NDA Mathematics 17 April 2016 Official Paper (Previous Year Paper)

Concept:

• Three cube root of unity are 1,$;{\rm{;}} - \frac{1}{2} + \frac{{i\sqrt 3 }}{2}$,${\rm{;;}} - \frac{1}{2} - \frac{{i\sqrt 3 }}{2}$
• The sum of three cube roots of unity is zero. ⇔  $1 + \omega + {\omega ^2} = 0$
• Square of any imaginary cube root of unity is equal to the other imaginary cube root of unity.

i.e., if ${\rm{\omega }} = - \frac{1}{2} + \frac{{i\sqrt 3 }}{2};{\rm{then;}}{{\rm{\omega }}^2} = - \frac{1}{2} - \frac{{i\sqrt 3 }}{2}$ 

• Slope of the line if two points are given:

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Let AB is the line such that ${\rm{A}} = ({{\rm{x}}_1},{{\rm{y}}_1}){\rm{;and;B}} = \left( {{{\rm{x}}_2},{\rm{;}}{{\rm{y}}_2}} \right)$

Then slope of line AB

$\Rightarrow \left( {\bf{m}} \right) = ;\frac{{{{\bf{y}}_2} - {{\bf{y}}_1}}}{{{{\bf{x}}_2} - {{\bf{x}}_1}}}$

• Angle between the two lines if their slopes are known:

$\tan {\rm{θ }} = \left( {{\frac{{{{\rm{m}}_2} - {{\rm{m}}_1}}}{{1 + {{\rm{m}}_1}{{\rm{m}}_2}}}} } \right)$  

$\Rightarrow {\bf{θ }} = {\tan ^{ - 1}}\left( { {\frac{{{{\bf{m}}_2} - {{\bf{m}}_1}}}{{1 + {{\bf{m}}_1}{{\bf{m}}_2}}}} } \right)$

Where, ${\rm{;}}{{\rm{m}}_1}{\rm{;and;}}{{\rm{m}}_2}$ are slope of the lines.

Calculation:

We have, points P and Q on complex plane which are ${\rm{\omega ;and;}}{{\rm{\omega }}^2}$

We know ${\rm{\omega }} = - \frac{1}{2} + \frac{{i\sqrt 3 }}{2};{\rm{and;}}{{\rm{\omega }}^2} = - \frac{1}{2} - \frac{{i\sqrt 3 }}{2}$

For point P, ${\rm{\omega }} = - \frac{1}{2} + \frac{{i\sqrt 3 }}{2}$, Here real part is $- \frac{1}{2};;{\rm{and;imaginary;part;is;}}\frac{{\sqrt 3 }}{2}$

Similarly, for point Q, ${\rm{;}}{{\rm{\omega }}^2} = - \frac{1}{2} - \frac{{i\sqrt 3 }}{2}$, Here real part is $- \frac{1}{2};;{\rm{and;imaginary;part;is;}}\frac{{ - \sqrt 3 }}{2}$

On complex plane point P and Q will look something like (shown in image)

$∴ {\rm{P;}}\left( { - \frac{1}{2},;\frac{{\sqrt 3 }}{2}} \right){\rm{and;Q;}}\left( { - \frac{1}{2}, - ;\frac{{\sqrt 3 }}{2}} \right)$ And origin O (0, 0) 

Let’s find out slope of OP and slope of OQ:

Slope of OP (m₁) = $\frac{{\frac{{\sqrt 3 }}{2}; - ;0}}{{ - \frac{1}{2} - 0}} = - \sqrt 3$                                … ( ${\rm{slope;}}\left( {\bf{m}} \right) = ;\frac{{{{\bf{y}}_2} - {{\bf{y}}_1}}}{{{{\bf{x}}_2} - {{\bf{x}}_1}}}$ )

Slope of OP (m₂) = $\frac{{ - \frac{{\sqrt 3 }}{2}; - ;0}}{{ - \frac{1}{2} - 0}} = \sqrt 3$

Now we have slope of the lines, so we can estimate the angle between the lines OP and OQ

We have,

$\Rightarrow {\bf{θ }} = {\tan ^{ - 1}}\left( { {\frac{{{{\bf{m}}_2} - {{\bf{m}}_1}}}{{1 + {{\bf{m}}_1}{{\bf{m}}_2}}}} } \right)$

$\Rightarrow {\rm{θ }} = {\tan ^{ - 1}}\left( {{\frac{{\sqrt 3 - \left( { - \sqrt 3} \right)}}{{1 + \sqrt3\left( { - \sqrt3} \right)}}} } \right)$

$\Rightarrow {\rm{θ }} = {\tan ^{ - 1}}\left( {{\frac{{2\sqrt3}}{{1 - 3}}} } \right)$

$\Rightarrow {\rm{θ }} = {\rm{ta}}{{\rm{n}}^{ - 1}}\left( -{\sqrt3} \right)$

∴ θ = 120° 

Hence, option 1 is correct.

Additional points:

1. ${\bf{\omega }} \times {{\bf{\omega }}^2} = \left( { - \frac{1}{2} + \frac{{i\sqrt 3 }}{2}} \right) \times \left( { - \frac{1}{2} - \frac{{i\sqrt 3 }}{2}} \right)$ = 1.
2. Reciprocal of each imaginary cube roots of unity is the other.

$\Rightarrow \omega = \frac{1}{{{\omega ^2}}}$  And ${\omega ^2} = \frac{1}{\omega }$

Concept:

Let R be a binary relation on a set A.

1. Reflexive: Each element is related to itself.

• R is reflexive if for all x ∈ A, xRx.

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2. Symmetric: If any one element is related to any other element, then the second element is related to the first.

• R is symmetric if for all x, y ∈ A, if xRy, then yRx.

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3. Transitive: If any one element is related to a second and that second element is related to a third, then the first element is related to the third.

• R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

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Calculation:

Given, x, y and z are positive numbers i.e 1, 2, 3 and so on

1. Reflexive:

To show x ≤ x²

 x² – x ≥ 0

x ( x – 1 ) ≥ 0

So, for x ≥ 1 or x ≤ 0, the above relation is true. It is given that the x > 0 so the acceptable value will be,

x ≥ 1 

So, for all x ≥ 1, the given relation will be reflexive but for the value of xϵ (0, 1), it will not be reflexive.

So, the relation is not reflexive for all the positive values of x.

2. Symmetric:

To show when x ≤ y² leads to y ≤ x²

Considering x = 1 and y = 2

x² = 1 and y² = 4

Here x ≤ y² as 1 ≤ 4 but y ≤ x² is not true as y = 2 is greater than x² = 1

Thus relation * is not symmetric

3. Transitive:  To show when xRy or x ≤ y² and yRz or y ≤ z² leads to xRz or x ≤ z²

Considering x = 5, y = 3 and z = 2

⇒ x² = 25 , y² = 9 and z² = 4

Here x ≤ y² as 5 ≤ 9, y ≤ z² as 3 ≤ 4, but x ≤ z² is not true as  x = 5 is greater than z² = 4

Thus relation * is not transitive

Concept:

Consider a quadratic equation: ${\rm{a}}{{\rm{x}}^2} + {\rm{bx;}} + {\rm{;c}} = {\rm{;}}0$

Roots of quadratic equation are given by,

⇔ ${\rm{x}} = \frac{{ - {\rm{b}} \pm \sqrt {{{\rm{b}}^2} - 4{\rm{ac}}} }}{{2{\rm{a}}}}$ 

The term  b²- 4ac is called discriminant, it is important as it tells about the nature of roots.

 

A parabola graphically represents a quadratic equation with a vertex located at the origin, below the x-axis, or above the x-axis. Therefore, a quadratic function may have one, two, or zero roots.

Calculation:

Given: ${{\rm{x}}^2}{\rm{;}}-{\rm{;px;}} + {\rm{;}}4{\rm{;}} > {\rm{;}}0{\rm{;;}}\forall ;{\rm{real;values;of;x}}$ 

On comparing the given equation with the standard quadratic equation

${\rm{a}}{{\rm{x}}^2} + {\rm{bx;}} + {\rm{;c}} = {\rm{;}}0$, we get

a = 1, b = -p and c = 4

Given quadratic equation is more than zero that means it has no real roots If roots are not real (Then only the quadratic equation will not intersect the x-axis)

∴ ${{\rm{b}}^2} - 4{\rm{ac}} < 0$ 

$\Rightarrow {\left( { - {\rm{p}}} \right)^2} - 4\left( 1 \right)\left( 4 \right) < 0$

$\Rightarrow {\left( {\rm{p}} \right)^2} - 16 < 0$

$\Rightarrow {\left( {\rm{p}} \right)^2} - {4^2} < 0$

$\Rightarrow \left( {{\rm{p}} + 4} \right)\left( {{\rm{p}} - 4} \right) < 0$                     (Using a²- b² = (a + b). (a - b))

This inequality holds if p ranges from -4 to 4 i.e., p = (-4, 4)

Only |p| < 4 satisfy the above condition

Hence, option 1 is correct.

Important Points

If the quadratic function is always more than zero, that means, when graphed on the coordinate system, the function never intersects the X-axis. That means that the quadratic equation has no real roots.

Concept:

• Using Euler form of complex number: ${\rm{cos;\theta ;}} + {\rm{;i;sin;\theta ;}} = {\rm{;}}{{\rm{e}}^{{\rm{i\theta }}}}$
• For a complex number z = a + ib , the amplitude θ is given by:${\rm{\theta }} = {\tan ^{ - 1}}\frac{{\rm{b}}}{{\rm{a}}}$

Calculation:

${\rm{z}} = {\rm{x}} + {\rm{iy}} = {\left( {\frac{1}{{\sqrt 2 }} - \frac{{\rm{i}}}{{\sqrt 2 }}} \right)^{ - 25}} = {\left( {\cos \left( {\frac{{\rm{\pi }}}{4}} \right) - {\rm{i;sin}}\left( {\frac{{\rm{\pi }}}{4}} \right)} \right)^{ - 25}} = {\left( {\cos \left( { - \frac{{\rm{\pi }}}{4}} \right) + {\rm{i;sin}}\left( { - \frac{{\rm{\pi }}}{4}} \right)} \right)^{ - 25}}$

$\Rightarrow {\rm{z}} = {\rm{;}}{\left( {{{\rm{e}}^{ - {\rm{i}}\frac{{\rm{\pi }}}{4}}}} \right)^{ - 25}} = {{\rm{e}}^{{\rm{i}}\frac{{25{\rm{\pi }}}}{4}}} = \cos \left( {\frac{{25{\rm{\pi }}}}{4}} \right) + {\rm{i;sin}}\left( {\frac{{25{\rm{\pi }}}}{4}} \right) = \frac{1}{{\sqrt 2 }} + \frac{{\rm{i}}}{{\sqrt 2 }}$

Calculating for next part:

$\frac{{{\rm{z}} - \sqrt 2 }}{{{\rm{z}} - {\rm{i}}\sqrt 2 }} = \frac{{\frac{1}{{\sqrt 2 }} + \frac{{\rm{i}}}{{\sqrt 2 }} - \sqrt 2 }}{{\frac{1}{{\sqrt 2 }} + \frac{{\rm{i}}}{{\sqrt 2 }} - {\rm{i}}\sqrt 2 }} = {\rm{;}}\frac{{\frac{{\sqrt 2 }}{2} + \frac{{{\rm{i}}\sqrt 2 }}{2} - \sqrt 2 }}{{\frac{{\sqrt 2 }}{2} + \frac{{{\rm{i}}\sqrt 2 }}{2} - {\rm{i}}\sqrt 2 }} = \frac{{ - \frac{{\sqrt 2 }}{2} + \frac{{{\rm{i}}\sqrt 2 }}{2}}}{{\frac{{\sqrt 2 }}{2} - \frac{{{\rm{i}}\sqrt 2 }}{2}}} = \frac{{ - 1 + {\rm{i}}}}{{{\rm{;}}1 - {\rm{i;}}}}$

$\Rightarrow {\rm{z}} = \frac{{ - \left( {1 - {\rm{i}}} \right)}}{{{\rm{;}}1 - {\rm{i;}}}} = - 1 = - 1 + 0{\rm{i}}$

${\rm{Amplitude}}\left( {\rm{\theta }} \right) = {\tan ^{ - 1}}\frac{0}{{ - 1}} = {\rm{;}}{\tan ^{ - 1}}\frac{0}{{ - 1}} = {\rm{;}}{\tan ^{ - 1}}0 = {\rm{;}}0{\rm{;}},{\rm{;\pi }}$

Here x is negative so amplitude (θ) is equal to π.

Calculation:

Given: ${\rm{f}}\left( {{{\rm{x}}_1}} \right) - {\rm{f}}\left( {{{\rm{x}}_2}} \right) = {\rm{f}}\left( {\frac{{{{\rm{x}}_1} - {{\rm{x}}_2}}}{{1 - {{\rm{x}}_1}{{\rm{x}}_2}}}} \right)$ for x₁, x₂ ∈ (-1, 1)

Put x₁ = x₂ = 0

${\rm{f}}\left( 0 \right) - {\rm{f}}\left( 0 \right) = {\rm{f}}\left( {\frac{{0 - 0}}{{1 - 0}}} \right)$

⇒ 0 = f(0)

So, f(x) at x = 0 is zero.

For option (1), ${\rm{f}}\left( {\rm{x}} \right) = {\rm{;In;}}\left( {\frac{{1 - {\rm{x}}}}{{1 + {\rm{x}}}}} \right)$

f(0) = ln (1) = 0.

For option (2), ${\rm{f}}\left( {\rm{x}} \right) = {\rm{In;}}\left( {\frac{{2 + {\rm{x}}}}{{1 - {\rm{x}}}}} \right)$

f(0) = ln (2)

For option, ${\rm{f}}\left( {\rm{x}} \right) = {\tan ^{ - 1}}\left( {\frac{{1 - {\rm{x}}}}{{1 + {\rm{x}}}}} \right)$

f(0) = tan^-1 (1) = π/4

For option, ${\rm{f}}\left( {\rm{x}} \right) = {\tan ^{ - 1}}\left( {\frac{{1 + {\rm{x}}}}{{1 - {\rm{x}}}}} \right)$

f(0) = tan^-1 (1) = π/4

Hence, option (1) is correct.

Concept:

• For a quadratic equation ax² + bx +c = 0 to have real roots: D = b² – 4ac ≥ 0

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Calculation:

${\rm{y}} = \frac{{{{\rm{x}}^2}}}{{1 + {{\rm{x}}^2}}}$, Where x ∈ R

On solving,

y + yx² = x²

⇒ (y - 1) x² + y = 0

⇒(y - 1) x² + y = 0

Also for the quadratic equation to exist y – 1 = 0 is not possible

Which is a quadratic equation in x , on comparing ax² + bx +c = 0 we get a = (y - 1) , b = 0 , c = y.

For x ∈ R considering D ≥ 0 so b² – 4ac ≥ 0

0 – 4 y ( y – 1 ) ≥ 0

⇒ y ( y - 1 ) ≤ 0

Which gives the solution y ∈ [0, 1)

Concept:

Let A(x₁, y₁) and B(x₂, y₂) be any two points on the X-Y plane. Suppose point C is the mid-point of the line segment AB, then the co-ordinates of point C is:

 

• ${\rm{Midpoint;}} = {\rm{;}}\left( {\frac{{{{\rm{x}}_1}{\rm{;}} + {\rm{;;}}{{\rm{x}}_2}}}{2},{\rm{;}}\frac{{{{\rm{y}}_1}{\rm{;}} + {\rm{;;}}{{\rm{y}}_2}}}{2}} \right)$

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Calculation:

Let the intersection on x axis (P) be (x, 0)

Let the intersection on y axis (Q) be (0, y)

Given mid-point of PQ: (3 , 5 )

Using midpoint formula:

$\frac{{0; + ;{\rm{x}}}}{2} = 3$

∴ x = 6

$\frac{{0; + ;4}}{2} = 5$

∴ y = 10

So, P (6, 0) and Q (0, 10)

Since the triangle is formed with sides as both the coordinate axes it is right angled at O.

PO = 6

QO = 10

${\rm{Area}} = \frac{1}{2} \times {\rm{PO}} \times {\rm{QO}} = \frac{{6: \times :10{\rm{;}}}}{2} = 30$

Concept:

To find distance between the straight line ax + by + c = 0 and point (x₁, y₁), we use

${\bf{d}} = \left| {\frac{{{\bf{a}}{{\bf{x}}_1} + {\bf{b}}{{\bf{y}}_1} + {\bf{c}}}}{{\sqrt {{{\bf{a}}^2} + {{\bf{b}}^2};} }}} \right|$

Calculation:

Given:

Equation of line, y = x - √2

Radius of circle = b units

Center of circle, (0, b)

To find: Value of b =?

Let’s find out the distance of a line from the center point (0, b),

We know that ${\bf{d}} = \left| {\frac{{{\bf{a}}{{\bf{x}}_1} + {\bf{b}}{{\bf{y}}_1} + {\bf{c}}}}{{\sqrt {{{\bf{a}}^2} + {{\bf{b}}^2};} }}} \right|$ 

Here line y = x - √2 can be written as, x - y - √2 = 0

On comparing the equation with ax + by + c = 0, we get

a = 1, b = -1, and c = -√2

Points are (x_1, y₁) = (0, b)

$\Rightarrow {\bf{d}} = \left| {\frac{{{\bf{a}}{{\bf{x}}_1} + {\bf{b}}{{\bf{y}}_1} + {\bf{c}}}}{{\sqrt {{{\bf{a}}^2} + {{\bf{b}}^2};} }}} \right|$

$\Rightarrow {\rm{d}} = \left| {\frac{{\left( 1 \right)\left( 0 \right) - \left( 1 \right)\left( {\rm{b}} \right) - \surd 2}}{{\sqrt {{1^2} + {1^2}{\rm{;}}} }}} \right|$

$\Rightarrow {\rm{d}} = \left| {\frac{{\left( {{\rm{-b}} -\sqrt 2 } \right)}}{{\sqrt {2{\rm{;}}} }}} \right|$

Here line touches the circle

Hence Distance = Radius

∴ d = b

$\Rightarrow {\rm{b}} = \frac{{\left( {{\rm{b}} + \sqrt 2 } \right)}}{{\sqrt {2{\rm{;}}} }}$

$\Rightarrow \sqrt {2{\rm{;}}} {\rm{b}} = {\rm{;b}} + \sqrt 2$

$\Rightarrow \sqrt {2{\rm{;}}} {\rm{b}} - {\rm{;b}} = \sqrt 2$

$\Rightarrow {\rm{b}}\left( {\sqrt {2{\rm{;}}} - {\rm{;}}1} \right) = \sqrt 2$

$\Rightarrow {\rm{b}} = \frac{{\surd 2}}{{\sqrt {2{\rm{;}}} - {\rm{;}}1}}$

$\Rightarrow {\rm{b}} = \frac{{\surd 2}}{{\sqrt {2{\rm{;}}} - {\rm{;}}1}} \times \frac{{\sqrt {2{\rm{;}}} + {\rm{;}}1}}{{\sqrt {2{\rm{;}}} + {\rm{;}}1}}$(\Rightarrow {\rm{b}} = ;2 + \surd 2\)

Hence, option 1 is correct

Concept:

• Using the domain of trigonometric identities: -1 ≤ sin2θ ≤ 1

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Calculation:

Given:

f(θ) = 4(sin² θ + cos⁴ θ)

⇒ f(θ) = 4(1 - cos² θ + cos⁴ θ)

Which is a quadratic equation in cos²θ

$\Rightarrow {\rm{f}}\left( {\rm{\theta }} \right) = 4\left[ {1 - {{\cos }^2}{\rm{\theta }}(1 - {{\cos }^2}{\rm{\theta }}} \right)] = 4\left( {1 - {{\sin }^2}{\rm{\theta }}{{\cos }^2}{\rm{\theta }}} \right){\rm{;}}$

$\Rightarrow {\rm{f}}\left( {\rm{\theta }} \right) = {\rm{;}}4\left( {1 - {{\left( {\sin {\rm{\theta }}\cos {\rm{\theta }}} \right)}^2}} \right) = 4 - {\left( {2\sin {\rm{\theta }}\cos {\rm{\theta }}} \right)^2} = 4 - {\rm{si}}{{\rm{n}}^2}2{\rm{\theta ;;}}$

We know the range of sin 2θ:

-1 ≤ sin2θ ≤ 1

⇒ 0 ≤ sin² 2θ ≤ 1

⇒ -1 ≤ -sin²2θ ≤ 0

⇒ 3 ≤ 4 -sin²2θ ≤ 4

⇒ 3 ≤ f(θ) ≤ 4

Maximum value = 4

Calculation:

Given:

f(θ) = 4(sin² θ + cos⁴ θ)

⇒ f(θ) = 4(1 - cos² θ + cos⁴ θ)

Which is a quadratic equation in cos²θ

$\Rightarrow {\rm{f}}\left( {\rm{\theta }} \right) = 4\left[ {1 - {{\cos }^2}{\rm{\theta }}(1 - {{\cos }^2}{\rm{\theta }}} \right)] = 4\left( {1 - {{\sin }^2}{\rm{\theta }}{{\cos }^2}{\rm{\theta }}} \right){\rm{;}}$

$\Rightarrow {\rm{f}}\left( {\rm{\theta }} \right) = {\rm{;}}4\left( {1 - {{\left( {\sin {\rm{\theta }}\cos {\rm{\theta }}} \right)}^2}} \right) = 4 - {\left( {2\sin {\rm{\theta }}\cos {\rm{\theta }}} \right)^2} = 4 - {\rm{si}}{{\rm{n}}^2}2{\rm{\theta ;;}}$

We know the range of sin 2θ:

-1 ≤ sin2θ ≤ 1

⇒ 0 ≤ sin² 2θ ≤ 1

⇒ -1 ≤ -sin²2θ ≤ 0

⇒ 3 ≤ 4 -sin²2θ ≤ 4

⇒ 3 ≤ f(θ) ≤ 4

Minimum value = 3

Concept:

1. ${\sin ^2}{\rm{\theta }} + {\cos ^2}{\rm{\theta }} = 1$
2. ${\rm{sin}}2{\rm{\theta }} = 2{\rm{sin\theta cos\theta }}$
3. Value of $\sin {\rm{\theta }}$ cannot be less than -1 and greater than 1.

i.e., -1 < ${\bf{sin}};{\bf{\theta }}$ < 1

Calculation:

Given: f (θ) = 4 (${\sin ^2}\theta + {\cos ^4}\theta$)

= $4\left( {{{\sin }^2}\theta + {{\cos }^2}{\rm{\theta }}(1 - {{\sin }^2}{\rm{\theta }}} \right)$)               $\because ({\sin ^2}{\rm{\theta }} + {\cos ^2}{\rm{\theta }} = 1;)$

= $4\left( {{{\sin }^2}\theta + {{\cos }^2}{\rm{\theta }} - {{\cos }^2}\theta {{\sin }^2}\theta } \right))$

= $4\left( {1 - {{\left( {{\rm{sin\theta ;cos\theta }}} \right)}^2}} \right)$                          $\because({\sin ^2}{\rm{\theta }} + {\cos ^2}{\rm{\theta }} = 1;)$

= $4\left( {1 - {{\left( {\frac{1}{2}{\rm{sin}}2{\rm{\theta }}} \right)}^2}} \right)$                            $\because \left( {{\rm{sin}}2{\rm{\theta }} = 2{\rm{sin\theta cos\theta }}} \right)$

= $4\left( {1 - \frac{1}{4}{{\sin }^2}2{\rm{\theta }}} \right)$

Now,

1. f (θ) = 2

$\Rightarrow 4\left( {1 - \frac{1}{4}{{\sin }^2}2{\rm{\theta }}} \right) = 2$  

$\Rightarrow \left( {1 - \frac{1}{4}{{\sin }^2}2{\rm{\theta }}} \right) = \frac{2}{4}$

$\Rightarrow 1 - \frac{1}{4}{\sin ^2}2{\rm{\theta }} = \frac{1}{2}$

$\Rightarrow \frac{1}{4}{\sin ^2}2{\rm{\theta }} = 1 - \frac{1}{2}$

$\Rightarrow \frac{1}{4}{\sin ^2}2{\rm{\theta }} = \frac{1}{2}$

$\Rightarrow {\sin ^2}2{\rm{\theta }} = 2$

$\Rightarrow {\rm{sin}}2{\rm{\theta }} = \pm \surd 2$

We know that value of ${\rm{sin\theta }}$ cannot be less than -1 and greater than 1.

Hence, f (θ) = 2 has no solution.

2. ${\rm{f;}}\left( {\rm{\theta }} \right){\rm{;}} = {\rm{;}}\frac{7}{2}$

$\Rightarrow 4\left( {1 - \frac{1}{4}{{\sin }^2}2{\rm{\theta }}} \right) = \frac{7}{2}$

$\Rightarrow \left( {1 - \frac{1}{4}{{\sin }^2}2{\rm{\theta }}} \right) = \frac{7}{8}$

$\Rightarrow - \frac{1}{4}{\sin ^2}2{\rm{\theta }} = \frac{7}{8} - 1$

$\Rightarrow \frac{1}{4}{\sin ^2}2{\rm{\theta }} = \frac{1}{8}$

$\Rightarrow {\sin ^2}2{\rm{\theta }} = \frac{1}{2}$

$\Rightarrow \sin 2{\rm{\theta }} = \pm \sqrt {\frac{1}{2}}$

$\sin 45^\circ = \frac{1}{{\surd 2}} \Rightarrow \pm \sin 45^\circ = \pm \frac{1}{{\surd 2}}$

$\therefore \sin 2{\rm{\theta }} = \pm {\rm{sin}}45^\circ$

$\Rightarrow 2\theta = \pm 45^\circ$

$\Rightarrow {\rm{\theta }} = \pm 22.5^\circ$

Hence, ${\rm{;f;}}\left( {\rm{\theta }} \right){\rm{}} = {\rm{}}\frac{7}{2}$ has a solution.

From above analysis we can say that both the statements are correct.

Hence, option 3 is correct.

Concept:

Modulus:

If f(x) = |x| then f(x) is defined as,

\({\rm{f}}\left( {\rm{x}} \right) = \left{ {\begin{array}{*{20}{c}} {x,;x \ge 0}\ { - x,;x < 0} \end{array}} \right.\)

 

Calculation:

Given:

\({\rm{f}}\left( {\rm{x}} \right) = {\rm{x}}\left| {\rm{x}} \right| - 1{\rm{;and;g}}\left( {\rm{x}} \right) = \left{ {\begin{array}{*{20}{c}} {\frac{{3{\rm{x}}}}{2},{\rm{;x}} > 0}\ {2{\rm{x}},{\rm{;x}} \le 0} \end{array}} \right.\)

To find: Intersection point of f(x) & g(x)

\({\rm{f}}\left( {\rm{x}} \right) = \left{ {\begin{array}{{20}{c}} {{\rm{x}}\left( {\rm{x}} \right) - 1,{\rm{;x}} \ge 0}\ {{\rm{x}}\left( { - {\rm{x}}} \right) - 1,{\rm{;x}} < 0} \end{array}} \right. = \left{ {\begin{array}{{20}{c}} {{{\rm{x}}^2} - 1,{\rm{;x}} \ge 0}\ { - {{\rm{x}}^2} - 1,{\rm{;x}} < 0} \end{array}} \right.\)

For x > 0,

f(x) = g(x)

⇒ x² – 1 = 3x/2

⇒ 2x² – 3x – 2 = 0

⇒ 2x² – 4x + x – 2 = 0

⇒ 2x (x - 2) + 1 (x – 2) = 0

⇒ (2x + 1) (x – 2) = 0

⇒ x = 2 or x = -1/2

For x > 0.

∴ x = 2

At x = 2, f (x) = g(x) = 3

Therefore Intersection point is (2, 3)

For x < 0,

F(x) = g(x)

⇒ -x² – 1 = 2x

⇒ x² + 2x + 1 = 0

⇒ (x + 1)² = 0

⇒ x = -1

At x = -1, f(x) = g(x) = -2

Therefore Intersection point is (-1, -2)

Hence the intersection points are (2, 3) & (-1, -2).

Concept:

Area between Two Curves: Let curves are f(x) and g(x)

\({\rm{Area}} = \mathop \smallint \nolimits_{\rm{a}}^{\rm{b}} \left[ {{\rm{f}}\left( {\rm{x}} \right) - {\rm{g}}\left( {\rm{x}} \right)} \right]{\rm{dx}} = \mathop \smallint \nolimits_{\rm{a}}^{\rm{b}} \left[ {{\rm{Top}} - {\rm{bottom}}} \right]{\rm{dx}}\)

Calculation:

We already calculated intersection point: Intersection points are (2, 3) & (-1, -2).

Plotting the graph,

\({\rm{Area}} = {\rm{;}}\mathop \smallint \nolimits_{\rm{a}}^{\rm{b}} \left[ {{\rm{Top}} - {\rm{bottom}}} \right]{\rm{dx}}\)

\({\rm{Area}} = \mathop \smallint \nolimits_{ - 1}^2 \left[ {{\rm{g}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{x}} \right)} \right]{\rm{dx}}\)

\(= \left| {\mathop \smallint \limits_{ - 1}^0 \left{ {2{\rm{x}} - \left( { - {{\rm{x}}^2} - 1} \right)} \right}{\rm{dx}}} \right| + \left| {\mathop \smallint \limits_0^2 \left{ {\frac{{3{\rm{x}}}}{2} - \left( {{{\rm{x}}^2} - 1} \right)} \right}{\rm{dx}}} \right|\)

\(= \left| {\mathop \smallint \limits_{ - 1}^0 \left{ {2{\rm{x}} + {\rm{;}}{{\rm{x}}^2} + 1} \right}{\rm{dx}}} \right| + \left| {\mathop \smallint \limits_0^2 \left{ {\frac{{3{\rm{x}}}}{2} - {{\rm{x}}^2} + 1} \right}{\rm{dx}}} \right|\)

$= \left| {{{\rm{x}}^2} + {\rm{;}}\frac{{{{\rm{x}}^3}}}{3} + {\rm{x}}} \right|_{ - 1}^0 + \left| {\frac{{3{{\rm{x}}^2}}}{4} - \frac{{{{\rm{x}}^3}}}{3} + {\rm{x}}} \right|_0^2$

 

= 8/3 square units

Calculation:

Given: ${\rm{f}}\left( {\rm{x}} \right) = \frac{{27\left( {{{\rm{x}}^{\frac{2}{3}}} - {\rm{x}}} \right)}}{4}$ 

Let x^1/3 be t

f(x) = 1

${\rm{f}}\left( {\rm{x}} \right) = \frac{{27\left( {{{\rm{t}}^2} - {{\rm{t}}^3}} \right)}}{4} = 1$

$\Rightarrow {{\rm{t}}^2} - {{\rm{t}}^3} = \frac{4}{{27}}$

$\Rightarrow 27{{\rm{t}}^2} - 27{{\rm{t}}^3} = 4$

$\Rightarrow 27{{\rm{t}}^3} - 27{{\rm{t}}^2} + 4 = 0$

$\Rightarrow 27{{\rm{t}}^2}\left( {{\rm{t}} - \frac{2}{3}} \right) - 9{\rm{t}}\left( {{\rm{t}} - \frac{2}{3}} \right) - 6\left( {{\rm{t}} - \frac{2}{3}} \right) = 0$

$\Rightarrow 3\left( {9{{\rm{t}}^2} - 3{\rm{t}} - 2} \right)\left( {{\rm{t}} - \frac{2}{3}} \right) = 0$

$\Rightarrow \left( {3{\rm{t}} - 2} \right)\left( {3{\rm{t}} + 1} \right)\left( {{\rm{t}} - \frac{2}{3}} \right) = 0$

$\Rightarrow \left( {{\rm{t}} - \frac{2}{3}} \right)\left( {{\rm{t}} + \frac{1}{3}} \right)\left( {{\rm{t}} - \frac{2}{3}} \right) = 0$

${{\rm{x}}^{\frac{1}{3}}} = {\rm{t}} = - \frac{1}{3},\frac{2}{3}$

${{\rm{x}}^{\frac{1}{3}}} = \frac{2}{3}$

$\Rightarrow {\rm{x}} = {\left( {\frac{2}{3}} \right)^3} = \frac{8}{{27}}$

$\Rightarrow {\rm{x}} = {\left( {\frac{{ - 1}}{3}} \right)^3} = \frac{{ - 1}}{{27}}$

Therefore 2 solutions

Calculation:

f(x) = -1

Let x^1/3 be t

${\rm{f}}\left( {\rm{x}} \right) = \frac{{27\left( {{{\rm{x}}^{\frac{2}{3}}} - {\rm{x}}} \right)}}{4} = \frac{{27\left( {{{\rm{t}}^2} - {{\rm{t}}^3}} \right)}}{4} = - 1$

$\Rightarrow {{\rm{t}}^2} - {{\rm{t}}^3} = - \frac{4}{{27}}$

$\Rightarrow 27{{\rm{t}}^2} - 27{{\rm{t}}^3} = - 4$

$\Rightarrow {\rm{g}}\left( {\rm{t}} \right):27{{\rm{t}}^3} - 27{{\rm{t}}^2} - 4 = 0$

$\Rightarrow {\rm{g'}}\left( {\rm{t}} \right):81{{\rm{t}}^2} - 54{\rm{t}} = 27{\rm{t}}\left( {3{\rm{t}} - 2} \right) = 0$

Which gives:

${\rm{t;}} = {\rm{;}}0{\rm{;and;t}} = \frac{2}{3}$

${\rm{at;t}} = 0,{\rm{;g}}\left( {\rm{t}} \right) = - 4$

${\rm{at;t}} = \frac{2}{3},{\rm{;g}}\left( {\rm{t}} \right) = - 8$

Since at both values of t for which g’(t) = 0 ; g(t) is negative and coefficient of x³ > 0

So, g(t) = 0 will have only 1 solution which is positive

There t will have 1 positive solution.

Hence x^1/3 will have 1 solution.

Concept:

Greatest Integer Function: Greatest Integer Function [x] indicates an integral part of the real number x which is nearest and smaller integer to x. It is also known as floor of x

• In general, If n≤ x ≤ n+1 Then [x] = n     (n ∈ Integer)
• Means if x lies in [n, n+1) then the Greatest Integer Function of x will be n.

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Example:

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Calculation:

Given: ${\rm{g}}\left( {\rm{x}} \right) = \left[ {\frac{1}{{\rm{x}}}} \right]$

\({\rm{Let;I;}} = {\rm{;}}\mathop \smallint \limits_{\frac{1}{3}}^{\frac{1}{2}} {\rm{g}}\left( {\rm{x}} \right){\rm{dx}}\)

Here,

$\frac{1}{3} < {\rm{;x;}} < \frac{1}{2}$

$\Rightarrow 3 > \frac{1}{{\rm{x}}} > 2{\rm{;or;}}2 < \frac{1}{{\rm{x}}} < 3$

$\therefore \left[ {\frac{1}{{\rm{x}}}} \right] = 2$

Now,

\({\rm{;I;}} = {\rm{;}}\mathop \smallint \limits_{\frac{1}{3}}^{\frac{1}{2}} {\rm{g}}\left( {\rm{x}} \right){\rm{dx}} = {\rm{;}}\mathop \smallint \limits_{\frac{1}{3}}^{\frac{1}{2}} \left[ {\frac{1}{{\rm{x}}}} \right]{\rm{dx}} = {\rm{;}}\mathop \smallint \limits_{\frac{1}{3}}^{\frac{1}{2}} 2{\rm{dx}} = {\rm{;}}\left[ {2{\rm{x}}} \right]_{\frac{1}{3}}^{\frac{1}{2}} = 2\left[ {\frac{1}{2} - \frac{1}{3}} \right] = \frac{1}{3}\)

Calculation:

Given:

${\rm{f}}\left( {\rm{x}} \right) = {\rm{xg}}\left( {\rm{x}} \right){\rm{;and;g}}\left( {\rm{x}} \right) = \left[ {\frac{1}{{\rm{x}}}} \right]$

$\therefore {\rm{f}}\left( {\rm{x}} \right) = {\rm{x}}\left[ {\frac{1}{{\rm{x}}}} \right]$

\({\rm{Let;I}} = \mathop \smallint \limits_{\frac{1}{3}}^1 {\rm{f}}\left( {\rm{x}} \right){\rm{dx}}\)

\( = \mathop \smallint \limits_{\frac{1}{3}}^1 {\rm{x}}\left[ {\frac{1}{{\rm{x}}}} \right]{\rm{dx}}\)

\({\rm{g}}\left( {\rm{x}} \right) = \left[ {\frac{1}{{\rm{x}}}} \right] = {\rm{;}}\left{ {\begin{array}{{20}{c}} {\left[ {\frac{1}{{\rm{x}}}} \right],\frac{1}{3} < ;x; < \frac{1}{2}}\ {\left[ {\frac{1}{{\rm{x}}}} \right],\frac{1}{2} < ;x; < 1} \end{array}} \right. = \left{ {\begin{array}{{20}{c}} {\left[ {\frac{1}{{\rm{x}}}} \right] = 2,;;;;2 < \frac{1}{{\rm{x}}} < 3}\ {\left[ {\frac{1}{{\rm{x}}}} \right] = 1,;;;;1 < \frac{1}{{\rm{x}}} < 2} \end{array}} \right.\)

\({\rm{I}} = \mathop \smallint \limits_{\frac{1}{3}}^1 {\rm{x}}\left[ {\frac{1}{{\rm{x}}}} \right]{\rm{dx}} = \mathop \smallint \limits_{\frac{1}{3}}^{\frac{1}{2}} 2{\rm{xdx}} + {\rm{;}}\mathop \smallint \limits_{\frac{1}{2}}^1 {\rm{xdx}}\)

\({\rm{I}} = {\rm{;}}\left[ {\frac{{2{{\rm{x}}^2}}}{2}} \right]{\frac{1}{3}}^{\frac{1}{2}} + {\rm{;}}\left[ {\frac{{{{\rm{x}}^2}}}{2}} \right]{\frac{1}{2}}^1 = {\rm{;}}\left[ {\frac{1}{4} - \frac{1}{9}} \right] + \frac{1}{2}\left[ {1 - \frac{1}{4}} \right] = {\rm{;}}\left[ {\frac{{9 - 4}}{{36}}} \right] + \frac{1}{2}\left[ {\frac{{4 - 1}}{4}} \right] = \frac{5}{{36}} + \frac{3}{8} = \frac{{37}}{{72}}\)

Concept:

1. A function f(x) is said to be continuous at a point x = a, in its domain if $\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} {\rm{f}}\left( {\rm{x}} \right) = {\rm{f}}\left( {\rm{a}} \right)$ exists or its graph is a single unbroken curve.

• If f(x) is Continuous at x = a then ${\rm{;}}\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{a}}^ + }} {\rm{f}}\left( {\rm{x}} \right){\rm{;}} = {\rm{;}}\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{a}}^ - }} {\rm{f}}\left( {\rm{x}} \right){\rm{;}} = {\rm{;}}\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} {\rm{f}}\left( {\rm{x}} \right)$
• If f(x) is differentiable at x = a then ${\rm{;}}\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{a}}^ + }} {\rm{f'}}\left( {\rm{x}} \right){\rm{;}} = {\rm{;}}\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{a}}^ - }} {\rm{f'}}\left( {\rm{x}} \right){\rm{;}}$

2. Let Standard form of quadratic Functions: ⇒ f(x) = ax² + bx + c

Steps to Graphing Quadratic Equations

1. Check the sign of a

• If a > 0 the graph will be a convex parabola.
• If a < 0 The graph will be a concave parabola

1. Vertex = $\left( {\frac{{ - {\rm{b}}}}{{2{\rm{a}}}},\frac{{ - {\rm{D}}}}{{4{\rm{a}}}}} \right)$
2. Parabola is symmetric about the line that intersects x = $\frac{{ - {\rm{b}}}}{{2{\rm{a}}}}$
3. Find the coordinates of the points of intersection with the x axis, we must solve the equation f(x) = 0

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Calculation:

Given: f(x) = |x – 1| + x² Where x ∈ R

\({\rm{f}}\left( {\rm{x}} \right) = \left| {{\rm{x;}}-{\rm{;}}1} \right| + {\rm{;}}{{\rm{x}}^2} = \left{ {\begin{array}{{20}{c}} {1 - x + {{\rm{x}}^2},;;x < 1}\ {x - 1 + {{\rm{x}}^2},;;x \ge 1} \end{array}} \right. = \left{ {\begin{array}{{20}{c}} {{{\rm{x}}^2} - x + 1,;;x < 1}\ {{{\rm{x}}^2} + x - 1,;;x \ge 1} \end{array}} \right.\)

Differentiating with respect to x, we get

\( \Rightarrow {\rm{f'}}\left( {\rm{x}} \right) = \left{ {\begin{array}{*{20}{c}} {2x - 1,;;x < 1}\ {2x + 1,;;x \ge 1} \end{array}} \right.\)

Let’s check continuity at x = 1

${\rm{LHL}} = \mathop {\lim }\limits_{{\rm{x}} \to {1^ - }} {\rm{f}}\left( {\rm{x}} \right) = {\rm{;;}}\mathop {\lim }\limits_{{\rm{x}} \to {1^ - }} \left( {{{\rm{x}}^2} - {\rm{x}} + 1} \right) = 1 - 1 + 1 = 1$

${\rm{RHL}} = \mathop {\lim }\limits_{{\rm{x}} \to {1^ + }} {\rm{f}}\left( {\rm{x}} \right) = {\rm{;;}}\mathop {\lim }\limits_{{\rm{x}} \to {1^ + }} \left( {{{\rm{x}}^2} + {\rm{x}} - 1} \right) = 1 + 1 - 1 = 1$

LHL = RHL

Hence f(x) is continuous at x = 1

Let’s check differentiability at x = 1

${\rm{LHD}} = \mathop {\lim }\limits_{{\rm{x}} \to {1^ - }} {\rm{f'}}\left( {\rm{x}} \right) = {\rm{;;}}\mathop {\lim }\limits_{{\rm{x}} \to {1^ - }} \left( {2{\rm{x}} - 1} \right) = 2 - 1 = 1$

${\rm{RHD}} = \mathop {\lim }\limits_{{\rm{x}} \to {1^ + }} {\rm{f'}}\left( {\rm{x}} \right) = {\rm{;;}}\mathop {\lim }\limits_{{\rm{x}} \to {1^ + }} \left( {2{\rm{x}} + 1} \right) = 2 + 1 = 3$

LHD ≠ RHD

Hence f(x) is not differentiable at x = 1

∴ Option 2 is correct answer.

Alternative method:

Graphical Approach:

\({\rm{f}}\left( {\rm{x}} \right) = \left{ {\begin{array}{*{20}{c}} {{{\rm{x}}^2} - x + 1,;;x < 1}\ {{{\rm{x}}^2} + x - 1,;;x \ge 1} \end{array}} \right.\)

 

Dark line represent the function f(x) graph.

As we can see in graph f(x) is unbroken curve so f(x) is continuous.

At x = 1 graph has sharp corner so f(x) is not differentiable at x = 1

Calculation:

We get the graph of f(x) in previous solution,

Dark line represent the function f(x) graph.

We can see that graph is decreasing up to x= 1/2 after that graph is increasing.

∴ f(x) is decreasing in $\left( { - \infty ,\frac{1}{2}} \right)$ and increasing in $\left( {\frac{1}{2},{\rm{;}}\infty } \right)$

Concept:

Let f(x) be a real-valued function and c be an interior point in the domain of f such that f’(c) = 0

a) If f’(x) > 0 at every point close to the left of c and f’(x) < 0 at every point right of c, then it is a local maximum

b) If f’(x) > 0 at every point close to the right of c and f’(x) < 0 at every point left of c, then it is a local minimum

Calculation:

As we can see at f(x) gives local maxima at x = -∞ and ∞.

∴ f(x) has local maxima at more than one point in (-∞, ∞)

Concept:

Area bounded by function f(x) and g(x) is given as,

\({\rm{Area;}} = {\rm{;}}\mathop \smallint \nolimits_{\rm{a}}^{\rm{b}} \left[ {{\rm{f}}\left( {\rm{x}} \right) - {\rm{g}}\left( {\rm{x}} \right)} \right]{\rm{dx}} = {\rm{;}}\mathop \smallint \nolimits_{\rm{a}}^{\rm{b}} \left[ {{\rm{Top}} - {\rm{bottom}}} \right]{\rm{dx}}\)

Calculation:

Given:

\({\rm{f}}\left( {\rm{x}} \right) = \left| {{\rm{x;}}-{\rm{;}}1} \right| + {\rm{;}}{{\rm{x}}^2} = \left{ {\begin{array}{{20}{c}} {1 - x + {{\rm{x}}^2},;;x < 1}\ {x - 1 + {{\rm{x}}^2},;;x \ge 1} \end{array}} \right. = \left{ {\begin{array}{{20}{c}} {{{\rm{x}}^2} - x + 1,;;x < 1}\ {{{\rm{x}}^2} + x - 1,;;x \ge 1} \end{array}} \right.\)

For x < 1

⇒ f(x) = x² – x + 1

\({\rm{Area;}} = {\rm{;}}\mathop \smallint \nolimits_{\frac{1}{2}}^1 ({{\rm{x}}^2} - {\rm{x}} + 1){\rm{dx}}\)

\(= {\rm{;}}\left[ {\frac{{{{\rm{x}}^3}}}{3}} \right]{\frac{1}{2}}^1 - {\rm{;}}\left[ {\frac{{{{\rm{x}}^2}}}{2}} \right]{\frac{1}{2}}^1 + {\rm{;}}\left[ {\rm{x}} \right]_{\frac{1}{2}}^1 = \frac{1}{3}\left[ {1 - \frac{1}{8}} \right] - \frac{1}{2}\left[ {1 - \frac{1}{4}} \right] + \left[ {1 - \frac{1}{2}} \right] = \frac{7}{{24}} - \frac{3}{8} + \frac{1}{2} = \frac{5}{{12}}\)

Hence Area is 5/12 square unit.

Calculation:

Given:

\({\rm{f}}\left( {\rm{x}} \right) = \left| {{\rm{x;}}-{\rm{;}}1} \right| + {\rm{;}}{{\rm{x}}^2} = \left{ {\begin{array}{{20}{c}} {1 - x + {{\rm{x}}^2},;;x < 1}\ {x - 1 + {{\rm{x}}^2},;;x \ge 1} \end{array}} \right. = \left{ {\begin{array}{{20}{c}} {{{\rm{x}}^2} - x + 1,;;x < 1}\ {{{\rm{x}}^2} + x - 1,;;x \ge 1} \end{array}} \right.\)

For x > 1

⇒ f(x) = x² + x – 1

\({\rm{Area;}} = {\rm{;}}\mathop \smallint \nolimits_1^{\frac{3}{2}} ({{\rm{x}}^2} + {\rm{x}} - 1){\rm{dx}}\)

$= {\rm{;}}\left[ {\frac{{{{\rm{x}}^3}}}{3}} \right]_1^{\frac{3}{2}} + {\rm{;}}\left[ {\frac{{{{\rm{x}}^2}}}{2}} \right]_1^{\frac{3}{2}} - {\rm{;}}\left[ {\rm{x}} \right]_1^{\frac{3}{2}} = \frac{1}{3}\left[ {\frac{{27}}{8} - 1} \right] + \frac{1}{2}\left[ {\frac{9}{4} - 1} \right] - \left[ {\frac{3}{2} - 1} \right] = \frac{{19}}{{24}} + \frac{5}{8} - \frac{1}{2} = \frac{{11}}{{12}}$

Hence Area is 11/12 square unit.

Concept:

• sin (2π – x) = -sin x
• sin {(2n + 1)π  – x} = -sin x
• sin {(2n + 2)π  – x} = -sin x
• \(\mathop \smallint \nolimits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = \mathop \smallint \nolimits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {{\rm{a}} + {\rm{b}} - {\rm{;x}}} \right){\rm{dx;}}\)

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Calculation:

Given:

\({{\rm{a}}_{\rm{n}}} = \mathop \smallint \limits_0^{\rm{\pi }} \frac{{{{\sin }^2}\left{ {\left( {{\rm{n}} + 1} \right){\rm{x}}} \right}}}{{\sin 2{\rm{x}}}}{\rm{dx}}\)               

Replace n → 2n

\({{\rm{a}}_{2{\rm{n}}}} = \mathop \smallint \limits_0^{\rm{\pi }} \frac{{{{\sin }^2}\left{ {\left( {2{\rm{n}} + 1} \right){\rm{x}}} \right}}}{{\sin 2{\rm{x}}}}{\rm{dx;}}\)               

Apply property f(x) = f(a + b - x)

\({{\rm{a}}_{2{\rm{n}}}} = \mathop \smallint \limits_0^{\rm{\pi }} \frac{{{{\sin }^2}\left( {2{\rm{n}} + 1} \right)\left( {{\rm{\pi }} - {\rm{x}}} \right)} }}{{\sin 2\left( {{\rm{\pi }} - {\rm{x}}} \right)}}{\rm{dx}} = \mathop \smallint \limits_0^{\rm{\pi }} \frac{{{{\sin }^2}\left{ {\left( {2{\rm{n}} + 1} \right){\rm{\pi }} - \left( {2{\rm{n}} + 1} \right){\rm{x}}} \right}}}{{\sin \left( {2{\rm{\pi }} - 2{\rm{x}}} \right)}}{\rm{dx}}\)

\({{\rm{a}}{2{\rm{n}}}} = \mathop \smallint \limits_0^{\rm{\pi }} \frac{{{{\sin }^2}\left{ {\left( {2{\rm{n}} + 1} \right){\rm{x}}} \right}}}{{ - \sin 2{\rm{x}}}}{\rm{dx}} = {\rm{;}} - {{\rm{a}}{2{\rm{n}}}}\)

⇒ 2a_2n = 0

∴ a_2n = 0

Sequence is independent of n so the sequence {a_2n} is in AP with common difference zero.

Hence statement 1 is correct.

Again,

Given:

\({{\rm{a}}_{\rm{n}}} = \mathop \smallint \limits_0^{\rm{\pi }} \frac{{{{\sin }^2}\left{ {\left( {{\rm{n}} + 1} \right){\rm{x}}} \right}}}{{\sin 2{\rm{x}}}}{\rm{dx}}\)               

Replace n → 2n + 1

\({{\rm{a}}_{2{\rm{n}} + 1}} = \mathop \smallint \limits_0^{\rm{\pi }} \frac{{{{\sin }^2}\left{ {\left( {2{\rm{n}} + 1 + 1} \right){\rm{x}}} \right}}}{{\sin 2{\rm{x}}}}{\rm{dx}} = \mathop \smallint \limits_0^{\rm{\pi }} \frac{{{{\sin }^2}\left{ {\left( {2{\rm{n}} + 2} \right){\rm{x}}} \right}}}{{\sin 2{\rm{x}}}}{\rm{dx}}\)              

Apply property f(x) = f(a + b - x)

\({{\rm{a}}_{2{\rm{n}} + 1}} = \mathop \smallint \limits_0^{\rm{\pi }} \frac{{{{\sin }^2}\left( {2{\rm{n}} + 2} \right)\left( {{\rm{\pi }} - {\rm{x}}} \right)} }}{{\sin 2\left( {{\rm{\pi }} - {\rm{x}}} \right)}}{\rm{dx}} = \mathop \smallint \limits_0^{\rm{\pi }} \frac{{{{\sin }^2}\left{ {\left( {2{\rm{n}} + 2} \right){\rm{\pi }} - \left( {2{\rm{n}} + 2} \right){\rm{x}}} \right}}}{{\sin \left( {2{\rm{\pi }} - 2{\rm{x}}} \right)}}{\rm{dx}}\)

\({{\rm{a}}{2{\rm{n}} + 1}} = \mathop \smallint \limits_0^{\rm{\pi }} \frac{{{{\sin }^2}\left{ {\left( {2{\rm{n}} + 2} \right){\rm{x}}} \right}}}{{ - \sin 2{\rm{x}}}}{\rm{dx}} = {\rm{;}} - {{\rm{a}}{2{\rm{n}} + 1}}\)                                     (∵ sin (2π – x) = -sin x)

⇒ 2a_2n+1 = 0

∴ a_2n+! = 0

Sequence is independent of n so the sequence {a_2n+1} is in AP with common difference zero.

Hence statement 2 is also correct.

Concept:

If a sequence is given in the form of Arithematic progression (A.P.) and all its terms are equal, i.e. 

a, a, a, . . .

then such A.P. are called as Constant A.P. and has a common difference = 0.

\( \smallint sin2xdx=-\frac{cos~2x}{2}\)

Calculation:

We are given the sequence \(a_n= \mathop \smallint \limits_0^{\rm{\pi }} \frac{{{{\sin }^2}\left{ {\left( {{\rm{n}} + 1} \right){\rm{x}}} \right}}}{{\sin 2{\rm{x}}}}{\rm{dx}}\)

putting n = 1,

$a_1= \smallint \limits_0^{\rm{\pi }} \frac{sin^22x}{sin2x}dx$

$a_1= \smallint \limits_0^{\rm{\pi }}sin2xdx$

$a_1=-[\frac{cos~2x}{2}]^\pi _0$

\(a_1=-[\frac{cos2 \pi}{2}-\frac{cos2 \times 0}{2}]\)

$a_1=-[\frac12-\frac12]$

a₁ = 0

putting n = 2, $a_2= \smallint \limits_0^{\rm{\pi }}\frac{sin^23x}{sin2x}dx=0$

Similarly, we get all the terms equal to 0.

Thus, all the terms will have 0 differences.

Hence, a_n-1 – a_n-4 = 0

Concept:

1. A function f(x) is said to be continuous at a point x = a, in its domain if $\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} {\rm{f}}\left( {\rm{x}} \right) = {\rm{f}}\left( {\rm{a}} \right)$ exists or its graph is a single unbroken curve.

• If f(x) is Continuous at x = a then ${\rm{;}}\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{a}}^ + }} {\rm{f}}\left( {\rm{x}} \right){\rm{;}} = {\rm{;}}\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{a}}^ - }} {\rm{f}}\left( {\rm{x}} \right){\rm{;}} = {\rm{;}}\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} {\rm{f}}\left( {\rm{x}} \right)$
• If f(x) is differentiable at x = a then ${\rm{;}}\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{a}}^ + }} {\rm{f'}}\left( {\rm{x}} \right){\rm{;}} = {\rm{;}}\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{a}}^ - }} {\rm{f'}}\left( {\rm{x}} \right){\rm{;}}$ 2. \({\rm{f}}\left( {\rm{x}} \right) = \left| {\rm{x}} \right| = \left{ {\begin{array}{*{20}{c}} { - x,;;x < 0}\ {x,;;x \ge 0} \end{array}} \right.\)

Calculation:

Given equation x + |y| = 2y

We know that, \(\left| {\rm{y}} \right| = \left{ {\begin{array}{*{20}{c}} { - y,;;y < 0}\ {y,;;y \ge 0} \end{array}} \right.\) 

For y ≥ 0

⇒ x + y = 2y

∴ y = x     … (1)

For y < 0

⇒ x – y = 2y

∴ y = x/3      …. (2)

\( \Rightarrow {\rm{y}} = {\rm{f}}\left( {\rm{x}} \right) = \left{ {\begin{array}{*{20}{c}} {\frac{{\rm{x}}}{3},;;x < 0,;y < 0}\ {x,;;x \ge 0,;y \ge 0} \end{array}} \right.\)

Hence we can say that y as a function of x is defined for all real x.

Let’s check continuity at x = 0

${\rm{LHL}} = \mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {0^ - }} \left( {\frac{{\rm{x}}}{3}} \right) = 0{\rm{;}}$

${\rm{RHL}} = \mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {0^ + }} {\rm{x}} = 0$

LHL = RHL

Hence function is continuous at x = 0

Let’s check differentiability

\({\rm{f}}\left( {\rm{x}} \right) = \left{ {\begin{array}{*{20}{c}} {\frac{{\rm{x}}}{3},;;x < 0}\ {x,;;x \ge 0} \end{array}} \right.\)

Differentiate with respect to x, we get

\( \Rightarrow {\rm{f'}}\left( {\rm{x}} \right) = \left{ {\begin{array}{*{20}{c}} {\frac{1}{3},;;x < 0}\ {1,;;x \ge 0} \end{array}} \right.\)

$\mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} {\rm{f'}}\left( {\rm{x}} \right) \ne {\rm{;}}\mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} {\rm{f'}}\left( {\rm{x}} \right)$

Therefore function is not differentiable at x = 0.