Quantitative Aptitude PYQs

Concept:

• Area of a triangle whose vertices are (x1, y1), (x2, y2) and (x3, y3), is given by the expression

Area = (1/2) [x₁ (y₂- y₃) + x₂ (y₃-y₁) + x₃(y₁-y₂)]

• Whose expression can be written in the form of a determinant as

\(Area = ;\frac{1}{2}\left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\ {{x_2}}&{{y_2}}&1\ {{x_3}}&{{y_3}}&1 \end{array}} \right|\)

Calculation:

We have to calculate area of the triangle with vertices \(\left( {{x_1},\frac{1}{{{x_1}}}} \right),\left( {{x_2},\frac{1}{{{x_2}}}} \right),\left( {{x_3},\frac{1}{{{x_3}}}} \right)\)

\(Area = ;\frac{1}{2}\left| {\begin{array}{*{20}{c}} {{x_1}}&{\frac{1}{{{x_1}}}}&1\ {{x_2}}&{\frac{1}{{{x_2}}}}&1\ {{x_3}}&{{{\frac{1}{{{x_3}}}}_3}}&1 \end{array}} \right|\)

Now, Applying R₁ → R₁ – R₂ and R₂ → R₂ – R₃

\(Area = ;\frac{1}{2}\left| {\begin{array}{*{20}{c}} {{x_1} - {x_2}}&{\frac{1}{{{x_1}}} - \frac{1}{{{x_2}}}}&0\ {{x_2} - {x_3}}&{\frac{1}{{{x_2}}} - \frac{1}{{{x_3}}}}&0\ {{x_3}}&{\frac{1}{{{x_3}}}}&1 \end{array}} \right|;\) 

\(= ;\frac{1}{2};\left| {\begin{array}{*{20}{c}} {{x_1} - {x_2}}&{\frac{{ - ;\left( {{x_1} - {x_2}} \right)}}{{{x_1}{x_2}}}}&0\ {{x_2} - {x_3}}&{\frac{{ - ;\left( {{x_2} - {x_3}} \right)}}{{{x_2}{x_3}}}}&0\ {{x_3}}&{\frac{1}{{{x_3}}}}&1 \end{array}} \right|\)

\(= ;\frac{1}{2};\left( {{x_1} - {x_2}} \right)\left( {{x_2} - {x_3}} \right);\left| {\begin{array}{*{20}{c}} 1&{\frac{{ - ;1}}{{{x_1}{x_2}}}}&0\ 1&{\frac{{ - ;1}}{{{x_2}{x_3}}}}&0\ {{x_3}}&{\frac{1}{{{x_3}}}}&1 \end{array}} \right|\)

Now, expand c₃

\(= \frac{1}{2};\left( {{x_1} - {x_2}} \right)\left( {{x_2} - {x_3}} \right); \times 1 \times \left[ {\frac{{ - ;1}}{{{x_2}{x_3}}} - ;\frac{{ - ;1}}{{{x_1}{x_2}}}} \right]\)

\(= ;\frac{1}{2};\left( {{x_1} - {x_2}} \right)\left( {{x_2} - {x_3}} \right); \times 1 \times \left[ {\frac{{ - ;{x_1}}}{{{x_1}{x_2}{x_3}}} - ;\frac{{ - ;{x_3}}}{{{x_1}{x_2}{x_3}}}} \right]\)

\(= \frac{1}{2};\left( {{x_1} - {x_2}} \right)\left( {{x_2} - {x_3}} \right); \times 1 \times \frac{{\left( {{x_3} - {x_1}} \right)}}{{{x_1}{x_2}{x_3}}}\)

∴ \(Area = ;\left| {\frac{{\left( {{x_1} - {x_2}} \right)\left( {{x_2} - {x_3}} \right)\left( {{x_3} - {x_1}} \right)}}{{2{x_1}{x_2}{x_3}}}} \right|\)

Concept:

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 =\(\sqrt {{g^2} + {f^2} - c} \), where g, f and c are three constants

Note:

• If circle touches x axis then radius = |- f| = |y coordinate of centre|
• If circle touches y axis then radius = |- g| = |x coordinate of centre|

Calculation:

If the circle touches the y axis, it means y axis is tangent then normal intersects the circle at points A and B.

Equation of circle is x² + y² + gx + fy + \(\frac{e}{4}\) = 0

Centre of circle = (- g/2, - f/2)

If the circle touches y-axis

Then, radius = AC = -g/2

AC = CB = radius

So, coordinate of point B = (2 × -g/2, -f/2) = (- g, -f/2)

∴ Option 2 is correct.

Concept:

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

Calculation:

Given:

\(\left( {\vec a + \vec b} \right).\left( {\vec a + \vec b} \right) = {\left| {\vec a} \right|^2} + {\left| {\vec b} \right|^2}\)

\( \Rightarrow ;{\left| {\vec a} \right|^2} + {\left| {\vec b} \right|^2} + 2\vec a.{\rm{;}}\vec b = ;{\left| {\vec a} \right|^2} + {\left| {\vec b} \right|^2};\)

\(\Rightarrow 2\vec a.{\rm{;}}\vec b = 0\)

\(\Rightarrow \vec a.{\rm{;}}\vec b = 0\)

∴ \(\vec a\) and \(\vec b\) are perpendicular.

Concept:

Dot Product: it is also called the inner product or scalar product

Let the two vectors be \(\vec a\) and \(\vec b\)

Then, dot Product of two vector = \(\vec a.{\rm{;}}\vec b\) = |a||b| cos θ

Where |\(\vec a\)| = Magnitudes of vectors a

|\(\vec b\) | = Magnitudes of vectors b and θ is angle between a and b

2.  \(\vec i.\vec i\)= \(\vec j.\vec j\) = \(\vec k.\vec k\) = 1 and

\(\overrightarrow {;i} \cdot \vec j\) = \(\vec j \cdot \overrightarrow {i;} \) = \(\vec i \cdot \vec k\)

= \(\overrightarrow {;k} \cdot \overrightarrow {i;} \) = \(\vec j \cdot \vec k\) = \(\vec k \cdot \overrightarrow {j;} \) = 0

Calculation:

Given: \(\vec r\) = xî + yĵ + zk̂,

We have to find the value of \(\vec r\). (î + ĵ + k̂)

\(\vec r\). (î + ĵ + k̂) = (xî + yĵ + zk̂) . (î + ĵ + k̂) = x + y + z

∴ Option 4 is correct.

Concept:

• Let \(\vec a = x;\vec i + y;\vec j + z;\vec k\)

Magnitude of vector of a = \(\left| {\vec a} \right| = ;\sqrt {{x^2} + ;{y^2} + {z^2}} \)

Unit vector = \(\hat a = ;\frac{{\vec a}}{{\left| {\vec a} \right|}}\)

• Let \(\vec a\) and \(\vec b\) be the two vectors, then the vector \(\vec c\) perpendicular to both \(\vec a\) and \(\vec b\)

\(\vec a = ;{a_1}\vec i + {b_1}\vec j + ;{c_1}\vec k;\) and \(\vec b = ;{a_2}\vec i + {b_2}\vec j + ;{c_2}\vec k\)

\(\therefore ;\vec c; = ;\vec a; \times ;\vec b = ;\left| {\begin{array}{*{20}{c}} {\vec i}&{\vec j}&{\vec k;}\ {{a_1}}&{{b_1}}&{{c_1}}\ {{a_2}}&{{b_2}}&{{c_2}} \end{array}} \right|\)

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

 

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

Calculation:

Let vector \(\vec a\)= = 2î - ĵ + k̂ and \(\vec b\) = 3î - 4ĵ - k̂ and vector \(\vec c\)perpendicular to both \(\vec a\) and \(\vec b\)

\(\therefore ;\vec c; = ;\vec a; \times ;\vec b = ;;\left| {\begin{array}{*{20}{c}} {\vec i}&{\vec j}&{\vec k;}\ 2&{ - 1}&1\ 3&{ - 4}&{ - 1} \end{array}} \right|\)

= \(\hat i;\left( {1 + 4} \right) - \hat j;\left( { - 2 - 3} \right) + \hat k\left( { - 8 + 3} \right)\)

= \(5\hat i + 5\hat j - 5\hat k;\)

∴ Unit vector = \(\hat c = ;\frac{{\vec c}}{{\left| c \right|}} = ;\frac{{5\hat {i;} + ;5\hat {j;} - ;5\hat k}}{{\sqrt {{5^2} + {5^2} + {{\left( { - 5} \right)}^2}} }} = ;\frac{{5\hat {i;} + ;5\hat {j;} - ;5\hat k}}{{5\sqrt 3 }} = ;\frac{{\hat {i;} + ;\hat {j;} - ;\hat k}}{{\sqrt 3 }}\)

Concept:

• A negative correlation means that there is an inverse relationship between two variables - when one variable decreases, the other increases.

Calculation:

Insurance companies’ profits and the number at claims they have to pay

The above statement is true by the definition of negative correlation because it gives inverse relationship.

∴ Option 3 is correct

Concept:

1. Moment: It will be equal to the cross product of a position vector r from the point to anywhere on the line of action of the force and the force vector itself.⇔ \(\vec M = \vec r; \times \vec F\)
2. Cross product of two vectors: The cross product can be written as a determinant.

Let \(\vec a\) and \(\vec b\) be the two vectors. ⇒ \(\vec a = ;{a_1}\vec i + {b_1}\vec j + ;{c_1}\vec k;\)and \(\vec b = ;{a_2}\vec i + {b_2}\vec j + ;{c_2}\vec k\)

\(\therefore ;\vec a; \times ;\vec b = ;\left| {\begin{array}{*{20}{c}} {\vec i}&{\vec j}&{\vec k;}\ {{a_1}}&{{b_1}}&{{c_1}}\ {{a_2}}&{{b_2}}&{{c_2}} \end{array}} \right|\)
3. If  \(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}}\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}}\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}} \end{array}} \right]\) then determinant of A is given by:

|A| = a₁₁ × {(a₂₂ × a₃₃) – (a₂₃ × a₃₂)} - a₁₂ × {(a₂₁ × a₃₃) – (a₂₃ × a₃₁)} + a₁₃ × {(a₂₁ × a₃₂) – (a₂₂ × a₃₁)}
4. Magnitude of vector: Let \(\vec a = x;\vec i + y;\vec j + z;\vec k\) 

Magnitude of vector of a = \(\left| {\vec a} \right| = ;\sqrt {{x^2} + ;{y^2} + {z^2}} \)

Calculation:

Given: \({\rm{\vec r}} = {\rm{;;\hat i}} + {\rm{;}}2{\rm{\hat j}} + {\rm{;}}3{\rm{\hat k;and;\vec F}} = {\rm{;\lambda ;\hat k}}\)

We have to find magnitude of moment,

We know that, \({\rm{\vec M}} = {\rm{\vec r;}} \times {\rm{\vec F}}\)

\(\therefore ;{\rm{\vec M}} = ;\left| {\begin{array}{*{20}{c}} {\vec i}&{\vec j}&{\vec k;}\ 1&2&3\ 0&0&{\rm{\lambda }} \end{array}} \right|\) 

\( \Rightarrow ;{\rm{\vec M}} = ;\vec i\left( {2{\rm{\lambda }} - 0} \right) - ;\vec j;\left( {{\rm{\lambda }} - 0} \right) + ;\vec k;\left( {0 - 0} \right) = ;2{\rm{\lambda }}\vec i - ;{\rm{\lambda }}\vec j;\)

Now,

Magnitude of moment = \(\left| {{\rm{\vec M}}} \right| = ;\sqrt {{{\left( {2{\rm{\lambda }}} \right)}^2} + ;{{\left( { - {\rm{\lambda }}} \right)}^2}} \) = √5 λ

Concept:

• A function f(x) is said to be continuous at a point x = a, in its domain if \(\mathop {\lim }\limits_{x \to {\rm{a}}} f\left( x \right) = f\left( a \right)\) exists
• \(\mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^{ - 1}}x}}{x} = 1\)
• \(\mathop {\lim }\limits_{x \to 0} \frac{{{{\tan }^{ - 1}}x}}{x} = 1\)

 

Calculation:

Given: Function f(x) = \(\frac{{2x - {{\sin }^{ - 1}}x}}{{2x + ta{n^{ - 1}}x}}\) is continuous at each point in its domain.

We have to find the value of f(0),

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

\( \Rightarrow {\rm{f}}\left( 0 \right) = ;\mathop {\lim }\limits_{x \to 0} \frac{{x;\left( {2;--{\rm{;}}\frac{{{{\sin }^{ - 1}}x}}{x}} \right)}}{{x;\left( {2{\rm{;}} + {\rm{;}}\frac{{ta{n^{ - 1}}x}}{x}} \right)}} = ;\mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{\left( {2;--{\rm{;}}\frac{{{{\sin }^{ - 1}}x}}{x}} \right)}}{{\left( {2{\rm{;}} + {\rm{;}}\frac{{ta{n^{ - 1}}x}}{x}} \right)}} = ;\frac{{\left( {2 - 1} \right)}}{{\left( {2 + 1} \right)}} = ;\frac{1}{3}\) 

∴ f(0) = 1/3

Concept:

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

Suppose that we have one of the following cases,

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

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

Note: 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: \(f\left( x \right) = \sqrt {25 - {x^2}} \)

We have to find the value of \(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}}\)

\( \Rightarrow \mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}} = ;\frac{0}{0};\left( {Form} \right)\)

Apply L-Hospital Rule,

\( \Rightarrow \mathop {\lim }\limits_{x \to 1} \frac{{f'\left( x \right) - 0}}{{1 - 0}} = f'\left( 1 \right)\)

 Now, \(f\left( x \right) = \sqrt {25 - {x^2}} \)

\( \Rightarrow {{\rm{f}}^{\rm{'}}}\left( {\rm{x}} \right) = {\rm{;}}\frac{{1{\rm{;}} \times \left( {0 - 2{\rm{x}}} \right)}}{{2\sqrt {25 - {{\rm{x}}^2}} }} = {\rm{;}}\frac{{ - {\rm{x}}}}{{\sqrt {25 - {{\rm{x}}^2}} }}\)

\( \Rightarrow {{\rm{f}}^{\rm{'}}}\left( 1 \right) = {\rm{;}}\frac{{ - 1}}{{\sqrt {25 - {1^2}} }} = ;\frac{{ - 1}}{{\sqrt {24} }}\)

∴ \(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}} = ;f'\left( 1 \right) = ;\frac{{ - 1}}{{\sqrt {24} }}\)

Concept:

• \({\tan ^{ - 1}}x - ;{\tan ^{ - 1}}y = ;{\tan ^{ - 1}}\left[ {\frac{{x; - ;y}}{{1; + ;xy}}} \right]\)
• \({\tan ^{ - 1}}x + ;{\tan ^{ - 1}}y = ;{\tan ^{ - 1}}\left[ {\frac{{x; + ;;y}}{{1 - ;xy}}} \right]\)
• \({\tan ^{ - 1}}\tan x = x\)

Calculation:

We have to find the value of dy/dx

Given: \(y = {\tan ^{ - 1}}\left( {\frac{{5 - 2\tan \sqrt x }}{{2 + 5\tan \sqrt x }}} \right)\)

\( \Rightarrow y = {\tan ^{ - 1}}\left[ {\frac{{2{\rm{;}}\left( {\frac{5}{2} - \tan \sqrt x } \right)}}{{2{\rm{;}}\left( {1 + \frac{5}{2}\tan \sqrt x } \right)}}} \right] = ;{\tan ^{ - 1}}\left[ {\frac{{\left( {\frac{5}{2} - \tan \sqrt x } \right)}}{{\left( {1 + \frac{5}{2}\tan \sqrt x } \right)}}} \right] = ;{\tan ^{ - 1}}\left( {\frac{5}{2}} \right) - ;{\tan ^{ - 1}}\tan \sqrt x \)

⇒ y = tan^-1 (5/2) - √x                          (∵ tan^-1 (tan x) = x)

Differentiating, both sides, we get

\( \Rightarrow \frac{{dy}}{{dx}} = 0 - ;\frac{1}{{2\sqrt x }} = ;\frac{{ - 1}}{{2\sqrt x }}\)  

∴ Option 1 is correct

Concepts:

1. Let Standard form of quadratic Functions: ⇒ f(x) = ax² + bx + c

Steps to Graphing Quadratic Equations

1. Check the sign of a

• If a > 0 the graph will be a convex parabola (Upward)
• If a < 0 the graph will be a concave parabola (Downward)

1. Vertex = \(\left( {\frac{{ - b}}{{2a}},\frac{{ - D}}{{4a}}} \right)\), Where D is Discriminants (D = b² – 4ac)
2. Parabola is symmetric about the line that intersects x = \(\frac{{ - b}}{{2a}}\)
3. Find the coordinates of the points of intersection with the x axis, we must solve the equation f(x) = 0

 

Calculation:

Given: f(x) = x² – 4x + 5

Compare above equation with ax² + bx + c,

So, a = 1, b = - 4 and c = 5

Here a > 0 so graph will be a convex parabola (upward)

Now, Discriminants = D = b² – 4ac = (- 4)² – 4 × 1 × 5 = 16 – 20 = - 4 < 0

Hence no real roots are possible.

Vertex = \(\left( {\frac{{ - b}}{{2a}},\frac{{ - D}}{{4a}}} \right) = ;\left( {\frac{{ - ;\left( { - 4} \right)}}{{2; \times 1}},;\frac{{ - ;\left( { - 4} \right)}}{{4; \times 1}}} \right) = \left( {2,;1} \right)\) 

Minimum value of function is one (At vertex)

Now, Maximum value of f(x) = x² – 4x + 5 in domain (1, 4)

⇒ f(1) = 1² – 4 × 1 + 5 = 2

⇒ f(4) = 4² – 4 × 4 + 5 = 5

So, maximum value of function is 5

∴ Range = [1, 5)      

Note: The domain of the function is given in open interval (1, 4) so range at extreme position will not consider.

Concept:

• \(\mathop \smallint \nolimits_a^b f\left( x \right)dx = ;\mathop \smallint \nolimits_a^c f\left( x \right)dx + ;\mathop \smallint \nolimits_c^b f\left( x \right)dx\)
• \(\mathop \smallint \nolimits_a^b f'\left( x \right)dx = ;\left[ {f\left( x \right)} \right]_a^b = \left[ {f\left( b \right) - f\left( a \right)} \right]\)
• \(\smallint {x^n}dx = ;\frac{{{x^{n; + ;1}}}}{{n + 1}} + c\)

Calculation:

We have to find the value of \(\mathop \smallint \nolimits_2^8 \left| {x - 5} \right|dx\)

Now, f(x) = \(\left| {x - 5} \right|; ⇒ ;f\left( x \right) = \left{ {\begin{array}{*{20}{c}} { - ;\left( {x - 5} \right),;;;2 \le x < 5}\ {x - 5,;;;5 \le x \le 8} \end{array}} \right.\)

\( ⇒ \mathop \smallint \nolimits_2^8 \left| {x - 5} \right|dx = ;\mathop \smallint \nolimits_2^5 - \left( {{\rm{x}} - 5} \right);dx + ;\mathop \smallint \nolimits_5^8 \left( {{\rm{x}} - 5} \right);dx\)

\( ⇒ \mathop \smallint \nolimits_2^8 \left| {x - 5} \right|dx = ;\mathop \smallint \nolimits_2^5 \left( {5 - x} \right);dx + ;\mathop \smallint \nolimits_5^8 \left( {{\rm{x}} - 5} \right);dx\)

\( ⇒;\left[ {5x - ;\frac{{{x^2}}}{2}} \right]_2^5; + ;\left[ {\frac{{{x^2}}}{2} - 5x} \right]_5^8;\)

\( ⇒ ;\left[ {\left( {25 - ;\frac{{25}}{2}} \right) - ;\left( {10 - 2} \right)} \right] + ;\left[ {\left( {32 - 40} \right) - ;\left( {\frac{{25}}{2} - 25} \right)} \right]\)

\( ⇒ ;\frac{{25}}{2} - 8 - 8 + ;\frac{{25}}{2}\)

⇒ 25 – 16 = 9

∴ The value of the integral \(\mathop \smallint \nolimits_2^8 \left| {x - 5} \right|dx\) is 9.

Concept:

• \({e^{\ln x}}; = ;x\)
• \(\smallint \tan x;dx = ;\ln \left| {\sec x} \right| + c\)

Calculation:

We have to find the value of \(\smallint {e^{{\rm{ln}}(\tan x)}}dx\)

We know that \({e^{\ln x}}; = ;x\) 

So, \({e^{{\rm{ln}}(\tan x)}} = \tan x\)

\(\smallint {e^{{\rm{ln}}(\tan x)}}dx = ;\smallint \tan x;dx = ;\ln \left| {\sec x} \right| + c\)

∴ Option 2 is correct.

Concept:

• \({\tan ^{ - 1}}\frac{1}{x} = ;{\cot ^{ - 1}}x\)
• \(\frac{{d({{\cot }^{ - 1}}x)}}{{dx}} = ;\frac{{ - 1}}{{1; + ;;{x^2}}}\)
• \(\smallint \frac{1}{{1; + ;;{x^2}}};dx = ;{\tan ^{ - 1}}x + c\)
• \(\mathop \smallint \limits_{ - {\rm{a}}}^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = \left{ {\begin{array}{*{20}{c}} {2\mathop \smallint \limits_0^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}},;;;;;If;f\left( {\rm{x}} \right);is;even;function}\ {0,;;;;;If;f\left( {\rm{x}} \right);is;odd;function} \end{array}} \right.\)
• If f(x) is even function then f(-x) = f(x)
• If f(x) is odd function then f(-x) = - f(x)

Calculation:

\({\rm{Let;I}} = \mathop \smallint \limits_{ - 1}^1 \left{ {\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\tan }^{ - 1}}\frac{1}{{\rm{x}}}} \right)} \right}{\rm{dx}}\)

\( = {\rm{;}}\mathop \smallint \limits_{ - 1}^1 \left( {\frac{{\rm{d}}}{{{\rm{dx}}}}{{\cot }^{ - 1}}{\rm{x}}} \right){\rm{dx}} \)

\(= {\rm{;}}\mathop \smallint \limits_{ - 1}^1 \frac{{ - 1}}{{1{\rm{;}} + {\rm{;;}}{{\rm{x}}^2}}}{\rm{dx;}}\)

Let f(x) = \(\frac{{ - 1}}{{1; + ;;{x^2}}}\)

Now, f(-x) = \(\frac{{ - 1}}{{1; + ;;{{\left( { - x} \right)}^2}}} = ;\frac{{ - 1}}{{1; + ;;{x^2}}}\)

⇒ f(-x) = f(x)

⇒ f(x) is even function

According to the concept used

\({\rm{;I;}} = {\rm{;}}\mathop \smallint \limits_{ - 1}^1 \frac{{ - 1}}{{1{\rm{;}} + {\rm{;;}}{{\rm{x}}^2}}}{\rm{dx}}\)

\( = {\rm{;}}2\mathop \smallint \limits_0^1 \frac{{ - 1}}{{1{\rm{;}} + {\rm{;;}}{{\rm{x}}^2}}}{\rm{dx}} \)

\(= {\rm{;}} - 2\mathop \smallint \limits_0^1 \frac{1}{{1{\rm{;}} + {\rm{;;}}{{\rm{x}}^2}}}{\rm{dx;}}\)

\( \Rightarrow {\rm{;I}} = {\rm{;}} - 2{\rm{;}}\left[ {{{\tan }^{ - 1}}x} \right]_0^1\)

⇒ I = -2 [tan^-1 (1) - tan^-1 (0)]

⇒ I = - 2 (π/4) = - π/2

∴ The value of the integral \(\mathop \smallint \limits_{ - 1}^1 \left{ {\frac{d}{{dx}}\left( {{{\tan }^{ - 1}}\frac{1}{x}} \right)} \right}dx\) is -π/2.

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) = [x(x – 1) + 1]^1/3

[x(x – 1) + 1]^1/3 is minimum when x(x – 1) + 1 is minimum.

Let g(x) = x(x – 1) + 1 = x² – x + 1

Differentiating, both sides, we get

⇒ g’(x) = 2x – 1

For maxima and minima, g’(x) = 0

⇒ 2x – 1 = 0

∴ x = 1/2

Now we have to find second derivative,

⇒ g”(x) = 2 > 0

So, function gives minimum value at x = 1/2

\( \Rightarrow ;f{\left( x \right)_{min}};; = ;{\left[ {\frac{1}{2};\left( {\frac{1}{2};-;1} \right); + ;1} \right]^{\frac{1}{3}}}\)

\( \Rightarrow ;f{\left( x \right)_{min}};; = ;{\left[ {\frac{1}{2};\left( {\frac{{ - 1}}{2};} \right); + ;1} \right]^{\frac{1}{3}}} = ;{\left[ {\frac{{ - 1}}{4} + 1} \right]^{\frac{1}{3}}} = ;{\left( {\frac{3}{4}} \right)^{\frac{1}{3}}};\)

∴ Option 1 is correct.

Concept:

• \(\left| {{\rm{sin;x}}} \right| = \left{ {\begin{array}{*{20}{c}} { - \sin {\rm{x}},;; - \pi ; \le x < 0}\ {\sin {\rm{x}},;;0; \le x \le ;\pi } \end{array}} \right.\)
• \(\left| x \right| = \left{ {\begin{array}{*{20}{c}} { - x,;;x < 0}\ {x,;;x \ge 0} \end{array}} \right.\)
• \(\frac{{{\rm{d}}\left( {{\rm{uv}}} \right)}}{{{\rm{dx}}}} = {\rm{v}}\frac{{{\rm{du}}}}{{{\rm{dx}}}} + {\rm{u}}\frac{{{\rm{dv}}}}{{{\rm{dx}}}}\) , Where u and v are function of x.
• log (m)ⁿ = n log (m)
• log (1/x) = - log (x)

Calculation:

We have to find the value of \(\frac{{dy}}{{dx}}{\rm{;}}at{\rm{;}}x = {\rm{;}} - \frac{\pi }{6}\)  (x < 0)

y = |sin x|^|x| = (- sin x) ^–x             (∵ x < 0)

Taking log both sides,

⇒ log y = log (- sin x) ^–x = (- x) log (- sin x)                        [∵ log (m)ⁿ = n log (m) ]

Differentiating, both sides, we get

\( \Rightarrow ;\frac{1}{y};\frac{{dy}}{{dx}} = ;\left[ {\left{ {\frac{{ - x}}{{ - \sin x}} \cdot \left( { - \cos x} \right)} \right} + ;\log \left( { - \sin x} \right) \cdot \left( { - 1} \right)} \right]\)

 \( \Rightarrow {\rm{;;}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = {\rm{;y}}\left[ {\left( {\frac{{ - {\rm{x}}\cos {\rm{x}}}}{{\sin {\rm{x}}}}} \right) - {\rm{;}}\log \left( { - \sin {\rm{x}}} \right)} \right] = {\rm{;}}{\left( { - \sin {\rm{x}}} \right)^{ - {\rm{x}}}}\left[ {\left( { - {\rm{x;}}\cot {\rm{x}}} \right) - {\rm{;}}\log \left( { - \sin {\rm{x}}} \right)} \right]\) 

\( \Rightarrow {\rm{;;}}{\left[ {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right]_{{\rm{x;}} = {\rm{;}}\left( {\frac{{ - {\rm{;\pi }}}}{6}} \right){\rm{;}}}} = {\rm{;}}{\left( { - \sin \left( {\frac{{ - {\rm{;\pi }}}}{6}} \right)} \right)^{ - {\rm{;}}\left( {\frac{{ - {\rm{;\pi }}}}{6}} \right)}}\left[ {\left( { - {\rm{;}}\left( {\frac{{ - {\rm{;\pi }}}}{6}} \right)\cot \left( {\frac{{ - {\rm{;\pi }}}}{6}} \right)} \right) - {\rm{;}}\log \left( { - \sin \left( {\frac{{ - {\rm{;\pi }}}}{6}} \right)} \right)} \right]\)

We know that sin (-θ) = - sin θ and cot (-θ) = - cot θ

\( \Rightarrow {\rm{;;}}{\left[ {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right]_{{\rm{x;}} = {\rm{;}}\left( {\frac{{ - {\rm{;\pi }}}}{6}} \right){\rm{;}}}} = {\rm{;}}{\left( {\sin \left( {\frac{{\rm{\pi }}}{6}} \right)} \right)^{\left( {\frac{{\rm{\pi }}}{6}} \right)}}\left[ { - \left( {\left( {\frac{{\rm{\pi }}}{6}} \right)\cot \left( {\frac{{\rm{\pi }}}{6}} \right)} \right) - {\rm{;}}\log \left( {\sin \left( {\frac{{\rm{\pi }}}{6}} \right)} \right)} \right]\)

\( \Rightarrow {\rm{;;}}{\left[ {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right]_{{\rm{x;}} = {\rm{;}}\left( {\frac{{ - {\rm{;\pi }}}}{6}} \right){\rm{;}}}} = {\rm{;}}{\left( {\frac{1}{2}} \right)^{\left( {\frac{{\rm{\pi }}}{6}} \right)}}\left[ { - \left( {\left( {\frac{{\rm{\pi }}}{6}} \right)\sqrt 3 } \right) - {\rm{;}}\log \left( {\frac{1}{2}} \right)} \right] = ;{2^{ - \frac{{\rm{\pi }}}{6}}};\left[ {\frac{{ - \pi \sqrt 3 }}{6} + ;\log 2} \right]\)      [∵ log (1/x) = - log (x)]

\(\therefore {\left[ {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right]_{{\rm{x;}} = {\rm{;}}\left( {\frac{{ - {\rm{;\pi }}}}{6}} \right){\rm{;}}}} = ;\frac{{{2^{ - \frac{\pi }{6}}}\left( {6\ln 2 - \sqrt {3 }\pi } \right)}}{6}\)

Concept:

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

Calculation:

Given:

\(\frac{{dx}}{{dy}} = \frac{{x; + {\rm{;}}y; + {\rm{;}}1}}{{x; + {\rm{;}}y; - {\rm{;}}1}}\)

Let x + y = t

Differentiating, both sides with respect to y, we get

\( \Rightarrow ;1; + ;\frac{{dx}}{{dy}} = ;\frac{{dt}}{{dy}}\)

\( \Rightarrow \frac{{dx}}{{dy}} = ;\frac{{dt}}{{dy}} - 1\)

Now,

\( \Rightarrow \frac{{{\rm{dt}}}}{{{\rm{dy}}}} - 1 = {\rm{;}}\frac{{{\rm{t;}} + {\rm{;}}1}}{{{\rm{t;}} - {\rm{;}}1}}\)

\( \Rightarrow \frac{{{\rm{dt}}}}{{{\rm{dy}}}} = {\rm{;}}\frac{{{\rm{t;}} + {\rm{;}}1}}{{{\rm{t;}} - {\rm{;}}1}} + 1\)

\( \Rightarrow \frac{{{\rm{dt}}}}{{{\rm{dy}}}} = {\rm{;}}\frac{{{\rm{t;}} + {\rm{;}}1{\rm{;}} + {\rm{;t;}} - 1}}{{{\rm{t}} - 1}}\)

\( \Rightarrow \frac{{{\rm{dt}}}}{{{\rm{dy}}}} = {\rm{;}}\frac{{2{\rm{t}}}}{{{\rm{t}} - 1}}\)

\( \Rightarrow \smallint \frac{{t - 1}}{t}{\rm{dt;}} = \smallint 2{\rm{;dy}}\)

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

⇒ t – log t = 2y + c₁

⇒ x + y – log (x + y) = 2y + c₁

⇒ 0 = y – x + log (x + y) + c₁

⇒ y – x + log (x + y) = -c₁

⇒ y – x + log (x + y) = c

∴ Option 3 is correct.

Calculation:

We have to find the value of \(\mathop {\lim }\limits_{x \to \frac{\pi }{6}} ;\frac{{2{{\sin }^2}x; + {\rm{;}}\sin x; - {\rm{;}}1}}{{2{{\sin }^2}x; - {\rm{;}}3\sin x; + {\rm{;}}1}}\)

\( \Rightarrow \mathop {\lim }\limits_{x \to \frac{\pi }{6}} ;\frac{{2{{\sin }^2}x; + {\rm{;}}\sin x; - {\rm{;}}1}}{{2{{\sin }^2}x; - {\rm{;}}3\sin x; + {\rm{;}}1}} = ;\mathop {{\rm{lim}}}\limits_{x \to \frac{\pi }{6}} ;\frac{{2{{\sin }^2}x; + {\rm{;}}2\sin x - \sin x - {\rm{;}}1}}{{2{{\sin }^2}x; - {\rm{;}}2\sin x - \sin x + {\rm{;}}1}}\)

\( = ;\mathop {{\rm{lim}}}\limits_{x \to \frac{\pi }{6}} ;\frac{{2\sin x(\sin x + 1) - 1;(\sin x + 1)}}{{2\sin x(\sin x - 1) - 1;(\sin x - 1)}}\)

\( = ;\mathop {{\rm{lim}}}\limits_{x \to \frac{\pi }{6}} ;\frac{{\left( {2\sin x - 1} \right);(\sin x + 1)}}{{\left( {2\sin x - 1} \right);(\sin x - 1)}}\)

\( = ;\mathop {{\rm{lim}}}\limits_{x \to \frac{\pi }{6}} ;\frac{{(\sin x + 1)}}{{(\sin x - 1)}} = ;\frac{{\left( {\frac{1}{2} + 1} \right)}}{{\left( {\frac{1}{2} - 1} \right)}} = ;\frac{{\left( {\frac{3}{2}} \right)}}{{\left( {\frac{{ - 1}}{2}} \right)}} = ; - 3\)

∴ Option 4 is correct.

Concept:

• If A is the arithmetic mean of numbers a and b and is given by ⇔ \(A = ;\frac{{a; + ;b}}{2}\)
• If G is the geometric mean of the numbers a and b and is given by  ⇔ \({\rm{G}} = {\rm{;}}\sqrt {{\rm{ab}}} \)
• If H is the Harmonic mean of numbers a and b and is given by ⇔ \({\rm{H}} = {\rm{;}}\frac{{2{\rm{ab}}}}{{{\rm{a}} + {\rm{b}}}}\)
• Relation between AM, GM and HM

1. G² = AH
2. \({\rm{AM;}} \ge {\rm{;GM;}} \ge {\rm{;HM}}\)

• sin (A + B) = sin A cos B + cos A sin B
• cos (A – B) = cos A cos B + sin A sin B

 

Calculation:

Given:

sin β is the harmonic mean of sin α and cos α

\( \Rightarrow ;\sin \beta = ;\frac{{2\sin \alpha \cos \alpha }}{{\sin \alpha + ;\cos \alpha }}\)

\(\therefore \sin \alpha + ;\cos \alpha = ;\frac{{2\sin \alpha \cos \alpha }}{{\sin \beta }}\)       ---(1)

And sin θ is the arithmetic mean of sin α and cos α

\( \Rightarrow ;\sin \theta = ;\frac{{\sin \alpha + ;\cos \alpha }}{2}\)         ---(2)

Now, Statement 1,

\({\rm{LHS}} = \sqrt 2 \sin \left( {{\rm{\alpha }} + \frac{{\rm{\pi }}}{4}} \right)\sin {\rm{\beta }} = {\rm{;}}\sqrt 2 {\rm{;}}\left( {\sin {\rm{\alpha }}\cos \frac{{\rm{\pi }}}{4} + {\rm{;}}\cos {\rm{\alpha }}\sin \frac{{\rm{\pi }}}{4}} \right){\rm{;}}\sin {\rm{\beta }}\)

\( = {\rm{;}}\sqrt 2 ;\left( {\frac{{\sin \alpha + ;\cos \alpha }}{{\sqrt 2 }}} \right);\sin \beta \)         (∵ sin (π/4) = cos (π/4) = 1/√2)  

\( = {\rm{;}}\left( {\sin \alpha + ;\cos \alpha } \right);\sin \beta \)

Put the value of (sin α + cos α) from equation 1^st

\( = ;\left( {\frac{{2\sin \alpha \cos \alpha }}{{\sin \beta }}} \right)\sin \beta \)

= 2 sin α cos α = sin 2α

∴ Statement 1 is correct.

Now, Statement 2,

LHS = √2 sin θ

 Put the value of sin θ from equation 2^nd

\({\rm{LHS;}} = {\rm{;}}\sqrt 2 \left( {\frac{{\sin \alpha + ;\cos \alpha }}{2}} \right) = ;\frac{{\sin \alpha + ;\cos \alpha }}{{\sqrt 2 }}\)

\( = ;\sin \alpha \left( {\frac{1}{{\sqrt 2 }}} \right) + ;\cos \alpha ;\left( {\frac{1}{{\sqrt 2 }}} \right) = ;\sin {\rm{\alpha }}\sin \frac{{\rm{\pi }}}{4} + {\rm{;}}\cos {\rm{\alpha }}\cos \frac{{\rm{\pi }}}{4};\)

= cos (α – π/4)

∴ Statement 2 is correct.

Concept:

• A, B and C are mutually exclusive and exhaustive events.
• P (A ∪ B ∪ C) = P (A) + P (B) + P (C)
• If A ∪ B ∪ C = S = sample space ⇔ P (A ∪ B ∪ C) = 1

Calculation:

Given:

P (B) = 1.5 P (A) and P (C) = 0.5 P (B) and A, B and C be three mutually exclusive and exhaustive events.

⇒ P (A ∪ B ∪ C) = 1

⇒ P (A) + P (B) + P (C) = 1

⇒ P (A) + 1.5 P (A) + 0.5 P (B) = 1

⇒ 2.5 P (A) + 0.5 × 1.5 P (A) = 1

⇒ 2.5 P (A) + 0.75 P (A) = 1

⇒ 3.25 P (A) = 1

⇒ P (A) = 1/3.25 = 4/13

∴ Option 2 is correct.