Quantitative Aptitude PYQs

Concept:

Perpendicular distance from point to straight line. Where point (h, k) and line ax + by + c = 0

• \({\rm{D\;}} = \frac{{\left| {{\rm{ah}} + {\rm{bk}} + {\rm{c}}} \right|}}{{\sqrt {{{\rm{a}}^2} + {{\rm{b}}^2}} }}\)

 

Calculation:

Given that,

Point on 2x + 11y = 5

Let’s take an point on line is (5/2, 0)

Perpendicular fall on line 24x + 7y = 20

⇒ D₁ = \(\frac{{\left| {{\rm{\;}}24 \times \frac{5}{2} + 7 \times 0 - 20} \right|}}{{\sqrt {{{24}^2} + {7^2}} }}\)

⇒ D₁ = \(\frac{{\left| {{\rm{\;}}60{\rm{\;}} - {\rm{\;}}20} \right|}}{{\sqrt {625} }}\) 

⇒ D₁ = 40/25

⇒ D₁ = 8/5      ….(1)

Perpendicular fall on line 4x – 3y = 2,

⇒ D₂ = \(\frac{{\left| {4 \times \frac{5}{2} - 3 \times 0 - 2} \right|}}{{{\rm{\;}}\sqrt {{4^2} + {3^2}} }}\)

⇒ D₂ = \(\frac{{\left| {10{\rm{\;}} - {\rm{\;}}2} \right|}}{{{\rm{\;}}\sqrt {{4^2} + {3^2}} }}\)

⇒ D₂ = 8/5 …. (2) From equation 1 and 2 we can say that perpendicular fall same on both lines from any point of the give straight line.

Concept:

1. Let’s take two lines ax + by + c = 0 and px + qy + r = 0 ,

• If a/p = b/q then both lines will be parallel to each other.

1. Two parallel line are ax + by + c = 0 and ax + by + r = 0 than

• Distance between two parallel lines D = \(\frac{{\left| {{\rm{c}} - {\rm{r}}} \right|}}{{\sqrt {{{\rm{a}}^2} + {{\rm{b}}^2}} }}\)

Calculation:

Given that,

Two lines are 3x + 4y = 9 and 6x + 8y = 15,

a = 3, b = 4 and p = 6, q = 8

If a/p = b/q then both lines will be parallel to each other

\(\Rightarrow \frac{3}{6} = \frac{4}{8} = \frac{1}{2}\)

So both lines are parallel to each other

First line 3x + 4y – 9 = 0

Second line 6x + 8y – 15 = 0 divide whole equation by 2

⇒ 3x + 4y – 15/2 = 0

So c = -9 and r = -15/2

⇒ Distance D = \(\frac{{\left| { - 9 + 15/2} \right|}}{{\sqrt {{3^2} + {4^2}} }}\left[ {D = {\rm{\;}}\frac{{\left| {{\rm{c}} - {\rm{r}}} \right|}}{{\sqrt {{{\rm{a}}^2} + {{\rm{b}}^2}} }}} \right]\)

⇒ D = \(\frac{{\left| { - \frac{3}{2}} \right|}}{{\sqrt {25} }}\)

⇒ D = 3/10

Concept:

1. If two vector a and b then \(\left| {{\rm{\vec a\;}} \times {\rm{\;\vec b}}} \right| = \left| {{\rm{\vec a}}} \right|{\rm{\;}}\left| {{\rm{\vec b}}} \right|\sin {\rm{\theta }}\)

 Where θ is angle between vector a and b.

1. Vector A = \({\rm{x\hat i}} + {\rm{y\hat j}} + {\rm{z\hat k}}\) then |A| = \(\sqrt {{{\rm{x}}^2} + {{\rm{y}}^2} + {{\rm{z}}^2}} \)

 

Calculation:

Given that,

\(\overrightarrow {|{\rm{a|}}} { = 2,{\rm{\;}}\overrightarrow {|{\rm{b}|}} }= 7​​​​\:and\:\overrightarrow {{\rm{a\;}}} \times {\rm{\;\vec b}} = 3{\rm{\hat i}} + 2{\rm{\hat j}} + 6{\rm{\hat k}},\)

Now

\(\begin{array}{l} \Rightarrow \left| {{\rm{\vec a\;}} \times {\rm{\;\vec b}}} \right| = \sqrt {{3^2} + {2^2} + {6^2}} \\ \Rightarrow \left| {{\rm{\vec a\;}} \times {\rm{\;\vec b}}} \right| = \sqrt {49} = 7\\ \Rightarrow \left| {{\rm{\vec a\;}} \times {\rm{\;\vec b}}} \right| = \left| {{\rm{\vec a}}} \right|{\rm{\;}}\left| {{\rm{\vec b}}} \right|\sin {\rm{\theta }} \end{array}\)

Put all values in above equation

⇒ 7 = 2 × 7 × sin θ

⇒ sin θ = 1/2

⇒ θ = 30°

Concept:

1. Let the given two position vector p and q respectively and r be the vector dividing the segment PQ internally in the ratio m: n
2. Internal Section Formula: When the segment PQ is divided internally in the ration m: n, we use this formula. ⇔ \({\rm{r}} = \left( {\frac{{{\rm{mq\;}} + {\rm{\;np}}}}{{{\rm{m}} + {\rm{\;n}}}}} \right)\)
3. External Section Formula: When the segment PQ is divided externally in the ratio m: n, we use this formula. ⇔ \({\rm{s}} = \left( {\frac{{{\rm{mq}} - {\rm{np}}}}{{{\rm{m}} - {\rm{\;n}}}}} \right)\)
4. \({\rm{\vec a\;and\;\vec b}}\) are two vectors perpendicular to each other ⇔ \({\rm{\vec a}}.{\rm{\vec b}} = 0\)

 

 

Calculation:

Given that,

Vector R divide segment PQ internally in ration 2 : 3 so,

Form the diagram,

\(\Rightarrow {\rm{R}} = \left( {\frac{{2{\rm{q\;}} + {\rm{\;}}3{\rm{p}}}}{{2 + {\rm{\;}}3}}} \right)\)

\(\Rightarrow {\rm{R}} = \frac{{\left( {2{\rm{q}} + 3{\rm{p}}} \right)}}{5}\)       …(1)

Given Vector S divide segment PQ internally in ration 2 : 3 so,

\(\Rightarrow {\rm{S}} = \left( {\frac{{2{\rm{q}} - 3{\rm{p}}}}{{2 - {\rm{\;}}3}}} \right)\)

\(\Rightarrow {\rm{S}} = \frac{{3{\rm{p}} - 2{\rm{q}}}}{1}\)       …(2)

Note:

given vectors are position vectors with respect to O so vector OR = R and OS = s

Given OR and OS perpendicular to each other so,

⇒ R.S = 0

From equation 1 and 2

\(\Rightarrow \frac{{\left( {2{\rm{q}} + 3{\rm{p}}} \right)}}{5}.\frac{{3{\rm{p}} - 2{\rm{q}}}}{1} = 0\)

⇒ 6q.p - 4q² + 9p² - 6p.q = 0

⇒ 4q² = 9p²

Concept:

Moment of a force, M = r X F

Where, M Torque, r position vector, and F force.

If two vector A = a₁i + a₂j + a₃k and B = b₁i + b₂j + b₃k then \({\rm{A}} \times {\rm{B}} = \left[ {\begin{array}{*{20}{c}} {\rm{i}}&{\rm{j}}&{\rm{k}}\\ {{{\rm{a}}_1}}&{{{\rm{a}}_2}}&{{{\rm{a}}_3}}\\ {{{\rm{b}}_1}}&{{{\rm{b}}_2}}&{{{\rm{b}}_3}} \end{array}} \right]\) 

Calculation:

Given that,

F = \(3\hat i + \hat k\), position a vector = \(\hat i + 2\hat j - \hat k\) and position b vector = \(2\hat i - \hat j + 3\hat k\)

⇒ Position vector L = b – a = \(2\hat i - \hat j + 3\hat k - \left( {\hat i + 2\hat j - \hat k} \right) = {\rm{\hat i}} - 3{\rm{\hat j}} + 4{\rm{k}}\)

So the moment of force,

⇒ M = r X F

⇒ M = \(\left( {{\rm{\hat i}} - 3{\rm{\hat j}} + 4{\rm{\hat k}}} \right)\; \times \left( {3\hat i + \hat k} \right)\) 

⇒ M = \(\left[ {\begin{array}{*{20}{c}} {\rm{i}}&{\rm{j}}&{\rm{k}}\\ 1&{ - 3}&4\\ 3&0&1 \end{array}} \right]\)

⇒ M = i (-3 -0) – j (1 – 12)) + k (0 – (-9))

⇒ M = \(- {\rm{\;}}3{\rm{\hat i}} + 11{\rm{\hat j}} + 9{\rm{\hat k}}\)

Concept:

\({\rm{\vec a\;and\;\vec b}}\) are two vectors parallel to each other ⇔ \({\rm{\vec a}} \times {\rm{\vec b}} = 0\)

Calculation:

Given: \({\rm{\vec k}}\) and \({\rm{\vec A}}\) are parallel to each other

∴ \({\rm{\vec k}} \times {\rm{\vec A}} = 0\)

Calculation:

Given that,

⇒ f(x) = |x + 1|

first option

⇒ f(x²) = |x² + 1|

⇒ f(x)² = |x + 1|²

so we can say that

⇒ f(x²) = |f(x)|²

⇒ |x + 1|² ≠ |x² + 1|

First option not correct.

Second option,

⇒ f(|x|) = |f(x)|

⇒ f(|x|) = ||x| + 1|

⇒ |f(x)| = ||x + 1|| = |x + 1|

f(|x|) ≠ |f(x)| because ||x| + 1| ≠ |x + 1| for real values of x.

Third option,

⇒ f(x + y) = f(X) + f(y)

⇒ f(x + y) = |(x + y) + 1|

⇒ f(y) = |y + 1|

so,

⇒ |(x + y) + 1| ≠ |x + 1| + |y + 1|

So f(x + y) ≠ f(x) + f(y)

Concept:

Below are steps to find out range of function:

1. Write down y = f(x) and then solve the equation for x, giving something of the form x=g(y).

Find the domain of g(y), and this will be the range of f(x).

If you can't seem to solve for x, then try graphing the function to find the range.

 

Calculation:

Given that,

\({\rm{f}}\left( {\rm{x}} \right) = \frac{{{{\rm{x}}^2}}}{{1 + {{\rm{x}}^2}}}\)

Put f(x) = y

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

Cross multiply

⇒ y(1 + x² ) = x²

⇒ y + y . x² = x²

⇒ x² (1 -y) = y

⇒ x² = y/(1 -y)

⇒ \({\rm{x}} = \sqrt {\frac{{\rm{y}}}{{1 - {\rm{y}}}}}\)

⇒ x = g(y) = \(\sqrt {\frac{{\rm{y}}}{{1 - {\rm{y}}}}} \)

Now we find out domain of g(y) which will be range of f(x)

So as we know inside root always positive values comes so,

⇒ y ≠ 1 and \(\frac{{\rm{y}}}{{{\rm{y}} - 1}} > 0\)

By showing on number line we will get range [0, 1)

Concept:

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

Suppose that we have one of the following cases,

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

 

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

1. 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:

Given that,

\({\rm{f}}\left( {\rm{x}} \right) = \left\{ {\begin{array}{*{20}{c}} {{{\rm{x}}^2}{\rm{ln}}\left| {\rm{x}} \right|}&{{\rm{x}} \ne 0}\\ 0&{{\rm{x}} = 0} \end{array}} \right.\)

f’(x) = \(\left\{ {{\rm{\;}}\begin{array}{*{20}{c}} {2x\ln {\rm{\;}}\left| {\rm{x}} \right| + {{\rm{x}}^2} \times \frac{1}{{\rm{x}}}{\rm{\;x}} \ne 0}\\ {0\;x = 0} \end{array}} \right. = \left\{ {{\rm{\;}}\begin{array}{*{20}{c}} {\frac{{2{\rm{\;lnx}}}}{{\frac{1}{{\rm{x}}}}} + {\rm{x\;x}} \ne 0}\\ {0\;x = 0} \end{array}} \right.\)

\({\rm{f'}}\left( 0 \right)\mathop { = \lim }\limits_{{\rm{x}} \to 0} {\rm{f'}}\left( {\rm{x}} \right) = {\rm{\;}}\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{2{\rm{\;lnx}}}}{{\frac{1}{{\rm{x}}}}} + {\rm{x\;}}\) (It become \(\frac{\infty }{\infty }\) form. So we can use L hospital rule)

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

\({\rm{f'}}\left( 0 \right) = \mathop {\lim }\limits_{{\rm{x}} \to 0} - 2{\rm{\;x}} + {\rm{x\;\;}}\)

f’(0) = 0

Concept:

1. Standard Equation of Parabola

 

(y - y₁)² = 4a(x - x₁)

Where (x₁,y₁) is vertex and focus is (x₁ + a,y₁)

Calculation:

y² = 6 (x – 1) and y² = 3x

From the diagram

Area bounded by both parabola is

⇒ A = 2 × \(\left[ {\mathop \smallint \nolimits_0^{\sqrt 6 } \left( {\frac{{{{\rm{y}}^2}}}{6} + 1} \right){\rm{dy\;}} - {\rm{\;\;}}\mathop \smallint \nolimits_0^{\sqrt 6 } \frac{{{{\rm{y}}^2}}}{3}{\rm{\;dy}}} \right]\)

⇒ A = 2 × \(\left[ {\left[ {\frac{{{{\rm{y}}^3}}}{{18}} + {\rm{y}}} \right]\begin{array}{*{20}{c}} {\sqrt 6 }\\ 0 \end{array} - \left[ {\frac{{{{\rm{y}}^3}}}{9}} \right]\begin{array}{*{20}{c}} {\sqrt 6 }\\ 0 \end{array}} \right]\)

⇒ A = 2 × \(\left[ {\left[ {\frac{{6\sqrt 6 }}{{18}} + \sqrt 6 - 0{\rm{\;}}} \right] - \frac{{6\sqrt 6 }}{9} - 0{\rm{\;}}} \right]\)

⇒ A = 2 × \(\left[ { - {\rm{\;}}\frac{{6\sqrt 6 }}{{18}} + \sqrt 6 {\rm{\;}}} \right]\)

⇒ A = \(\frac{{4\sqrt 6 }}{3}\)

Concept:

Area of trapezium = \(\frac{1}{2} \times \left( {{\rm{sum\;of\;parallel\;side}}} \right) \times {\rm{height}}\)

1. 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.

1. f``(x) is less than 0 then the given function is said to be maxima
2. If f``(x) Is greater than 0 then the function is said to be minima

 

Calculation:

Given that,

Side of trapezium = 6 cm.

From above diagram,

⇒ Area = A = \(\frac{1}{2} \times \left( {6 + 6 + 2{\rm{x}}} \right) \times \sqrt {36 - {{\rm{x}}^2}} \)

\(\Rightarrow \frac{{{\rm{dA}}}}{{{\rm{dx}}}} = \frac{1}{2} \times \left( {2 \times \sqrt {36 - {{\rm{x}}^2}} {\rm{\;}} + \left( {12 + 2{\rm{x}}} \right).\frac{1}{2}\frac{{\left( { - 2{\rm{x}}} \right)}}{{\sqrt {36 - {{\rm{x}}^2}} }}} \right)\)

To find maximum area,

\(\Rightarrow \frac{{{\rm{dA}}}}{{{\rm{dx}}}} = 0\)

\(\Rightarrow \frac{1}{2} \times \left( {2 \times \sqrt {36 - {{\rm{x}}^2}} {\rm{\;}} + \left( {12 + 2{\rm{x}}} \right).\frac{1}{2}\frac{{\left( { - 2{\rm{x}}} \right)}}{{\sqrt {36 - {{\rm{x}}^2}} }}} \right) = 0\)

\(\Rightarrow 2 \times \sqrt {36 - {{\rm{x}}^2}} {\rm{\;}} = {\rm{\;}}\left( {12 + 2{\rm{x}}} \right).\frac{{\left( {\rm{x}} \right)}}{{\sqrt {36 - {{\rm{x}}^2}} }}\)

By cross multiplication

⇒ 36 – x² = (6 + x). x

⇒ 36 – x² = 6x + x²

⇒ x² + 3x - 18 = 0

⇒ x² + 6x – 3x -18 = 0

⇒ x(x + 6) - 3(x + 6) = 0

⇒ (x -3)(x + 6) = 0

⇒ x = 3

x is length which can’t be negative so x = -6 not possible.

So from the diagram

\(\cos {\rm{\alpha }} = \frac{{\rm{x}}}{6}\)

\(\cos {\rm{\alpha }} = \frac{1}{2}\)

\(\Rightarrow {\rm{\alpha }} = 60^\circ = \frac{{\rm{\pi }}}{3}\)

Calculation:

From the given information, Figure can be made as:

Now,

we know that Area of trapezium = A = \(\frac{1}{2}\)×(sum of parallel sides)× height.

By applying Pythagoras Theorem, 6² = x² + h² 

Height, \(h=\sqrt{36-x^2}\)

Thus, \(A=\frac{1}{2}\times (6 + 6 + x +x)\times\sqrt{36-x^2}\) 

\(\Rightarrow A=(6+x)\sqrt{36-x^2}\)

Differentiating with respect to x, we get,

\(\frac{dA}{dx}= (6+x)\frac{-2x}{2\sqrt{36-x^2}}+\sqrt{36-x^2}\)

Putting  \(\frac{dA}{dx}=0\),

i.e. \(\frac{dA}{dx}= (6+x)\frac{-2x}{2\sqrt{36-x^2}}+\sqrt{36-x^2}=0\)  

⇒ x = 3

thus, the fourth side will be = 6 + 2x

⇒ 6 + 2(3)

⇒ 12

Calculation:

Area at x = 3 will be maximum so,

⇒ Area = A = \(\frac{1}{2}{\rm{\;}} \times \left( {6 + 6 + 2{\rm{x}}} \right) \times \sqrt {36 - {{\rm{x}}^2}} \)

⇒ A = \(\frac{1}{2}{\rm{\;}} \times \left( {12 + 3 \times 2} \right) \times \sqrt {36 - {3^2}} {\rm{\;}}\)

⇒ A = 27√3 cm²

Concept:

1. Integration by parts: Integration by parts is a method to find integrals of products

 The formula for integrating by parts is given by;

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

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

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

 

Calculation:

Given that,

I = \(\mathop \smallint \nolimits_0^{\rm{\pi }} {{\rm{e}}^{\rm{x}}}\sin {\rm{x}}\) dx

By using integration by part,

\(\begin{array}{l} {\rm{I}} = \mathop \smallint \nolimits_0^{\rm{\pi }} {{\rm{e}}^{\rm{x}}}\sin {\rm{x\;dx}} = {\rm{sinx\;}}\smallint {{\rm{e}}^{\rm{x}}}{\rm{\;dx}} - {\rm{\;}}\mathop \smallint \nolimits_0^{\rm{\pi }} (\frac{{{\rm{d}}\left( {{\rm{sinx}}} \right)}}{{{\rm{dx}}}}\smallint {{\rm{e}}^{\rm{x}}}{\rm{\;dx}}){\rm{\;dx}}){\rm{\;dx}}\\ \Rightarrow {\rm{I}} = \left[ {{\rm{sinx}}.{{\rm{e}}^{\rm{x}}}} \right]\begin{array}{*{20}{c}} {\rm{\pi }}\\ 0\end{array} - \mathop \smallint \nolimits_0^{\rm{\pi }} {\rm{cosx}}.{{\rm{e}}^{\rm{x}}}{\rm{\;dx}}\\ \Rightarrow {\rm{I}} = 0 - \mathop \smallint \nolimits_0^{\rm{\pi }} {\rm{cosx}}{{\rm{e}}^{\rm{x}}}{\rm{dx}}\\ \Rightarrow - {\rm{I}} = \mathop \smallint \nolimits_0^{\rm{\pi }} {\rm{cosx}}{{\rm{e}}^{\rm{x}}}{\rm{dx}} \end{array}\)

Apply integration by part,

\(\begin{array}{l} \Rightarrow - {\rm{\;I}} = {\rm{\;cosx\;}}\mathop \smallint \nolimits_0^{\rm{\pi }} {{\rm{e}}^{\rm{x}}}{\rm{\;dx}} - {\rm{\;}}\mathop \smallint \nolimits_0^{\rm{\pi }} (\frac{{{\rm{d}}\left( {{\rm{cosx}}} \right)}}{{{\rm{dx}}}}{\rm{\;}}\mathop \smallint \nolimits_0^{\rm{\pi }} {{\rm{e}}^{\rm{x}}}{\rm{\;dx}}){\rm{\;\;dx}}\\ \Rightarrow - {\rm{I}} = \left[ {{\rm{cosx\;}}.{{\rm{e}}^{\rm{x}}}} \right]\begin{array}{*{20}{c}} {\rm{\pi }}\\ 0 \end{array} + {\rm{\;}}\mathop \smallint \nolimits_0^{\rm{\pi }} {\rm{sinx\;}}{{\rm{e}}^{\rm{x}}}{\rm{\;dx}} \end{array}\)

Given

⇒ I = \(\mathop \smallint \nolimits_0^{\rm{\pi }} {{\rm{e}}^{\rm{x}}}\sin {\rm{x}}\) dx

So,

\(\Rightarrow - {\rm{I}} = \left[ {{\rm{cosx\;}}.{{\rm{e}}^{\rm{x}}}} \right]\begin{array}{*{20}{c}} {\rm{\pi }}\\ 0 \end{array} + {\rm{\;I}}\)

\(\begin{array}{l} \Rightarrow - 2{\rm{I}} = \left( { - {{\rm{e}}^{\rm{\pi }}} - 1} \right)\\ \Rightarrow {\rm{\;I}} = \frac{{{{\rm{e}}^{\rm{\pi }}} + 1}}{2} \end{array}\)

Concept:

a² - b² = (a - b) (a + b)

Calculation:

Given that,

\(\Rightarrow {\rm{f}}\left( {\rm{x}} \right) = \frac{{{{\rm{x}}^2} - 9}}{{{{\rm{x}}^2} - 2{\rm{x}} - 3}}\)

\(\Rightarrow {\rm{f}}\left( {\rm{x}} \right) = \frac{{\left( {{\rm{x}} - 3} \right)\left( {{\rm{x}} + 3} \right)}}{{{{\rm{x}}^2} - 3{\rm{x}} + {\rm{x}} - 3}}\) [∵ a² - b² = (a-b) (a+b)]

⇒ \({\rm{f}}\left( {\rm{x}} \right) = \frac{{\left( {{\rm{x}} - 3} \right)\left( {{\rm{x}} + 3} \right)}}{{{\rm{x}}\left( {{\rm{x}} - 3} \right) + 1\left( {{\rm{x}} - 3} \right)}}\)

\(\Rightarrow {\rm{f}}\left( {\rm{x}} \right) = \frac{{\left( {{\rm{x}} - 3} \right)\left( {{\rm{x}} + 3} \right)}}{{\left( {{\rm{x}} - 3} \right)\left( {{\rm{x}} + 1} \right)}}\)

⇒ \({\rm{f}}\left( {\rm{x}} \right) = \frac{{\left( {{\rm{x}} + 3} \right)}}{{{\rm{x}} + 1}}\)

Given f(x) is continuous at x = 3

\(\therefore {\rm{f}}\left( 3 \right) = \mathop {\lim }\limits_{{\rm{x}} \to 3} {\rm{f}}\left( {\rm{x}} \right) = \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to 3} \frac{{\left( {{\rm{x}} + 3} \right)}}{{{\rm{x}} + 1}} = \frac{{\left( {3 + 3} \right)}}{{3 + 1}} = \frac{6}{4} = 1.5\)

Concept:

• If f(x) = a sin x

 b cos x then range of f(x) is \( - {\rm{\;}}\sqrt {{{\rm{a}}^2} + {{\rm{b}}^2}} \le {\rm{f}}\left( {\rm{x}} \right) \le \sqrt {{{\rm{a}}^2} + {{\rm{b}}^2}} {\rm{\;}}\).

• If f : R → S then R denotes domain of f(x) and S define as range of f(x).

 

Calculation:

Given that,

⇒ f(x) = 4 sin x – 3 cos x + 1

When we compare 4 sin x – 3 cos x to a sin x

 b cos x we find that,

a = 4 and b = - 3 so range of 4 sin x – 3 cos x is

\(\Rightarrow - \sqrt {{4^2} + {3^2}{\rm{\;}}} {\rm{\;}} \le {\rm{\;}}4{\rm{\;sin\;x\;}}-{\rm{\;}}3{\rm{\;cos\;x\;}} \le \sqrt {{4^2} + {3^2}} \)

⇒- 5 ≤ 4 sin x – 3 cos x ≤ 5

Now add 1

⇒ - 5 + 1 ≤ 4 sin x – 3 cos x + 1 ≤ 5 + 1

⇒ - 4 ≤ f(x) ≤ 6

So S = [-4, 6]

Concept:

Domain of function f(x) is define as values of x for which function f(x) exist.

Function f(x) = |x| which is define as

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

Calculation:

Given that,

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

⇒ |x| - x > 0

Which can be possible only when x < 0. If x > 0 then |x| - x become zero which is not possible.

So domain of f(x) is (-∞, 0).

Concept:

\(\smallint \frac{1}{{\rm{x}}}{\rm{\;dx}} = {\rm{lnx}} + {\rm{c}}\)

ln a – ln b = ln(a/b)

Calculation:

given that,

⇒ x dy – y dx = 0

⇒ x dy = y dx

\(\Rightarrow \smallint \frac{{{\rm{dy}}}}{{\rm{y}}} = \smallint \frac{{{\rm{dx}}}}{{\rm{x}}}\)

⇒ ln y = ln x + c

⇒ ln y – ln x = lnc

⇒ ln (y/x) = ln c

[∵ we can write constantly as ln c because it makes calculation easy]

⇒ y/x = c

Concept:

Derivative of some standard function are given below:

• \(\frac{{{\rm{d}}\left( {{{\rm{e}}^{\rm{x}}}} \right)}}{{{\rm{dx}}}} = {{\rm{e}}^{\rm{x}}}\)
• \(\frac{{{\rm{d}}\left( {{\rm{tanx}}} \right)}}{{{\rm{dx}}}} = {\sec ^2}{\rm{x\;}}\)
• \(\frac{{{\rm{d}}\left( {{\rm{secx}}} \right)}}{{{\rm{dx}}}} = {\rm{secx}}.{\rm{tanx}}\)
• \(\frac{{{\rm{d}}\left( {{\rm{lnx}}} \right)}}{{{\rm{dx}}}} = \frac{1}{{\rm{x}}}\)

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Calculation: 

Given that,

f(x) = e^{tan x} + ln(sec x) – e^{ln x}

⇒ f(x) = e^{tan x} + ln(sec x) – x [∵ e^lnx = x]

\(\Rightarrow \frac{{{\rm{d}}\left( {{\rm{f}}\left( {\rm{x}} \right)} \right)}}{{{\rm{dx}}}} = {{\rm{e}}^{{\rm{tanx}}}}.{\sec ^2}{\rm{x\;}} + \frac{1}{{{\rm{secx}}}}{\rm{\;}}.{\rm{secx}}.{\rm{tanx}} - 1\)

Put x = \({\frac{{\rm{\pi }}}{4}}\) in derivative

\(\Rightarrow {\rm{f}}\left( {\frac{{\rm{\pi }}}{4}} \right) = {\rm{\;}}{{\rm{e}}^1}{\rm{\;}}.{\left( {\sqrt 2 } \right)^2} + 1 - 1\) [ ∵ tan (π/4) = 1, sec (π/4) = √2]

⇒ f’(π/4) = 2e

Where μ > 0 

Concept:

SHM differential equation is \(\frac{{{{\rm{d}}^2}{\rm{x}}}}{{{\rm{d}}{{\rm{t}}^2}}} = - {{\rm{\omega }}^2}{\rm{x}}\), Where ω is angular frequency.

Time period of SHM = T = \(\frac{{2{\rm{\pi }}}}{{\rm{\omega }}}\)

 

Calculation:

If we compare all differential equation with SHM equation we find that,

\(\Rightarrow \frac{{{{\rm{d}}^2}{\rm{x}}}}{{{\rm{d}}{{\rm{t}}^2}}} + {\rm{\mu x}} = 0\)

\(\Rightarrow \frac{{{{\rm{d}}^2}{\rm{x}}}}{{{\rm{d}}{{\rm{t}}^2}}} = - {\rm{\;\mu x}}\)

Where \({\rm{\mu }} = {{\rm{\omega }}^2}\)

So \({\rm{\omega }} = \sqrt {\rm{\mu }} \)

Or time period of equation \({\rm{T}} = \frac{{2{\rm{\pi }}}}{{\sqrt {\rm{\mu }} }}\) 

So \(\frac{{{{\rm{d}}^2}{\rm{x}}}}{{{\rm{d}}{{\rm{t}}^2}}} + {\rm{\mu x}} = 0\) is a periodic differential equation with time period \({\rm{T}} = \frac{{2{\rm{\pi }}}}{{\sqrt {\rm{\mu }} }}\)