Quantitative Aptitude PYQs

Concept:

• Order: The order of a differential equation is the order of the highest derivative appearing in it.
• Degree: The degree of a differential equation is the power of the highest derivative occurring in it, after the Equation has been expressed in a form free from radicals as far as the derivatives are concerned.

 

Calculation:

Given: \({\left[ {1 + {{\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)}^2}} \right]^3} = {{\rm{\rho }}^2}{\left[ {\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}}} \right]^2}\)

Here, highest derivative = \(\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}}\)

So order = 2,

And Power of the highest derivative = 2

Hence, option (2) is correct.

Concept:

• \({\sin ^{ - 1}}{\rm{x}} + {\cos ^{ - 1}}{\rm{x}} = \frac{{\rm{\pi }}}{2} \Rightarrow {\cos ^{ - 1}}{\rm{x}} = \frac{{\rm{\pi }}}{2} - {\sin ^{ - 1}}{\rm{x}}\)
• \(2{\tan ^{ - 1}}{\rm{x}} = {\sin ^{ - 1}}\left( {\frac{{2{\rm{x}}}}{{1 + {{\rm{x}}^2}}}} \right), - 1 < {\rm{x}} < 1\)
• \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\tan }^{ - 1}}{\rm{x}}} \right) = \frac{1}{{1 + {{\rm{x}}^2}}}\)

 

Calculation:

Given: \({\rm{y}} = {\cos ^{ - 1}}\left( {\frac{{2{\rm{x}}}}{{1 + {{\rm{x}}^2}}}} \right)\)

We know, \({\cos ^{ - 1}}{\rm{x}} = \frac{{\rm{\pi }}}{2} - {\sin ^{ - 1}}{\rm{x}}\)

So, \({\rm{y}} = {\cos ^{ - 1}}\left( {\frac{{2{\rm{x}}}}{{1 + {{\rm{x}}^2}}}} \right)\)

\(\Rightarrow {\rm{y}} = \frac{{\rm{\pi }}}{2} - {\sin ^{ - 1}}\left( {\frac{{2{\rm{x}}}}{{1 + {{\rm{x}}^2}}}} \right)\)

We also know that, \({\sin ^{ - 1}}\left( {\frac{{2{\rm{x}}}}{{1 + {{\rm{x}}^2}}}} \right) = 2{\tan ^{ - 1}}{\rm{x}},{\rm{\;}} - 1 < {\rm{x}} < 1\)

\(\therefore {\rm{y}} = \frac{{\rm{\pi }}}{2} - {\sin ^{ - 1}}\left( {\frac{{2{\rm{x}}}}{{1 + {{\rm{x}}^2}}}} \right)\)

\(\Rightarrow {\rm{y}} = \frac{{\rm{\pi }}}{2} - 2{\tan ^{ - 1}}{\rm{x}}\) , |x| < 1

Now, \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\frac{{\rm{\pi }}}{2} - 2{{\tan }^{ - 1}}{\rm{x}}} \right)\)

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

\(= \frac{{ - 2}}{{1 + {{\rm{x}}^2}}}\) |x| < 1

Hence, option (1) is correct.

Concept:

• A function is differentiable at a point when there's a defined derivative at that point.
• Denominator should not be zero
• Chain Rule: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right)} \right] = {\rm{\;f'}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right){\rm{g'}}\left( {\rm{x}} \right)\)

 

Calculation:

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

\({\rm{f'}}\left( {\rm{x}} \right) = \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{f}}\left( {\rm{x}} \right)} \right)\)

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

\(= \frac{1}{{2\sqrt {1 - {{\rm{e}}^{ - {{\rm{x}}^2}}}} }} \times \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {1 - {{\rm{e}}^{ - {{\rm{x}}^2}}}} \right)\)      \(\left( \because {\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\sqrt {\rm{x}} } \right) = \frac{1}{{2\sqrt {\rm{x}} }}} \right)\)

\(\Rightarrow {\rm{f'}}\left( {\rm{x}} \right) = \frac{1}{{2\sqrt {1 - {{\rm{e}}^{ - {{\rm{x}}^2}}}} }}\left( { - {{\rm{e}}^{ - {{\rm{x}}^2}}}} \right)\frac{{\rm{d}}}{{{\rm{dx}}}}\left( { - {{\rm{x}}^2}} \right)\)

\(= \frac{{\left( { - {{\rm{e}}^{ - {{\rm{x}}^2}}}} \right)}}{{2\sqrt {1 - {{\rm{e}}^{ - {{\rm{x}}^2}}}} }}\left( { - 2{\rm{x}}} \right)\)

\(= \frac{{{\rm{x}}{{\rm{e}}^{ - {{\rm{x}}^2}}}}}{{\sqrt {1 - {{\rm{e}}^{ - {{\rm{x}}^2}}}} }}\)

Now, at x = 0, f’(x) = 0/0

So, f’(x) is defined for all values of x except x = 0

⇒ f(x) is differentiable on (-∞, 0) ∪ (0, ∞)

Hence, option (3) is correct.

Concept:

Following steps to finding maxima and minima using derivatives.

• Find the derivative of the function.
• Set the derivative equal to 0 and solve. This gives the values of the maximum and minimum points.
• Now we have find second derivative.

I. f``(x) is less than 0 then the given function is said to be maxima

II. If f``(x) Is greater than 0 then the function is said to be minima

• sin x = cos x when x = π/4

 

Short trick:

Maximum value of trigonometric function

\({\bf{a}}\;{\bf{sin}}\;{\bf{x}}\; + \;{\bf{b}}\;{\bf{cos}}\;{\bf{x}}\) is \(\sqrt {{{\bf{a}}^2} + {{\bf{b}}^2}} \)  

**Calculation:**​

A) f(x) = sin x + cos x

f’(x) = cos x - sin x = 0 ⇒ sin x = cos x

∴ x = π/4

f’’(x) = -sin x - cos x

⇒ \({\rm{f''}}\left( {\frac{{\rm{\pi }}}{4}} \right) = - \sin \left( {\frac{{\rm{\pi }}}{4}} \right) - \cos \left( {\frac{{\rm{\pi }}}{4}} \right)\)

\(= - \frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 2 }}\)    \(= - \sqrt 2 < 0\)

Maximum value gets at x = π/4

\({\rm{f}}\left( {\frac{{\rm{\pi }}}{4}} \right){\rm{\;}} = \sin \left( {\frac{{\rm{\pi }}}{4}} \right) + \cos \left( {\frac{{\rm{\pi }}}{4}} \right)\)

\(= \frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }}\)  \(= \frac{2}{{\sqrt 2 }}\)  \(= \sqrt 2 \)

Alternative way:

Use short trick to find maximum value of sin x + cos x

We know maximum value of \({\rm{a\;sin\;x\;}} + {\rm{\;b\;cos\;x}}\) is \(\sqrt {{{\rm{a}}^2} + {{\rm{b}}^2}} \)

∴ Maximum value of sin x + cos x is

\(\sqrt {{1^2} + {1^2}} \)   \(= \sqrt 2 \)

A → (2)

B) f(x) = 3 sin x + 4 cos x

Use short trick to find maximum value of 3 sin x + 4 cos x

We know maximum value of \({\rm{a\;sin\;x\;}} + {\rm{\;b\;cos\;x}}\) is \(\sqrt {{{\rm{a}}^2} + {{\rm{b}}^2}} \)

∴ Maximum value of 3 sin x + 4 cos x is

\(\sqrt {{3^2} + {4^2}} = \sqrt {9 + 16} = \sqrt {25} \)   = 5 

B → (3)

C) f (x) = 2 sin x + cos x

Use short trick to find maximum value of 2 sin x + cos x

We know maximum value of \({\rm{a\;sin\;x\;}} + {\rm{\;b\;cos\;x}}\) is \(\sqrt {{{\rm{a}}^2} + {{\rm{b}}^2}} \)

∴ Maximum value of 2 sin x + cos x is

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

C → (4)

D) f(x) = Sin x + 3 cos x

Use short trick to find maximum value of sin x + 3 cos x

We know maximum value of \({\rm{a\;sin\;x\;}} + {\rm{\;b\;cos\;x}}\) is \(\sqrt {{{\rm{a}}^2} + {{\rm{b}}^2}} \)

∴ Maximum value of sin x + 3 cos x is

\(\sqrt {{1^2} + {3^2}} = \sqrt {1 + 9} \)   = √10

D → (1)

Hence, option (2) is correct.

Concept:

We say f(x) is continuous at x = c if

LHL = RHL = value of f(c)

i.e., \(\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{c}}^ - }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {{\rm{c}}^ + }} {\rm{f}}\left( {\rm{x}} \right) = {\rm{f}}\left( {\rm{c}} \right)\)

We say f(x) is differentiable at x = c, if

LHD = RHD

\({\rm{LHD}} = \mathop {{\rm{lim}}}\limits_{{\rm{h}} \to {0^ - }} \frac{{{\rm{f}}\left( {{\rm{c}} - {\rm{h}}} \right) - {\rm{f}}\left( {\rm{c}} \right)}}{{ - {\rm{h}}}}\)

\({\rm{RHD}} = \mathop {{\rm{lim}}}\limits_{{\rm{h}} \to {0^ + }} \frac{{{\rm{f}}\left( {{\rm{c}} + {\rm{h}}} \right) - {\rm{f}}\left( {\rm{c}} \right)}}{{\rm{h}}}\)

Calculation:

Given:

\({\rm{f}}\left( {\rm{x}} \right) = {\rm{x}}\left( {\sqrt {\rm{x}} - \sqrt {{\rm{x}} + 1} } \right)\)

Let’s check continuity of f(x) at x = 0

\({\rm{LHL}} = {\rm{\;}}\mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {0^ - }} {\rm{x}}\left( {\sqrt {\rm{x}} - \sqrt {{\rm{x}} + 1} } \right) = 0\)

\(\begin{array}{l} {\rm{RHL}} = \mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {0^ + }} {\rm{x}}\left( {\sqrt {\rm{x}} - \sqrt {{\rm{x}} + 1} } \right) = 0\\ {\rm{f}}\left( 0 \right) = 0\left( {\sqrt 0 - \sqrt {\left( {0 + 1} \right)} } \right) = 0 \end{array}\)

Here, LHL = RHL = f (0) = 0

⇒ f(x) is continuous at x = 0

Now, let’s check differentiability of f(x) at x = 0

\(\begin{array}{l} {\rm{LHD}} = {\rm{f'}}\left( {{0^ - }} \right)= \mathop {{\rm{lim}}}\limits_{{\rm{h}} \to {0^ - }} \frac{{{\rm{f}}\left( {0 - {\rm{h}}} \right) - {\rm{f}}\left( 0 \right)}}{{ - {\rm{h}}}}\\ \Rightarrow \mathop {{\rm{lim}}}\limits_{{\rm{h}} \to {0^ - }} \frac{{ - {\rm{h}}\left( {\sqrt {\left( { - {\rm{h}}} \right)} - \sqrt { - {\rm{h}} + 1} } \right)}}{{ - {\rm{h}}}} \end{array}\)

\(\begin{array}{l} \Rightarrow \mathop {{\rm{lim}}}\limits_{{\rm{h}} \to {0^ - }} \sqrt {\left( { - {\rm{h}}} \right)} - \sqrt { - {\rm{h}} + 1} = \sqrt 0 - \sqrt {0 + 1} \\ \Rightarrow {\rm{LHD}} = {\rm{\;}} - 1 \end{array}\)

Now, \({\rm{RHD}} = {\rm{\;f'}}\left( {{0^ + }} \right)\)

\(\begin{array}{l} = \mathop {{\rm{lim}}}\limits_{{\rm{h}} \to {0^ + }} \frac{{{\rm{f}}\left( {0 + {\rm{h}}} \right) - {\rm{f}}\left( 0 \right)}}{{\rm{h}}}\\ \Rightarrow \mathop {{\rm{lim}}}\limits_{{\rm{h}} \to {0^ + }} \frac{{{\rm{h}}\left( {\sqrt {\left( {\rm{h}} \right)} - \sqrt {{\rm{h}} + 1} } \right)}}{{\rm{h}}} \end{array}\)

\(\begin{array}{l} \Rightarrow\mathop {{\rm{lim}}}\limits_{{\rm{h}} \to {0^ + }} \sqrt {\left( {\rm{h}} \right)} - \sqrt {{\rm{h}} + 1} = \sqrt 0 - \sqrt {0 + 1} \end{array}\)

⇒ RHD = -1

Here, LHD = RHD = -1

⇒ f(x) is differentiable at x = 0

Hence, option (2) is correct.

Calculation:

Here, f(x) = y = x/x and x ≠ 0

Now, at y = 1, x ≠ 0

But at x = 0, we get 0/0 (indeterminate form)

So, f(x) = x/x = 1 for all point except x = 0

Hence, option (3) is correct.

Concept:

• The value of ⌊x⌋ is the largest integer that is less than or equal to x.

 

Example: [0.23] = 0, [4.2] = 4, [-3.2] = -4

Calculation:

Given: \({\rm{f}}\left( {\rm{n}} \right) = \left[ {\frac{1}{4} + \frac{{\rm{n}}}{{1000}}} \right]\)

To find: \(\mathop \sum \limits_{{\rm{n}} = 1}^{1000} {\rm{f}}\left( {\rm{n}} \right)\)

\(\begin{array}{l} \mathop \sum \limits_{{\rm{n}} = 1}^{1000} {\rm{f}}\left( {\rm{n}} \right) = \left[ {\frac{1}{4} + \frac{1}{{1000}}} \right] + \left[ {\frac{1}{4} + \frac{2}{{1000}}} \right] + \left[ {\frac{1}{4} + \frac{3}{{1000}}} \right] + \ldots + \left[ {\frac{1}{4} + \frac{{1000}}{{1000}}} \right]\\ \Rightarrow \mathop \sum \limits_{{\rm{n}} = 1}^{1000} {\rm{f}}\left( {\rm{n}} \right) = \left[ {0.25 + 0.001} \right] + \left[ {0.25 + 0.002} \right] + \left[ {0.25 + 0.003} \right] + \ldots \left[ {0.25 + 1} \right] \end{array}\)

Now, \(\left[ {0.25 + 0.001} \right] = \left[ {0.251} \right] = 0\)       (∵ greatest integer [0.25] = 0)

Similarly, \(\left[ {0.25 + 0.002} \right] = 0\)

\(\left[ {0.25 + 0.003} \right] = 0\)

.

.

.

\(\left[ {0.25 + 0.75} \right] = \left[ 1 \right] = 1\)

It means, till 750^th term the value will come zero

\(\therefore \mathop \sum \limits_{{\rm{n}} = 1}^{1000} {\rm{f}}\left( {\rm{n}} \right) = \left[ {\frac{1}{4} + \frac{1}{{1000}}} \right] + \left[ {\frac{1}{4} + \frac{2}{{1000}}} \right] + \left[ {\frac{1}{4} + \frac{3}{{1000}}} \right] + \ldots + \left[ {\frac{1}{4} + \frac{{750}}{{1000}}} \right] + \left[ {\frac{1}{4} + \frac{{752}}{{1000}}} \right] + \ldots + \left[ {\frac{1}{4} + \frac{{1000}}{{1000}}} \right]\)

Now, last term \(\left[ {0.25 + 1} \right] = \left[ {1.25} \right] = 1\)

So from 750^th term till end the values will be 1

\(\therefore \mathop \sum \limits_{{\rm{n}} = 1}^{1000} {\rm{f}}\left( {\rm{n}} \right) = 0 + 0 + 0 + \ldots 1 + 1 + 1 + \ldots + 1\)

So 1 will come: (1000 - 750) + 1(750^th term) = 251 times

\(\therefore \mathop \sum \limits_{{\rm{n}} = 1}^{1000} {\rm{f}}\left( {\rm{n}} \right) = 251\)

Hence, option (1) is correct.

Concept:

• \(\smallint {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} - \smallint {\rm{g}}\left( {\rm{x}} \right){\rm{dx}} = \smallint \left[ {{\rm{f}}\left( {\rm{x}} \right) - {\rm{g}}\left( {\rm{x}} \right)} \right]{\rm{dx}}\)
• \({\log _{\rm{a}}}{\rm{b}} = {\rm{k}} \Rightarrow {\rm{b}} = {{\rm{a}}^{\rm{k}}}\)
• Integration by parts:

 

​\(\Rightarrow \smallint {\rm{u\;vdx}} = {\rm{u\;}}\smallint {\rm{vdx}} - \smallint \left( {\frac{{{\rm{du}}}}{{{\rm{dx}}}}{\rm{\;}}\smallint {\rm{vdx}}} \right){\rm{dx}}\)

Where u is the function u(x) and v is the function v(x)

ILATE Rule: Usually, the preference order of this rule is based on some functions such as Inverse, Logarithm, Algebraic, Trigonometric and Exponent.

 

Calculation:

To find: \(\smallint \frac{1}{{\ln {\rm{x}}}}{\rm{dx}} - \smallint \frac{1}{{{{\left( {\ln {\rm{x}}} \right)}^2}}}{\rm{dx}} = ?\)

Let, \({\rm{I}} = \smallint \frac{1}{{\ln {\rm{x}}}}{\rm{dx}} - \smallint \frac{1}{{{{\left( {\ln {\rm{x}}} \right)}^2}}}{\rm{dx}}\)

\(\Rightarrow {\rm{I}} = \smallint \left[ {\frac{1}{{\ln {\rm{x}}}} - \frac{1}{{{{\left( {\ln {\rm{x}}} \right)}^2}}}} \right]{\rm{dx}}\)

Put, ln x = t

⇒ x = e^t                 \(\left( {{{\log }_{\rm{e}}}{\rm{x}} = {\rm{t}} \Rightarrow {\rm{x}} = {{\rm{e}}^{\rm{t}}}} \right)\)

dx = e^t dt                              (taking derivative)

\({\rm{I}} = \smallint \left[ {\frac{1}{{\ln {\rm{x}}}} - \frac{1}{{{{\left( {\ln {\rm{x}}} \right)}^2}}}} \right]{\rm{dx}}\)

\(= \smallint {{\rm{e}}^{\rm{t}}}\left[ {\left( {\frac{1}{{\rm{t}}}} \right) - \left( {\frac{1}{{{{\rm{t}}^2}}}} \right)} \right]{\rm{dt}}\)

\(= \smallint \left( {\frac{1}{{\rm{t}}}} \right){{\rm{e}}^{\rm{t}}}{\rm{dt}} - \smallint \left( {\frac{1}{{{{\rm{t}}^2}}}} \right){{\rm{e}}^{\rm{t}}}{\rm{dt}}\)        ….

(1) Now, \(\smallint \left( {\frac{1}{{\rm{t}}}} \right){{\rm{e}}^{\rm{t}}}{\rm{dt}} = \frac{1}{{\rm{t}}}\smallint {{\rm{e}}^{\rm{t}}}{\rm{dt}} - \smallint \frac{{\rm{d}}}{{{\rm{dt}}}}\left( {\frac{1}{{\rm{t}}}} \right)\smallint {{\rm{e}}^{\rm{t}}}{\rm{dt\;}}\)         (integration by part)

\(\Rightarrow \smallint \left( {\frac{1}{{\rm{t}}}} \right){{\rm{e}}^{\rm{t}}}{\rm{dt}} = \frac{{{{\rm{e}}^{\rm{t}}}}}{{\rm{t}}} - \smallint - \frac{{{{\rm{e}}^{\rm{t}}}}}{{{{\rm{t}}^2}}}{\rm{dt}}\)

\(= \frac{{{{\rm{e}}^{\rm{t}}}}}{{\rm{t}}} + \smallint \frac{{{{\rm{e}}^{\rm{t}}}}}{{{{\rm{t}}^2}}}{\rm{dt}}\)     …. (2) \({\rm{I}} = \smallint \left( {\frac{1}{{\rm{t}}}} \right){{\rm{e}}^{\rm{t}}}{\rm{dt}} - \smallint \left( {\frac{1}{{{{\rm{t}}^2}}}} \right){{\rm{e}}^{\rm{t}}}{\rm{dt}}\)         (from (1))

\(\Rightarrow \frac{{{{\rm{e}}^{\rm{t}}}}}{{\rm{t}}} + \smallint \frac{{{{\rm{e}}^{\rm{t}}}}}{{{{\rm{t}}^2}}}{\rm{dt}} - \smallint \left( {\frac{1}{{{{\rm{t}}^2}}}} \right){{\rm{e}}^{\rm{t}}}{\rm{dt}}\)          (from (2))

\(= \frac{{{{\rm{e}}^{\rm{t}}}}}{{\rm{t}}} + {\rm{C}}\)

Now, t = ln x

\(\therefore {\rm{I}} = \frac{{{{\rm{e}}^{{\rm{lnx}}}}}}{{{\rm{lnx}}}} + {\rm{c}}\)

\(\Rightarrow {\rm{I}} = \frac{{\rm{x}}}{{\ln {\rm{x}}}} + {\rm{c}}\)        (e^{ln x} = x)

Hence, option (1) is correct.

Concept:

• Surface area of cylinder without a lid (A) = 2πrh + πr²
• Volume of cylinder (V) = πr²h
• If volume is maximum, then dV/dr = 0

Here, r = radius, h = height

 

Calculation:

Given: d = k × h,

Where, d = diameter, h = height

To find: k =?

A = 2πrh + πr²

⇒ 2πrh = A - πr²

\(\Rightarrow {\rm{h}} = \frac{{\rm{A}}}{{2{\rm{\pi r}}}} - \frac{{\rm{r}}}{2}\)        (1)

We have,

V = πr² h

\(= {\rm{\pi }}{{\rm{r}}^2}\left( {\frac{{\rm{A}}}{{2{\rm{\pi r}}}} - \frac{{\rm{r}}}{2}{\rm{\;}}} \right)\)         (from (1))

\(= \frac{{{\rm{Ar}}}}{2} - \frac{{{\rm{\pi }}{{\rm{r}}^3}}}{2}\)

Capacity/volume is maximum so \(\frac{{{\rm{dV}}}}{{{\rm{dr}}}} = 0\),

\(\frac{{{\rm{dV}}}}{{{\rm{dr}}}} = \frac{{\rm{d}}}{{{\rm{dr}}}}\left( {\frac{{{\rm{Ar}}}}{2} - \frac{{{\rm{\pi }}{{\rm{r}}^3}}}{2}} \right)\)

\(= \frac{{\rm{A}}}{2} - \frac{3}{2}{\rm{\pi }}{{\rm{r}}^2}\)

\(= \frac{1}{2}\left( {{\rm{A}} - 3{\rm{\pi }}{{\rm{r}}^2}} \right) = 0\)

⇒ A = 3πr²

Put, A = 3πr² in equation (1),

\({\rm{h}} = \frac{{3{\rm{\pi }}{{\rm{r}}^2}}}{{2{\rm{\pi r}}}} - \frac{{\rm{r}}}{2}\)

\(\Rightarrow {\rm{h}} = \frac{3}{2}{\rm{r}} - \frac{1}{2}{\rm{r}}\)

⇒ h = r

We know diameter (d) = 2 × r

And given that, diameter (d) = k × h

∴ 2 × r = k × h

⇒ k = 2                 (∵ h = r)

Hence, option (2) is correct.

Concept:

Formulas used:

• 2sin x cos x = 1 - (sin x - cos x)²
• \(\smallint \frac{{{\rm{dx}}}}{{\sqrt {{{\rm{a}}^2} - {{\rm{x}}^2}} }} = {\sin ^{ - 1}}\left( {\frac{{\rm{x}}}{{\rm{a}}}} \right) + {\rm{c}}\)

 

Calculation:

\(\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} \sqrt {\tan {\rm{x\;}}} {\rm{dx}} + \mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} \sqrt {\cot {\rm{x}}} {\rm{dx}} = \mathop \smallint \limits_0^{{\rm{\pi }}/4} \left\{ {\sqrt {\tan {\rm{x}}} + \frac{1}{{\sqrt {\tan {\rm{x}}} }}} \right\}{\rm{dx}}\)

\(= \mathop \smallint \limits_0^{{\rm{\pi }}/4} \frac{{\tan {\rm{x}} + 1}}{{\sqrt {\tan {\rm{x}}} }}{\rm{\;dx}}\)

\(= \mathop \smallint \limits_0^{{\rm{\pi }}/4} \frac{{\sin {\rm{x}} + \cos {\rm{x}}}}{{\sqrt {\sin {\rm{x}}\cos {\rm{x}}} }}{\rm{\;dx}}\)

\(= \mathop \smallint \limits_0^{{\rm{\pi }}/4} \sqrt 2 \times \frac{{\sin {\rm{x}} + \cos {\rm{x}}}}{{\sqrt {2\sin {\rm{x}}\cos {\rm{x}}} }}{\rm{\;dx}}\)

We know that, 2sin x cos x = 1 - (sin x - cos x) ²

\(= \mathop \smallint \limits_0^{{\rm{\pi }}/4} \sqrt 2 \times \frac{{\sin {\rm{x}} + \cos {\rm{x}}}}{{\sqrt {1 - {{\left( {\sin {\rm{x}} - \cos {\rm{x}}} \right)}^2}} }}{\rm{\;dx}}\)

Let (sin x - cos x) = t

Differentiating both sides, we get,

⇒ (cos x + sin x) dx = dt

\(\mathop \smallint \limits_0^{{\rm{\pi }}/4} \sqrt 2 \times \frac{{\sin {\rm{x}} + \cos {\rm{x}}}}{{\sqrt {1 - {{\left( {\sin {\rm{x}} - \cos {\rm{x}}} \right)}^2}} }}{\rm{\;dx}} = \mathop \smallint \limits_0^1 \sqrt 2 \times \frac{{{\rm{dt}}}}{{\sqrt {1 - {{\rm{t}}^2}} }}\)

\(= \sqrt 2 \left[ {{{\sin }^{ - 1}}{\rm{x}}} \right]_0^1\)

\(= \frac{{\rm{\pi }}}{{\sqrt 2 }}\)

Concept:

Greatest integer function: It is denoted as [x], it indicates the integral part of real number x which is nearest and smaller integer to x.

For example: [2.4] = 2, [4] = 4, [-3.2] = -4

Note:

• If n is any integer, then:

 

\(\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{n}}^ - }} \left[ {\rm{x}} \right] = {\rm{n}} - 1\)

\(\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{n}}^ + }} \left[ {\rm{x}} \right] = {\rm{n}}\)

Continuity: 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.

f(x) is Continuous at x = a ⇔ \(\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{a}}^ + }} {\rm{f}}\left( {\rm{x}} \right){\rm{\;}} = {\rm{\;}}\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{a}}^ - }} {\rm{f}}\left( {\rm{x}} \right){\rm{\;}} = {\rm{\;}}\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} {\rm{f}}\left( {\rm{x}} \right)\)

 

Calculation:

Given: g is a greatest integer function

f(x) = (g(x))² - g(x)

 

So, let g(x) = [x]

∴ f(x) = [x]² - [x]

⇒ f(x) = [x]{[x] - 1}

Calculating the limit at x = c, where c is an integer

Left hand limit:

\(\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{c}}^ - }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {{\rm{c}}^ - }} \left[ {\rm{x}} \right]\left\{ {\left[ {\rm{x}} \right] - 1} \right\}\)

\(= \mathop {\lim }\limits_{{\rm{h}} \to 0} \left[ {{\rm{c}} - {\rm{h}}} \right]\left\{ {\left[ {{\rm{c}} - {\rm{h}}} \right] - 1} \right\}\)

\(= \mathop {\lim }\limits_{{\rm{h}} \to 0} \left( {{\rm{c}} - 1} \right)\left\{ {\left( {{\rm{c}} - 1} \right) - 1} \right\}\)

\(= \mathop {\lim }\limits_{{\rm{h}} \to 0} \left( {{\rm{c}} - 1} \right)\left( {{\rm{c}} - 2} \right)\)

= (c - 1)(c - 2)

Right hand limit:

\(\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{c}}^ + }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {{\rm{c}}^ + }} \left[ {\rm{x}} \right]\left\{ {\left[ {\rm{x}} \right] - 1} \right\}\)

\(= \mathop {\lim }\limits_{{\rm{h}} \to 0} \left[ {{\rm{c}} + {\rm{h}}} \right]\left\{ {\left[ {{\rm{c}} + {\rm{h}}} \right] - 1} \right\}\)

\(= \mathop {\lim }\limits_{{\rm{h}} \to 0} \left( {\rm{c}} \right)\left\{ {\left( {\rm{c}} \right) - 1} \right\}\)

\(= \mathop {\lim }\limits_{{\rm{h}} \to 0} {\rm{c}}\left( {{\rm{c}} - 1} \right)\)

= c(c - 1)

 

For f(x) to be continuous left hand limit = right hand limit and that is only when c = 1.

So, f(x) is discontinuous at all integers except at x = 1.

Concept:

Eliminating arbitrary constant: Let there be n arbitrary constants in a given differential equation. So differentiate the given equation up to n^th order. And from those n equations try to eliminate the arbitrary constants.

 

Calculation:

Given:

y = A [sin (x + C) + cos (x + C)] ….

(1) Since, we have to eliminate two arbitrary constant, thus we have to get a 2^nd order differential equation.

Differentiating equation (1) w.r.t x

⇒ y’ = A [cos (x + C) - sin (x + C)]

Now, again differentiating w.r.t x we get,

⇒ y’’ = A [-sin (x + C) - cos (x + C)]

⇒ y’’ = -A [sin (x + C) + cos (x + C)]

⇒ y’’ = -y

⇒ y’’ + y = 0

Concept:

Monotonic function: If a function is differentiable on the interval (a, b) and if the function is increasing/strictly increasing or decreasing/strictly decreasing, then the function is known as monotonic function.

First derivative test:

• If \(\frac{{{\rm{df}}\left( {\rm{x}} \right)}}{{{\rm{dx}}}} \ge 0\) for all x in (a, b) then, f(x) is an increasing function in (a, b).
• If \(\frac{{{\rm{df}}\left( {\rm{x}} \right)}}{{{\rm{dx}}}} \le 0\) for all x in (a, b) then, f(x) is an decreasing function in (a, b).

For strictly increasing or decreasing:

• \(\frac{{{\rm{df}}\left( {\rm{x}} \right)}}{{{\rm{dx}}}} > 0\) for strictly increasing.
• \(\frac{{{\rm{df}}\left( {\rm{x}} \right)}}{{{\rm{dx}}}} < 0\) for strictly decreasing.

 

Calculation:

Let f(x) = x - sin x

Now, differentiating both sides w.r.t x

⇒ f’(x) = 1 - cos x

We know, -1 ≤ cos x ≤ 1 for all x > 0

⇒ 1 ≥ -cos x ≥ -1

⇒ 2 ≥ 1 - cos x ≥ 0

∴ f’(x) ≥ 0 for all x > 0

Hence, f(x) is an increasing function for all x > 0.

And since, f(x) is an increasing function for all x > 0, then x > sin x for all x > 0.

Hence, both statement I & II are correct and statement II is correct explanation of statement I.

Calculation:

Given: \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{{{\rm{y}}\phi '\left( {\rm{x}} \right) - {{\rm{y}}^2}}}{{\phi \left( {\rm{x}} \right)}}\)

\(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{{{\rm{y}}\phi '\left( {\rm{x}} \right) - {{\rm{y}}^2}}}{{\phi \left( {\rm{x}} \right)}} = \frac{{{\rm{y}}\phi '\left( {\rm{x}} \right)}}{{\phi \left( {\rm{x}} \right)}} - \frac{{{\rm{y}}.{\rm{y}}}}{{\phi \left( {\rm{x}} \right)}}\)

\({\rm{Let}},{\rm{\;}}\frac{{\rm{y}}}{{\phi \left( {\rm{x}} \right)}} = {\rm{z}}\)     ...1)

\(\therefore \frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{{{\rm{y}}\phi '\left( {\rm{x}} \right)}}{{\phi \left( {\rm{x}} \right)}} - \frac{{{\rm{y}}.{\rm{y}}}}{{\phi \left( {\rm{x}} \right)}} = \phi '\left( {\rm{x}} \right){\rm{z}} - \phi \left( {\rm{x}} \right){{\rm{z}}^2}\)

Now, \({\rm{y}} = \phi \left( {\rm{x}} \right){\rm{z}}\) 

\(\Rightarrow \frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \phi \left( {\rm{x}} \right)\frac{{{\rm{dz}}}}{{{\rm{dx}}}} + \phi '\left( {\rm{x}} \right){\rm{z}}\)

\(\therefore \phi '\left( {\rm{x}} \right){\rm{z}} - \phi \left( {\rm{x}} \right){{\rm{z}}^2} = \phi \left( {\rm{x}} \right)\frac{{{\rm{dz}}}}{{{\rm{dx}}}} + \phi '\left( {\rm{x}} \right){\rm{z}}\)

\(\Rightarrow - {{\rm{z}}^2} = \frac{{{\rm{dz}}}}{{{\rm{dx}}}}\)

\(\Rightarrow - {\rm{dx}} = \frac{{{\rm{dz}}}}{{{{\rm{z}}^2}}}\)

Integrating both sides, we get

\(\Rightarrow {\rm{x}} + {\rm{c}} = \frac{1}{{\rm{z}}}\)      

\({\rm{But\;z}} = \frac{{\rm{y}}}{{\phi \left( {\rm{x}} \right)}}\)     ...(From (1)

\(\therefore {\rm{x}} + {\rm{c}} = \frac{{\phi \left( {\rm{x}} \right)}}{{\rm{y}}}\)

\( \Rightarrow {\rm{y}} = \frac{{\phi \left( {\rm{x}} \right)}}{{{\rm{x}} + {\rm{c}}}}\)

Hence, option (4) is correct.

Concept:

Composite function: Let f: X → Y₁ and g: Y₂ → Z be two functions and the set D = { x ∈ X: f(x) ∈ Y₂}. If D ∉ ∅ then function h is defined as:

h(x) = g{f(x)} = g o f (x)

 

Calculation:

Given: \({\rm{f}}\left( {\rm{x}} \right) = \frac{{4{\rm{x}} + {{\rm{x}}^4}}}{{1 + 4{{\rm{x}}^3}}}\) and \({\rm{g}}\left( {\rm{x}} \right) = {\rm{In\;}}\left( {\frac{{1 + {\rm{x}}}}{{1 - {\rm{x}}}}} \right)\)

To find: \({\rm{f\;o\;g\;}}\left( {\frac{{{\rm{e}} - 1}}{{{\rm{e}} + 1}}} \right)\)

So, we know that, \({\rm{f\;o\;g\;}}\left( {\rm{x}} \right) = {\rm{f}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right)\)

∴ \({\rm{f\;o\;g\;}}\left( {\frac{{{\rm{e}} - 1}}{{{\rm{e}} + 1}}} \right) = {\rm{f}}\left( {{\rm{g}}\left( {\frac{{{\rm{e}} - 1}}{{{\rm{e}} + 1}}} \right)} \right)\)

So, \({\rm{g}}\left( {\frac{{{\rm{e}} - 1}}{{{\rm{e}} + 1}}} \right) = \ln \left( {\frac{{1 + \frac{{{\rm{e}} - 1}}{{{\rm{e}} + 1}}}}{{1 - \frac{{{\rm{e}} - 1}}{{{\rm{e}} + 1}}}}} \right)\)

\(= \ln \left( {\frac{{2{\rm{e}}}}{2}} \right)\)

= ln e = 1

\({\rm{f\;o\;g\;}}\left( {\frac{{{\rm{e}} - 1}}{{{\rm{e}} + 1}}} \right) = {\rm{f}}\left( {{\rm{g}}\left( {\frac{{{\rm{e}} - 1}}{{{\rm{e}} + 1}}} \right)} \right)\)

= f(1)

\(= \frac{{4\left( 1 \right) + {{\left( 1 \right)}^4}}}{{1 + 4{{\left( 1 \right)}^3}}}\)

= 5/5

= 1

Concept:

Determinant: If \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} {{{\rm{a}}_{11}}}&{{{\rm{a}}_{12}}}&{{{\rm{a}}_{13}}}\\ {{{\rm{a}}_{21}}}&{{{\rm{a}}_{22}}}&{{{\rm{a}}_{23}}}\\ {{{\rm{a}}_{31}}}&{{{\rm{a}}_{32}}}&{{{\rm{a}}_{33}}} \end{array}} \right]\) then determinant of A is given by:

|A| = a₁₁ × {(a₂₂ × a₃₃) - (a₂₃ × a₃₂)} - a₁₂ × {(a₂₁ × a₃₃) - (a₂₃ × a₃₁)} + a₁₃ × {(a₂₁ × a₃₂) - (a₂₂ × a₃₁)}

Note:

• If row or column operations are applied, then the value of determinant doesn’t change.

 

Calculation:

Given:

\(\left| {\begin{array}{*{20}{c}} {1 - {\rm{\alpha }}}&{{\rm{\alpha }} - {{\rm{\alpha }}^2}}&{{{\rm{\alpha }}^2}}\\ {1 - {\rm{\beta }}}&{{\rm{\beta }} - {{\rm{\beta }}^2}}&{{{\rm{\beta }}^2}}\\ {1 - {\rm{\gamma }}}&{{\rm{\gamma }} - {{\rm{\gamma }}^2}}&{{{\rm{\gamma }}^2}} \end{array}} \right|\)

Let’s apply row or column operations to simplify.

Replace C₁ → C₁ + C₂ + C₃

\(= \left| {\begin{array}{*{20}{c}} 1&{{\rm{\alpha }} - {{\rm{\alpha }}^2}}&{{{\rm{\alpha }}^2}}\\ 1&{{\rm{\beta }} - {{\rm{\beta }}^2}}&{{{\rm{\beta }}^2}}\\ 1&{{\rm{\gamma }} - {{\rm{\gamma }}^2}}&{{{\rm{\gamma }}^2}} \end{array}} \right|\)

Replace C₂ → C₂ + C₃

\(= \left| {\begin{array}{*{20}{c}} 1&{\rm{\alpha }}&{{{\rm{\alpha }}^2}}\\ 1&{\rm{\beta }}&{{{\rm{\beta }}^2}}\\ 1&{\rm{\gamma }}&{{{\rm{\gamma }}^2}} \end{array}} \right|\)

Replace R₁ → R₁ - R₂ & R₃ → R₃ - R₂

\(= \left| {\begin{array}{*{20}{c}} 0&{{\rm{\alpha }} - {\rm{\beta }}}&{{{\rm{\alpha }}^2} - {{\rm{\beta }}^2}}\\ 1&{\rm{\beta }}&{{{\rm{\beta }}^2}}\\ 0&{{\rm{\gamma }} - {\rm{\beta }}}&{{{\rm{\gamma }}^2} - {{\rm{\beta }}^2}} \end{array}} \right|\)

= -(α - β)(γ² - β²) + (α² - β²)(γ - β)

= (α - β)(γ - β){ - γ - β + α + β}

= (α - β)(γ - β)(α - γ)

= (α - β)(β - γ)(γ - α)

Concept:

Steps to find adjoint of a matrix:

• Find the minors of the matrix: Ignore the value of current row and column and calculate the determinant of remaining vales.
• For m₁₁ = a₂₂ × a₃₃ - a₂₃ × a₃₂. Similarly, for every element.
• Find the co-factor of matrix formed: C = [c_ij]_{3 × 3} = (-1)^{(i + j)} × m_ij
• Take the transpose of the matrix i.e., C^T: Swapping the (ij)^th element with (ji)^th element.

 

 

Calculation:

Given: \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} 1&0&2\\ 2&1&0\\ 0&3&1 \end{array}} \right]\)

To find: adj(A)

 

Minors for matrix:

m₁₁ = 1 × 1 - 0 × 3 = 1

m₁₂ = 2 × 1 - 0 = 2

m₁₃ = 2 × 3 - 0 = 6

m₂₁ = 0 × 1 - 2 × 3 = -6

m₂₂ = 1 - 0 = 1

m₂₃ = 3 - 0 = 3

m₃₁ = 0 - 2 = -2

m₃₂ = 0 - 2 × 2 = -4

m₃₃ = 1 - 0 = 1

Co-factors: C = [c_ij]_{3 × 3} = (-1)^{(i + j)} × m_ij

\({\rm{C}} = \left[ {\begin{array}{*{20}{c}} 1&{ - 2}&6\\ 6&1&{ - 3}\\ -2&{ 4}&{ 1} \end{array}} \right]\)

So, adj(A) = C^T

\(\therefore {\rm{adj}}\left( {\rm{A}} \right){\rm{\;}} = \left[ {\begin{array}{*{20}{c}} 1&6&-2\\ { - 2}&1&{ 4}\\ 6&{ - 3}&{ 1} \end{array}} \right]\)

Concept:

Matrix Multiplication:

• First make sure the number of columns of first matrix is equal to number of rows of second matrix.
• Multiply elements of each row of first matrix to elements of each column of second matrix.
• Let A = [a_ij]_{m × p} & B = [b_ij]_{p × n} then AB = [c_ij]_{m × n} where \({{\rm{c}}_{{\rm{ij}}}} = \mathop \sum \limits_{{\rm{k}} = 1}^{\rm{p}} \left( {{{\rm{a}}_{{\rm{ik}}}}{{\rm{b}}_{{\rm{kj}}}}} \right)\).

 

Formulas used:

• Let A = [a_ij]_{m × n} be a matrix then kA = [b_ij]_{m × n} where b_ij = ka_ij

 

Calculation:

Given: \({\rm{A}} = \left( {\begin{array}{*{20}{c}} { - 2}&2\\ 2&{ - 2} \end{array}} \right)\)

\({{\rm{A}}^2} = \left( {\begin{array}{*{20}{c}} { - 2}&2\\ 2&{ - 2} \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} { - 2}&2\\ 2&{ - 2} \end{array}} \right)\)

\(\Rightarrow {{\rm{A}}^2} = \left( {\begin{array}{*{20}{c}} {\left\{ {\left( { - 2} \right) \times \left( { - 2} \right) + 2 \times 2} \right\}}&{\left\{ {\left( { - 2} \right) \times 2 + 2 \times \left( { - 2} \right)} \right\}}\\ {\left\{ {2 \times \left( { - 2} \right) + \left( { - 2} \right) \times 2} \right\}}&{\left\{ {2 \times 2 + \left( { - 2} \right) \times \left( { - 2} \right)} \right\}} \end{array}} \right)\)

\(\Rightarrow {{\rm{A}}^2} = \left( {\begin{array}{*{20}{c}} 8&{ - 8}\\ { - 8}&8 \end{array}} \right)\)

\(\Rightarrow {{\rm{A}}^2} = - 4 \times \left( {\begin{array}{*{20}{c}} { - 2}&2\\ 2&{ - 2} \end{array}} \right)\)

⇒ A² = -4A

Concept:

• z = x + iy where x and y are real values and z is complex number.
• Re (z) = x, imaginary part of z = y.
• Equation of circle x² + y² = r²
• Equation of ellipse \(\frac{{{x^2}}}{{{a^2}}}\; + \frac{{{y^2}}}{{{b^2}}} = 1\)
• Equation of Parabola y² = 4ax, x² = 4ay
• Equation of hyperbola \(\frac{{{x^2}}}{{{a^2}}} - \;\frac{{{y^2}}}{{{b^2}}} = 1\)
• If a² = b² then hyperbola is called rectangular hyperbola and x² − y² = a² is the general form of rectangular hyperbola

 

Calculation:

⇒ Re (z² – i) = 2

Put z = x + iy

Let’s find the value of z² – i

⇒ (x + iy)² - i

⇒ x² + 2xyi – y² – i     ...\([{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}]\)

Now take real part and imaginary part collectively

⇒ (x² – y²) + i (2xy – 1)

So the real part (z² – i)

⇒ Re (z² – i) = x² – y² = 2

⇒ x² – y² = 2

So it is a rectangular hyperbola.

Concept:

Determinant: If \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} {{{\rm{a}}_{11}}}&{{{\rm{a}}_{12}}}&{{{\rm{a}}_{13}}}\\ {{{\rm{a}}_{21}}}&{{{\rm{a}}_{22}}}&{{{\rm{a}}_{23}}}\\ {{{\rm{a}}_{31}}}&{{{\rm{a}}_{32}}}&{{{\rm{a}}_{33}}} \end{array}} \right]\) then determinant of A is given by:

|A| = a₁₁ × {(a₂₂ × a₃₃) - (a₂₃ × a₃₂)} - a₁₂ × {(a₂₁ × a₃₃) - (a₂₃ × a₃₁)} + a₁₃ × {(a₂₁ × a₃₂) - (a₂₂ × a₃₁)}

 

Formulas used:

• a³ + b³ + c³ = 3abc, if a + b + c = 0.

 

Calculation:

Given: p + q + r = a + b + c = 0

To find: determinant \(\left| {\begin{array}{*{20}{c}} {{\rm{pa}}}&{{\rm{qb}}}&{{\rm{rc}}}\\ {{\rm{qc}}}&{{\rm{ra}}}&{{\rm{pb}}}\\ {{\rm{rb}}}&{{\rm{pc}}}&{{\rm{qa}}} \end{array}} \right|\)

\(\left| {\begin{array}{*{20}{c}} {{\rm{pa}}}&{{\rm{qb}}}&{{\rm{rc}}}\\ {{\rm{qc}}}&{{\rm{ra}}}&{{\rm{pb}}}\\ {{\rm{rb}}}&{{\rm{pc}}}&{{\rm{qa}}} \end{array}} \right|\)

Expanding R₁,

= pa (ra × qa - pb × pc) - qb (qc × qa - pb × rb) + rc (qc × pc - ra × rb)

= pa(a²qr - p²bc) - qb(q²ac - b²pr) + rc(c²pq - r²ab)

= a³pqr - p³abc - q³abc + b³pqr + c³pqr - r³abc

= pqr(a³ + b³ + c³) - abc(p³ + q³ + r³)

Now we know that, a³ + b³ + c³ = 3abc, if a + b + c = 0.

= pqr (3abc) - abc (3pqr)

= 0