∴ LHS = \(\frac { \pi }{ 4 } +\frac { 3\pi }{ 4 } =\frac { 4\pi }{ 4 } =\pi \)
RHS = arg[(1+i) (-1+i)]
= arg[-1-i + i + i²]
= (-1-i + i-1) = arg(-2)
= arg(2) - (1) = 2 arg(-1)
= 2 (cos π + isin π) = π
∴ LHS = RHS

Given F₁P + F₂P = 8
By the focal property of ellipse
F₁P + F₂P = 2a
∴ 2a = 8 ⇒ a = 4
and distance between the foci = F₁F₂ = 6
2ae = 6 ⇒ ae = 3
∴ 4(e) = 3 ⇒ e \(\frac34\)

∴ b² = a²(1- e²)
= \(16\left( { 1-\left( \frac { 3 }{ 4 } \right) }^{ 2 } \right) =16\left( 1-\frac { 9 }{ 10 } \right) =16\left( \frac { 7 }{ 16 } \right) =7\)
∴ The path traced by him is an ellipse and its equation is \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
⇒ \(\frac { { x }^{ 2 } }{ 16 } -\frac { { y }^{ 2 } }{ 7 } \) = 1

Given \({ tan }^{ -1 }\left( \frac { \sqrt { 1+{ x }^{ 2 } } -\sqrt { 1-{ x }^{ 2 } } }{ \sqrt { 1+{ x }^{ 2 } } +\sqrt { 1-{ x }^{ 2 } } } \right) =a\)
\(\Rightarrow \frac { \left( \sqrt { 1+{ x }^{ 2 } } -\sqrt { 1-{ x }^{ 2 } } \right) \left( \sqrt { 1+{ x }^{ 2 } } +\sqrt { 1-{ x }^{ 2 } } \right) }{ \left( \sqrt { 1+{ x }^{ 3 } } -\sqrt { 1-{ x }^{ 2 } } \right) \left( \sqrt { 1+{ x }^{ 2 } } +\sqrt { 1-{ x }^{ 2 } } \right) } \)
= \(\frac { tan\alpha +1 }{ tan\alpha -1 } \)
\(\Rightarrow \frac { 2\sqrt { 1+{ x }^{ 2 } } }{ -2\sqrt { 1-{ x }^{ 2 } } } =\frac { tan\alpha +1 }{ tan\alpha -1 } \)
\(\Rightarrow \sqrt { \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } } =\frac { 1-tan\alpha }{ 1+tan\alpha } \)
\(\Rightarrow \sqrt { \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } } =\frac { cos\alpha -sin\alpha }{ cos\alpha +sin\alpha } \)
\(\Rightarrow \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \left( \frac { cos\alpha -sin\alpha }{ cos\alpha +sin\alpha } \right) ^{ 2 }\)
\(\Rightarrow \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } =\frac { 1-sin2\alpha }{ 1+sin2\alpha } \Rightarrow { x }^{ 2 }=sin2\alpha \)

Given \(\overset { \rightarrow }{ AB } =\overset { \rightarrow }{ \alpha } \), \(\overset { \rightarrow }{ AD } =\overset { \rightarrow }{ \beta } \) and \(\overset { \rightarrow }{ AC } =2\overset { \rightarrow }{ \alpha } +3\overset { \rightarrow }{ \beta } \)
Area of the quadrilateral ABCD 
∴ = are of ∆ ABC + area of ∆ ACD
\(=\frac { 1 }{ 2 } \left| \overset { \rightarrow }{ AB } \times \overset { \rightarrow }{ AC } \right| +\frac { 1 }{ 2 } \left| \overset { \rightarrow }{ AC } \times \overset { \rightarrow }{ AD } \right| \)
\(=\frac { 1 }{ 2 } \left| \overset { \rightarrow }{ \alpha } \times \left( 2\overset { \rightarrow }{ \alpha } +3\overset { \rightarrow }{ \beta } \right) \right| +\frac { 1 }{ 2 } \left| \left( 2\overset { \rightarrow }{ \alpha } +3\overset { \rightarrow }{ \beta } \right) \times \overset { \rightarrow }{ \beta } \right| \)
\(=\frac { 1 }{ 2 } \left| 2\left( \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \alpha } \right) +3\left( \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \beta } \right) \right| +\frac { 1 }{ 2 } \left| 2\left( \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \beta } \right) +3\left( \overset { \rightarrow }{ \beta } \times \overset { \rightarrow }{ \beta } \right) \right| \)
\(=\frac { 1 }{ 2 } \left| 3\left( \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \beta } \right) \right| +\frac { 1 }{ 2 } \left| 2\left( \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \beta } \right) \right| \quad \quad \quad \left[ \because \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \alpha } =\overset { \rightarrow }{ \beta } \times \overset { \rightarrow }{ \beta } =0 \right] \)
\(=\left( \frac { 3 }{ 2 } +\frac { 2 }{ 2 } \right) \left( \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \beta } \right) =\left( \frac { 5 }{ 2 } \right) \left| \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \beta } \right| \quad \quad (1)\)
Now, Area of the parallelogram with \(\overset { \rightarrow }{ AB } \) and \(\overset { \rightarrow }{ AD } \) as
adjacent sides = \(\left| \overset { \rightarrow }{ AB } \times \overset { \rightarrow }{ AD } \right| =\left| \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \beta } \right| .... (2)\)
From (1) & (2), \(\frac { 5 }{ 2 } \left| \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \beta } \right| =\lambda \left| \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \beta } \right| \) [Given]
\(\lambda =\frac { 5 }{ 2 } \)

Put \(\frac { 1 }{ x } \) = a, \(\frac { 1 }{ y } \) = b, \(\frac { 1 }{ z } \) = c
∴ 2a + 3b + 10c = 4  ....(1)
4a- 6b - 5c = 1 .....(2)
6a + 9b -20c = 2 ...(3)
Δ = \(\left| \begin{matrix} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{matrix} \right| \)
= \(2\left| \begin{matrix} -6 & 5 \\ 9 & -20 \end{matrix} \right| -3\left| \begin{matrix} 4 & 5 \\ 6 & -20 \end{matrix} \right| +10\left| \begin{matrix} 4 & -6 \\ 6 & 9 \end{matrix} \right| \)
= 2 (120 - 45) -3 (-80 - 30) + 10 (36 + 36)
= 150 + 330 + 720 = 1200
Δ₁ = \(\left| \begin{matrix} 4 & 3 & 10 \\ 1 & -6 & 5 \\ 2 & 9 & -20 \end{matrix} \right| \)
= \(4\left| \begin{matrix} -6 & 5 \\ 9 & -20 \end{matrix} \right| -3\left| \begin{matrix} 1 & 5 \\ 2 & -20 \end{matrix} \right| +10\left| \begin{matrix} 1 & -6 \\ 2 & 9 \end{matrix} \right| \)
= 4 (120 - 45) -3 (-20 - 10) + 10 (9 + 12)
= 300 + 90 + 120 = 600
Δ₂ = \(\left| \begin{matrix} 2 & 4 & 10 \\ 4 & 1 & 5 \\ 6 & 2 & -20 \end{matrix} \right| =2\left| \begin{matrix} 1 & 5 \\ 2 & -20 \end{matrix} \right| -4\left| \begin{matrix} 4 & 5 \\ 6 & -20 \end{matrix} \right| +10\left| \begin{matrix} 4 & 1 \\ 6 & 2 \end{matrix} \right| \)
= 2 (-2 - 10) - 4 (-80 - 30) + 10 (8 - 6)
= -60 + 440 + 20 = 400
Δ₃ = \(\left| \begin{matrix} 2 & 3 & 4 \\ 4 & -6 & 1 \\ 6 & 9 & 2 \end{matrix} \right| 2\left| \begin{matrix} -6 & 1 \\ 9 & 2 \end{matrix} \right| -3\left| \begin{matrix} 4 & 1 \\ 6 & 2 \end{matrix} \right| +10\left| \begin{matrix} 4 & -6 \\ 6 & 9 \end{matrix} \right| \)
= 2 (-12 - 9) -3 (8 - 6) + 4 (36 + 36)
= - 42 - 6 + 288 = 240
∴ a = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { 600 }{ 1200 } =\frac { 1 }{ 2 } \Rightarrow \frac { 1 }{ x } =\frac { 1 }{ 2 } \) ⇒ x = 2
∴ b = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { 400 }{ 1200 } =\frac { 1 }{ 3 } \Rightarrow \frac { 1 }{ y } =\frac { 1 }{ 3 } \) ⇒ y = 1 
∴ c = \(\frac { { \triangle }_{ 3 } }{ \triangle } =\frac { 240 }{ 1200 } =\frac { 1 }{ 5 } \Rightarrow \frac { 1 }{ z } =\frac { 1 }{ 5 } \) ⇒ z = 5
∴ Solution set is {2, 3, 5}

Given equation is (a²+b²+c²) p²-2
(ab+bc+cd) p+(b²+c²+d²) ≤ 0...(1)
(1) can be rewritten as
(a²p² - 2abp + b²) + (b²p² - 2bcp + c²) + (c²p2 - 2cdp + d2) ≤ 0
Since a, b, c, d, p∈ R
(ap-b)² ≥ 0, (bp-c)² ≥ 0 and (cp-d)² ≥ 0
∴ (2) will be satisfied only if
ap - b = 0, bp - c = 0, cp - d = 0
\(\Rightarrow \frac { b }{ a } =\frac { c }{ b } =\frac { d }{ c } =p\)
⇒ a, b, c, d are in G.P and ad = bc