Quantitative Aptitude PYQs

Calculation:

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

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

\(\Rightarrow {{\rm{A}}^2} = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} {0 + 1}&{0 + 0}\\ {0 + 0}&{1 + 0} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]\)

Now,

\(\Rightarrow {{\rm{A}}^4} = {{\rm{A}}^2}{{\rm{A}}^2} = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right] = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]\)

Hence Option 1^st is correct answer.

Concept:

• \(1 + \cos {\rm{A}} = 2{\cos ^2}\left( {\frac{{\rm{A}}}{2}} \right)\)

Calculation:

Given: sin A = 3/5

\(\cos {\rm{A}} = {\rm{\;}}\sqrt {1 - {\rm{\;}}{{\sin }^2}{\rm{A}}} = {\rm{\;}}\sqrt {1 - {\rm{\;}}{{\left( {\frac{3}{5}} \right)}^2}} = {\rm{\;}}\sqrt {\frac{{16}}{{25}}} = {\rm{\;}} \pm \frac{4}{5}\)

450° < A < 540°, 

360°+ 90° < A < 360°+180° 

∴ 90° < A < 180° 

So A lies in 2^nd quad

 

We know that in the 2^nd Quadrant cos A is negative.

∴ cos A = -4/5

\(1 + \cos {\rm{A}} = 2{\cos ^2}\left( {\frac{{\rm{A}}}{2}} \right)\)

\(\Rightarrow \cos \left( {\frac{{\rm{A}}}{2}} \right) = {\rm{\;}}\sqrt {\frac{{1 + \cos {\rm{A}}}}{2}} = {\rm{\;}}\sqrt {\frac{{1 - \frac{4}{5}}}{2}} = {\rm{\;}}\sqrt {\frac{1}{{10}}} {\rm{\;}}\)

Concept:

• 2 sin θ cos θ = sin 2θ
• sin (x - y) = sin x cos y - cos x sin y
• sin (x + y) = sin x cos y + cos x sin y

 

Calculation:

We have to find the value of \(\frac{1}{{\sin 10^\circ }} - \frac{{\sqrt 3 }}{{\cos 10^\circ }}\) 

\(\frac{1}{{\sin 10^\circ }} - \frac{{\sqrt 3 }}{{\cos 10^\circ }} = {\rm{\;}}\frac{{\cos 10^\circ - {\rm{\;}}\sqrt 3 \sin 10^\circ }}{{\sin 10^\circ \cos 10^\circ }} = {\rm{\;}}\frac{{2\; \times \;\left( {\cos 10^\circ - {\rm{\;}}\sqrt 3 \sin 10^\circ } \right)}}{{2\; \times \;\sin 10^\circ \cos 10^\circ }}\)

\( = {\rm{\;}}\frac{{2\; \times \;\left( {\cos 10^\circ - {\rm{\;}}\sqrt 3 \sin 10^\circ } \right)}}{{\sin 20^\circ }}\)

\(= {\rm{\;}}\frac{{4\; \times \;\left( {\frac{1}{2}\cos 10^\circ - {\rm{\;}}\frac{{\sqrt 3 }}{2}\sin 10^\circ } \right)}}{{\sin 20^\circ }}\)

\(= {\rm{\;}}\frac{{4\; \times \;\left( {\sin 30^\circ \cos 10^\circ - {\rm{\;}}\cos 30^\circ \sin 10^\circ } \right)}}{{\sin 20^\circ }}\)

\(= {\rm{\;}}\frac{{4\; \times {\rm{\;}}\sin (30^\circ - 10^\circ )}}{{\sin 20^\circ }}\)

\(= {\rm{}}\frac{{4\; \times {\rm{\;}}\sin 20^\circ }}{{\sin 20^\circ }}\)

= 4

Concept:

Angle of depression: The angle of depression is formed when the observer is higher than the object he/she is looking at.

• When an observer looks at an object that is situated at a distance lower than the observer, an angle is formed below the horizontal line drawn with the level of the eye of the observer and line joining object with the observer’s eye.

 

Calculation:

Given: Angle of depression = \({\tan ^{ - 1}}\left( {\frac{5}{{12}}} \right)\)

Let Angle of depression = θ

\(\Rightarrow {\rm{\theta }} = {\rm{\;}}{\tan ^{ - 1}}\left( {\frac{5}{{12}}} \right)\)

\(\Rightarrow \tan {\rm{\theta }} = {\rm{\;}}\frac{5}{{12}}\)

In triangle ABC,

\(\tan {\rm{\theta }} = {\rm{\;}}\frac{{{\rm{AB}}}}{{{\rm{BC}}}}\)

\(\Rightarrow \frac{5}{{12}} = {\rm{\;}}\frac{{100}}{{{\rm{BC}}}}\)

∴ BC = 12 × 20 = 240 m

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.

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:

Let f(x) = \(\sin \left( {{\rm{x}} + \frac{{\rm{\pi }}}{6}} \right) + \cos \left( {{\rm{x}} + \frac{{\rm{\pi }}}{6}} \right)\)

Differentiate with respect to x, we get

\(\Rightarrow {\rm{f'}}\left( {\rm{x}} \right) = \cos \left( {{\rm{x}} + \frac{{\rm{\pi }}}{6}} \right) - {\rm{\;}}\sin \left( {{\rm{x}} + \frac{{\rm{\pi }}}{6}} \right){\rm{\;\;}}\)

Again differentiate with respect to x,

\(\Rightarrow {\rm{f''}}\left( {\rm{x}} \right) = - \sin \left( {{\rm{x}} + \frac{{\rm{\pi }}}{6}} \right) - \cos \left( {{\rm{x}} + \frac{{\rm{\pi }}}{6}} \right) = {\rm{\;}} - \left[ {\sin \left( {{\rm{x}} + \frac{{\rm{\pi }}}{6}} \right) + \cos \left( {{\rm{x}} + \frac{{\rm{\pi }}}{6}} \right)} \right]{\rm{\;}}\)

For maximum value f’(x) = 0

\(\Rightarrow \cos \left( {{\rm{x}} + \frac{{\rm{\pi }}}{6}} \right) - {\rm{\;}}\sin \left( {{\rm{x}} + \frac{{\rm{\pi }}}{6}} \right) = 0\)

\(\Rightarrow \cos \left( {{\rm{x}} + \frac{{\rm{\pi }}}{6}} \right) = {\rm{\;}}\sin \left( {{\rm{x}} + \frac{{\rm{\pi }}}{6}} \right)\)

\(\Rightarrow \tan \left( {{\rm{x}} + \frac{{\rm{\pi }}}{6}} \right) = 1\)

∴ x + (π/6) = (π/4)

⇒ x = π/12

At x = π/12, f”(x) < 0

So function is maximum at x = π/12

Concept:

• 2sin x sin y = cos (x - y) – cos (x + y)

Calculation:

We know that π = 180°

∴ π/18 = 10°, 5π/18 = 50° and 7π/18 = 70°

K = sin (π/18) × sin (5π/18) × sin (7π/18)

= sin 10 × sin 50 × sin 70

= (1/2) × (2 sin 10 × sin 50) × sin 70

= (1/2) × (cos (50-10) - cos (10+50)) × sin 70

= - (1/2) × (cos 60 - cos 40) × sin 70

= - (1/2) × (1/2 - cos 40) × sin 70

= - (1/4) × (1 - 2cos 40) × sin 70

= -1/4 (sin 70 - 2 cos 40 × sin 70)

= -1/4 × (sin 70 - 2 sin (90 - 50) × sin 70)

= -1/4 × (sin 70 - 2 sin 50 × sin 70)

= -1/4 × (sin 70 + (cos (50 + 70) – cos (70 - 50)))

= -1/4 × (sin 70 + (cos 120 - cos 20))

= -1/4 × (sin 70 + (cos 120 - sin (90 - 70)))

= -1/4 × (sin 70 + (cos 120 - sin 70))

= -1/4 × (sin 70 + cos 120 - sin 70)

= -1/4 × (cos 120)

= -1/4 × (-1/2)

= 1/8

Concept:

• \(\sin {\rm{x}} + {\rm{}}\sin {\rm{y}} = 2\sin \left( {\frac{{{\rm{x}}\; +\; {\rm{y}}}}{2}} \right)\cos \left( {\frac{{{\rm{x}} - {\rm{y}}}}{2}} \right)\)
• \(\cos {\rm{x}} + {\rm{}}\cos {\rm{y}} = 2\cos \left( {\frac{{{\rm{x}}\; +\; {\rm{y}}}}{2}} \right)\cos \left( {\frac{{{\rm{x}}\; - \;{\rm{y}}}}{2}} \right)\)

 

Calculation:

We have to find the value of \(\frac{{\sin {\rm{\alpha }}\; +\; \sin {\rm{\beta }}}}{{\cos {\rm{\alpha }}\; + \;\cos {\rm{\beta }}}}\)

\(\Rightarrow \frac{{\sin {\rm{\alpha }}\; + \;\sin {\rm{\beta }}}}{{\cos {\rm{\alpha }}\; + \;\cos {\rm{\beta }}}} = {\rm{\;}}\frac{{2\sin \left( {\frac{{{\rm{\alpha }}\; + \;{\rm{\beta }}}}{2}} \right)\cos \left( {\frac{{{\rm{\alpha }}\; - \;{\rm{\beta }}}}{2}} \right)}}{{2\cos \left( {\frac{{{\rm{\alpha }}\; + \;{\rm{\beta }}}}{2}} \right)\cos \left( {\frac{{{\rm{\alpha }}\; - \;{\rm{\beta }}}}{2}} \right)}}\)

\(= \tan \left( {\frac{{{\rm{\alpha }}\; + \;{\rm{\beta }}}}{2}} \right)\)

Concept:

• sin (x - y) = sin x cos y - cos x sin y
• sin (x + y) = sin x cos y + cos x sin y

Calculation:

Given:

sin θ = 3 sin (θ + 2α)

⇒ sin (θ + α −  α) = 3 sin (θ + α + α)

⇒ sin (θ + α) cos α – cos (θ + α) sin α = 3 sin (θ + α) cos α + 3 cos (θ + α) sin α

⇒ −2 sin (θ + α) cos α = 4 cos (θ + α) sin α

⇒ 2 tan (θ + α) = -4 tan α

⇒ tan (θ + α) = -2 tan α

∴ tan (θ + α) + 2 tan α = 0

Concept:

Let A(x₁, y_1, z₁) and B(x₂, y₂, z₂)

• \({\rm{Distance\;between\;A\;and\;B\;}} = \sqrt {{\rm{\;}}{{\left( {{{\rm{x}}_2} - {\rm{\;}}{{\rm{x}}_1}} \right)}^2} + {\rm{\;}}{{\left( {{{\rm{y}}_2} - {\rm{\;}}{{\rm{y}}_1}} \right)}^2} + \;{{\left( {{{\rm{z}}_2} - {\rm{\;}}{{\rm{z}}_1}} \right)}^2}} \)

Calculation:

Let A(0, 0, 0), B(a, 0, 0), C(0, b, 0) and 0(0, 0, c) be the four distinct points.

Let P(x, y, z) be the required point.

P is equidistance from A, B, C and O.

∴ AP = BP = CP = OP

Take AP = OP

Squaring both sides, we get

⇒ AP² = BP²

⇒ (x - 0)² + (y - 0)² + (z - 0)² = (x - a) ² + (y - 0)² + (z - 0)²

⇒ x² + y² + z² = x² - 2ax + a² + y² + z²

⇒ 2ax = a²

∴ x = a/2

Similarly, y = b/2 and z = c/2

So, P(x, y, z) = \({\rm{P}}\left( {\frac{{\rm{a}}}{2},\frac{{\rm{b}}}{2},\frac{{\rm{c}}}{2}} \right)\)

Concept:

Let A(x₁, y_1, z₁) and B(x₂, y₂, z₂)

• \({\rm{Distance\;between\;A\;and\;B\;}} = \sqrt{{{\rm{\;}}{{\left( {{{\rm{x}}_2} - {\rm{\;}}{{\rm{x}}_1}} \right)}^2} + {\rm{\;}}{{\left( {{{\rm{y}}_2} - {\rm{\;}}{{\rm{y}}_1}} \right)}^2} + \;{{\left( {{{\rm{z}}_2} - {\rm{\;}}{{\rm{z}}_1}} \right)}^2}} }\)

Calculation:

Given points are P(3, 2, 4), Q(4, 5, 2), R(5, 8, 0) and S(2 -1, 6).

\({\rm{PQ\;}} = \sqrt{ {{\rm{\;}}{{\left( {4 - {\rm{\;}}3} \right)}^2} + {\rm{\;}}{{\left( {5 - {\rm{\;}}2} \right)}^2} + \;{{\left( {2 - {\rm{\;}}4} \right)}^2}}} = \;\sqrt{{1 + 9 + 4} }= \;\sqrt{ {14} }\)

\({\rm{QR\;}} = ​​\sqrt{ {{\rm{\;}}{{\left( {5 - {\rm{\;}}4} \right)}^2} + {\rm{\;}}{{\left( {8 - {\rm{\;}}5} \right)}^2} + \;{{\left( {0 - {\rm{\;}}2} \right)}^2}} }= \;\sqrt{ {1 + 9 + 4}} = \;\sqrt{ {14} }\) 

\({\rm{RS\;}} = ​​\sqrt{ {{\rm{\;}}{{\left( {2 - {\rm{\;}}5} \right)}^2} + {\rm{\;}}{{\left( { - 1 - {\rm{\;}}8} \right)}^2} + \;{{\left( {6 - {\rm{\;}}0} \right)}^2}} }= \;\sqrt{ {9 + 81 + 36} }= \;\sqrt{ {126}} = 3\sqrt{ {14}} \)

\({\rm{PS\;}} = \sqrt{{{\rm{\;}}{{\left( {2 - {\rm{\;}}3} \right)}^2} + {\rm{\;}}{{\left( { - 1 - {\rm{\;}}2} \right)}^2} + \;{{\left( {6 - {\rm{\;}}4} \right)}^2}}} = \;\sqrt{ {1 + 9 + 4}} = \;\sqrt{ {14} }\) 

Here, We can write 3√14 = √14 + √14 + √14

⇒ RS = PQ + QR + PS

So, points are collinear.

Concept:

• The equation of the line that passes through the two points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂)  is given by the formula: \(\frac{{{\rm{x}} - {{\rm{x}}_1}}}{{{{\rm{x}}_2} - {{\rm{x}}_1}}} = \frac{{{\rm{y}} - {{\rm{y}}_1}}}{{{{\rm{y}}_2} - {{\rm{y}}_1}}} = \frac{{{\rm{z}} - {{\rm{z}}_1}}}{{{{\rm{z}}_2} - {{\rm{z}}_1}}}\)

​​

Calculation:

The line passing through the points (1, 2, -1) and (3, -1, 2) is,

\(\frac{{{\rm{x}} - 1}}{{3 - 1}} = \frac{{{\rm{y}} - 2}}{{ - 1 - 2}} = \frac{{{\rm{z}} + 1}}{{2 + 1}}\) 

⇒ \( \frac{{{\rm{x}} - 1}}{2} = \frac{{{\rm{y}} - 2}}{{ - 3}} = \frac{{{\rm{z}} + 1}}{3} = {\rm{\lambda }}\) 

⇒ x = 2λ + 1, y = -3λ + 2 and z = 3λ – 1

Line meets the yz-plane,

∴ x = 0

⇒ 2λ + 1 = 0

So, λ = -1/2

Now, y = -3λ + 2 = 7/2 and z = 3λ – 1 = -5/2

∴ Point (x, y, z) is \(\left( {0,\frac{7}{2},\; - \frac{5}{2}} \right)\)

Concept:

• If two lines are perpendicular to each other’s then product of direction ratio of these lines is zero.

Calculation:

Given lines are x = ay + b and z = cy + d

⇒ x – b = ay and z – d = cy

\(\therefore {\rm{\;}}\frac{{{\rm{x}} - {\rm{b}}}}{{\rm{a}}} = {\rm{\;}}\frac{{\rm{y}}}{1} = {\rm{\;}}\frac{{{\rm{z}} - {\rm{d}}}}{{\rm{c}}}\)            ….

(1) Direction ration of line is (a, 1, c)

Again, Lines are x = ey + f and z = gy + h

⇒ x – f = ey and z – h = gy

\(\therefore {\rm{\;}}\frac{{{\rm{x}} - {\rm{f}}}}{{\rm{e}}} = {\rm{\;}}\frac{{\rm{y}}}{1} = {\rm{\;}}\frac{{{\rm{z}} - {\rm{h}}}}{{\rm{g}}}\)            …. (2) Direction ration of line is (e, 1, g)

Given both line are perpendicular

∴ (a × e) + (1 × 1) + (c × g) = 0

⇒ ae + 1 + cg = 0

Or, ae + cg + 1 = 0

Concept:

• Diagonals of a parallelogram bisect each other.

Calculation:

Since, the diagonals of a parallelogram bisect each other. Therefore, P is the middle point of AC and BD both.

\(\Rightarrow \frac{{\overrightarrow {{\rm{OA}}} + \overrightarrow {{\rm{OC}}} }}{2} = \overrightarrow {{\rm{OP}}} \) 

\(\therefore \overrightarrow {{\rm{OA}}} + \overrightarrow {{\rm{OC}}} = 2 \times \overrightarrow {{\rm{OP}}} \)         ….

(1) Now,

\( \Rightarrow \frac{{\overrightarrow {{\rm{OB}}} + \overrightarrow {{\rm{OD}}} }}{2} = \overrightarrow {{\rm{OP}}} \)  

\(\therefore \overrightarrow {{\rm{OB}}} + \overrightarrow {{\rm{OD}}} = 2 \times \overrightarrow {{\rm{OP}}} \)         …. (2) Adding equation 1 and 2, we get

\(\overrightarrow {{\rm{OA}}} + \overrightarrow {{\rm{OB}}} + \overrightarrow {{\rm{OC}}} + \overrightarrow {{\rm{OD}}} = 4{\rm{\;}}\overrightarrow {{\rm{OP}}} {\rm{\;}}\)

Concept:

Triangle Law of Vector Addition: Triangle law of vector addition states that when two vectors are represented as two sides of the triangle with the order of magnitude and direction, then the third side of the triangle represents the magnitude and direction of the resultant vector.

\(\Rightarrow {\rm{\vec R}} = {\rm{\vec A}} + {\rm{\vec B}}\)

Calculation:

In triangle ABC,

\( \Rightarrow \overrightarrow {{\rm{BA}}} {\rm{\;}} + \overrightarrow {{\rm{AC}}} {\rm{\;}} = \overrightarrow {{\rm{BC}}} \)               ….

(1) In triangle BCD,

\(\Rightarrow \overrightarrow {{\rm{BC}}} + \overrightarrow {{\rm{CD}}} \; = \;\overrightarrow {{\rm{BD}}} \;\)            …. (2) Put the value of BC vector in equation 2^nd,

\(\Rightarrow \overrightarrow {{\rm{BA}}} {\rm{\;}} + \overrightarrow {{\rm{AC}}} {\rm{\;}} + \overrightarrow {{\rm{CD}}} \; = \;\overrightarrow {{\rm{BD}}} \)

\(\Rightarrow \overrightarrow {{\rm{BA}}} - \overrightarrow {{\rm{CA}}} {\rm{\;}} + \overrightarrow {{\rm{CD}}} \; = \;\overrightarrow {{\rm{BD}}}\)                           \( \left(\because {\overrightarrow {{\rm{AC}}} = \; - \overrightarrow {{\rm{CA}}} } \right)\) 

\(\therefore \overrightarrow {{\rm{BA}}} + \overrightarrow {{\rm{CD}}} = \overrightarrow {{\rm{BD}}} + \overrightarrow {{\rm{CA}}}\)

Concept:

Cross product of vectors: The Cross Product \({\rm{\vec a}} \times {\rm{\vec b}}\) of two vectors is another vector that is at right angles to both,

 

\( \Rightarrow {\rm{\vec a}} \times {\rm{\vec b}}\) = |a| |b| sin (θ) n

Where, |a| is the magnitude of vector a, |b| is the magnitude of vector b, θ is the angle between a and b and n is the unit vector at right angles to both a and b

Orthogonal Vectors: When you take the cross product of two vectors a and b, the resultant vector, (a x b), is orthogonal to both a and b.

Calculation:

Given:

\({\rm{\vec a}} \times {\rm{\vec b}} = {\rm{\vec c}}\) 

∴ \({\rm{\vec c}}\) is the resultant vector of vector a and vector b and it will be perpendicular to both vectors.

Now,

\({\rm{\vec b}} \times {\rm{\vec c}} = {\rm{\vec a}}\) 

∴ \({\rm{\vec a}}\) is the resultant vector of vector b and c and it will be perpendicular to both vectors.

Hence \({\rm{\vec a}},{\rm{\;\vec b}},{\rm{\;\vec c}}\) are orthogonal to each other.

Now, \({\rm{\vec a}} \times {\rm{\vec b}} = {\rm{\vec c}}\)

Taking mod both sides

|a × b| = |c|

⇒ |a| |b| sin θ = |c|                        (∵ |n| = 1)

Here θ = 90° then sin θ = 1

⇒ |a| |b| = |c|                       ….

(1) Now, \({\rm{\vec b}} \times {\rm{\vec c}} = {\rm{\vec a}}\)

Taking mod both sides

|b × c| = |a|

⇒ |b| |c| sin θ = |a|                        (∵ |n| = 1)

Here θ = 90° then sin θ = 1

⇒ |b| |c| = |a|                       …. (2) Put the value of |a| in equation 1^st,

⇒ |b| |c||b| = |c|

⇒ |b|² = 1

∴ |b| = 1

Put the value of |b| in equation 1^st, we get

|a| = |c|

Hence option 1^st is correct answer,

Concept:

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

 

Calculation:

Given \({\rm{\vec a}} = 2{\rm{\hat i}} + 3{\rm{\hat j}} + 4{\rm{\hat k}}\) 1) and \({\rm{\vec b}} = 3{\rm{\hat i}} + 2{\rm{\hat j}} - {\rm{\lambda \hat k}}\) perpendicular

∴ \({\rm{\vec a}} \cdot {\rm{\vec b}} = 0\)

\(\Rightarrow \left( {2{\rm{\hat i}} + 3{\rm{\hat j}} + 4{\rm{\hat k}}} \right)\; \cdot \left( {3{\rm{\hat i}} + 2{\rm{\hat j}} - {\rm{\lambda \hat k}}} \right) = 0\) 

⇒ 6 + 6 – 4λ = 0

⇒ 12 – 4λ = 0

∴ λ = 3

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:

\(\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - \;\left( {1 + x} \right)}}{{{x^2}}}\)                 [Form 0/0]

Apply L-Hospital Rule,

\( = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - \;\left( {0 + 1} \right)}}{{2x}} = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - \;1}}{{2x}}\)              [Form 0/0]

Again apply L-Hospital Rule,

\( = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - \;0}}{2} = \frac{{{e^0}}}{2} = \frac{1}{2}\)

Concept:

• \(\smallint {\sec ^2}{\rm{x\;dx}} = {\rm{tan\;x}} + {\rm{c}}\)

Calculation:

Let I = \(\mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} \frac{{{\rm{d\theta }}}}{{1 + \cos {\rm{\theta }}}}\)

\( \Rightarrow {\rm{I}} = \mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} \frac{{{\rm{d\theta }}}}{{2{{\cos }^2}\left( {\frac{{\rm{\theta }}}{2}} \right)}}\) 

\( \Rightarrow {\rm{I}} = \frac{1}{2}\mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} {\sec ^2}\left( {\frac{{\rm{\theta }}}{2}} \right){\rm{d\theta }}\) 

Let θ/2 = t

⇒ (dθ)/2 = dt

 

\( \Rightarrow {\rm{I}} = \mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{4}} {\sec ^2}{\rm{tdt}}\) 

\(\Rightarrow {\rm{I}} = {\rm{\;}}\left[ {\tan {\rm{t}}} \right]_0^{\frac{{\rm{\pi }}}{4}} = {\rm{\;}}\tan \frac{{\rm{\pi }}}{4} - {\rm{\;}}\tan 0 = 1 - 0 = 1\)

Concept:

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

Calculation:

Let I = \(\smallint \frac{{{\rm{dx}}}}{{{\rm{x}}\left( {{{\rm{x}}^7} + 1} \right)}} = \;\smallint \frac{{{\rm{dx}}}}{{{{\rm{x}}^8}\left( {1{\rm{\;}} + \frac{1}{{{x^7}}}} \right)}}\)

Let \(1{\rm{\;}} + \frac{1}{{{{\rm{x}}^7}}} = {\rm{t}}\)

Differentiating both sides, we get

\(\Rightarrow \left( {0{\rm{\;}} - \frac{7}{{{{\rm{x}}^8}}}} \right){\rm{dx}} = {\rm{dt}}\) 

\(\therefore \frac{1}{{{x^8}}}{\rm{dx}} = - \frac{{{\rm{dt}}}}{7}\) 

Now,

\({\rm{I}} = {\rm{}} - \frac{1}{7}\smallint \frac{{{\rm{dt}}}}{{\rm{t}}} = {\rm{}} - \frac{1}{7}\ln \left| {\rm{t}} \right| + {\rm{c}}\) 

\(\Rightarrow {\rm{I\;}} = -{\rm{\;}}\frac{1}{7}\ln \left| {1{\rm{\;}} + \frac{1}{{{{\rm{x}}^7}}}} \right| + {\rm{c}} = -{\rm{\;}}\frac{1}{7}{\rm{\;In\;}}\left| {\frac{{{{\rm{x}}^7} + 1}}{{{{\rm{x}}^7}}}} \right| + {\rm{c}}=\frac{1}{7}{\rm{In\;}}\left| {\frac{{{{\rm{x}}^7}}}{{{{\rm{x}}^7} + 1}}} \right| + {\rm{c}}\)

Concept:

One to one function:

• If for each x ∈ A there exist only one image y ∈ B and each y ∈ B has a unique pre-image x ∈ A (i.e. no two elements of A have the same image in B), then f is said to be one-one function
• Graphically, if a line parallel to x axis cuts the graph of f(x) at more than one point then f(x) is many-to-one function.
• Functions that are both one-to-one and onto are referred to as bijective.

Onto Function:

• The mapping of 'f' is said to be onto if every element of Y is the f-image of at least one element of X.
• A surjective function is a function whose image is equal to its codomain.

One-to-one and onto: A function f from A (the domain) to B (the range) is BOTH one-to-one and onto when no element of B is the image of more than one element in A, and all elements in B are used.

Calculation:

Observe the cos x graph in the figure.

It is clear that, function is one-one and onto when x and y are [0, π] and [-1, 1]

Explanation:

Graphically, if a line parallel to x axis cuts the graph of f(x) at more than one point then f(x) is many-to-one function. But in [0, π] if we draw a line parallel to x axis cut the graph of f(x) at one point so f(x) is one-one.

As we can see in the graph, range of f(x) is [-1, 1]

∴ Co-domain = Range

Hence function is onto.