Quantitative Aptitude PYQs

Calculation:

Given: \({\rm{\;f}}\left( {\rm{x}} \right) = \frac{{\rm{x}}}{{{\rm{x}} - 1}},\)

\( \Rightarrow {\rm{f}}\left( {\rm{a}} \right) = \frac{{\rm{a}}}{{{\rm{a}} - 1}}\)              

\( \Rightarrow {\rm{f}}\left( {{\rm{a}} + 1} \right) = \frac{{({\rm{a}} + 1)}}{{\left( {{\rm{a}} + 1} \right) - 1}} = \frac{{{\rm{a}} + 1}}{{\rm{a}}}\) 

Now,

\(\frac{{{\rm{F}}\left( {\rm{a}} \right)}}{{{\rm{f}}\left( {{\rm{a}} + 1} \right)}} = {\rm{\;}}\frac{{\left( {\frac{{\rm{a}}}{{{\rm{a}} - 1}}} \right)}}{{\left( {\frac{{{\rm{a}} + 1}}{{\rm{a}}}} \right)}}{\rm{\;}} = {\rm{\;}}\frac{{{{\rm{a}}^2}}}{{{{\rm{a}}^2} - 1}} \)

As we already known as \({\rm{\;f}}\left( {\rm{x}} \right) = \frac{{\rm{x}}}{{{\rm{x}} - 1}},\)

So, put a² instead of x

The equation become, \(\frac{{{\rm{f}}\left( {\rm{a}} \right)}}{{{\rm{f}}\left( {{\rm{a}} + 1} \right)}}={\rm{f}}\left( {{{\rm{a}}^2}} \right)\)

∴ Option 2 is correct answer.

Concept:

• \(\smallint \frac{{{\rm{dx}}}}{{\rm{x}}} = \ln x + c\)

Calculation:

Let I = \(\smallint \frac{{{{\rm{x}}^{{\rm{e}} - 1}} + {{\rm{e}}^{{\rm{x}} - 1}}}}{{{{\rm{x}}^{\rm{e}}} + {{\rm{e}}^{\rm{x}}}}}{\rm{dx}}\)

Let x^e + e^x = t

Differentiating both sides, we get

⇒ (ex^e-1 + e^x) dx = dt

⇒ e (x^e-1 + e^x-1) dx = dt

∴ (x^e-1 + e^x-1) dx = dt/e

Now,

\({\rm{I}} = \frac{1}{{\rm{e}}}\smallint \frac{{{\rm{dt}}}}{{\rm{t}}} = {\rm{\;}}\frac{1}{{\rm{e}}}\ln {\rm{t}} + {\rm{c}} = {\rm{\;}}\frac{1}{{\rm{e}}}{\rm{\;In\;}}\left( {{{\rm{x}}^{\rm{e}}} + {{\rm{e}}^{\rm{x}}}} \right) + {\rm{c}}\)

Calculation:

Given: f(x) = x² – 3

⇒ (f o f) (x) = (f(x)) ² – 3 = (x² – 3)² – 3 = x⁴ – 6x² + 9 – 3 = x⁴ – 6x² + 6

∴ (f o f) (x) = x⁴ – 6x² + 6        ….

(1) Now,

⇒ (f o f 0 f) (x) = (x⁴ – 6x² + 6)² – 3

Statement-1:

LHS = (f ∘ f ∘ f) (-1) = [(-1)⁴ – 6(-1)² + 6]² – 3 = (1 – 6 + 6)² – 3 = -2

RHS = (f ∘ f ∘ f) (1) = [(1)⁴ – 6(1)² + 6]² – 3 = (1 – 6 + 6)² – 3 = -2

LHS = RHS

Hence statement 1 is correct.

Statement-2:

LHS = (f ∘ f ∘ f) (-1) – 4 (f ∘ f ∘ f) (1) = -2 – 4 × (-2) = -2 + 8 = 6

RHS = (f ∘ f) (0) = (0)⁴ – 69(0)² + 6 = 6

Hence statement 2 is correct.

∴ Both 1 and 2 are correct.

Calculation:

Given: f(x) = px + q and g(x) = mx + n

f (g(x)) = g (f(x))

⇒ f (mx + n) = g (px + q)

⇒ p (mx + n) + q = m (px + q) + n

⇒ pmx + pn + q = pmx + mq + n

⇒ pn + q = mq + n

⇒ f (n) = g (q)

∴ Option 3 is correct answer.

Concept:

L-Hospital Rule: Let f(x) and g(x) be two functions

Suppose that we have one of the following cases,

I. \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = {\rm{\;}}\frac{0}{0}\)

II. \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = {\rm{\;}}\frac{\infty }{\infty }\)

Then we can apply L-Hospital Rule ⇔ \(\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}\left( {\bf{x}} \right)}}{{{\bf{g}}\left( {\bf{x}} \right)}} = \;\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}'\left( {\bf{x}} \right)}}{{{\bf{g}}'\left( {\bf{x}} \right)}}\) 

Note: We have to differentiate both the numerator and denominator with respect to x unless and until \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = {\rm{l}} \ne \frac{0}{0}\) where l is a finite value.

Calculation:

Let L = \(\mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{{\rm{F}}\left( {\rm{x}} \right) - {\rm{F}}\left( 1 \right)}}{{{\rm{x}} - 1}}\)             (Form 0/0)

Apply L-Hospital rule,

\({\rm{L}} = \mathop {\lim }\limits_{{\rm{x}} \to 1} \frac{{{\rm{F'}}\left( {\rm{x}} \right){\rm{\;}} - 0)}}{{1 - 0}}\) 

L = F’(1)

Given: \({\rm{F}}\left( {\rm{x}} \right) = \sqrt {9 - {{\rm{x}}^2}} \)

Differentiating with respect to x, we get

\( \Rightarrow {\rm{F'}}\left( {\rm{x}} \right) = {\rm{\;}}\frac{{1 \times \left( {0 - 2{\rm{x}}} \right)}}{{2\sqrt {9 - {{\rm{x}}^2}} }} = {\rm{\;}}\frac{{ - {\rm{x}}}}{{\sqrt {9 - {{\rm{x}}^2}} }}\) 

Put x= 1

\( \Rightarrow {\rm{F'}}\left( 1 \right) = \frac{{ - 1}}{{\sqrt {9 - {1^2}} }} = \frac{{ - 1}}{{\sqrt 8 }} = - \frac{1}{{2\sqrt 2 }}\) 

So, L \(= - \frac{1}{{2\sqrt 2 }}\)

Concept:

• \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{x}}^{ - 1}}} \right) = {\rm{\;}} - {{\rm{x}}^{ - 2}}\)

Calculation:

We have to find the value of \(\frac{{{{\rm{d}}^2}{\rm{x}}}}{{{\rm{d}}{{\rm{y}}^2}}}\)

\( \Rightarrow \frac{{{{\rm{d}}^2}{\rm{x}}}}{{{\rm{d}}{{\rm{y}}^2}}} = {\rm{\;}}\frac{{\rm{d}}}{{{\rm{dy}}}}\left( {\frac{{{\rm{dx}}}}{{{\rm{dy}}}}} \right)\) 

\( = {\rm{\;}}\frac{{{\rm{d}}{{\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)}^{ - 1}}}}{{{\rm{dy}}}} = {\rm{\;}}\frac{{{\rm{d}}{{\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)}^{ - 1}}}}{{{\rm{dx}}}} \times \frac{{{\rm{dx}}}}{{{\rm{dy}}}}\) 

\( = {\rm{}} - {\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^{ - 2}}\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}} \times \frac{{{\rm{dx}}}}{{{\rm{dy}}}}\)                                  

\(= {\rm{}} - {\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^{ - 2}}\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}} \times {\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^{ - 1}} = {\rm{}} - \left( {\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}}} \right){\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^{ - 3}}\)

Concept:

• cos 3θ = 4 cos³ θ - 3 cos θ
• sin 2θ = 2 sin θ cos θ
• cos 2θ = cos² θ – sin² θ = 1 – 2 sin² θ = 2 cos² θ - 1

Calculation:

Let us consider θ = 18°

So, 5θ = 90°

⇒ 2θ + 3θ = 90˚

⇒ 2θ = 90˚ - 3θ

By taking sine on both sides, we get

Sin 2θ = sin (90˚ - 3θ) = cos 3θ

⇒ 2 sin θ cos θ = 4 cos³ θ - 3 cos θ

⇒ 2 sin θ cos θ - 4 cos³ θ + 3 cos θ = 0

⇒ cos θ (2 sin θ - 4 cos² θ + 3) = 0 

⇒ (2 sin θ - 4 cos² θ + 3) = 0 and cos θ = cos 18˚ ≠ 0

⇒ 2 sin θ - 4 (1 - sin² θ) + 3 = 0

⇒ 4 sin² θ + 2 sin θ - 1 = 0

This is a quadratic equation in Sine,

So, \(= {\rm{\;}}\frac{{ - 2{\rm{\;}} \pm \sqrt {4 + 16} }}{{2 \times 4}} = \frac{{ - 1 \pm \sqrt 5 }}{4}\)

Now since 18° is first quadrant so sin 18° is positive

Therefore, \(= \frac{{ - 1 + \sqrt 5 }}{4} = {\rm{\;}}\frac{{\sqrt 5 - 1}}{4}\)

Now, cos θ = \(\sqrt {1 - {\rm{\;}}{{\sin }^2}{\rm{\theta }}} = {\rm{\;}}\sqrt {1 - {\rm{\;}}{{\left( {\frac{{\sqrt 5 - 1}}{4}} \right)}^2}} = {\rm{\;}}\sqrt {\frac{{16{\rm{\;}} - {\rm{\;}}5{\rm{\;}} - 1{\rm{\;}} + 2\sqrt 5 }}{{16}}} = \frac{{\sqrt {10 + 2\sqrt 5 } }}{4}\)

∴ \(\tan {\rm{\theta }} = {\rm{\;}}\frac{{\sin {\rm{\theta }}}}{{\cos {\rm{\theta }}}} = {\rm{\;}}\frac{{\left( {\frac{{\sqrt 5 - 1}}{4}} \right)}}{{\left( {\frac{{\sqrt {10 + 2\sqrt 5 } }}{4}} \right)}} = \frac{{\sqrt 5 - 1}}{{\sqrt {10 + 2\sqrt 5 } }}{\rm{\;}}\)

Concept:

• If x, y, z are in A.P ⇔ 2y = x + z
• If x, y, z are in G.P ⇔  y² = xz
• If x, y, z are in H.P ⇔ \({\rm{y}} = {\rm{\;}}\frac{{2{\rm{xz}}}}{{{\rm{x\;}} + {\rm{z}}}}\)
• \({\tan ^{ - 1}}{\rm{x}} + {\rm{\;}}{\tan ^{ - 1}}{\rm{y}} = {\tan ^{ - 1}}\left( {\frac{{{\rm{x}} + {\rm{y}}}}{{1 - {\rm{\;xy}}}}} \right)\)
• \(2{\tan ^{ - 1}}{\rm{x}} = {\tan ^{ - 1}}\left( {\frac{{2{\rm{x}}}}{{1 - {\rm{\;}}{{\rm{x}}^2}}}} \right)\)

Calculation:

Given:

x, y, z are in GP.

∴ y² = xz       ….

(1) tan^-1 x, tan^-1 y and tan^-1 z are in AP.

∴ 2 tan^-1 y = tan^-1 x + tan^-1 z

\( \Rightarrow {\tan ^{ - 1}}\left( {\frac{{2{\rm{y}}}}{{1 - {\rm{\;}}{{\rm{y}}^2}}}} \right) = {\rm{\;}}{\tan ^{ - 1}}\left( {\frac{{{\rm{x}} + {\rm{z}}}}{{1 - {\rm{\;xz}}}}} \right)\) 

\( \Rightarrow \left( {\frac{{2{\rm{y}}}}{{1 - {\rm{\;}}{{\rm{y}}^2}}}} \right) = {\rm{\;}}\left( {\frac{{{\rm{x}} + {\rm{z}}}}{{1 - {\rm{\;xz}}}}} \right)\) 

\(\Rightarrow \left( {\frac{{2{\rm{y}}}}{{1 - {\rm{\;}}{{\rm{y}}^2}}}} \right) = {\rm{\;}}\left( {\frac{{{\rm{x}} + {\rm{z}}}}{{1 - {\rm{\;}}{{\rm{y}}^2}}}} \right)\)                          (∵ y² = xz)

⇒ 2y = x + z

So, x, y, z are in AP

Here x, y, z are in AP as well in GP, it is only possible when x = y = z

Concept:

• \(\tan \left( {{\rm{x}} + {\rm{y}}} \right) = {\rm{\;}}\frac{{\tan {\rm{x}} + {\rm{\;}}\tan {\rm{y}}}}{{1 - {\rm{\;}}\tan {\rm{x}}\tan {\rm{y}}}}\)

Calculation:

Given: tan (α + β) = 2 and tan (α - β) = 1

Now,

\(\Rightarrow \tan 2{\rm{\alpha }} = {\rm{tan\;}}\left[ {\left( {{\rm{\alpha \;}} + {\rm{\beta }}} \right) + \left( {{\rm{\alpha \;}}-{\rm{\beta }}} \right)} \right]{\rm{\;}} = {\rm{\;}}\frac{{\tan \left( {{\rm{\alpha \;}} + {\rm{\;\beta }}} \right) + {\rm{\;}}\tan \left( {{\rm{\alpha \;}}-{\rm{\;\beta }}} \right)}}{{1 - {\rm{\;}}\tan \left( {{\rm{\alpha \;}} + {\rm{\;\beta }}} \right)\tan \left( {{\rm{\alpha \;}}-{\rm{\;\beta }}} \right)}}\) 

\(\Rightarrow \tan 2{\rm{\alpha }} = {\rm{\;}}\frac{{2\; + \;1}}{{1 - 2\; \times \;1}} = {\rm{\;}} - 3\)

Concept:

• sin (90° - θ) = cos θ
• tan (90° - θ) = cot θ
• sin (180° - θ) = sin θ
• tan (180° - θ) = tan θ

Calculation:

We know that A + B + C = 180°    (Triangle ABC)

⇒ B + C = 180° - A

Statement 1:

\({\rm{LHS}} = {\rm{}}\sin \left( {\frac{{{\rm{B}} + {\rm{C}}}}{2}} \right) = \sin \left( {\frac{{180^\circ - {\rm{A}}}}{2}} \right) = \sin \left( {90^\circ - {\rm{\;}}\frac{{\rm{A}}}{2}} \right) = {\rm{}}\cos \left( {\frac{{\rm{A}}}{2}} \right)\) 

∴ Statement 1 is true.

Statement 2:

\({\rm{LHS}} = {\rm{}}\tan \left( {\frac{{{\rm{B}} + {\rm{C}}}}{2}} \right) = \tan \left( {\frac{{180^\circ - {\rm{A}}}}{2}} \right) = \tan \left( {90^\circ - {\rm{\;}}\frac{{\rm{A}}}{2}} \right) = {\rm{}}\cot \left( {\frac{{\rm{A}}}{2}} \right)\) 

∴ Statement 2 is true.

Statement 3:

LHS = sin (B + C) = sin (180° - A) = sin A

∴ Statement 3 is wrong.

Statement 4:

LHS = tan (B + C) = tan (180° - A) = -tan A

∴ Statement 4 is wrong.

Hence option 2 is correct.

Concept:

Calculation:

Given: \(\sec {\rm{\theta }} - {\rm{cosec\;\theta }} = \frac{4}{3}\)

\(\Rightarrow \frac{1}{{\cos {\rm{\theta }}}} - {\rm{\;}}\frac{1}{{\sin {\rm{\theta }}}} = \frac{4}{3}\) 

\(\Rightarrow \frac{{\sin {\rm{\theta }} - {\rm{\;}}\cos {\rm{\theta \;}}}}{{\sin {\rm{\theta }}\cos {\rm{\theta }}}} = \frac{4}{3}\)            ….

(1) Squaring both sides, we get

\( \Rightarrow {\left( {\frac{{\sin {\rm{\theta }} - {\rm{\;}}\cos {\rm{\theta \;}}}}{{\sin {\rm{\theta }}\cos {\rm{\theta }}}}} \right)^2} = {\left( {\frac{4}{3}} \right)^2}\) 

\( \Rightarrow {\rm{\;}}\frac{{{{\sin }^2}{\rm{\theta }} + {{\cos }^2}{\rm{\theta }} - 2\sin {\rm{\theta }}\cos {\rm{\theta }}}}{{{{\left( {\sin {\rm{\theta }}\cos {\rm{\theta }}} \right)}^2}}} = {\rm{\;}}\frac{{16}}{9}\) 

\(\Rightarrow {\rm{\;}}\frac{{1 - 2\sin {\rm{\theta }}\cos {\rm{\theta }}}}{{{{\left( {\sin {\rm{\theta }}\cos {\rm{\theta }}} \right)}^2}}} = {\rm{\;}}\frac{{16}}{9}\) 

Let sin θ cos θ = x

\(\Rightarrow {\rm{\;}}\frac{{1 - 2{\rm{x}}}}{{{{\rm{x}}^2}}} = {\rm{\;}}\frac{{16}}{9}\) 

⇒ 9 – 18x = 16x²

⇒ 16x² + 18x – 9 = 0

⇒ (8x – 3) (2x + 3) = 0

∴ x = 3/8, x = -3/2

So, sin θ cos θ = x = 3/8

Put the value of sin θ cos θ in equation 1^st,

⇒ (sin θ – cos θ) = 4/3 × sin θ cos θ = (4/3) × (3/8) = 1/2

Now, x = -3/2

⇒ sin θ cos θ = -3/2

We know that -1 ≤ sin θ ≤ 1 and -1 ≤ cos θ ≤ 1

∴ -1 ≤ sin θ cos θ ≤ 1

Hence sin θ cos θ = -3/2 is not possible.

So correct answer is 1/2 only.

Concept:

Centroid of a Triangle: The point where the three medians of the triangle intersect.

Let’s consider a triangle. If the three vertices of the triangle are A(x₁, y₁), B(x₂, y₂), C(x₃, y₃).

Centroid of a triangle = \(\left( {\frac{{{{\rm{x}}_1} + {\rm{\;}}{{\rm{x}}_2} + {\rm{\;}}{{\rm{x}}_3}}}{3},\frac{{{{\rm{y}}_1} + {\rm{\;}}{{\rm{y}}_2} + {\rm{\;}}{{\rm{y}}_3}}}{3}} \right)\)

Calculation:

Let’s consider a triangle. If the three vertices of the triangle are A(x₁, y₁), B(x₂, y₂), C(x₃, y₃).

Given: A (x₁, y₁) = A (1, 1)

Mid-Point of AB is given (-1, 2)

\(\frac{{{{\rm{x}}_1} + {\rm{\;}}{{\rm{x}}_2}}}{2} = {\rm{\;}} - 1{\rm{\;and\;}}\frac{{{{\rm{y}}_1} + {\rm{\;}}{{\rm{y}}_2}}}{2} = {\rm{\;}}2{\rm{\;\;}}\) 

⇒ x₂ = -2 – 1 = -3 and y₂ = 4 – 1 = 3

So, B(x₂, y₂) = B (-3, 3)

Mid-Point of AC is given (3, 2)

\(\frac{{{{\rm{x}}_1} + {\rm{\;}}{{\rm{x}}_3}}}{2} = 3{\rm{\;and\;}}\frac{{{{\rm{y}}_1} + {\rm{\;}}{{\rm{y}}_3}}}{2} = {\rm{\;}}2{\rm{\;\;}}\) 

⇒ x₃ = 6 – 1 = 5 and y₃ = 4 – 1 = 3

C(x₃, y₃) = (5, 3)

Centroid of a triangle = \(\left( {\frac{{{{\rm{x}}_1} + {\rm{\;}}{{\rm{x}}_2} + {\rm{\;}}{{\rm{x}}_3}}}{3},\frac{{{{\rm{y}}_1} + {\rm{\;}}{{\rm{y}}_2} + {\rm{\;}}{{\rm{y}}_3}}}{3}} \right) = {\rm{}}\left( {\frac{{1 - 3 + {\rm{\;}}5}}{3},\frac{{1 + {\rm{\;}}3 + {\rm{\;}}3}}{3}} \right) = {\rm{}}\left( {1,\frac{7}{3}} \right)\)

Concept:

Incentre: The incentre is the point where the angle bisectors intersect.

• For a triangle with side lengths a, b, c with vertices at the points (x₁, y₁), (x₂, y₂), (x₃, y₃) the incentre lies at \(\left( {\frac{{{\rm{a}}{{\rm{x}}_1} + {\rm{b}}{{\rm{x}}_2} + {\rm{c}}{{\rm{x}}_3}}}{{{\rm{a}} + {\rm{b}} + {\rm{c}}}},{\rm{\;}}\frac{{{\rm{a}}{{\rm{y}}_1} + {\rm{b}}{{\rm{y}}_2} + {\rm{c}}{{\rm{y}}_3}}}{{{\rm{a}} + {\rm{b}} + {\rm{c}}}}} \right)\)
• For equilateral triangle incentre coincide with the centroid.

Calculation:

OA = 2

\({\rm{OB\;}} = {\rm{\;}}\sqrt {{{\left( {1 - 0} \right)}^2} + {\rm{\;}}{{\left( {\sqrt 3 - 0} \right)}^2}} = 2{\rm{\;}}\)

\({\rm{AB\;}} = {\rm{\;}}\sqrt {{{\left( {1 - 2} \right)}^2} + {\rm{\;}}{{\left( {\sqrt 3 - 0} \right)}^2}} = 2\)

Here OA = OB = AB,

Since, the given triangle is an equilateral triangle.

Therefore, incentre coincide with the centroid.

Incentre of triangle = \(\left( {\frac{{0\; + \;2\; +{\rm{\;}}1}}{3},\frac{{0\; + {\rm{\;}}0\; + {\rm{\;}}\sqrt 3 }}{3}} \right) = {\rm{}}\left( {1,\frac{1}{{\sqrt 3 }}} \right)\)

Concept:

Properties of a parallelogram

• The diagonals of a parallelogram bisect each other.
• Opposite sides of a parallelogram are congruent.
• Opposite angles of a parallelogram are congruent.

Calculation:

Diagonals of a parallelogram bisect each other,

Let the point of intersection of diagonals be P(x, y)

So, P is mid-point of AC,

\(\left( {{\rm{x}},{\rm{\;y}}} \right) = \left( {\frac{{ - 2\; + \;4}}{2},\frac{{ - 1\; + \;3}}{2}} \right) = \left( {1,{\rm{\;}}1} \right)\) 

Let the Fourth vertex be D (x_D, y_D)

P is mid-point of BD,

\(\therefore \left( {{\rm{x}},{\rm{\;y}}} \right) = \left( {1,{\rm{\;}}1} \right) = \left( {\frac{{1\; + \;{{\rm{x}}_{\rm{D}}}}}{2},\frac{{0\; + \;{{\rm{y}}_{\rm{D}}}}}{2}} \right)\)  ⇒ (x_D, y_D) = (1, 2)

Fourth vertex of the parallelogram is D = (1, 2)

Concept:

Let center of two circles are C₁ and C₂ and radius are r₁ and r₂.

Case-1: Circles touching each other externally: If the two circles are touching each other externally then the distance between their centers is sum of their radii.

⇒ C₁C₂ = r₁ + r₂

Case-2: Circles touching each other internally: If the two circles are touching each other internally then the distance between their centers is differences of their radii.

⇒ C₁C₂ = r₁ - r₂

Case-2: Two circles will intersect at two points

⇒ r₁ - r₂ < C₁C₂ < r₁ + r₂

General form of the equation of a circle: x² + y² + 2gx + 2fy + c = 0

• Centre is (-g, -f) or \(\left( {\frac{{ - \;{\rm{coefficient\;of\;x}}}}{2},\;\frac{{ - \;{\rm{coefficient\;of\;y}}\;}}{2}} \right)\) and radius = \(\sqrt {{{\rm{g}}^2} + {{\rm{f}}^2} - {\rm{c}}}\), where g, f and c are three constants

Calculation:

Given circles are x² + y² = r² and x² + y² – 10x + 16 = 0

Let center of two circles are C₁ and C₂ and radius are r₁ and r₂.

C₁ = (0, 0) and r₁ = r

C₂ = (5, 0) and r₂ = \(\sqrt {{5^2} + {0^2} - 16} = 3\)

Now,

\({{\rm{C}}_1}{{\rm{C}}_2} = {\rm{\;}}\sqrt {{{\left( {5 - 0} \right)}^2} + {\rm{\;}}{{\left( {0 - 0} \right)}^2}} = 5\) 

When two circles will intersect at two points

⇒ r₁ - r₂ < C₁C₂ < r₁ + r₂

⇒ r – 3 < 5 < r + 3

∴ 2 < r < 8

Concept:

Equation of circle having centre (h, k) and radius r is (x – h) ² + (y – k) ² = r²

Calculation:

Equation of circle with centre (h, k) is

(x − h)² + (y − k)² = r²

Since the circle passes through (3, −2)

(3 − h) ² + (-2 − k) ² = r²        ….

(1) Since the circle passes through (−2, 0)

(-2 − h) ² + (0 − k) ² = r²        …. (2) From equation 1 and 2,

(3 − h) ² + (-2 − k) ² = (-2 − h) ² + (0 − k) ²

⇒ 9 + h² − 6h + 4 + K² + 4k = 4 + h² + 4h + k²

⇒ 10h − 4k = 9         …. (3) Given: centre (h, k) lies on the line 2x – y – 3 = 0

⇒ 2h – k = 3          …. (4) Solving equation 3 and 4, we get

h = -3/2 and k = -6

Put the value of h and k in equation 1^st,

⇒ (3 + (3/2)) ² + (−2 + 6)² = r²

⇒ r² = 145​/4

So, The equation of the circle is (x + 3/2​)² + (y + 6)² = 145​/4

∴ The equation of the circle is x² + y² + 3x + 12y + 2 = 0

Concept:

Section Formula: Section formula is used to determine the coordinate of a point that divides a line into two parts such that ratio of their length is m : n

⇒ Let A and B be the given two points (x₁, y₁) and (x₂, y₂) respectively and C(x, y) be the point dividing the line- segment AB internally in the ratio m: n

I. Internal Section Formula: When the line segment is divided internally in the ration m: n, we use this formula. ⇔ \(\left( {{\rm{x}},{\rm{\;y}}} \right) = \left( {\frac{{{\rm{m}}{{\rm{x}}_2}{\rm{\;}} + {\rm{\;n}}{{\rm{x}}_1}}}{{{\rm{m}} + {\rm{\;n}}}},\frac{{{\rm{m}}{{\rm{y}}_2}{\rm{\;}} + {\rm{\;n}}{{\rm{y}}_1}}}{{{\rm{m}} + {\rm{n}}}}} \right)\)

II. External Section Formula: When the point C lies on the external part of the line segment. ⇔ \(\left( {{\rm{x}},{\rm{\;y}}} \right) = \left( {\frac{{{\rm{m}}{{\rm{x}}_2} - {\rm{n}}{{\rm{x}}_1}}}{{{\rm{m}} - {\rm{\;n}}}},\frac{{{\rm{m}}{{\rm{y}}_2} - {\rm{n}}{{\rm{y}}_1}}}{{{\rm{m}} - {\rm{\;n}}}}} \right)\)

Calculation:

Given points are A (-2, -2) and B (2, -4)

Let, point \({\rm{C}}\left( { - \frac{2}{7},{\rm{\;}} - \frac{{20}}{7}} \right)\) ) be the point dividing the line- segment AB in the ratio k : 1

Applying section formula,

\(\left( { - \frac{2}{7},{\rm{\;}} - \frac{{20}}{7}} \right) = \left( {\frac{{{\rm{k\;}} \times {\rm{\;}}2{\rm{\;}} + {\rm{\;}}1{\rm{\;}} \times - 2}}{{{\rm{k\;}} + {\rm{\;}}1}},\frac{{{\rm{k\;}} \times - 4{\rm{\;}} + {\rm{\;}}1{\rm{\;}} \times - 2}}{{{\rm{k\;}} + {\rm{\;}}1}}} \right)\)

\( \Rightarrow - \frac{2}{7} = {\rm{\;}}\frac{{{\rm{k\;}} \times {\rm{\;}}2{\rm{\;}} + {\rm{\;}}1{\rm{\;}} \times - 2}}{{{\rm{k\;}} + {\rm{\;}}1}}\)  

⇒ -2 (k + 1) = 7 × (2k – 2)

⇒ -k – 1 = 7k – 7

⇒ 8k = 6

∴ k = 3/4

So, k : 1 = 3 : 4

Concept:

Standard equation of an ellipse: \(\frac{{{\rm{\;}}{{\bf{x}}^2}}}{{{{\bf{a}}^2}}} + \frac{{{{\bf{y}}^2}}}{{{{\bf{b}}^2}}} = 1\) (a > b)

• Coordinates of foci = (± ae, 0)
• Eccentricity (e) = \(\sqrt {1 - {\rm{\;}}\frac{{{{\rm{b}}^2}}}{{{{\rm{a}}^2}}}} \) ⇔ a²e² = a² – b²

Calculation:

Given: Focus: (±2, 0) and eccentricity = 1/4

⇒ Focus = (±ae, 0) = (±2, 0)

⇒ ae = 2

⇒ a × (1/4) = 2

∴ a = 8

⇒ a² = 64

Now,

a²e² = a² – b²

⇒ 2² = 8² - b²

∴ b² = 60

Equation of an ellipse = \(\frac{{{\rm{\;}}{{\rm{x}}^2}}}{{64}} + \frac{{{{\rm{y}}^2}}}{{60}} = 1\)

Concept:

• Equation of a line parallel to a given line ax + by + c = 0 is ax + by + λ = 0, where λ is a constant

Calculation:

Equation of a line parallel to a given line 2x + 3y + 1 = 0 is 2x + 3y + λ = 0

It is passing through point (-1, 2)

⇒ (2 × -1) + (3 × 2) + λ = 0

⇒ -2 + 6 + λ = 0

∴ λ = -4

So, Equation of line is 2x + 3y – 4 = 0

Concept:

• Angle between two lines: The angle θ between the lines having slope m₁ and m₂ is given by

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

Calculation:

Let Slope of √2 x + √3 y = 1 is m₁,

\(\Rightarrow {{\rm{m}}_1} = -{\rm{}}\frac{{\sqrt 2 }}{{\sqrt 3 }}\) 

Let Slope of √3 x + √2 y = 2 is m₂,

\(\Rightarrow {{\rm{m}}_2} = -{\rm{\;}}\frac{{\sqrt 3 }}{{\sqrt 2 }}\) 

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

\(\Rightarrow \tan {\rm{\theta }} = {\rm{\;}}\left| {\frac{{\frac{{-\sqrt 3 }}{{\sqrt 2 }} - (\frac{{-\sqrt 2 }}{{\sqrt 3 }})}}{{1 + \left( {\frac{{\sqrt 2 }}{{\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 2 }}} \right)}}} \right| = \;\left| {\frac{{-3 + 2}}{{2 \times \sqrt 3 \times \sqrt 2 }}} \right| = \;\frac{1}{{2\sqrt 6 }}\) 

\(\therefore {\rm{\theta }} = \;{\tan ^{ - 1}}\left( {\frac{1}{{2\sqrt 6 }}} \right)\)