Quantitative Aptitude PYQs

Concept:

If A (x₁, y₁, z₁), B (x₂, y₂, z₂) and C (x₃, y₃, z₃) are the vertices of ΔABC. Then the centroid of ΔABC is given by: \(\left( {\frac{{{x_1} + {x_2} + {x_3}}}{3},\frac{{{y_1} + {y_2} + {y_3}}}{3},\frac{{{z_1} + {z_2} + {z_3}}}{3}} \right)\)

Calculation:

Given: The vertices of ΔABC are: A(2, -3, 3), B(5, -3, -4) and C(2, -3, -2)

As we know that, if A (x₁, y₁, z₁), B (x₂, y₂, z₂) and C (x₃, y₃, z₃) are the vertices of ΔABC. Then the centroid of ΔABC is given by: \(\left( {\frac{{{x_1} + {x_2} + {x_3}}}{3},\frac{{{y_1} + {y_2} + {y_3}}}{3},\frac{{{z_1} + {z_2} + {z_3}}}{3}} \right)\)

\(\Rightarrow ;\left( {\frac{{2 + 5 + 2}}{3},\frac{{ - ;3 - 3 - 3}}{3},\frac{{3 - 4 - 2}}{3}} \right) = \left( {3,; - 3,; - 1} \right)\)

Concept:

The general equation of sphere is x² + y² + z² + 2ux + 2vy + 2wz + d = 0 with centre at (-u, -v, -w) and radius \(r = \sqrt {{u^2} + {v^2} + {w^2} - d}\)

Calculation:

Given: x² + y² + z² – 6x + 8y – 10z + 1 = 0 represents a sphere.

Now, by comparing the given equation of sphere with the general equation of sphere x² + y² + z² + 2ux + 2vy + 2wz + d = 0, we get: u = -3, v = 4, w = -5 and d = 1.

As we know that, for a sphere x² + y² + z² + 2ux + 2vy + 2wz + d = 0 the centre is (- u, - v, - w) and radius \(r = \sqrt {{u^2} + {v^2} + {w^2} - d}\)

\(r = \sqrt {{u^2} + {v^2} + {w^2} - d} = \sqrt {{{\left( { - ;3} \right)}^2} + {4^2} + {{\left( { - ;5} \right)}^2} - 1} = 7\)

Concept:

If a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 represents two different planes, then equation of plane passing through the intersection of these planes is given by:

(a₁x + b₁y + c₁z + d₁) + λ × (a₂x + b₂y + c₂z + d₂) = 0.

Calculation:

Given: Two planes 2x + y + 2z – 9 = 0 and 4x – 5y – 4z – 1 = 0.

As we know that, if a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 represents two different planes, then equation of plane passing through the intersection of these planes is given by:

(a₁x + b₁y + c₁z + d₁) + λ × (a₂x + b₂y + c₂z + d₂) = 0.

So, the plane passing through the intersection of two given planes is:

⇒ (2x + y + 2z – 9) + λ × (4x – 5y – 4z – 1) = 0     ...1)

∵ it is given that the plane passing through the intersection of two given plane also passes through the point (3, 2, 1).

⇒ The point (3, 2, 1) will satisfy the equation (1)

⇒ (6 + 2 + 2 - 9) + λ × (12 – 10 – 4 - 1) = 0 ⇒ λ = 1/3.

So, by substituting the value of λ in equation (1), we get

⇒ 10x – 2y + 2z = 28

Concept:

The distance between two parallel planes ax + by + cz + d₁ = 0 and ax + by + cz + d₂ = 0 is given by: \(\left| {\frac{{{d_1} - {d_2}}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right|\)

Calculation:

Given: 4x – 2y + 4z + 9 = 0 and 8x – 4y + 8z + 21 = 0

Now, we can rewrite 8x – 4y + 8z + 21 = 0 as 4x – 2y + 4z + (21/2) = 0

So, let P₁ = 4x – 2y + 4z + 9 = 0 and P₂ = 4x – 2y + 4z + (21/2) = 0

As we know that, the distance between two parallel planes ax + by + cz + d₁ = 0 and ax + by + cz + d₂ = 0 is given by: \(\left| {\frac{{{d_1} - {d_2}}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right|\)

Here, a = 4, b = - 2, c = 4, d₁ = 9 and d₂ = 21/2

\(\Rightarrow ;\left| {\frac{{{d_1} - {d_2}}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right| = \left| {\frac{{9 - \frac{{21}}{2}}}{{\sqrt {{4^2} + {{\left( { - ;2} \right)}^2} + {4^2}} }}} \right| = \frac{1}{4}\)

Concept:

If L is a line in a 3D plane makes angles α, β and γ with the positive directions of x – axis, Y – axis and z – axis respectively, then cos α, cos β and cos γ are called its direction cosines. The direction cosines are denoted as l, m and n.

i.e l = cos α, m = cos β and n = cos γ

Calculation:

As we know that, the z – axis makes 90°, 90° and 0° with the positive directions of x – axis, Y – axis and z – axis respectively.

So, the direction cosines of z – axis are: l = cos 90°, m = cos 90° and n = cos 0°

Hence, the direction cosines of z – axis are: < 0, 0, 1 >

Concept:

If \(\vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k;and;\vec b = {b_1}\hat i + {b_2}\hat j + {b_3}\hat k\) then \(\vec a \cdot ;\vec b = {a_1}{b_1} + {a_2}{b_2} + {a_3}{b_3}\)

If \(\vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k\) and λ is a scalar then \(\lambda \times \vec a = \lambda {a_1}\hat i + \lambda {a_2}\hat j + \lambda {a_3}\hat k\)

Calculation:

Given: \(\vec a = \hat i - 2\hat j + 5\hat k\) and \(\vec b = 2\hat i + \hat j - 3\hat k\)

\( \Rightarrow ;\left( {\vec b - \vec a} \right) = \left( {2\hat i + \hat j - 3\hat k} \right) - \left( {\hat i - 2\hat j + 5\hat k{\rm{;}}} \right) = ;\hat i + 3\hat j - 8\hat k\)

\(\Rightarrow ;\left( {3\vec a + \vec b} \right) = \left{ {3;\left( {\hat i - 2\hat j + 5\hat k} \right)} \right} + ;2\hat i + \hat j - 3\hat k = 5\hat i - 5\hat j + 12\hat k\)

\(\Rightarrow ;\left( {\vec b - \vec a} \right) \cdot \left( {3\vec a + \vec b} \right) = \left( {\hat i + 3\hat j - 8\hat k} \right) \cdot \left( {5\hat i - 5\hat j + 12\hat k} \right) = 5 - 15 - 96 = ; - 106\)

Concept:

If A and B are points with position vectors \(\vec a;and;\vec b\) respectively then \(\overrightarrow {AB} = ;\vec b - ;\vec a\)

If \(\vec a = ;{a_1}\hat i + {a_2}\hat j + {a_3}\hat k\) is a vector then the magnitude of the vector is given by

\(\left| {\vec a} \right| = \sqrt {a_1^2 + a_2^2 + a_3^2}\)

Calculation:

Given: The position vectors of points A and B are \(3\hat i - 2\hat j + \hat k\) and \(2\hat i + 4\hat j - 3\hat k\) respectively

As we know that, If A and B are points with position vectors \(\vec a;and;\vec b\) respectively then \(\overrightarrow {AB} = ;\vec b - \vec a\)

\(\Rightarrow ;\overrightarrow {AB} = \left( {2\hat i + 4\hat j - 3\hat k} \right) - \left( {3\hat i - 2\hat j + \hat k} \right) = ; - ;\hat i + 6\hat j - 4\hat k\)

As we know that, If \(\vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k\) is a vector then the magnitude of the vector is given by

\(\left| {\vec a} \right| = \sqrt {a_1^2 + a_2^2 + a_3^2}\)

\( \Rightarrow ;\left| {\overrightarrow {AB} } \right| = \sqrt {{{\left( { - 1} \right)}^2} + {6^2} + {{\left( { - ;4} \right)}^2}} = \sqrt {53} \)

Calculation:

Given: In the right-angled triangle ABC, hypotenuse AC = p

As we can see from the above diagram angle between vector \(\overrightarrow {BC} ;and;\overrightarrow {BA}\) is 90° ⇒ \(\overrightarrow {BC} ; \cdot \overrightarrow {BA} = 0\)

\(\Rightarrow ;\overrightarrow {AB} \cdot \overrightarrow {AC} + \overrightarrow {BC} ; \cdot \overrightarrow {BA} + \overrightarrow {CA} \cdot \overrightarrow {CB} = ;\overrightarrow {AB} \cdot \overrightarrow {AC} + ;\overrightarrow {CA} \cdot \overrightarrow {CB}\)

\(\Rightarrow ;\overrightarrow {AB} \cdot \overrightarrow {AC} + \overrightarrow {BC} ; \cdot \overrightarrow {BA} + \overrightarrow {CA} \cdot \overrightarrow {CB} = \overrightarrow {AB} \cdot \overrightarrow {AC} + ;\overrightarrow {AC} \cdot \overrightarrow {BC}\)

\( = ;\overrightarrow {AC} \cdot \left( {\overrightarrow {AB} + ;\overrightarrow {BC} } \right) = \overrightarrow {AC} \cdot ;\overrightarrow {AC} = {\left| {\overrightarrow {AC} } \right|^2}\)

\(\Rightarrow ;\overrightarrow {AB} \cdot \overrightarrow {AC} + \overrightarrow {BC} ; \cdot \overrightarrow {BA} + \overrightarrow {CA} \cdot \overrightarrow {CB} = {p^2}\)

Concept:

If \(\vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k;and;\vec b = {b_1}\hat i + {b_2}\hat j + {b_3}\hat k\) then \(\vec a \cdot ;\vec b = \left| {\vec a} \right| \times \left| {\vec b} \right|\cos \theta\)

Calculation:

Given: \(\vec a = 2\hat i - 6\hat j - 3\hat k\) and \(\vec b = 4\hat i + 3\hat j - \hat k\)

\(\left| {\vec a} \right| = 7,;\left| {\vec b} \right| = \sqrt {26} ;and;\vec a \cdot ;\vec b = - 7\)

\(\Rightarrow ;\cos \theta = \frac{{\vec a \cdot ;\vec b}}{{\left| {\vec a} \right| \times \left| {\vec b} \right|}} = \frac{{ - ;7}}{{7 \times \sqrt {26} }} = - \frac{1}{{\sqrt {26} }}\)

\( \Rightarrow ;{\sin ^2}\theta = 1 - {\cos ^2}\theta = 1 - \frac{1}{{26}} = \frac{{25}}{{26}}\)

\(\Rightarrow ;\sin \theta = \frac{5}{{\sqrt {26} }}\)

Concept:

If vectors \(\vec a;and;\vec b\) are perpendicular then \(\vec a \cdot ;\vec b = 0\)

Calculation:

Given: \(3\hat i + 4\hat j - \hat k\) and \(- 2\hat i + \lambda \hat j + 10\hat k\) are perpendicular

\(\Rightarrow \vec a \cdot ;\vec b = \left( {3\hat i + 4\hat j - \hat k} \right) \cdot \left( { - 2\hat i + \lambda \hat j + 10\hat k} \right) = ; - 6 + 4\lambda - 10 = 4\lambda - 16\)

∵ \(3\hat i + 4\hat j - \hat k\) and \(- 2\hat i + \lambda \hat j + 10\hat k\) are perpendicular

⇒ 4λ – 16 = 0

⇒ λ = 4

Concept:

• sec2θ = 1 + tan2θ

Calculation:

Let y = sec² (tan^-1 x)

Since, tan-1 x is an angle 

Thus, Substitute tan-1 x by any angle θ.

i.e., let tan^-1x = θ ...........(equation 1)

⇒ x = tanθ 

⇒ y = sec²θ 

⇒ y = 1 + tan2θ 

⇒ y = 1 + x²  (Using equation 1)

Now, y = sec2 (tan-1 x) becomes y = 1 + x2

Differentiating y with respect to x,

We get,

\(\frac{dy}{dx}=2x\)

Calculation:

Given: f(x) = log₁₀ (1 + x)

⇒ f(4) = log₁₀ (5)

⇒ f(1) = log₁₀ (2)

⇒ 4f(4) + 5f(1) – log₁₀ 2 = log₁₀ 625 + log₁₀ 32 - log₁₀ 2

⇒ 4f(4) + 5f(1) – log₁₀ 2 = log₁₀ (625 × 32) - log₁₀ 2

⇒ 4f(4) + 5f(1) – log₁₀ 2 = log₁₀ (2 × 10⁴) - log₁₀ 2

⇒ 4f(4) + 5f(1) – log₁₀ 2 = log₁₀ 2 + 4 log₁₀ 10 - log₁₀ 2

⇒ 4f(4) + 5f(1) – log₁₀ 2 = 4

Concept:

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

Calculation:

Given: f(x) = In \(\left( {\sqrt {{x^2} + 1} - x} \right)\)

⇒ f(- x) = ln \(\left( {\sqrt {{x^2} + 1} + x} \right)\)

Let’s calculate - f(x) = - ln \(\left( {\sqrt {{x^2} + 1} - x} \right)\)

⇒ - f(x) = ln \(\left[ {{{\left( {\sqrt {{x^2} + 1} - x} \right)}^{ - 1}}} \right] = \ln \frac{1}{{\left( {\sqrt {{x^2} + 1} - x} \right)}}\)

\(\Rightarrow - f\left( x \right) = \ln \left[ {\frac{{\sqrt {{x^2} + 1} + x}}{{\left( {\sqrt {{x^2} + 1} - x} \right) \times ;\left( {\sqrt {{x^2} + 1} + x} \right)}}} \right]\)

\(\Rightarrow - f\left( x \right) = \ln \left( {\sqrt {{x^2} + 1} + x} \right)\)

Hence, f(- x) = - f(x) so, f is an odd function

Calculation:

Given: f(x) = log_x 10

We can re write given f(x) as \(f\left( x \right) = \frac{{\log 10}}{{\log x}}\)

We know that, f(x) will exist only if x > 0 and x ≠ 1.

So, the domain of the given function f is x > 0 excluding x = 1

Concept:

L’ Hospital’s rule:

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

Calculation:

Given: \(\mathop {\lim }\limits_{x \to 0} \frac{{1 - {{\cos }^3}4x}}{{{x^2}}}\)

\(\Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{1 - {{\cos }^3}4x}}{{{x^2}}} = \frac{0}{0}\)

By applying L hospital’s rule we get

\(\Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{1 - {{\cos }^3}4x}}{{{x^2}}} = ;\mathop {\lim }\limits_{x; \to 0} \frac{{12{{\cos }^2}4x\sin 4x}}{{2x}}\)

\(\Rightarrow \mathop {\lim }\limits_{x; \to 0} \frac{{12{{\cos }^2}4x\sin 4x}}{{2x}} = ;\mathop {\lim }\limits_{x \to 0} \frac{{24;\left( {{{\cos }^2}4x\sin 4x} \right)}}{{4x}}\)

\(\Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{24;\left( {{{\cos }^2}4x\sin 4x} \right)}}{{4x}} = 24 \times ;\mathop {\lim }\limits_{x \to 0} {\cos ^2}4x \times ;\mathop {\lim }\limits_{x \to 0} \frac{{\sin 4x}}{{4x}}\)

\(\Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{1 - {{\cos }^3}4x}}{{{x^2}}} = 24 \times 1 \times 1 = 24\)

Concept:

Area of the circle = π × r² 

The perimeter of the circle = 2πr 

Calculation:

We are given that f(r) is the ratio of perimeter to the area of a circle of radius r,

for r > 0.(r > 0 because r represents the radius of the circle and it should be more than 0 as the radius of the circle is the magnitude of the distance between the center of the circle to its boundary.)

that is, f(r) = \(\frac{perimeter~of ~circle}{area~of~circle}\)

Since, 

Area of the circle = π × r2 

The perimeter of the circle = 2πr

Thus,

⇒ f(r) =  \( \frac{{2;π r;}}{{π {r^2}}} = \frac{2}{r};\)

⇒ f(1) = \(\frac{2}{1}\) = 2 and f(2) = \(\frac{2}{2}\) = 1

⇒ f(1) + f(2) = 2 + 1 = 3

Calculation:

Given: f(x) = 3^{1+ x}

Then,

⇒ f(y) = 3^{1+ y}

⇒ f(z) = 3^{1+ z}

We have to find out value of f(x) f(y) f(z),

⇒ f(x) f(y) f(z) = 3^{1+ x} × 3^{1+ y} × 3^{1+ z}

⇒ f(x) f(y) f(z) = 3^{3+ x + y + z} = 3^{1 + 2 + x + y + z} = f(x + y + z + 2)

Calculation:

Given: x² + 9 |x| + 20 = 0

Case – 1: If x ≥ 0

⇒ x² + 9x + 20 = 0

⇒ x² + 4x + 5x + 20 = 0

⇒ (x + 4) (x + 5) = 0

⇒ x = - 4 or – 5

∵ x ≥ 0

So, f(x) has no real roots for this case.

Case-2: If x < 0

⇒ x² - 9x + 20 = 0

⇒ x² - 4x - 5x + 20 = 0

⇒ (x - 4) (x - 5) = 0

⇒ x = 4 or 5

∵ x < 0

So, f(x) has no real roots for this case.

Concept:

• Chain rule of Differentiation is given by:

\(\frac{df(g(x))}{dx}=\frac{d(f(g(x))}{dx}\times \frac{dg(x)}{dx}\)........(equation 1)

• \(\frac{d}{dx}sinx=cosx\)
• \(\frac{d}{dx}cosx=-sinx\)

Calculation:

We are given that f(x) = sin (cos x)

On differentiating both sides with respect to x, we get,

⇒\( f'\left( x \right) = \cos \left( {\cos x} \right) \times \frac{{d\left( {\cos x} \right)}}{{dx}}\) (Using equation 1)

⇒ \( f'\left( x \right) = \cos \left( {\cos x} \right) \times (-sin ~x)\)

⇒ \( f'\left( x \right) = \left( { - \sin x} \right)\cos \left( {\cos x} \right)\)

Concept:

Domain:

• Let f : A → B be a function, then the set A is called as the domain of the function f.

Co-domain:

• Let f : A → B be a function, then the set B is called as the co-domain of the function f.

Calculation:

Given: \(f\left( x \right) = \sqrt {\left( {2 - x} \right)\left( {x - 3} \right)} \)

∵ The given function is  square root so the expression inside the square root will always be ≥ 0 

In order to define the domain of the given function,

(2 - x) (x - 3)  ≥  0

⇒ (x - 2) (x - 3) ≤ 0

⇒ x ∈ [2, 3] 

So, when x ∈ [2, 3] then f(x) ≥ 0.

Hence, the domain of a given function is [2, 3]