Quantitative Aptitude PYQs

Concept:

Consider a triangle ABC inscribed in a circle with centre O.

So, the angle subtended at the centre of a circle is twice the angle subtended at its circumference.

Hence, β = 2α

Formula used:

\(\cos 2{\rm{A}} = \frac{{1 - {{\tan }^2}{\rm{A}}}}{{1 + {{\tan }^2}{\rm{A}}}}\)

Calculation:

Given: α = ∠BAC & β = ∠BOC

Now we know, that the angle subtended at the centre of a circle is twice the angle subtended at its circumference.

So, β = 2α

cos β = cos 2α

We know that, 

\(\cos 2{\rm{A}} = \frac{{1 - {{\tan }^2}{\rm{A}}}}{{1 + {{\tan }^2}{\rm{A}}}}\)

\(\Rightarrow \cos {\rm{\beta }} = \frac{{1 - {{\tan }^2}{\rm{\alpha }}}}{{1 + {{\tan }^2}{\rm{\alpha }}}}\)

Concept:

Formulas used:

• tan θ = Hypotenuse/Base

 

Calculation:

Let H be height of tower, θ be the angle of elevation, L be length of shadow.

Now, \(\rm \tan θ=\frac{h}{x} \)

Also, \(\rm \frac{h+6}{x+2√3}=\tan θ \)

∴ \(\rm \frac{h}{x}=\frac{h+6}{x+2√3}⇒ hx+2√3h=hx+6x \)

\(\rm ⇒ 2√3h=6x \)

\(\rm ⇒ \frac{h}{x}=\frac{6}{2√3}=√3 \)

⇒ tan θ = √3. ∴ θ = 60°.

Concept:

Inverse trigonometry formulas:

\({\tan ^{ - 1}}{\rm{A}} + {\tan ^{ - 1}}{\rm{B}} = {\tan ^{ - 1}}\frac{{{\rm{A\;}} + {\rm{\;B}}}}{{1{\rm{\;}} - {\rm{\;AB}}}}\) where A > 0, B > 0 & AB < 1

Calculation:

According to the concept used

\(\tan ^{- 1}\left( {\frac{1}{4}} \right) + {\tan ^{ - 1}}\left( {\frac{3}{5}} \right) = {\tan ^{ - 1}}\frac{{\frac{1}{4}{\rm{\;}} + \frac{3}{5}}}{{1{\rm{\;}} - \frac{1}{4} \times \frac{3} {5}}}\)

\(⇒ {\tan ^{ - 1}}\frac{{\frac{{17}}{{20}}}}{{1 - \frac{3}{{20}}}}\)

⇒ tan^-1 (1)

⇒ tan^-1 (tan π/4)

⇒ π/4 

∴ The value of \(\tan ^{- 1}\left( {\frac{1}{4}} \right) + {\tan ^{ - 1}}\left( {\frac{3}{5}} \right)\) is π/4.

Calculation:

Let O be the observer (neglecting height of observer), C be the centre of balloon of radius r. The rough diagram as shown below:

Given: ∠AOB = α; ∠COD = β

To find: length of CD

Now we know that, OC bisects ∠AOB.

So, ∠COB = α/2

 

In ∆COB,

\(\sin \frac{{\rm{\alpha }}}{2} = \frac{{{\rm{BC}}}}{{{\rm{OC}}}}\)

\(\Rightarrow \sin \frac{{\rm{\alpha }}}{2} = \frac{{\rm{r}}}{{{\rm{OC}}}}\)

\(\Rightarrow {\rm{OC}} = \frac{{\rm{r}}}{{\sin \frac{{\rm{\alpha }}}{2}}}\)

In ∆COD,

\(\sin {\rm{\beta }} = \frac{{{\rm{CD}}}}{{{\rm{OC}}}}\)

⇒ CD = OC × sin β

\(\Rightarrow {\rm{CD}} = \frac{{{\rm{r}}\sin {\rm{\beta }}}}{{\sin \frac{{\rm{\alpha }}}{2}}}\)

Concept:

Formulas used:

• cos (A - B) = cos A × cos B + sin A × sin B
• sin² θ + cos² θ = 1

 

Calculation:

Given: sin α + sin β = 0 & cos α + cos β = 0

Now, (sin α + sin β)² + (cos α + cos β)² = 0

⇒ sin² α + sin² β + 2sin α × sin β + cos² α + cos² β + 2cos α × cos β = 0

⇒ 1 + 1 + 2{sin α × sin β + cos α × cos β} = 0 [∵sin² θ + cos² θ = 1]

We know that, cos (A - B) = cos A × cos B + sin A × sin B

⇒ 2 + 2{cos (α - β)} = 0

⇒ cos (α - β) = -1

⇒ cos (α - β) = cos π

⇒ α - β = π

⇒ α = π + β

Concept:

We know that cos θ ≥ 0 for θ ∈ [0, π/2].

Calculation:

Given: cos α + cos β + cos γ = 0, where \(0 < {\rm{\alpha }} \le \frac{{\rm{\pi }}}{2},0 < {\rm{\beta }} \le \frac{{\rm{\pi }}}{2}\:\&\: 0 < {\rm{\gamma }} \le \frac{{\rm{\pi }}}{2}\)

To find: sin α + sin β + sin γ

 

We know that cos θ ≥ 0 for θ ∈ [0, π/2], i.e., cos θ is not negative for a defined of interval θ.

So, cos α + cos β + cos γ = 0, only when individually they are zero, i.e., cos α = 0, cos β = 0 & cos γ = 0.

∴ α = β = γ = π/2

So, sin α + sin β + sin γ = 3sin π/2

= 3

Concept:

Maxima: Let f(x) be a function defined on an interval x. For finding the maximum:

1. Find the stationary points \(\frac{{{\rm{df}}\left( {\rm{x}} \right)}}{{{\rm{dx}}}} = 0\).
2. \({\left( {\frac{{{{\rm{d}}^2}{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{d}}{{\rm{x}}^2}}}} \right)_{{\rm{x\;}} = {\rm{\;a}}}} < 0\), for the stationary point (x = a) to be point of maxima.

 

Note:

If f(x) = a(sin x) + b(cos x) then:

• \({\rm{f}}\left( {\rm{x}} \right) = \sqrt {{{\rm{a}}^2} + {{\rm{b}}^2}} \times \sin \left\{ {{\rm{x}} + \frac{{\rm{\pi }}}{4}} \right\}\) and for f(x)_max function of sin should be maximum i.e., .

 

Calculation:

Given: \({\rm{f}}\left( {\rm{x}} \right) = \sin \left( {{\rm{x}} + \frac{{\rm{\pi }}}{5}} \right) + \cos \left( {{\rm{x}} + \frac{{\rm{\pi }}}{5}} \right)\)

To find: point of maxima

\({\rm{f}}\left( {\rm{x}} \right) = \sin \left( {{\rm{x}} + \frac{{\rm{\pi }}}{5}} \right) + \cos \left( {{\rm{x}} + \frac{{\rm{\pi }}}{5}} \right)\)

Now, differentiating both sides w.r.t x

\(\Rightarrow {\rm{f'}}\left( {\rm{x}} \right) = \cos \left( {{\rm{x}} + \frac{{\rm{\pi }}}{5}} \right) - \sin \left( {{\rm{x}} + \frac{{\rm{\pi }}}{5}} \right)\)

Put f’(x) = 0

\(\Rightarrow 0 = \cos \left( {{\rm{x}} + \frac{{\rm{\pi }}}{5}} \right) - \sin \left( {{\rm{x}} + \frac{{\rm{\pi }}}{5}} \right)\)

⇒ sin (x + π/5) = cos (x + π/5)

⇒ tan (x + π/5) = 1

⇒ tan (x + π/5) = tan π/4

⇒ x + π/5 = π/4

⇒ x = π/4 - π/5

⇒ x = π/20

 

Alternate method:

\({\rm{f}}\left( {\rm{x}} \right) = \sin \left( {{\rm{x}} + \frac{{\rm{\pi }}}{5}} \right) + \cos \left( {{\rm{x}} + \frac{{\rm{\pi }}}{5}} \right)\)

\(\Rightarrow {\rm{f}}\left( {\rm{x}} \right) = \sqrt {{1^2} + {1^2}} \times \sin \left\{ {{\rm{x}} + \frac{{\rm{\pi }}}{5} + \frac{{\rm{\pi }}}{4}} \right\}\)

\(\Rightarrow {\rm{f}}\left( {\rm{x}} \right) = \sqrt {{1^2} + {1^2}} \times \sin \left\{ {{\rm{x}} + \frac{{\rm{\pi }}}{5} + \frac{{\rm{\pi }}}{4}} \right\}\)

For, f(x)_max

\(\sin \left\{ {{\rm{x}} + \frac{{\rm{\pi }}}{5} + \frac{{\rm{\pi }}}{4}} \right\} = 1\)

⇒ x + π/5 + π/4 = π/2

⇒ x = π/20

Concept:

Section formula: Let P(x, y) divides AB in ratio (m : n), where A ≡ (x₁, y₁) and B ≡ (x₂, y₂) then:

\({\rm{x}} = \frac{{{\rm{m}}{{\rm{x}}_2}{\rm{\;}} \pm {\rm{\;n}}{{\rm{x}}_1}}}{{{\rm{m\;}} \pm {\rm{\;n}}}}\) & \({{\rm{y}}_1} = \frac{{{\rm{m}}{{\rm{y}}_2}{\rm{\;}} \pm {\rm{\;n}}{{\rm{y}}_1}}}{{{\rm{m\;}} \pm {\rm{\;n}}}}\)

+ve for internal division & -ve for external division.

 

Distance formula: Let A ≡ (x₁, y₁) and B ≡ (x₂, y₂), distance between A and B is given as:

\({\rm{AB}} = \sqrt {{{\left( {{{\rm{y}}_2} - {{\rm{y}}_1}} \right)}^2} + {{\left( {{{\rm{x}}_2} - {{\rm{x}}_1}} \right)}^2}} \)

 

Calculation:

Let A ≡ (4, 3) and B ≡ (5, 7)

Let P(x₁, y₁) divides AB externally and Q(x₂, y₂) divides AB internally.

Given: ratio of division 2 : 3

To find: length of PQ

External division:

Using section formula,

\({{\rm{x}}_1} = \frac{{2 \times 5{\rm{\;}} - {\rm{\;}}3 \times 4}}{{2{\rm{\;}} - {\rm{\;}}3}}\) & \({{\rm{y}}_1} = \frac{{2 \times 7{\rm{\;}} - {\rm{\;}}3 \times 3}}{{2{\rm{\;}} - {\rm{\;}}3}}\)

⇒ x₁ = 2 & y₁ = -5

Internal division;

Using section formula,

\({{\rm{x}}_2} = \frac{{2 \times 5{\rm{\;}} + {\rm{\;}}3 \times 4}}{{2{\rm{\;}} + {\rm{\;}}3}}\) & \({{\rm{y}}_2} = \frac{{2 \times 7{\rm{\;}} + {\rm{\;}}3 \times 3}}{{2{\rm{\;}} + {\rm{\;}}3}}\)

⇒ x₂ = 22/5 & y₂ = 23/5

So, P ≡ (2, -5) & Q ≡ (22/5, 23/5)

\({\rm{PQ}} = \sqrt {{{\left( {{{\rm{y}}_2} - {{\rm{y}}_1}} \right)}^2} + {{\left( {{{\rm{x}}_2} - {{\rm{x}}_1}} \right)}^2}} \)

\(= \sqrt {{{\left( {\frac{{22}}{5} - 2} \right)}^2} + {{\left( {\frac{{23}}{5} - \left( { - 5} \right)} \right)}^2}} \)

\(= \sqrt {{{\left( {\frac{{12}}{5}} \right)}^2} + {{\left( {\frac{{48}}{5}} \right)}^2}} \)

\(= \frac{{12}}{5}\sqrt {1 + {4^2}} \)

\(= \frac{{12\sqrt {17} }}{5}\)

Concept:

Angle between lines: Let m₁ and m₂ be the slope of the lines and let θ be the angle between the lines then:

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

 

Calculation:

Given:

(m² - mn)y = (mn + n²)x + n³       ----(1)

(mn + m²)y = (mn - n²)x + m³       ----(2)

 

Let the slope of line (1) be m₁ and the slope of the line (2) be m₂.

So, \({{\rm{m}}_1} = \frac{{{\rm{mn}} + {{\rm{n}}^2}}}{{{{\rm{m}}^2} - {\rm{mn}}}} = \frac{{{\rm{n}}\left( {{\rm{m}} + {\rm{n}}} \right)}}{{{\rm{m}}\left( {{\rm{m}} - {\rm{n}}} \right)}}\)

And, \({{\rm{m}}_2} = \frac{{{\rm{mn}} - {{\rm{n}}^2}}}{{{\rm{mn}} + {{\rm{m}}^2}}} = \frac{{{\rm{n}}\left( {{\rm{m}} - {\rm{n}}} \right)}}{{{\rm{m}}\left( {{\rm{m}} + {\rm{n}}} \right)}}\)

Now we know that angle between two lines is given as,

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

So, \({{\rm{m}}_2} - {{\rm{m}}_1} = \frac{{{\rm{n}}\left( {{\rm{m}} - {\rm{n}}} \right)}}{{{\rm{m}}\left( {{\rm{m}} + {\rm{n}}} \right)}} - \frac{{{\rm{n}}\left( {{\rm{m}} + {\rm{n}}} \right)}}{{{\rm{m}}\left( {{\rm{m}} - {\rm{n}}} \right)}}\)

\(\Rightarrow {{\rm{m}}_2} - {{\rm{m}}_1} = \frac{{\rm{n}}}{{\rm{m}}} \times \frac{{{{\left( {{\rm{m}} - {\rm{n}}} \right)}^2} - {{\left( {{\rm{m}} + {\rm{n}}} \right)}^2}}}{{\left( {{{\rm{m}}^2} - {{\rm{n}}^2}} \right)}}\)

\(\Rightarrow {{\rm{m}}_2} - {{\rm{m}}_1} = \frac{{\rm{n}}}{{\rm{m}}} \times \frac{{4{\rm{mn}}}}{{\left( {{{\rm{m}}^2} - {{\rm{n}}^2}} \right)}}\)

\(\Rightarrow {{\rm{m}}_2} - {{\rm{m}}_1} = \frac{{4{{\rm{n}}^2}}}{{\left( {{{\rm{m}}^2} - {{\rm{n}}^2}} \right)}}\)

 

Now, \(1 + {{\rm{m}}_1}{{\rm{m}}_2} = 1 + \frac{{{\rm{n}}\left( {{\rm{m}} + {\rm{n}}} \right)}}{{{\rm{m}}\left( {{\rm{m}} - {\rm{n}}} \right)}} \times \frac{{{\rm{n}}\left( {{\rm{m}} - {\rm{n}}} \right)}}{{{\rm{m}}\left( {{\rm{m}} + {\rm{n}}} \right)}}\)

\(\Rightarrow 1 + {{\rm{m}}_1}{{\rm{m}}_2} = 1 + \frac{{{{\rm{n}}^2}}}{{{{\rm{m}}^2}}}\)

\(\Rightarrow 1 + {{\rm{m}}_1}{{\rm{m}}_2} = \frac{{{{\rm{m}}^2} + {{\rm{n}}^2}}}{{{{\rm{m}}^2}}}\)

Angle between lines = θ

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

\(= {\tan ^{ - 1}}\left\{ {\frac{{\frac{{4{{\rm{n}}^2}}}{{\left( {{{\rm{m}}^2} - {{\rm{n}}^2}} \right)}}}}{{\frac{{{{\rm{m}}^2} + {{\rm{n}}^2}}}{{{{\rm{m}}^2}}}}}} \right\}\)

\(= {\tan ^{ - 1}}\left\{ {\frac{{4{{\rm{n}}^2}{{\rm{m}}^2}}}{{\left( {{{\rm{m}}^2} - {{\rm{n}}^2}} \right) \times \left( {{{\rm{m}}^2} + {{\rm{n}}^2}} \right)}}} \right\}\)

\(= {\tan ^{ - 1}}\left\{ {\frac{{4{{\rm{n}}^2}{{\rm{m}}^2}}}{{\left( {{{\rm{m}}^4} - {{\rm{n}}^4}} \right)}}} \right\}\)

Concept:

Line:

• Slope form of line:

y = mx + c, where m is the slope and c is y-intercept.

Calculation:

Given:

Intercept = c = -2

Slope = m = tan θ

⇒ m = tan 30°

⇒ m = 1/√3

Rough diagram:

Intercept form of a line is given as:

y = mx + c

where m is the slope and c is y-intercept.

⇒ y = (1/√3)x - 2

⇒ √3y = x - 2√3

⇒ x - √3y - 2√3 = 0

Concept:

Line: Line in slope form is given as,

y = mx + c

Where m is the slope and c is y-intercept.

Note:

• If a line (slope m₁) is parallel to any other line (slope m₂) then:
• m₁ = m₂
• If a line (slope m₁) is perpendicular to any other line (slope m₂) then:
• m₁ = -1/m₂

 

Calculation:

Given:

x + 2y - 3 = 0              ----(1)

2x - y + 5 = 0              ----(2)

y - x + 10 = 0              ----(3)

Let the required equation of line be y = mx + c              ----(4)

Where m is the slope and c is y-intercept.

∵ The required line is parallel to the line (3).

So, the slope of required line (4) will be equal to the slope of line (3).

∴ m = 1

Now, the required line (4) passes through the intersection of line (1) and (2),

Let’s find the intersection of line (1) and (2)

Adding (1) and 2 × (2) we get,

x + 2y - 3 + 2{2x - y + 5} = 0

⇒ x + 4x - 3 + 10 = 0

⇒ 5x = -7

⇒ x = -7/5­­

Put x = -7/5 in equation (2)

2(-7/5) - y + 5 = 0

⇒ y = 5 - 14/5

⇒ y = 11/5

 

Put m = 1, x = -7/5 & y = 11/5 in equation (4)

11/5 = -7/5 + c

⇒ c = 18/5

So, equation of line we get,

Y = x + 18/5

⇒ 5y = 5x + 18

⇒ 5x - 5y + 18 = 0

Concept:

Perpendicular distance from a point on a line**:**

Let a point be (x₁, y₁) and equation of line be ax + by + c = 0 then:

Perpendicular distance \(= \frac{{\left| {{\rm{a}}{{\rm{x}}_1}{\rm{\;}} + {\rm{\;b}}{{\rm{y}}_1}{\rm{\;}} + {\rm{\;c}}} \right|}}{{\sqrt {{{\rm{a}}^2}{\rm{\;}} + {\rm{\;}}{{\rm{b}}^2}} }}\)

 

Calculation:

For statement (1),

Point is (0,0) and equation of line is ax + by - c = 0.

So, \({\rm{p}} = \frac{{\left| { - {\rm{c}}} \right|}}{{\sqrt {{{\rm{a}}^{2{\rm{\;}}}} + {\rm{\;}}{{\rm{b}}^2}} }}\)

\(\Rightarrow {\rm{\;p}} = \frac{{\rm{c}}}{{\sqrt {{{\rm{a}}^{2{\rm{\;}}}} + {\rm{\;}}{{\rm{b}}^2}} }}\)

Squaring both sides

\(\Rightarrow {{\rm{p}}^2} = \frac{{{{\rm{c}}^2}}}{{{{\rm{a}}^2} + {{\rm{b}}^2}}}\)

So, first statement is correct.

For statement (2),

Point is (0,0) and equation of line is \(\frac{{\rm{x}}}{{\rm{a}}} + \frac{{\rm{y}}}{{\rm{b}}} = 1\)

Equation of line will be bx + ay = ab

⇒ bx + ay - ab = 0

So, \({\rm{p}} = \frac{{\left| { - {\rm{ab}}} \right|}}{{\sqrt {{{\rm{b}}^2}{\rm{\;}} + {\rm{\;}}{{\rm{a}}^2}} }}\)

\(\Rightarrow {\rm{p}} = \frac{{{\rm{ab}}}}{{\sqrt {{{\rm{b}}^2}{\rm{\;}} + {\rm{\;}}{{\rm{a}}^2}} }}\)

Squaring both sides

\(\Rightarrow {{\rm{p}}^2} = \frac{{{{\rm{a}}^2}{{\rm{b}}^2}}}{{{{\rm{a}}^2} + {{\rm{b}}^2}}}\)

Taking reciprocal of both sides,

\(\Rightarrow \frac{1}{{{{\rm{p}}^2}}} = \frac{{{{\rm{a}}^2} + {{\rm{b}}^2}}}{{{{\rm{a}}^2}{{\rm{b}}^2}}}\)

\(\Rightarrow \frac{1}{{{{\rm{p}}^2}}} = \frac{1}{{{{\rm{b}}^2}}} + \frac{1}{{{{\rm{a}}^2}}}\)

So, second statement is correct.

For statement (3),

Point is (0,0) and equation of line is y - mx - c = 0

So, \({\rm{p}} = \frac{{\left| { - {\rm{c}}} \right|}}{{\sqrt {{{\rm{m}}^2} + {1^2}} }}\)

Squaring both sides,

\(\Rightarrow {{\rm{p}}^2} = \frac{{{{\rm{c}}^2}}}{{{{\rm{m}}^2} + 1}}\)

Taking reciprocal of both sides,

\(\Rightarrow \frac{1}{{{{\rm{p}}^2}}} = \frac{{{{\rm{m}}^2} + 1}}{{{{\rm{c}}^2}}}\)

So, third statement is incorrect.

Concept:

Line: Line in slope form is given as,

\(\frac{{\rm{x}}}{{\rm{a}}} + \frac{{\rm{y}}}{{\rm{b}}} = 1\)

where a is x-intercept and b is y-intercept.

Calculation:

Let the equation of line be \(\frac{{\rm{x}}}{{\rm{a}}} + \frac{{\rm{y}}}{{\rm{b}}} = 1\)               ----(1)

Where a is x-intercept and b is y-intercept.

The line is passing through (2,3). So, (2,3) should satisfy the equation of line.

Put x = 2 & y = 3 in equation (1)

\(\frac{2}{{\rm{a}}} + \frac{3}{{\rm{b}}} = 1\)              ----(2)

 

Now, we have given that the intercept on the positive y-axis equal to twice its intercept on the positive x-axis.

So, b = 2a

Put b = 2a in equation 2

\(\frac{2}{{\rm{a}}} + \frac{3}{{2{\rm{a}}}} = 1\)

⇒ 2 + 3/2 = a

⇒ a = 7/2

Now, b = 2a = 7

So, equation of line will be,

\(\frac{{\rm{x}}}{{\frac{7}{2}}} + \frac{{\rm{y}}}{7} = 1\)

⇒ 2x + y = 7

Concept:

Line in 3D:

• Equation of line passing through a point P(x₁, y₁, z₁) with direction cosine l, m, n:
• \(\frac{{{\rm{x}} - {{\rm{x}}_1}}}{{\rm{l}}} = \frac{{{\rm{y}} - {{\rm{y}}_1}}}{{\rm{m}}} = \frac{{{\rm{z}} - {{\rm{z}}_1}}}{{\rm{n}}}\)
• Equation of line passing through two points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂):
• \(\frac{{{\rm{x}} - {{\rm{x}}_1}}}{{{{\rm{x}}_2} - {{\rm{x}}_1}}} = \frac{{{\rm{y}} - {{\rm{y}}_1}}}{{{{\rm{y}}_2} - {{\rm{y}}_1}}} = \frac{{{\rm{z}} - {{\rm{z}}_1}}}{{{{\rm{z}}_2} - {{\rm{z}}_1}}}\)
• General point on a line:

\(\frac{{{\rm{x}} - {{\rm{x}}_1}}}{{\rm{l}}} = \frac{{{\rm{y}} - {{\rm{y}}_1}}}{{\rm{m}}} = \frac{{{\rm{z}} - {{\rm{z}}_1}}}{{\rm{n}}} = {\rm{\lambda }}\), where λ is a constant.

Note:

• Let direction ratios of a line be (l₁, m₁, n₁) and for another line be (l₂, m₂, n₂) then:

If lines are perpendicular, l₁l₂ + m₁m₂ + n₁n₂ = 0

If lines are parallel, \(\frac{{{{\rm{l}}_1}}}{{{{\rm{l}}_2}}} = \frac{{{{\rm{m}}_1}}}{{{{\rm{m}}_2}}} = \frac{{{{\rm{n}}_1}}}{{{{\rm{n}}_2}}}\)

 

Calculation:

Given:

A ≡ (1, 8, 4)

B ≡ (0, -11, 4)

C ≡ (2, -3, 1)

Equation of BC,

\(\frac{{{\rm{x}} - 0}}{{2 - 0}} = \frac{{{\rm{y}} - \left( { - 11} \right)}}{{ - 3 - \left( { - 11} \right)}} = \frac{{{\rm{z}} - 4}}{{1 - 4}}\)

\(\Rightarrow \frac{{\rm{x}}}{2} = \frac{{{\rm{y}} + 11}}{8} = \frac{{{\rm{z}} - 4}}{{ - 3}}\)

General point on BC

\(\frac{{\rm{x}}}{2} = \frac{{{\rm{y}} + 11}}{8} = \frac{{{\rm{z}} - 4}}{{ - 3}} = {\rm{\lambda }}\)

⇒ x = 2λ; y = 8λ - 11; z = -3λ + 4

General point on BC be D ≡ (2λ, 8λ - 11, -3λ + 4) and let D be foot of perpendicular.

Direction ratio of line AD will be (2λ - 1, 8λ - 11 - 8, -3λ + 4 - 4)

⇒ (2λ - 1, 8λ - 19, -3λ)

Now, since the line BC and AD are perpendicular, l₁l₂ + m₁m₂ + n₁n₂ = 0

So, (2λ - 1) × 2 + (8λ - 19) × 8 + (-3λ) × (-3) = 0

⇒ 4λ - 2 + 64λ - 152 + 9λ = 0

⇒ 77λ = 154

⇒ λ = 2

So, D ≡ (4, 5, -2)

Concept:

Sphere: General equation of sphere is given as

• x² + y² + z² + 2ux + 2vy + 2wz + d = 0
• Radius, \({\rm{r}} = \sqrt {{{\rm{u}}^2} + {{\rm{v}}^2} + {{\rm{z}}^2} - {\rm{d}}} \)

Centroid of triangle: If A(x₁, y₁, z₁), B(x₂, y₂, z₂) & C(x₃, y₃, z₃) are vertices of triangle, then:

• \({\rm{G}} \equiv \left( {\frac{{{{\rm{x}}_1}{\rm{\;}} + {\rm{\;}}{{\rm{x}}_2}{\rm{\;}} + {\rm{\;}}{{\rm{x}}_3}}}{3},\frac{{{{\rm{y}}_1}{\rm{\;}} + {\rm{\;}}{{\rm{y}}_2}{\rm{\;}} + {\rm{\;}}{{\rm{y}}_3}}}{3},\frac{{{{\rm{z}}_1}{\rm{\;}} + {\rm{\;}}{{\rm{z}}_2}{\rm{\;}} + {\rm{\;}}{{\rm{z}}_3}}}{3}} \right)\)

 

Calculation:

Let A ≡ (a, 0, 0); B ≡ (0, b, 0); C ≡ (0, 0, c) where a, b & c are constants and a ≠ 0, b ≠ 0 & c ≠ 0

General equation of sphere, x² + y² + z² + 2ux + 2vy + 2wz + d = 0.

Let centroid of triangle be G ≡ (α, β, γ)

Since, it passes through origin, (0, 0, 0) should satisfy.

So, we get d = 0.

∴ Equation of sphere, x² + y² + z² + 2ux + 2vy + 2wz = 0

Sphere intersect co-ordinate axes at A, B & C. So these points should satisfy the equation.

For A, put x = a, y = 0 & z = 0 we get,

a² + 2ua = 0

⇒ a(a + 2u) = 0

Now, a ≠ 0, So, 2u = -a

Similarly, for B & C we get, 2v = -b & 2w = -c

So, equation of sphere becomes, x² + y² + z² - ax - by - cz = 0

Radius of sphere, \({\rm{r}} = \sqrt {{{\rm{u}}^2} + {{\rm{v}}^2} + {{\rm{z}}^2} - {\rm{d}}} \)

\(\Rightarrow {\rm{r}} = \sqrt {{{\left( {\frac{{\rm{a}}}{2}} \right)}^2} + {{\left( {\frac{{\rm{b}}}{2}} \right)}^2} + {{\left( {\frac{{\rm{c}}}{2}} \right)}^2}} \)

Squaring both sides,

⇒ 4r² = a² + b² + c²              ----(1)

Now, centroid of triangle ABC G is given as \(\left( {\frac{{{\rm{a\;}} + {\rm{\;}}0{\rm{\;}} + {\rm{\;}}0}}{3},\frac{{0{\rm{\;}} + {\rm{\;b\;}} + {\rm{\;}}0}}{3},\frac{{0{\rm{\;}} + {\rm{\;}}0{\rm{\;}} + {\rm{\;c}}}}{3}} \right)\)

\({\rm{G}} \equiv \left( {\frac{{\rm{a}}}{3},\frac{{\rm{b}}}{3},\frac{{\rm{c}}}{3}} \right)\)

So, α = a/3, β = b/3 & γ = c/3.

Now, α² + β² + γ² = a²/9 + b²/9 + c²/9

⇒ 9(α² + β² + γ²) = a² + b² + c²

⇒ 9(α² + β² + γ²) = 4r² [Using (1)]

Locus of centroid,

⇒ 9(x² + y² + z²) = 4r²

Concept:

Integral properties: Consider a function f(x) defined on x.

• \(\mathop \smallint \limits_{ - {\rm{a}}}^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{\;dx}} = \left\{ {\begin{array}{*{20}{c}} {2\mathop \smallint \limits_0^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}},\;\;\;f\left( {\rm{x}} \right) = f\left( { - x} \right)}\\ {0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;f\left( {\rm{x}} \right) = - f\left( { - x} \right)} \end{array}} \right.\)

Calculation:

Let f(x) = sin x – tan x

Checking the function is odd or even,

f(-x) = sin (-x) – tan (-x)

⇒ f(-x) = – sin x + tan x

⇒ f(-x) = –{sin x – tan x}

⇒ f(-x) = –f(x)

Hence, the function is odd.

And we know that, \(\mathop \smallint \limits_{ - {\rm{a}}}^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{\;dx}} = 0\)  if f(x) is odd.

∴ \(\mathop \smallint \nolimits_{ - {\rm{\pi }}/4}^{{\rm{\pi }}/4} \left( {\sin {\rm{x}} - \tan {\rm{x}}} \right){\rm{dx}} = 0\)

Concept:

Integral properties: Consider a function f(x) defined on x.

\(\mathop \smallint \limits_{ - {\rm{a}}}^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{\;dx}} = \left\{ {\begin{array}{*{20}{c}} {2\mathop \smallint \limits_0^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}},\;\;\;f\left( {\rm{x}} \right) = f\left( { - x} \right)}\\ {0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;f\left( {\rm{x}} \right) = - f\left( { - x} \right)} \end{array}} \right.\)

Calculation:

Given:

\(\mathop \smallint \nolimits_{\rm{a}}^{\rm{b}} {{\rm{x}}^2}{\rm{dx}} = \frac{2}{3}\)

\(\mathop \smallint \nolimits_{\rm{a}}^{\rm{b}} {{\rm{x}}^3}{\rm{dx}} = 0{\rm{\;}}\)

Let f(x) = x³

f(-x) = -x³

⇒ f(-x) = - f(x)

∴ f(x) is an odd function.

We know that, if f(x) is odd function then, \(\mathop \smallint \nolimits_{ - {\rm{a}}}^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = 0\)

So, b = -a      ----(1)

Now, \(\mathop \smallint \nolimits_{\rm{a}}^{\rm{b}} {{\rm{x}}^2}{\rm{dx}} = \frac{2}{3}\)

\(\Rightarrow \left[ {\frac{{{{\rm{x}}^3}}}{3}} \right]_{\rm{a}}^{\rm{b}} = \frac{2}{3}\)

⇒ b³ – a³ = 2

⇒ -a³ – a³ = 2      [Using (1)]

⇒ a³ = -1

⇒ a = -1

So, b = 1

Concept:

Definite Integral properties:

\(\mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{\;dx}} = \mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {{\rm{a}} + {\rm{b}} - {\rm{x}}} \right){\rm{\;dx}}\)

Calculation:

Let f(x) = x(1 – x)⁹

Now using property, \(\mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{\;dx}} = \mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {{\rm{a}} + {\rm{b}} - {\rm{x}}} \right){\rm{\;dx}}\)

\(\mathop \smallint \nolimits_0^1 {\rm{x}}{(1 - {\rm{x}})^9}{\rm{dx}} = \mathop \smallint \limits_0^1 \left( {1 - {\rm{x}}} \right){\left\{ {1 - \left( {1 - {\rm{x}}} \right)} \right\}^9}{\rm{dx}}\)

\(⇒ \mathop \smallint \limits_0^1 \left( {1 - {\rm{x}}} \right){{\rm{x}}^9}{\rm{dx}}\)

\(⇒ \mathop \smallint \limits_0^1 \left( {{{\rm{x}}^9} - {{\rm{x}}^{10}}} \right){\rm{dx}}\)

\(⇒\left[ {\frac{{{{\rm{x}}^{10}}}}{{10}} - \frac{{{{\rm{x}}^{11}}}}{{11}}} \right]_0^1\)

⇒ 1/10 – 1/11

⇒ 1/110

∴ The value of integral \(\mathop \smallint \nolimits_0^1 {\rm{x}}{(1 - {\rm{x}})^9}{\rm{dx}}\) is 1/110.

Concept:

Binomial expansion:

\({\left( {1 + {\rm{x}}} \right)^{\rm{n}}} = 1 + {\rm{nx}} + \frac{{{\rm{n}}\left( {{\rm{n}} - 1} \right){{\rm{x}}^2}}}{{2!}} + \frac{{{\rm{n}}\left( {{\rm{n}} - 1} \right)\left( {{\rm{n}} - 2} \right){{\rm{x}}^3}}}{{3!}} + \ldots + \frac{{{\rm{n}}\left( {{\rm{n}} - 1} \right) \ldots \left( {{\rm{n}} - {\rm{r}} + 1} \right){{\rm{x}}^{\rm{r}}}}}{{{\rm{r}}!}} + \ldots {\rm{up\;to\;infinite\;terms}}\)

Where n is negative or fraction value and -1< x < 1

Calculation:

\(\mathop {\lim }\limits_{{\rm{h}} \to 0} \frac{{\sqrt {2{\rm{x}} + 3{\rm{h}}} - \sqrt {2{\rm{x}}} }}{{2{\rm{h}}}} = \mathop {\lim }\limits_{{\rm{h}} \to 0} \frac{{\sqrt {2{\rm{x}}} \left( {\sqrt {1 + \frac{{3{\rm{h}}}}{{2{\rm{x}}}}} } \right) - \sqrt {2{\rm{x}}} {\rm{\;}}}}{{2{\rm{h}}}}\)

\(= \mathop {\lim }\limits_{{\rm{h}} \to 0} \frac{{\sqrt {2{\rm{x}}} \left\{ {{{\left( {1 + \frac{{3{\rm{h}}}}{{2{\rm{x}}}}} \right)}^{\frac{1}{2}}} - 1} \right\}{\rm{\;}}}}{{2{\rm{h}}}}{\rm{\;}}\)

Using binomial expansion,

\(= \mathop {\lim }\limits_{{\rm{h}} \to 0} \frac{{\sqrt {2{\rm{x}}} \left\{ {\left( {1 + \frac{1}{{2!}} \times \frac{{3{\rm{h}}}}{{2{\rm{x}}}} + \frac{1}{{3!}} \times {{\left( {\frac{{3{\rm{h}}}}{{2{\rm{x}}}}} \right)}^2} + \cdots } \right) - 1} \right\}{\rm{\;}}}}{{2{\rm{h}}}}{\rm{\;}}\)

Ignoring higher terms as they are very small. So, terms from \(\frac{1}{{3!}} \times {\left( {\frac{{3{\rm{h}}}}{{2{\rm{x}}}}} \right)^2}\)  to infinite terms, are all zero.

\(= \mathop {\lim }\limits_{{\rm{h}} \to 0} \frac{{\sqrt {2{\rm{x}}} \left\{ {\left( {1 + \frac{1}{{2!}} \times \frac{{3{\rm{h}}}}{{2{\rm{x}}}}} \right) - 1} \right\}{\rm{\;}}}}{{2{\rm{h}}}}\)

\(= \mathop {\lim }\limits_{{\rm{h}} \to 0} \frac{{\sqrt {2{\rm{x}}} \left\{ {\frac{1}{{2!}} \times \frac{{3{\rm{h}}}}{{2{\rm{x}}}}} \right\}{\rm{\;}}}}{{2{\rm{h}}}}\)

\(= \mathop {\lim }\limits_{{\rm{h}} \to 0} \frac{{3{\rm{h}}}}{{2 \times 2{\rm{h}} \times \sqrt {2{\rm{x}}} }}\)

\(= \frac{3}{{4\sqrt {2{\rm{x}}} }}\)

Concept:

Even & odd function: Let f(x) be a function defined on x.

• A function is said to be even if f(x) = f(-x).
• A function is said to be odd if f(x) = -f(-x).

Calculation:

Given: f(x) is an even function

So, f(x) = f(-x)

Now, differentiating to both sides w.r.t x

⇒ f’(x) = f’(-x) × (-1)

⇒ f’(x) = -f’(-x)

Hence, f’(x) is an odd function