We have, f (x) is continuous in \([\frac{1}{2},2 ]\) and differentiable in \((\frac{1}{2},2 )\) with  \(f(\frac{1}{2})=\frac{5}{2}=f(2) \). 
By  the Rolle’s theorem there must exist a \(c \in (\frac{1}{2},2 )\) such that, \(f'(c)=1-\frac{1}{c^{2}}=0 \Rightarrow c^{2}=1 \) gives \(\Rightarrow c=\pm1, \) As \(1\in(\frac{1}{2},2)\) we choose c = 1.

\(sin\left( { tan }^{ -1 }\left( \frac { 1 }{ 2 } \right) -{ cos }^{ -1 }\left( \frac { 4 }{ 5 } \right) \right) \)

Let \({ tan }^{ -1 }\left( \frac { 1 }{ 2 } \right) =x\Rightarrow \frac { 1 }{ 2 } =tanx\)
\(cosx=\frac { adj }{ hyp } =\frac { 2 }{ \sqrt { 5 } } \)
\(\therefore sinx= \frac { 1 }{ \sqrt { 5 } } \)

Let \({ cos }^{ -1 }\left( \frac { 4 }{ 5 } \right) =y\Rightarrow \frac { 4 }{ 5 } =cosy\)
\(siny=\frac { opp }{ hyp } =\frac { 3 }{ 5 } \)
\(\therefore sin\left( { tan }^{ -1 }\left( \frac { 1 }{ 2 } \right) \right) -{ cos }^{ -1 }\left( \frac { 4 }{ 5 } \right) \)
= sin (x - y) = sin x cos y - cos x sin y
= \(\frac { 1 }{ \sqrt { 5 } } .\frac { 4 }{ 5 } -\frac { 2 }{ \sqrt { 5 } } .\frac { 3 }{ 5 } \)
= \(\frac { 4 }{ 5\sqrt { 5 } } =\frac { 6 }{ 5\sqrt { 5 } } =\frac { -2 }{ 5\sqrt { 5 } }\)

|z-4|²-|z-1|² = 16
|x+iy-4|² - |x+iy-1|² = 16
⇒ |(x-4)+iy|² - |(x-1)+iy2|² = 16
⇒ [(x-4)²+y²] - [(x-1)²+y²] = 16
⇒ x²-8x+16+y²-[x²-2x+1+y²] = 16
\(\Rightarrow \not x^{2}-8 x+16+\not y^{2}-\not x^{2}+2 x-1-\not y^{2}=16\)
⇒ -6x+15-16 = 0
⇒ -6x-1 = 0
⇒ 6x+1 = 0 Which is the required Cartesian equation.
The locus of the point is a straight line.

2x-y = 8, 3x+2y+2 = -2
The matrix form of the system is
\(\left[ \begin{matrix} 2 & -1 \\ 3 & 2 \end{matrix} \right] \left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 8 \\ -2 \end{matrix} \right] \)
⇒ AX = B where A =\(\\ \left[ \begin{matrix} 2 & -1 \\ 3 & 2 \end{matrix} \right] \)
B =\(\left[ \begin{matrix} 8 \\ -2 \end{matrix} \right] \)
⇒ X = A^-1N
Now, |A| =\(\left[ \begin{matrix} 2 & -1 \\ 3 & 2 \end{matrix} \right] \)= 4 + 3 = 7
∴ A^-1= \(\frac { 1 }{ |A| } \)adj A
= \(\frac { 1 }{ 7 } \left[ \begin{matrix} 2 & 1 \\ -3 & 2 \end{matrix} \right] \)
∴ X = A^-1B = \(\frac { 1 }{ 7 } \left[ \begin{matrix} 2 & 1 \\ -3 & 2 \end{matrix} \right] \left[ \begin{matrix} 8 \\ -2 \end{matrix} \right] \)
= \(\frac { 1 }{ 7 } \left[ \begin{matrix} 16-2 \\ -24-4 \end{matrix} \right] \)
= \(\frac { 1 }{ 7 } \left[ \begin{matrix} 14 \\ -28 \end{matrix} \right] =\left[ \begin{matrix} \frac { 14 }{ 7 } \\ \frac { -28 }{ 7 } \end{matrix} \right] =\left[ \begin{matrix} 2 \\ -4 \end{matrix} \right] \)
∴ x = 2, y = -4
Hence, the solution set is {2, -4}

Let the given point be \(\vec { a } =5\hat { i } +2\hat { j } +8\hat { k } \)
Given lines are \(\vec { r } =(\hat { i } +\hat { j } -\hat { k } )+s(2\hat { i } -2\hat { j } +\hat { k } )\)
⇒ \(\vec { b } =2\hat { i } -2\hat { j } +\hat { k } \)
and \(\vec { r } =(2\hat { i } -\hat { j } -3\hat { k } )+t(\hat { i } +2\hat { j } +2\hat { k } )\)
⇒ \(\vec { d } =\hat { i } +2\hat { j } +2\hat { k } \)
Since the required line is perpendicular to both \(\vec { b } \) and \(\vec { d } \), it will be parallel to \(\vec { b } \times \vec { d } \)
\(\vec { b } \times \vec { d } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & -2 & 1 \\ 1 & 2 & 2 \end{matrix} \right| \)
= \(\hat { i } (-4-2)-\hat { j } (4-1)+\hat { k } (4+2)\)
= \(-6\hat { i } -3\hat { j } +6\hat { k } \)
= \(-3(2\hat { i } +\hat { j } -2\hat { k } )\)
∴ Equation of required straight line is
\(\vec { r } =\vec { a } +m(\vec { b } \times \vec { d } ),m\in R\)
\(\vec { r } =(5\hat { i } +2\hat { j } +8\hat { k } )-3m(2\hat { i } +\hat { j } -2\hat { k } )\)
\(\vec { r } =(5\hat { i } +2\hat { j } +8\hat { k } )+t(2\hat { i } +\hat { j } -2\hat { k } )\)
where t =\(-3m\in R\)
Cartesian equation of a straight line passing through (5, 2, 8) and is 丄 to the straight lines
\(\vec { r } =(\vec { i } +\vec { j } -\vec { k } )+s(2\vec { i } -2\vec { j } +\vec { k } )\)
\(\vec { r } =(2\vec { i } -\vec { j } -3\vec { k } )+t(\vec { i } +2\vec { j } +2\vec { k } )\)
\(\frac { x-{ x }_{ 1 } }{ b_{ 1 } } =\frac { y-{ y }_{ 1 } }{ { b }_{ 2 } } =\frac { z-{ z }_{ 1 } }{ { b }_{ 3 } } \)
\(\frac { x-5 }{ 2 } =\frac { y-2 }{ 1 } =\frac { z-8 }{ -2 } \).

Cartesian equation of straight line passing through two points (2, 1,4) and (a - 1,4, -1) is
\(\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { z-{ z }_{ 1 } }{ { z }_{ 2 }-{ z }_{ 1 } } \)
[∵ (x₁, y₁, z₁) = (2, 1, 4) (x₂, y₂, z₂) = (a, -1, 4, -1)]
⇒ \(\frac { x-2 }{ a-1-2 } =\frac { y-1 }{ 4-1 } =\frac { z-4 }{ -1-4 } \)
⇒ \(\frac { x-2 }{ a-3 } =\frac { y-1 }{ 3 } =\frac { z-4 }{ -5 } \)
Similarly, Cartesian equation of straight liens passing through two points (0, 2, b - 1) and (5, 3, -2) is
\(\frac { x-5 }{ 5-0 } =\frac { y-3 }{ 2-3 } =\frac { z+3 }{ b-1+2 } \)
[∵ (x₁,y₁,z₁)=(5,3,-2) (x₂,y₂,z₂) is (0,2,b)...(1)
⇒ \(\frac { x-5 }{ -5 } =\frac { y-3 }{ -1 } =\frac { z+2 }{ b+1 } \)
(1) and (2) are parallel if their direction cosines are equal
∴ Direction ratios ofline (1) are a - 3, 3, -5   ........... (3)
Direction ratios of line (2) are -5, -1, b + 1  ............ (4)
To make the direction ratios equal, multiply (4) by-3.
∴ (4) ⟶ +15, 3, -3b-3
(3) ⟶ a-3, 3, -5
∴ a-3 = +5 ⇒ a = 15+3 = 18
3 = 3
-3b-3 = -15 ⇒ -3b = -5+3 = -2
⇒ b = \(\frac { -2 }{ -3 } =\frac { 2 }{ 3 } \)
∴ a = 18 and b = \(\frac { 2 }{ 3 } \).

Let the vertex be (0, 0) .
The equation of the parabola is
y² = 4ax
(1) Since the diameter is 40cm and the depth is 30 cm , the point (30, 20) lies on the parabola.
20² = 4a × 30
4a = \(\frac { 400 }{ 30 } \) = \(\frac { 40 }{ 3 } \)
Equation is y² = \(\frac { 40 }{ 3 } \)x.
(2) The bulb is at focus (0, a).
Hence the bulb is at a distance of \(\frac { 10 }{ 3 } \)cm from the vertex.

 Passing through (5, -2) length of the transverse axis is a long x-axis and of length 8 units.
2a = 8 ⇒ a = 4
Since the transverse axis is along x-axis, centre is (0, 0)
Equation of the hyperbola is
\(\frac { { x }^{ 2 } }{ 16 } -\frac { { y }^{ 2 } }{ b^{ 2 } } =1\)
Since (5, -2)passes through the parabola,
\(\frac { 25 }{ 16 } -\frac { 4 }{ { b }^{ 2 } } \Rightarrow \frac { 4 }{ { b }^{ 2 } } =\frac { 25 }{ 16 } -1=\frac { 25-16 }{ 16 } =\frac { 9 }{ 16 } \)
∴ \({ b }^{ 2 }=\frac { 16\times 4 }{ 9 } =\frac { 64 }{ 9 } \)
∴ Equation of the hyperbola is
\(\frac { { x }^{ 2 } }{ 16 } -\frac { { y }^{ 2 } }{ \frac { 64 }{ 9 } } =1\Rightarrow \frac { { x }^{ 2 } }{ 16 } -\frac { 9{ y }^{ 2 } }{ 64 } =1\)

2cos²x - 9cosx + 4 = 0 ............ (1)
The left hand side of this equation is not a polynomial in x. But it looks like a polynomial. In fact, we can say that this is a polynomial in cos x. However, we can solve the equation (1) by using our knowledge on polynomial equations. If we replace cos x by y, then we get the polynomial equation 2y²- 9y + 4 = 0 for which 4 and \(\frac{1}{2}\) are solutions.
From this we conclude that x must satisfy cos x = 4 or cos x = \(\frac{1}{2}\).
But cos x = 4 is never possible, if we take cos x = \(\frac{1}{2}\), then we get infinitely many real numbers x satisfying cos x = \(\frac{1}{2}\); in fact, for all n\(\in \)Z, x = 2nπ ±\(\frac { \pi }{ 3 } \) are solutions for the given equation (1).
If we repeat the steps by taking the equation cos²x - 9 cosx + 20 = 0, we observe that this equation has no solution.

Given A =\(\left[ \begin{matrix} 1 & -1 \\ 2 & 0 \end{matrix} \right] \), B =\(\left[ \begin{matrix} 3 & -2 \\ 1 & 1 \end{matrix} \right] \) and C =\(\left[ \begin{matrix} 1 & 1 \\ 2 & 2 \end{matrix} \right] \)
Also,  A X B = C
Premultiply by A^-1 we get,
(A^-1 A) X B = A^-1 C
⇒ XB = A^-1C    [∵ A^-1a = 1]
Post Multiply by B^-1 we get
(X B) B^-1 = (A^-1 C) B^-1
⇒ X = (A^-1 C) B^-1
|A| = \(\left| \begin{matrix} 1 & -1 \\ 2 & 0 \end{matrix} \right| \) = 0 + 2 = 2 ≠ 0
∴ A^-1 = \(\frac { 1 }{ |A| } adjA=\frac { 1 }{ 2 } \left[ \begin{matrix} 0 & 1 \\ -2 & 1 \end{matrix} \right] \)
|B| = \(\left| \begin{matrix} 3 & -2 \\ 1 & 1 \end{matrix} \right| \) = 3 + 2 = 5 ≠ 0
∴ B^-1 = \(\frac { 1 }{ |B| } adjB=\frac { 1 }{ 5 } \left[ \begin{matrix} 1 & 2 \\ -1 & 3 \end{matrix} \right] \)
A^-1C = \(\\ \frac { 1 }{ 2 } \left[ \begin{matrix} 0 & 1 \\ -2 & 1 \end{matrix} \right] \left[ \begin{matrix} 1 & 1 \\ 2 & 2 \end{matrix} \right] \)
= \(\frac { 1 }{ 2 } \left[ \begin{matrix} 0+2 & 0+2 \\ -2+2 & -2+2 \end{matrix} \right] \)
=\(\frac { 1 }{ 2 } \left[ \begin{matrix} 2 & 2 \\ 0 & 0 \end{matrix} \right] =\frac { 1 }{ 2 } (2)\left[ \begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix} \right] \)
∴ X = Z(A^-1C).B^-1
= \(\left[ \begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix} \right] \frac { 1 }{ 5 } \left[ \begin{matrix} 1 & 2 \\ -1 & 3 \end{matrix} \right] =\frac { 1 }{ 5 } \left[ \begin{matrix} 1-1 & 2+3 \\ 0+0 & 0+0 \end{matrix} \right] \)
= \(\frac { 1 }{ 5 } \left[ \begin{matrix} 0 & 5 \\ 0 & 0 \end{matrix} \right] =\frac { 1 }{ 5 } (5)\left[ \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right] =\frac { 1 }{ 10 }\left[ \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right] \)
∴ X = \(\frac { 1 }{ 10 } \left[ \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right] \).

We know that the domain of sin⁻¹(x) is [-1, 1].
This leads to −1\(\le\)2 - 3x²\(\le\)1, Which implies -3\(\le\) -3x²\(\le\)-1
Now, -3\(\le\) -3x², gives x²\(\le\)1 and ............(1)
-3\(\le\)-3x²\(\le\)-1, gives x²\(\ge\)\(\frac{1}{2}\) .........(2)
Combining the equations (1) and (2), we get \(\frac{1}{3}\le x^2\le 1\). That is \(\frac{1}{\sqrt3}\le |x|\le1\), Which gives \(x\in[-1,-1\frac{1}{\sqrt3}]\cup[\frac{1}{\sqrt3},1]\)
since  a\(\le|x|\le b\) implies x \(\in[-b,-a]\cup[a,b]\).

\(log|x|=cos\left( \frac { y }{ x } \right) ,x\neq 0\)

c(x-y)2=|x|e-y/x,x≠0

y=log(x+1)+log(x+1)-x+5

\(\frac { ab\left( \pi -2 \right) }{ 4 } \)

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

2 sq.units

Let P_o be the initial population and the population after t year be P.
Given \(\frac { dp }{ dt } =\frac { 2p }{ 100 } \Rightarrow \frac { dp }{ dt } =\frac { p }{ 50 } \)
⇒ \(\frac { dp }{ p } =\frac { dt }{ 50 } \Rightarrow \int { \frac { dp }{ p } } =\int { \frac { dt }{ 50 } } \)
⇒ log p =\(\frac { t }{ 50 } \) + c ...(1)
when t = 0, p = 0
⇒ log p₀ = 0+c ⇒ log p₀  ....(1)
∴ (1) becomes, log p =\(\frac { t }{ 50 } \)+log P₀.
⇒ log\(\left( \frac { P }{ { p }_{ 0 } } \right) =\frac { t }{ 50 } \)
⇒ t = 50 log\(\left( \frac { P }{ { p }_{ 0 } } \right) \)
when p = 2p₀, t = 50 log\(\left( \frac { 2P_{ 0 } }{ { p }_{ 0 } } \right) \) = 50 log 2
= 50(0.3) = 15 years.
Hence the population doubles in 15 years.

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

Let r be the radius of the base, h be the height of the cone, I be the slant height and θ be the semi vertical angle.
In ΔOAB l² = h²+r²
⇒ r² =l²-h²
⇒ V =\(\frac { 1 }{ 3 } { \pi }r^{ 2 }h=\frac { \pi }{ 3 } \)(l²-h²)h
=\(\frac { \pi }{ 3 } \) (l²h - h³)
\(\frac { dv }{ dh } =\frac { \pi }{ 3 } \)(l²-3h²)

\(\frac { dv }{ dh } \)=0
⇒ \(\frac { \pi }{ 3 } \)(l² - 3h²) = 0
⇒ l² - 3h² = 0 ⇒ \(\frac { { l }^{ 2 } }{ { h }^{ 2 } } \) =3
⇒ \(\frac { l }{ h } \) = \(\sqrt { 3 } \)  ⇒ h=\(\frac { 1 }{ \sqrt { 3 } } \)
Now, \(\frac { d^{ 2 }V }{ dh^{ 2 } } =\frac { \pi }{ 3 } \)(-6h) = -2πh
at h =\(\frac { l }{ \sqrt { 3 } } ,\frac { d^{ 2 }V }{ dh^{ 2 } } =-2\pi \left( \frac { l }{ \sqrt { 3 } } \right) \) < 0
V is max at h =\(\frac { l }{ \sqrt { 3 } } \)
∴ r² =l²-\(\frac { { l }^{ 2 } }{ 3 } =\frac { 2{ l }^{ 2 } }{ 3 } \) =2h²
\(\\ \left[ \because h=\frac { l }{ \sqrt { 3 } } \Rightarrow h^{ 2 }=\frac { { l }^{ 2 } }{ 3 } \right] \)
∴ \(\frac { { r }^{ 2 } }{ { h }^{ 2 } } \) =2
⇒ \(\frac { r }{ h } =\sqrt { 2 } \)
⇒ tanθ =\(\sqrt { 2 } \)
⇒ θ = tan^-1(\(\sqrt { 2 } \))

\(tan\left[ \frac { 1 }{ 2 } { sin }^{ -1 }\left( \frac { 2a }{ 1+{ a }^{ 2 } } \right) +\frac { 1 }{ 2 } { cos }^{ -1 }\left( \frac { 1-{ a }^{ 2 } }{ 1+{ a }^{ 2 } } \right) \right] \)
Let a = tan 0
Now, \(tan\left[ \frac { 1 }{ 2 } { sin }^{ -1 }\left( \frac { 2a }{ 1+{ a }^{ 2 } } \right) +\frac { 1 }{ 2 } { cos }^{ -1 }\left( \frac { 1-{ a }^{ 2 } }{ 1+{ a }^{ 2 } } \right) \right] =tan\left[ \frac { 1 }{ 2 } { sin }^{ -1 }\left( \frac { 2tan\theta }{ 1+{ tan }^{ 2 }\theta } \right) +\frac { 1 }{ 2 } { cos }^{ -1 }\left( \frac { 1-{ tan }^{ 2 }\theta }{ 1+{ tan }^{ 2 }\theta } \right) \right] \)
\(tan\left[ \frac { 1 }{ 2 } { sin }^{ -1 }(sin2\theta )+\frac { 1 }{ 2 } { cos }^{ 1 }(cos2\theta ) \right] =tan[2\theta ]=\frac { 2tan\theta }{ 1-{ tan }^{ 2 }\theta } =\frac { 2a }{ 1-{ a }^{ 2 } } \)

Comparing the given equation \(\vec { r } =(\hat { i } -4\hat { j } +3\hat { k } )+t(2\hat { i } +3\hat { j } +\hat { k } )\) with \(\vec { r } =\hat { a } +t\hat { b } \),
we get \(\vec { a } =\hat { i } -4\hat { j } +3\hat { k } \) and \(\vec { b } =2\hat { i } +3\hat { j } +\hat { k } \). We denote the given point (-1, 2, 3) by D and the point (1, -4, 3) on the straight line by F.
If F is the foot of the perpendicular from to the straight line, then F is of the form (2t + 1, 3t - 4, t+3)  and \(\vec { DF } =\vec { OF } -\vec { OD } =(2t+2)\hat { i } +(3t-6)\hat { j } +t\hat { k } \)

Since \(\vec { b } \) is perpendicular to \(\vec { DF } \), we have
\(\vec { b } .\vec { DF } \) = 0 ⇒ 2(2t + 2)+ 3(3t - 6) + 1(t) ⇒ t= 1
Therefore, the coordinate of F is (3, -1, 4)
Now, the perpendicular distance from the given point to the given line is
\(DF=\left| \vec { DF } \right| =\sqrt { { 4 }^{ 2 }+(-{ 3) }^{ 2 }+{ 1 }^{ 2 } } =\sqrt { 26 } \)

Given cos α + cos β + cos \(\gamma\) = sin α + sin β + sin \(\gamma\)
∴ (cos α + cos β + cos \(\gamma\)) + i(sin α + sin β + sin \(\gamma\)) = 0
⇒ (cos α + i sin α) + (cos β + i sin β) + (cos \(\gamma\)+i sin \(\gamma\)) = 0
⇒ a + b + c = 0 where a = cos α + i sin α, b = cos β + i sin β, c = cos \(\gamma\) + i sin\(\gamma\)
If a + b + c = 0, then a³+b³+c³ = 3abc
∴ (cos α + i sin α)³ + (cos β + i sin β)³ + (cos \(\gamma\) + i sin \(\gamma\))³ = 3[ (cos α + i sin α) + (cos β + i sin β) + (cos \(\gamma\) + i sin \(\gamma\))
= 3[(cos(α + β + \(\gamma\)) + i sin(α + β + \(\gamma\))]
⇒ (cos 3α + cos β + cos \(\gamma\)) + i[sin 3α + sin 3β + sin 3\(\gamma\))]
= 3(cos(α + β + \(\gamma\)) + i sin(α + β + \(\gamma\))
Equating the real and imaginary parts, we get 
\( \cos 3 \alpha+\cos 3 \beta+\cos 3 \gamma=3 \cos (\alpha+\beta+\gamma) \)
\( \sin 3 \alpha+\sin 3 \beta+\sin 3 \gamma=3 \sin (\alpha+\beta+\gamma) \)

\(\left[ \begin{matrix} 2 & -1 \\ 5 & -2 \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 2 & -1 \\ 5 & -2 \end{matrix} \right] \)
Applying Gauss - Jordan method, we get
[A|I₂] =\(\left[ \begin{matrix} 2 & -1 \\ 5 & -2 \end{matrix}|\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\rightarrow { R }_{ 1 }\div 2 }{ \longrightarrow } \left[ \begin{matrix} 1 & -\frac { 1 }{ 2 } \\ 5 & -2 \end{matrix}|\begin{matrix} \frac { 1 }{ 2 } & 0 \\ 0 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-5{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -\frac { 1 }{ 2 } \\ 5 & -2 \end{matrix}|\begin{matrix} \frac { 1 }{ 2 } & 0 \\ -\frac { 5 }{ 2 } & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }\times 2 }{ \longrightarrow } \left[ \begin{matrix} 1 & -\frac { 1 }{ 2 } \\ 0 & 1 \end{matrix}|\begin{matrix} \frac { 1 }{ 2 } & 0 \\ -5 & 2 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\rightarrow { R }_{ 1 }+\frac { 1 }{ 2 } { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}|\begin{matrix} -2 & 1 \\ -5 & 2 \end{matrix} \right] \)
∴ We get A^-1=\(\left[ \begin{matrix} -2 & 1 \\ -5 & 2 \end{matrix} \right] \)

order – 2 : degree 2

\(\frac { { a }^{ 2 } }{ 6 } \)

\(\frac { 1 }{ 5 } \left( { e }^{ 5 }-2 \right) \)

f(x) is strictly decreasing on [0, 1]

f(x) is strictly increasing \(\left[ -\frac { 1 }{ 2 } ,\frac { 1 }{ 2 } \right] \)

(i) π
(ii) πX2^n-1
(iii) 2
(iv) 0

\(\left( \frac { 9 }{ 8 } ,\frac { 9 }{ 2 } \right) \)

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

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

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

f(x) = x³+ 2x²-1
f'(x) = 3x² + 4x = 0
⇒ x (3x + 4) = 0
⇒ x = 0 or \(\frac { 4 }{ 3 } \)
The possible intervals are \(\left( -\infty ,-\frac { 4 }{ 3 } \right) \left( -\frac { 4 }{ 3 } ,0 \right) \) and (0, ∞).

The equation of the required plane through the intersection of the given planes is
( x + y + z - 6 ) + λ (2x + 3y + 4z + 5) = 0 ......(1)
This passes through (1, 1, 1)
∴ ( 1+ 1 + z - 6) + λ (2 + 3+ 4 + 5) = 0
⇒ -3 +14λ = 0 \(\Rightarrow \lambda =\frac { 3 }{ 14 } \)
Substituting \(\lambda =\frac { 3 }{ 14 } \) in (1) we get
( x + y + z - 6 )+\(\frac { 3 }{ 14 } \) (2x + 3y + 4z + 5) = 0
⇒ 14( x + y + z - 6 ) +3 (2 + 3+ 4 + 5) = 0
⇒ 20x + 23y + 26z - 69 = 0