Quantitative Aptitude PYQs

Concept:

• Probability of exactly r success in n trials \( = {\rm{;p}}\left( {{\rm{x}} = {\rm{r}}} \right) = {\rm{;}}{{\rm{;}}^{\rm{n}}}{{\rm{C}}_{\rm{r}}}{\rm{;}}{{\rm{p}}^{\rm{r}}}{{\rm{q}}^{{\rm{n}} - {\rm{r}}}}\)

Where n = number of trials; p = probability of success; q = (1-p) = probability of failure

Calculation:

We have to find the probability of getting at least 6 heads

Given, n = 8

Here Success is getting a head on single trial.

We already know that there are exactly 2possible outcomes for each and every toss of coins

Probability of success = p = 1/2

Probability of failure = q = 1 – p = 1 – 1/2 = 1/2

The probability of getting at least 6 heads = P (x = 6) + P (x = 7) + P (x = 8)

\( = {{\rm{;}}^8}{{\rm{C}}_6}{\rm{;}}{\left( {\frac{1}{2}} \right)^6}{\left( {\frac{1}{2}} \right)^{8 - 6}} + {{\rm{;}}^8}{{\rm{C}}_7}{\rm{;}}{\left( {\frac{1}{2}} \right)^7}{\left( {\frac{1}{2}} \right)^{8 - 7}} + {{\rm{;}}^8}{{\rm{C}}_8}{\rm{;}}{\left( {\frac{1}{2}} \right)^8}{\left( {\frac{1}{2}} \right)^{8 - 8}}\)

\( = ;{\left( {\frac{1}{2}} \right)^8};\left( {{{\rm{;}}^8}{{\rm{C}}_6}{\rm{;}} + {{\rm{;}}^8}{{\rm{C}}_7}{\rm{;}} + {{\rm{;}}^8}{{\rm{C}}_8}{\rm{;}}} \right) = ;{\left( {\frac{1}{2}} \right)^8}\left( {28 + 8 + 1} \right) = ;\frac{{37}}{{256}}\)

Concept:

1) Combination: Selecting r objects from given n objects.

• The number of selections of r objects from the given n objects is denoted by \({;^n}{C_r};\)
• \({;^n}{C_r}; = ;\frac{{n!}}{{r!\left( {n; - ;r} \right)!}}\)

2). Probability of an event happening =  \(\frac{{{\rm{Number;of;ways;it;can;happen}}}}{{{\rm{Total;number;of;outcomes}}}}\) 

Note: Use combinations if a problem calls for the number of ways of selecting objects.

Calculation:

Given:

The combination of 1 girl and 2 boys can be selected in:

Case 1: P (girl from I group) and P (boy from II group) and P (boy from III group)

= (3/4) × (2/4) × (3/4) = 18/64

Case 2: P (boy from I group) and P (girl from II group) and P (boy from III group)

= (1/4) × (2/4) × (3/4) = 6/64

Case 3: P (boy from I group) and P (boy from II group) and P (girl from III group) 

= (1/4) × (2/4) × (1/4) = 2/64

Hence, required probability = (18/64) + (6/64) + (2/64) = 26/64 = 13/32

Concept:

• Coefficient of correlation \({\rm{;}} = {\rm{;r;}} = {\rm{;}}\sqrt {{{\rm{b}}_{{\rm{yx}}}} \times {{\rm{b}}_{{\rm{xy}}}}} \)

Where, b_yx and b_xy are regression coefficients or the slopes of the equation y on x and x on y are denoted as b_yx and b_xy

Calculation:

Given:

Regression coefficient of Y on X = b_yx = -6

And correlation coefficient between X and Y = r = -1/2

We have to find the value of regression coefficient of X on Y = b_xy

We know that Coefficient of correlation

\({\rm{;}} = {\rm{;r;}} = {\rm{;}}\sqrt {{{\rm{b}}_{{\rm{yx}}}} \times {{\rm{b}}_{{\rm{xy}}}}} \) 

Squaring both sides, we get

⇒ r² = b_yx × b_xy

\( \Rightarrow {\rm{;}}{\left( {\frac{{ - 1}}{2}} \right)^2} = {\rm{;}} - 6{\rm{;}} \times {\rm{;}}{{\rm{b}}_{{\rm{xy}}}}\)

∴ b_xy = -1/24

Concept:

Correlation Coefficient: The correlation coefficient is a statistical measure of the strength of the relationship between the relative movements of two variables.

• The values range between -1.0 and 1.0
• A correlation of -1.0 shows a perfect negative correlation, while a correlation of 1.0 shows a perfect positive correlation. A correlation of 0.0 shows no linear relationship between the movements of the two variables.
• Coefficient of correlation \({\rm{;}} = {\rm{;r;}} = {\rm{;}}\sqrt {{{\rm{b}}_{{\rm{yx}}}} \times {{\rm{b}}_{{\rm{xy}}}}} \) 

Where, b_yx and b_xy are regression coefficients or the slopes of the equation y on x and x on y are denoted as b_yx and b_xy

 

Calculation:

Given:

Straight line with non-zero slope,

⇒ b_yx ≠ 0 and b_xy ≠ 0

By definition.

Coefficient of correlation  \( = {\rm{;r;}} = {\rm{;}}\sqrt {{{\rm{b}}_{{\rm{yx}}}} \times {{\rm{b}}_{{\rm{xy}}}}} = ;\sqrt 1 \) 

∴ r = ± 1

Hence option 4 is correct.

Concept:

• Probability of an event happening =  \(\frac{{{\rm{Number;of;ways;it;can;happen}}}}{{{\rm{Total;number;of;outcomes}}}}\)

 

Calculation:

Number of elements in the set = {0, 1, 2.... 10}

Number of ordered pairs(x, y), from the elements of set, such that |x - y| > 5

Possible combinations of (x, y) are:

(0, 10), (0, 9), (0, 8), (0, 7), (0, 6)

(1, 10), (1, 9), (1, 8), (1, 7)

(2, 10), (2, 9), (2, 8)

(3, 10), (3, 9)

(4, 10)

And their reverse.

Thus, the number of combinations are = (5 + 4 + 3 + 2 + 1) × 2 = 30

Total no. of combinations (with replacement, hence numbers may be repeated) = 11 × 11 = 121

Hence, probability = 30/121​

Alternate solution:

Given:

Number of elements in the set = {0, 1, 2.... 10}

|x - y| > 5

⇒ -5 < x – y < 5

Hence, required probability = \(\frac{{2{\rm{;}} \times {\rm{Area;of;triangle}}}}{{{\rm{Area;of;square;}}}} = {\rm{;}}\frac{{2{\rm{;}} \times {\rm{;}}\frac{1}{2}{\rm{;}} \times {\rm{;}}6{\rm{;}} \times 5}}{{11{\rm{;}} \times 11}} = {\rm{;}}\frac{{30}}{{121}}\)

Concept:

Modulus:

If f(x) = |x – a| then f(x) is defined as,

\({\rm{f}}\left( {\rm{x}} \right) = \left{ {\begin{array}{*{20}{c}} {x - a,;;x \ge a}\ { - \left( {x - a} \right),;;x \le a} \end{array}} \right.\)

Continuity: A function is continuous if left hand limit is equal to right hand limit and both are equal to the value of function at that point.

\(\mathop {\lim }\limits_{{\rm{x}} \to {3^ + }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {3^ - }} {\rm{f}}\left( {\rm{x}} \right) = {\rm{f}}\left( {\rm{x}} \right)\)

Condition for Differentiability: A function is differentiable at x = a, if f’(a) exists.

\({\rm{f'}}\left( {\rm{a}} \right) = \mathop {\lim }\limits_{{\rm{h}} \to 0} \frac{{{\rm{f}}\left( {{\rm{a}} + {\rm{h}}} \right) - {\rm{f}}\left( {\rm{a}} \right)}}{{\rm{h}}}\)

Calculation:

Given: f(x) = |x – 3|

\({\rm{f}}\left( {\rm{x}} \right) = \left{ {\begin{array}{*{20}{c}} {x - 3,;;x \ge 3}\ {3 - x,;;x \le 3} \end{array}} \right.\)

\(\mathop {\lim }\limits_{{\rm{x}} \to {3^ + }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {3^ - }} {\rm{f}}\left( {\rm{x}} \right) = {\rm{f}}\left( {\rm{x}} \right)\)

\(\Rightarrow \mathop {\lim }\limits_{{\rm{x}} \to {3^ + }} \left( {{\rm{x}} - 3} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {3^ - }} \left( {3 - {\rm{x}}} \right) = 0\)

⇒ 0 = 0 = 0

Hence, f(x) is continuous at x = 3.

f(x) is continuous at x = 3. So, second statement is true and first statement is wrong.

∴ Option 1^st is wrong.

For third statement,

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

\(\Rightarrow {\rm{f'}}\left( 0 \right) = \mathop {\lim }\limits_{{\rm{h}} \to 0} \frac{{3 - \left( {0 + {\rm{h}}} \right) - 3}}{{\rm{h}}}\)

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

⇒ f’(0) = -1

So, third statement is true.

For fourth statement,

\({\rm{f'}}\left( { - 3} \right) = \mathop {\lim }\limits_{{\rm{h}} \to 0} \frac{{{\rm{f}}\left( { - 3 + {\rm{h}}} \right) - {\rm{f}}\left( { - 3} \right)}}{{\rm{h}}}\)

\(\Rightarrow {\rm{f'}}\left( { - 3} \right) = \mathop {\lim }\limits_{{\rm{h}} \to 0} \frac{{3 - \left( { - 3 + {\rm{h}}} \right) - 6}}{{\rm{h}}}\)

\(\Rightarrow {\rm{f'}}\left( { - 3} \right) = \mathop {\lim }\limits_{{\rm{h}} \to 0} \frac{{ - {\rm{h}}}}{{\rm{h}}}\)

⇒ f’(-3) = -1

So, fourth statement is true.

Concept:

Formulas used:

• Chain rule:

\(\frac{{{\rm{dy}}}}{{{\rm{dx}}}}\left( {{\rm{f}}\left( {\rm{x}} \right){\rm{g}}\left( {\rm{x}} \right)} \right) = {\rm{f'}}\left( {\rm{x}} \right){\rm{g}}\left( {\rm{x}} \right) + {\rm{g'}}\left( {\rm{x}} \right){\rm{f}}\left( {\rm{x}} \right)\)

• \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}}{{\rm{e}}^{{\rm{ax}}}} = {\rm{a}}{{\rm{e}}^{{\rm{ax}}}}\)

 

Calculation:

Given: u = e^ax sin bx and v = e^ax cos bx

To find: \({\rm{u}}\frac{{{\rm{du}}}}{{{\rm{dx}}}} + {\rm{v}}\frac{{{\rm{dv}}}}{{{\rm{dx}}}}\)

u = e^ax sinbx

Differentiating both sides w.r.t x

\(\Rightarrow \frac{{{\rm{du}}}}{{{\rm{dx}}}} = {\rm{a}}{{\rm{e}}^{{\rm{ax}}}}\sin {\rm{bx}} + {{\rm{e}}^{{\rm{ax}}}}\cos {\rm{bx}}\)

Multiplying u to both sides

\(\Rightarrow {\rm{u}}\frac{{{\rm{du}}}}{{{\rm{dx}}}} = {{\rm{e}}^{{\rm{ax}}}}\sin {\rm{bx}}\left{ {{\rm{a}}{{\rm{e}}^{{\rm{ax}}}}\sin {\rm{bx}} + {{\rm{e}}^{{\rm{ax}}}}\cos {\rm{bx}}} \right}\)

\(\Rightarrow {\rm{u}}\frac{{{\rm{du}}}}{{{\rm{dx}}}} = {{\rm{e}}^{2{\rm{ax}}}}\left{ {{\rm{a}}{{\sin }^2}{\rm{bx}} + \sin {\rm{bx}}\cos {\rm{bx}}} \right}\)

Now, v = e^ax cos bx

Differentiating both sides w.r.t x

\(\Rightarrow \frac{{{\rm{dv}}}}{{{\rm{dx}}}} = {\rm{a}}{{\rm{e}}^{{\rm{ax}}}}\cos {\rm{bx}} - {{\rm{e}}^{{\rm{ax}}}}\sin {\rm{bx}}\)

Multiplying v to both sides

\(\Rightarrow {\rm{v}}\frac{{{\rm{dv}}}}{{{\rm{dx}}}} = {{\rm{e}}^{{\rm{ax}}}}\cos {\rm{bx}}\left{ {{\rm{a}}{{\rm{e}}^{{\rm{ax}}}}\cos {\rm{bx}} - {{\rm{e}}^{{\rm{ax}}}}\sin {\rm{bx}}} \right}\)

\(\Rightarrow {\rm{v}}\frac{{{\rm{dv}}}}{{{\rm{dx}}}} = {{\rm{e}}^{2{\rm{ax}}}}\left{ {{\rm{a}}{{\cos }^2}{\rm{bx}} - \sin {\rm{bx}}\cos {\rm{bx}}} \right}\)

\({\rm{u}}\frac{{{\rm{du}}}}{{{\rm{dx}}}} + {\rm{v}}\frac{{{\rm{dv}}}}{{{\rm{dx}}}} = {{\rm{e}}^{2{\rm{ax}}}}\left{ {{\rm{a}}{{\sin }^2}{\rm{bx}} + \sin {\rm{bx}}\cos {\rm{bx}}} \right} + {{\rm{e}}^{2{\rm{ax}}}}\left{ {{\rm{a}}{{\cos }^2}{\rm{bx}} - \sin {\rm{bx}}\cos {\rm{bx}}} \right}\)

= e^2ax {a sin² bx + a cos² bx}

= ae^2ax

Concept:

Formulas used:

Perimeter of sector = (θ/360) × 2πr

Area of sector (A) = (θ/360) × πr²

Where, r is radius of sector, θ is the angle subtended and l is length of arc of sector.

Maxima: Let f(x) be a function defined on a interval x. For finding the maximum:

1. Find the stationary points \(\frac{{{\rm{df}}\left( {\rm{x}} \right)}}{{{\rm{dx}}}} = 0\).
2. \({\left( {\frac{{{{\rm{d}}^2}{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{d}}{{\rm{x}}^2}}}} \right)_{{\rm{r;}} = {\rm{;a}}}} < 0\), for the stationary point (x = a) to be point of maxima.

 

Calculation:

Let r be radius of sector and l be length of length of arc of the sector.

Total Perimeter = 2r + l = 40      ----(1)

Perimeter of sector = (θ/360) × 2πr      ----(2)

Area of sector (A) = (θ/360) × πr²

\(\Rightarrow {\rm{A}} = \frac{{\rm{\theta }}}{{360}} \times 2{\rm{\pi r}} \times \frac{{\rm{r}}}{2}\)

⇒ A = l × (r/2)      [Using (2)]

⇒ A = (40 – 2r) × (r/2)      [Using (1)]

For calculating area to be maximum:

1. Find the stationary points \(\frac{{{\rm{dA}}}}{{{\rm{dr}}}} = 0\), for r = a.
2. \({\left( {\frac{{{{\rm{d}}^2}{\rm{A}}}}{{{\rm{d}}{{\rm{r}}^2}}}} \right)_{{\rm{r;}} = {\rm{;a}}}} < 0\), for the stationary point (r = a) to be point of maxima.

So, \(\frac{{{\rm{dA}}}}{{{\rm{dr}}}} = \frac{{\rm{d}}}{{{\rm{dr}}}}{ \left( {40 - 2{\rm{r}}} \right)\left( {\frac{{\rm{r}}}{2}} \right)\)

\(\Rightarrow 0 = \frac{{40 - 2{\rm{r}}}}{2} + \frac{{\rm{r}}}{2} \times \left( { - 2} \right)\)

⇒ 0 = 40 – 2r – 2r

⇒ r = 10

Now, checking for r = 10 is point of maxima or not.

\({\left( {\frac{{{{\rm{d}}^2}{\rm{A}}}}{{{\rm{d}}{{\rm{r}}^2}}}} \right)_{{\rm{r;}} = {\rm{;a}}}} = \frac{{\rm{d}}}{{{\rm{dr}}}}\left{ {\frac{{40 - 2{\rm{r}}}}{2} + \frac{{\rm{r}}}{2} \times \left( { - 2} \right)} \right}\)

\(\Rightarrow {\left( {\frac{{{{\rm{d}}^2}{\rm{A}}}}{{{\rm{d}}{{\rm{r}}^2}}}} \right)_{{\rm{r;}} = {\rm{;}}10}} = - 1 - 1\)

\(\Rightarrow {\left( {\frac{{{{\rm{d}}^2}{\rm{A}}}}{{{\rm{d}}{{\rm{r}}^2}}}} \right)_{{\rm{r;}} = {\rm{;}}10}} = - 2\)

∴ \({\left( {\frac{{{{\rm{d}}^2}{\rm{A}}}}{{{\rm{d}}{{\rm{r}}^2}}}} \right)_{{\rm{r;}} = {\rm{;}}10}} < 0\)

Hence, at r = 10, area will be maximum.

Concept:

Formulas used:

• sin 2A = 2 × sin A × cos A
• 1 = sin² A + cos² A

Calculation:

We know that sin 2x = 2sin x cos x and sin² x + cos² x = 1

So, \(\frac{{{\rm{d}}\sqrt {1 - \sin 2{\rm{x}}} }}{{{\rm{dx}}}} = \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\sqrt {{{\sin }^2}{\rm{x}} + {{\cos }^2}{\rm{x}} - 2\sin {\rm{x}}\cos {\rm{x}}} } \right)\) 

Case 1:

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

Now, the value that comes out of root should not be negative. And for \(\frac{{\rm{\pi }}}{4} < {\rm{x}} < \frac{{\rm{\pi }}}{2}\), (sin x – cos x) > 0.

So, \(\sqrt {{{\left( {\sin {\rm{x}} - \cos {\rm{x}}} \right)}^2}} =( \sin {\rm{x}} - \cos {\rm{x;}})\)

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

= (cos x + sin x) 

Case 2: 

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

Now, the value that comes out of root should not be negative. And for \(\frac{{\rm{\pi }}}{4} < {\rm{x}} < \frac{{\rm{\pi }}}{2}\), (sin x – cos x) > 0.

So, \(\sqrt {{{\left( {\cos {\rm{x}} - \sin {\rm{x}}} \right)}^2}} =-( \cos {\rm{x}} - \sin {\rm{x;}})\)

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

= (sinx + cosx)

Option 1 is correct.

Concept:

Integral formula used:

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

 

Calculation:

Let I = \(\smallint \frac{{{\rm{dx}}}}{{{{\rm{a}}^2}{{\sin }^2}{\rm{x}} + {{\rm{b}}^2}{{\cos }^2}{\rm{x}}}}\)

Dividing with b² cos² x to both numerator and denominator.

\(= \smallint \frac{{\frac{1}{{{{\rm{b}}^2}{{\cos }^2}{\rm{x}}}}}}{{\frac{{{{\rm{a}}^2}{{\sin }^2}{\rm{x}}}}{{{{\rm{b}}^2}{{\cos }^2}{\rm{x}}}} + 1}}{\rm{dx}}\)

\(= \frac{1}{{{{\rm{b}}^2}}}\smallint \frac{{{{\sec }^2}{\rm{x}}}}{{\frac{{{{\rm{a}}^2}}}{{{{\rm{b}}^2}}}{{\tan }^2}{\rm{x}} + 1}}{\rm{dx}}\)

Take tan x = t

Differentiating both sides

⇒ sec² x dx = dt

Changing integral in form of t.

\(\frac{1}{{{{\rm{b}}^2}}}\smallint \frac{{{{\sec }^2}{\rm{x}}}}{{\frac{{{{\rm{a}}^2}}}{{{{\rm{b}}^2}}}{{\tan }^2}{\rm{x}} + 1}}{\rm{dx}} = \frac{1}{{{{\rm{b}}^2}}}\smallint \frac{{{\rm{dt}}}}{{\frac{{{{\rm{a}}^2}}}{{{{\rm{b}}^2}}} \times {{\rm{t}}^2} + 1}}\)

Multiplying \(\frac{{{{\rm{b}}^2}}}{{{{\rm{a}}^2}}}\) to both numerator and denominator.

\(= \frac{1}{{{{\rm{a}}^2}}}\smallint \frac{{{\rm{dt}}}}{{{{\rm{t}}^2} + \frac{{{{\rm{b}}^2}}}{{{{\rm{a}}^2}}}}}\)

Now, we know that \(\smallint \frac{{{\rm{dx}}}}{{{{\rm{a}}^2} + {{\rm{x}}^2}}} = \frac{1}{{\rm{a}}}{\tan ^{ - 1}}\left( {\frac{{\rm{x}}}{{\rm{a}}}} \right) + {\rm{c}}\)

So, \(\frac{1}{{{{\rm{a}}^2}}}\smallint \frac{{{\rm{dt}}}}{{{{\rm{t}}^2} + \frac{{{{\rm{b}}^2}}}{{{{\rm{a}}^2}}}}} = \frac{1}{{{{\rm{a}}^2}}} \times \frac{1}{{\frac{{\rm{b}}}{{\rm{a}}}}} \times {\tan ^{ - 1}}\left( {\frac{{\rm{t}}}{{\frac{{\rm{b}}}{{\rm{a}}}}}} \right) + {\rm{c}}\)

\(= \frac{1}{{{\rm{ab}}}}{\tan ^{ - 1}}\left( {\frac{{{\rm{at}}}}{{\rm{b}}}} \right) + {\rm{c}}\)

Put t = tan x

\(= \frac{1}{{{\rm{ab}}}}{\tan ^{ - 1}}\left( {\frac{{{\rm{a}}\tan {\rm{x}}}}{{\rm{b}}}} \right) + {\rm{c}}\)

Concept:

Differentiability: Let f(x) be a function defined on x.

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

Formulas used:

• Product rule of limit:

\(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} {\rm{f}}\left( {\rm{x}} \right){\rm{g}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} {\rm{f}}\left( {\rm{x}} \right) \times \mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} {\rm{g}}\left( {\rm{x}} \right)\)

Calculation:

Given:

f(x + y) = f(x)f(y)      ----(1)

f(x) = 1 + xg(x)φ(x)      ----(2)

\(\mathop {\lim }\limits_{{\rm{x}} \to 0} {\rm{g}}\left( {\rm{x}} \right) = {\rm{a}}\) and \(\mathop {\lim }\limits_{{\rm{x}} \to 0} \phi \left( {\rm{x}} \right) = {\rm{b}}\)      ----(3)

To find: f’(x)

We know that \({\rm{f'}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{h}} \to 0} \frac{{{\rm{f}}\left( {{\rm{x}} + {\rm{h}}} \right) - {\rm{f}}\left( {\rm{x}} \right)}}{{\rm{h}}}\)

\(= \mathop {\lim }\limits_{{\rm{h}} \to 0} \frac{{{\rm{f}}\left( {\rm{x}} \right){\rm{f}}\left( {\rm{h}} \right) - {\rm{f}}\left( {\rm{x}} \right)}}{{\rm{h}}}\)      [Using (1)]

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

\(= \mathop {\lim }\limits_{{\rm{h}} \to 0} \frac{{{\rm{f}}\left( {\rm{x}} \right)\left{ {1 + {\rm{hg}}\left( {\rm{h}} \right)\phi \left( {\rm{h}} \right) - 1} \right}}}{{\rm{h}}}\)

\(= \mathop {\lim }\limits_{{\rm{h}} \to 0} \frac{{{\rm{f}}\left( {\rm{x}} \right)\left{ {{\rm{hg}}\left( {\rm{h}} \right)\phi \left( {\rm{h}} \right)} \right}}}{{\rm{h}}}\)

\(= \mathop {\lim }\limits_{{\rm{h}} \to 0} {\rm{f}}\left( {\rm{x}} \right){\rm{g}}\left( {\rm{h}} \right)\phi \left( {\rm{h}} \right)\)

\(= {\rm{f}}\left( {\rm{x}} \right)\mathop {\lim }\limits_{{\rm{h}} \to 0} {\rm{g}}\left( {\rm{h}} \right)\phi \left( {\rm{h}} \right)\)

\(= {\rm{f}}\left( {\rm{x}} \right)\mathop {\lim }\limits_{{\rm{h}} \to 0} {\rm{g}}\left( {\rm{h}} \right) \times \mathop {\lim }\limits_{{\rm{h}} \to 0} \phi \left( {\rm{h}} \right)\)

= f(x) × a × b      [Using (3)]

= abf(x)

Concept:

Conditional probability: It is the measure of probability of an event A is occurring given that another event B is already occurred, then:

\({\rm{P}}\left( {\rm{B}} \right) = \frac{{{\rm{P}}\left( {{\rm{A}} \cap {\rm{B}}} \right)}}{{{\rm{P}}\left( {\rm{A}} \right)}}\)

Calculation:

Let

S: Event of rolling 2 dice.

A: Event of at least one die showing 5.

B: Event that the sum of the numbers appeared on both dice to be 10 or more.

So, S = n (6ⁿ) = n (6²) = n (36)

Where n is the number of dice rolled.

A = {(1,5), (2,5), (3,5), (4,5), (5,5), (6,5), (5,1), (5,2), (5,3), (5,4), (5,6)}

B = {(5,5), (4,6), (5,6), (6,4), (6,5), (6,6)}

(A ∩ B) = {(5,5), (5,6), (6,5)}

Hence, conditional probability of occurrence of B when A is already occurred = P(B|A)

\(= \frac{{{\rm{P}}\left( {{\rm{A}} \cap {\rm{B}}} \right)}}{{{\rm{P}}\left( {\rm{A}} \right)}}\)

\(= \frac{{3/36}}{{11/36}}\)

= 3/11

Concept:

Formulas used:

Correlation co-efficient = r

Co-efficient of determination = r²

Calculation

Number of observation, n = 30

Correlation co-efficient, r = 0.8

So, Co-efficient of determination = r²

= (0.8)²

= 0.64

= 64%

Percentage of variation not explained = 100 – 64

= 36 %

Concept:

Mean: It is sum of all observations upon number of observations.

If x₁, x₂, x₃, …., x_n are the observations, then mean is given as:

\({\rm{\bar X}} = \frac{{{{\rm{x}}_1} + {{\rm{x}}_2} + \ldots + {{\rm{x}}_{\rm{n}}}}}{{\rm{n}}}\)

Calculation:

Let number total men = m and m₁, m₂, … m_m be the ages of men.

Let number of total women = w and w₁, w₂, … w_w be the ages of women.

Given:

\(\frac{{{{\rm{m}}_1}{\rm{;}} + {\rm{;}}{{\rm{m}}_2}{\rm{;}} + {\rm{;}} \cdots {\rm{;}} + {\rm{;}}{{\rm{m}}_{\rm{m}}}}}{{\rm{m}}} = 26\)      ----(1)

\(\frac{{{{\rm{w}}_1}{\rm{;}} + {\rm{;}}{{\rm{w}}_2}{\rm{;}} + {\rm{;}} \cdots {\rm{;}} + {\rm{;}}{{\rm{w}}_{\rm{w}}}}}{{\rm{w}}} = 21\)      ----(2)

\(\frac{{{{\rm{m}}_1}{\rm{;}} + {\rm{;}}{{\rm{m}}_2} + {\rm{;}} \cdots {\rm{;}} + {\rm{;}}{{\rm{m}}_{\rm{m}}}{\rm{;}} + {\rm{;}}{{\rm{w}}_1} + {\rm{;}}{{\rm{w}}_2}{\rm{;}} + {\rm{;}} \cdots {\rm{;}} + {\rm{;}}{{\rm{w}}_{\rm{w}}}}}{{{\rm{m}} + {\rm{w}}}} = 25\)  

⇒ m₁ + m₂ + … + m_m + w₁ + w₂ + … + w_w = 25(m + w)

⇒ 26m + 21w = 25m + 25w

⇒ m = 4w

Percentage of men \(= \frac{{\rm{m}}}{{{\rm{m}} + {\rm{w}}}} \times 100\)  

= (4/5) × 100

= 80

Percentage of women \(= \frac{{\rm{w}}}{{{\rm{m}} + {\rm{w}}}} \times 100\) 

= (1/5) × 100

= 20

Concept:

Bayes’ theorem:

\({\rm{P}}\left( {{\rm{A|B}}} \right) = \frac{{{\rm{P}}({\rm{B}}|{\rm{A}}){\rm{;;P}}\left( {\rm{A}} \right)}}{{{\rm{P}}\left( {\rm{B}} \right)}}\)

Where A and B are events and P(B) 0.

• P(A|B) is a conditional probability: the likelihood of event of A occurring given that B is true.
• P(B|A) is a conditional probability: the likelihood of event of B occurring given that A is true.
• P(A) and P(B) are the probabilities of observing A and B respectively.

Calculation:

Let

X: bolt manufactured by machine X.

Y: bolt manufactured by machine Y.

Z: bolt manufactured by machine Z.

D: defective bolt.

Given:

P(X) = 0.25; P(Y) = 0.35; P(Z) = 0.40

Probability of defective bolt manufactured by X = P(D|X) = 0.02

Probability of defective bolt manufactured by Y = P(D|Y) = 0.04

Probability of defective bolt manufactured by Z = P(D|Z) = 0.05

We need to determine the probability that a bolt chosen, found to be defective and was manufactured by X = P(X|D)

Using Bayes’ theorem,

\({\rm{P}}\left( {{\rm{X|D}}} \right) = \frac{{{\rm{P}}\left( {{\rm{D|X}}} \right){\rm{P}}\left( {\rm{X}} \right)}}{{{\rm{P}}\left( {\rm{D}} \right)}}\)

P(D) = Probability of defective bolt

= P(X)P(D|X) + P(Y)P(D|Y) + P(Z)P(D|Z)

= 0.25 × 0.02 + 0.35 × 0.04 + 0.40 × 0.05

= (50 + 140 + 200) × 10^-4

= 390 × 10^-4

Now, \({\rm{P}}\left( {{\rm{X|D}}} \right) = \frac{{{\rm{P}}\left( {{\rm{D|X}}} \right){\rm{P}}\left( {\rm{X}} \right)}}{{{\rm{P}}\left( {\rm{D}} \right)}}\) 

\(= \frac{{0.02; \times ;0.25}}{{390; \times ;{{10}^{ - 4}}}}\)

\(= \frac{{50; \times ;{{10}^{ - 4}}}}{{390; \times ;{{10}^{ - 4}}}}\)

= 5/39

Concept:

Mean: It is the average of given observation. Let x₁, x₂, …… , x_n be n observations, then

Mean \(= {\rm{\bar X}} = \frac{{\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{n}} {{\rm{x}}_{\rm{i}}}}}{{\rm{n}}}\)

Properties:

• If C is added to each observation, then the new mean is also increased by C.
• If C is multiplied to each observation, then the new mean is also C times.

Standard deviation: Let x₁, x₂, …… , x_n be n observations of X, then:

Standard deviation (σ) = √{Var (X)}

\({{\rm{\sigma }}^2} = {\rm{;Var}}\left( {\rm{X}} \right) = \frac{{\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{n}} {{\left( {{{\rm{x}}_{\rm{i}}} - {\rm{\bar x}}} \right)}^2}}}{{\rm{n}}}\)

So, properties of standard deviation will depend on properties of variance. Properties of variance is given as:

• Var(X + C) = Var(X), where C is a constant.
• Var(CX) = C² × Var(X), where C is a constant.
• Var(aX + b) = a² × Var(X), where a and b are constants.

Calculation:

For first statement,

We know that if C is added to each observation, then the new mean is also increased by C.

Hence, first statement is correct.

For second statement,

We know that, adding fixed number to each observation doesn’t show any effect on variance and also standard deviation.

So, second statement is incorrect.

For third statement,

We know that if C is multiplied to each observation, then the new mean is also C times.

So, third statement is correct.

Concept:

Variance: It is a measure of how data points differ from the mean. Let x₁, x₂, …… , x_n be n observations of X, then:

\({\rm{Var}}\left( {\rm{X}} \right) = \frac{{\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{n}} {{\left( {{{\rm{x}}_{\rm{i}}} - {\rm{\bar x}}} \right)}^2}}}{{\rm{n}}}\)

Where, x_i (i = 1, 2, … , n) are the values of X

\({\rm{\bar x}}\) is the mean of X.

n is the number of observation.

Properties:

• Var(X + C) = Var(X), where C is a constant.
• Var(CX) = C² × Var(X), where C is a constant.
• Var(aX + b) = a² × Var(X), where a and b are constants.
• If X₁, X₂, … , X_n are n independent random variables, then

          Var(X₁ + X₂ + … + X_n) = Var(X₁) + Var(X₂) +……..+ Var(X_n).

Note:

Standard deviation (σ) = √{Var (X)}

Calculation:

Given:

Number of observation = 25.

Variance = 4

Now, we know that variance doesn’t change if a fixed value is added or subtracted from each observation.

Hence, New variance = variance = 4

Concept:

Geometric mean: It is the n^th root of product n numbers. Let x₁, x₂, …… , x_n be n values of X, then:

Geometric mean \({{\rm{\bar X}}_{{\rm{geom}}}} = \sqrt[{\rm{n}}]{{\mathop \prod \limits_{{\rm{i}} = 1}^{\rm{n}} {{\rm{x}}_{\rm{i}}}}}\)

Calculation:

Given: P is geometric mean of X and Q is geometric mean of Y.

So, P = (x₁ × x₂ × ⋯ × x_n)^1/n

Q = (y₁ × y­₂ × ⋯ × y_n)^1/n

Geometric mean of X/Y \(= \frac{{{{\left( {{{\rm{x}}_1}{\rm{;}} \times {\rm{;}}{{\rm{x}}_2}{\rm{;}} \times {\rm{;}} \cdots {\rm{;}} \times {\rm{;}}{{\rm{x}}_{\rm{n}}}} \right)}^{\frac{1}{{\rm{n}}}}}}}{{{{\left( {{{\rm{y}}_1}{\rm{;}} \times {\rm{;}}{{\rm{y}}_2}{\rm{;}} \times {\rm{;}} \cdots {\rm{;}} \times {\rm{;}}{{\rm{y}}_{\rm{n}}}} \right)}^{\frac{1}{{\rm{n}}}}}}}\) 

= P/Q

Concept:

De Morgan’s Law:

• \({\rm{P}}\left( {{\rm{\bar A}} \cap {\rm{\bar B}}} \right) = \) 1 - P(A ∪ B)

Formulas used:

• P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
• P(A̅ ) = 1 – P(A)

 

Calculation:

Probability of occurrence of A and B simultaneously = P(A ∩ B) = p

Probability of occurrence of exactly one of A, B = P(A ∪ B) – P(A ∩ B) = q

⇒ q = P(A) + P(B) – P(A ∩ B) – P(A ∩ B)

⇒ q = P(A) + P(B) – 2p

⇒ P(A) + P(B) = q + 2p

Now, P(A̅) + P(B̅) = 1 – P(A) + 1 – P(B)

= 2 – {P(A) + P(B)}

= 2 – q – 2p

\({\rm{P}}\left( {{\rm{\bar A}} \cap {\rm{\bar B}}} \right) \)      [Using De Morgan’s Law]

= 1 – P(A ∪ B)

= 1 – {P(A) + P(B) – P(A ∩ B)}

= 1 – {q + 2p – p}

= 1 – q – p

Concept:

Suppose we have two sets of data containing n₁ and n₂ observations with means \({{\rm{\bar X}}_1}\) and \({{\rm{\bar X}}_2}\) and variances S₁² and S₂²

• Combined mean = \({\rm{\bar X}} = {\rm{;}}\frac{{{{\rm{n}}_1}{\rm{;}}{{{\rm{\bar X}}}_1} + {\rm{;}}{{\rm{n}}_2}{\rm{;}}{{{\rm{\bar X}}}_2}}}{{{{\rm{n}}_1} + {\rm{;}}{{\rm{n}}_2}}}\)
• Combined variance = \({\rm{S}}_{\rm{C}}^2 = {\rm{;}}\frac{{{{\rm{n}}_1}\left[ {{\rm{S}}_1^2{\rm{;}} + {\rm{;}}{{\left( {{{{\rm{\bar X}}}_1} - {\rm{;\bar X}}} \right)}^2}{\rm{;}}} \right]{\rm{;}} + {\rm{;}}{{\rm{n}}_2}\left[ {{\rm{S}}_2^2{\rm{;}} + {\rm{;}}{{\left( {{{{\rm{\bar X}}}_2} - {\rm{;\bar X}}} \right)}^2}{\rm{;}}} \right]{\rm{;}}}}{{{{\rm{n}}_1} + {\rm{;}}{{\rm{n}}_2}}}\)

 

Calculation:

Given: n₁ = 500 and n₂ = 600

\({{\rm{\bar X}}_1} = 1860{\rm{;and;}}{{\rm{\bar X}}_2} = 1750{\rm{;}}\) 

\({\rm{S}}_1^2 = 81{\rm{;and;S}}_2^2 = 100{\rm{;}}\) 

Combined mean = \({\rm{\bar X}} = {\rm{;}}\frac{{{{\rm{n}}_1}{\rm{;}}{{{\rm{\bar X}}}_1} + {\rm{;}}{{\rm{n}}_2}{\rm{;}}{{{\rm{\bar X}}}_2}}}{{{{\rm{n}}_1} + {\rm{;}}{{\rm{n}}_2}}}\)

\(\Rightarrow {\rm{\bar X}} = {\rm{;}}\frac{{500{\rm{;}} \times 1860 + {\rm{;}}600{\rm{;}} \times 1750}}{{500 + {\rm{;}}600}} = ;\frac{{1980,000}}{{1100}} = 1800\) 

∴ Average of monthly wage = Rs. 1800

Now,

Combined variance = \({\rm{S}}_{\rm{C}}^2 = {\rm{;}}\frac{{{{\rm{n}}_1}\left[ {{\rm{S}}_1^2{\rm{;}} + {\rm{;}}{{\left( {{{{\rm{\bar X}}}_1} - {\rm{;\bar X}}} \right)}^2}{\rm{;}}} \right]{\rm{;}} + {\rm{;}}{{\rm{n}}_2}\left[ {{\rm{S}}_2^2{\rm{;}} + {\rm{;}}{{\left( {{{{\rm{\bar X}}}_2} - {\rm{;\bar X}}} \right)}^2}{\rm{;}}} \right]{\rm{;}}}}{{{{\rm{n}}_1} + {\rm{;}}{{\rm{n}}_2}}}\)

\(\Rightarrow {\rm{S}}_{\rm{C}}^2 = {\rm{;}}\frac{{500{\rm{;}} \times {\rm{;}}\left[ {81{\rm{;}} + {\rm{;}}{{\left( {1860 - 1800} \right)}^2}{\rm{;}}} \right]{\rm{;}} + {\rm{;}}600{\rm{;}} \times {\rm{;}}\left[ {100{\rm{;}} + {\rm{;}}{{\left( {1750{\rm{;}} - {\rm{;}}1800} \right)}^2}{\rm{;}}} \right]{\rm{;}}}}{{500; + {\rm{;}}600}}\)

\( \Rightarrow {\rm{S}}_{\rm{C}}^2 = {\rm{;}}\frac{{500{\rm{;}} \times {\rm{;}}\left[ {81{\rm{;}} + {\rm{;}}3600{\rm{;}}} \right]{\rm{;}} + {\rm{;}}600{\rm{;}} \times {\rm{;}}\left[ {100{\rm{;}} + {\rm{;}}2500{\rm{;}}} \right]{\rm{;}}}}{{500; + {\rm{;}}600}}\)  = 3092

∴ Variance of distribution of wages = 3092

Hence option 4 is correct.