NDA Mathematics 16 Dec 2015 Official Paper (Previous Year Paper)

Concept:

Covert Binary to Decimal:

For binary number with n digits;

D_n-1 …. D₃D₂D₁D₀

The decimal number is equal to the sum of binary digits (D_n) times their power of 2 (2ⁿ)

Decimal = D₀ × 2⁰ + D₁ × 2¹ + D₂ × 2² + ….

Calculation:

We have to find the value of (1000000001)₂ – (0.0101)­₂

(1000000001)₂ = 1 × 2⁰ + 1 × 2⁹ = 1 + 512 = 513

(0.0101)­₂ = 1 × 2^-2 + 1 × 2^-4 = (1/4) + (1/16) = 5/16 = 0.3125

Now,

(1000000001)₂ – (0.0101)­₂ = 513 - 0.3125 = (512.6875)₁₀

∴ Option 2 is correct.

Concept:

Steps for derivatives of functions expressed in the parametric form:

1. First of all, we write the given functions u and v in terms of the parameter x.
2. Using differentiation find out du/dx and dv/dx.
3. Then by using the formula used for solving functions in parametric form i.e.
4. Lastly substituting the values of du/dx and dv/dx and simplify to obtain the result.

Calculation:

Let u = In(x + sin x) and v = (x + cos x)

⇒ u = In (x + sin x)

Differentiating with respect to x, we get

$\Rightarrow \frac{{{\rm{du}}}}{{{\rm{dx}}}} = {\rm{;}}\frac{1}{{\left( {{\rm{x}} + \sin {\rm{x}}} \right)}} \times \left( {1 + \cos {\rm{x}}} \right)$

Now,

v = (x + cos x)

Differentiating with respect to x, we get

$\Rightarrow \frac{{{\rm{dv}}}}{{{\rm{dx}}}} = \left( {1 - \sin {\rm{x}}} \right)$

We have to find the derivative of In(x + sin x) with respect to (x + cos x)

Therefore,  $\frac{{{\rm{dIn}}\left( {{\rm{x;}} + {\rm{;sin;x}}} \right)}}{{{\rm{d}}\left( {{\rm{x;}} + {\rm{;cos;x}}} \right)}} = {\rm{;}}\frac{{{\rm{du}}}}{{{\rm{dv}}}}$

$\Rightarrow \frac{{{\rm{du}}}}{{{\rm{dv}}}} = \frac{{\left( {\frac{{{\rm{du}}}}{{{\rm{dx}}}}} \right)}}{{\left( {\frac{{{\rm{dv}}}}{{{\rm{dx}}}}} \right)}}{\rm{;}}$

$= \frac{{\frac{1}{{\left( {{\rm{x}} + \sin {\rm{x}}} \right)}} \times \left( {1 + \cos {\rm{x}}} \right){\rm{;}}}}{{\left( {1 - \sin {\rm{x}}} \right)}} = {\rm{;}}\frac{{1 + \cos {\rm{x}}}}{{\left( {{\rm{x}} + \sin {\rm{x}}} \right)\left( {1 - \sin {\rm{x}}} \right)}}$

Concept:

• $2;{\tan ^{ - 1}}x = {\tan ^{ - 1}}\left( {\frac{{2x}}{{1; - ;{x^2}}}} \right)$
• ${\tan ^{ - 1}}x - {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\frac{{x - y}}{{1 + xy}}} \right)$
• tan tan^-1 (x) = x

Calculation:

We have to find the value of ${\rm{tan}}\left( {2;{{\tan }^{ - 1}}\frac{1}{5} - \frac{\pi }{4}} \right)$

We know that $2;{\tan ^{ - 1}}x = ;{\tan ^{ - 1}}\left( {\frac{{2x}}{{1; - ;{x^2}}}} \right)$

${\rm{So}},{\rm{;}}2;{\tan ^{ - 1}}\frac{1}{5} = {\rm{;}}{\tan ^{ - 1}}\left[ {\frac{{2{\rm{}} \times {\rm{}}\frac{1}{5}}}{{1{\rm{}} - {\rm{}}{{\left( {\frac{1}{5}} \right)}^2}}}} \right] = {\rm{}}{\tan ^{ - 1}}\left[ {\frac{{\frac{2}{5}}}{{\frac{{24}}{{25}}}}} \right] = {\rm{}}{\tan ^{ - 1}}\frac{5}{{12}}$

We know that ${\tan ^{ - 1}}x - ;{\tan ^{ - 1}}y = ;{\tan ^{ - 1}}\left( {\frac{{x; - ;y}}{{1; + ;xy}}} \right)$

So, $2;{\tan ^{ - 1}}\frac{1}{5} - \frac{\pi }{4} = ;{\tan ^{ - 1}}\frac{5}{{12}} - {\tan ^{ - 1}}1$

$= {\tan ^{ - 1}}\left[ {\frac{{\frac{5}{{12}} - 1}}{{1 + \frac{5}{{12}} \times 1;}}} \right] = ;{\tan ^{ - 1}}\left[ {\frac{{\frac{{5 - 12}}{{12}}}}{{\frac{{12 + 5}}{{12}}}}} \right] = ;{\tan ^{ - 1}}\frac{{ - 7}}{{17}}$

Now,

${\rm{tan}}\left( {2{{\tan }^{ - 1}}\frac{1}{5} - \frac{\pi }{4}} \right) = ;\tan {\tan ^{ - 1}}\frac{{ - 7}}{{17}} = ;\frac{{ - 7}}{{17}}$     ...(∵tan tan^-1 (x) = x)

∴ Option 1 is correct.

Concept:

${\sin ^{ - 1}}x + ;{\sin ^{ - 1}}y = ;{\sin ^{ - 1}}\left[ {x√ {1 - ;{y^2}} ; + ;y√ {1 - ;{x^2}} } \right], x^2 + y^2 \le 1$

${\tan ^{ - 1}}x + ;{\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\frac{{x; + ;y}}{{1; - ;xy}}} \right), x > 1, y > 1 \ and \ xy > 1$

Calculation:

Statement 1:

L.H.S = ${\sin ^{ - 1}}\frac{4}{5} + {\sin ^{ - 1}}\frac{3}{5}$

Here, (4/5)² + (3/5)² = 1 

So, from the 1st concept 

${\rm{LHS}} = {\sin ^{ - 1}}\frac{4}{5} + {\sin ^{ - 1}}\frac{3}{5} = {\sin ^{ - 1}}\left[ {\frac{4}{5}√ {1 - ;{{\left( {\frac{3}{5}} \right)}^2}} + ;\frac{3}{5}√ {1 - ;{{\left( {\frac{4}{5}} \right)}^2}} } \right]$

$= {\rm{}}{\sin ^{ - 1}}\left[ {\frac{4}{5} × \frac{4}{5} + \frac{3}{5} × \frac{3}{5}} \right] = ;{\sin ^{ - 1}}\frac{{16 + 9}}{{25}} = ;{\sin ^{ - 1}}\frac{{25}}{{25}} = ;{\sin ^{ - 1}}1 = \frac{π }{2}$

Hence Statement 1 is correct.

Statement 2:

LHS = tan^-1 √3 + tan ^-1 1

Here, √3 > 0, 1 > 0 and (√3 × 1) > 0

So, from the 2nd concept we have

$⇒;{\tan ^{ - 1}}\left( {\frac{{√ 3 ; + ;1}}{{1; - ;√ 3 ; × 1}}} \right) = ;{\tan ^{ - 1}}\left[ {\frac{{1; + ;√ 3 }}{{1 - ;√ 3 }} × \frac{{1; + ;√ 3 }}{{1; + ;√ 3 }}} \right]$

$⇒ {\tan ^{ - 1}}\left[ {\frac{{{{\left( {1 + √ 3 } \right)}^2}}}{{{{\left( 1 \right)}^2} - {{\left( {√ 3 } \right)}^2}}}} \right] = π + {\tan ^{ - 1}}\left[ {\frac{{1 + 3 + 2√ 3 }}{{1 - 3}}} \right]$

$⇒ {\tan ^{ - 1}}\left[ {\frac{{4 + 2√ 3 }}{{ - 2}}} \right] = π + {\tan ^{ - 1}} - \left( {2 + √ 3 } \right)$

⇒ tan^-1 (2 + √3)

Hence Statement 2 is correct.

Additional Information

${\tan ^{ - 1}}x + ;{\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\frac{{x; + ;y}}{{1; - ;xy}}} \right), x > 1, y > 1 \ and \ xy > 1$

${\tan ^{ - 1}}x + ;{\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\frac{{x; + ;y}}{{1; - ;xy}}} \right), x < 1, y < 1 \ and \ xy < 1$

Concept:

• The triangle inequality states that for any triangle, the sum of the lengths of any two sides must be greater than or equal to the length of the remaining side.

Calculation:

Given: a, b, c are the sides of a triangle ABC

Since, a, b, c are positive numbers.

⇒ a^1/p, b^1/p, c^1/p are positive numbers.     ...(∵ p > 1)

We know that sum of any two sides must be greater than third side.

⇒ a^1/p + b^1/p > c^1/p

⇒ a^1/p + b^1/p - c^1/p > 0

∴ Option 2 is correct.

Concept:

• If f′(x) > 0 then the function is said to be increasing.
• If f′(x) < 0 then the function is said to be decreasing.

Calculation:

Given:

f(x) = -2x3 - 9x2 - 12x + 1

Differentiating, both sides, we get

⇒ f’(x) = -6x2 – 18x – 12

For f(x) to be increasing function,

∴ f’(x) > 0

⇒ -6x2 – 18x – 12 > 0

⇒ -6 (x2 + 3x + 2) > 0

⇒ (x2 + 3x + 2) < 0

⇒ (x + 2) (x + 1) < 0

∴ x ∈ (-2, -1)  

∴ Option 1 is correct.

Mistake Points(x + 2) (x + 1) < 0. This expression is satisfied in two cases:

1. When x + 2 < 0, and x + 1 > 0, i.e. when x < -2, and x > -1. Such an answer is not there in the given options.

2) When x + 2 > 0, and x + 1 < 0, i.e. when x > -2, and x < -1. Option (1) correctly defines this range.

Concept:

• If f′(x) > 0 then the function is said to be increasing.
• If f′(x) < 0 then the function is said to be decreasing.

Calculation:

Given:

f(x) = -2x3 - 9x2 - 12x + 1

Differentiating, both sides, we get

⇒ f’(x) = -6x2 – 18x – 12

For f(x) to be decreasing function,

∴ f’(x) < 0

⇒ -6x2 – 18x – 12 < 0

⇒ -6 (x2 + 3x + 2) < 0

⇒ (x2 + 3x + 2) > 0

⇒ (x + 2) (x + 1) > 0

∴ x ∈ (-∞, -2) ∪ (-1, ∞)   

∴ Option  4 is correct.

Concept:

• The intercept form of the plane is given by ⇔ $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$

Where ‘a’ is the x- intercept, ‘b’ is the y- intercept and ‘c’ is the z- intercept.

Calculation:

Given:

Equation of plane 5x + 2y + z – 13 = 0

⇒ 5x + 2y + z = 13

$\Rightarrow \frac{{5{\rm{x;}} + {\rm{;}}2{\rm{y;}} + {\rm{;z;}}}}{{13}} = 1$

$\Rightarrow \frac{x}{{\frac{{13}}{5}}} + \frac{y}{{\frac{{13}}{2}}} + \frac{z}{{\frac{{13}}{1}}} = 1$

∴ Lengths of intercepts are $\frac{{13}}{5},\frac{{13}}{2},;13$ unit

Concept:

Whenever the situations like finding the possible arrangements for m slots given to n people and there is no such restriction of giving the particular number of slots to any of their person, then a total number of possible arrangements is given by = m × m × m.....(n times) = mⁿ .

Calculation:

Since, In the given question, Each ticket can be given to 20 employees and there are 3 tickets in total.

Thus, each of the 3 tickets has 20 different options.

i.e. There are the following possible ways:

20 × 20 × 20 = 20³ = 8000.

Concept:

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

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 = 7

Here Success is getting a head on single trial.

We already know that there are exactly 2 possible 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 3 heads = 1- [P (x = 0) + P (x = 1) + P (x = 2)]

$= 1 - \left[ {^7{{\rm{C}}_0}{{\left( {\frac{1}{2}} \right)}^0}{{\left( {\frac{1}{2}} \right)}^{7 - 0}}{ + ^7}{{\rm{C}}_1}{{\left( {\frac{1}{2}} \right)}^1}{{\left( {\frac{1}{2}} \right)}^{7 - 1}}{ + ^7}{{\rm{C}}_2}{{\left( {\frac{1}{2}} \right)}^2}{{\left( {\frac{1}{2}} \right)}^{7 - 2}}} \right]$

$= 1 - \left[ {{{\left( {\frac{1}{2}} \right)}^7}\left( {^7{{\rm{C}}_0}{ + ^7}{{\rm{C}}_1}{ + ^7}{{\rm{C}}_2}} \right)} \right] = 1 - {\left( {\frac{1}{2}} \right)^7}\left( {1 + 7 + 21} \right) = 1 - \frac{{29}}{{128}} = \frac{{99}}{{128}}$

∴ For 128 times, probability of getting at least 3 heads = 128 × (99/128) = 99

Concept:

• det (kA) = kⁿ × det (A)    (Where n is the order and k is scalar constant)
• det (AB) = det (A) det (B)
• $\det \left( {{{\rm{A}}^{ - 1}}} \right) = \frac{1}{{{\rm{detA}}}}$

Calculation:

We have to find the value of [det(kA)]^-1 det A

⇒ [det(kA)]^-1 det A

$= \frac{1}{{\det \left( {{\rm{kA}}} \right)}} \times \det {\rm{A}}$------------(∵ x^-1 = 1/x)

$= \frac{1}{{{{\rm{k}}^{\rm{n}}}\det \left( {\rm{A}} \right)}} \times \det {\rm{A}}$-----------(∵ det (kA) = kⁿ × det (A))

= k^-n

∴ Option 1^st is correct.

Concept:

• x² = |x|²

Calculation:

Given:

x² - 3|x| + 2 = 0

⇒ |x|² - 3|x| + 2 = 0                (∵ x² = |x|²)

⇒ |x|² - 2|x| - |x| + 2 = 0               

⇒ |x| (|x| - 2) - 1 (|x| - 2) = 0

⇒ (|x| - 2) (|x| - 1) = 0

⇒ (|x| - 2) = 0 or (|x| - 1) = 0

⇒ |x| = 2 or 1

∴ x = ± 2 or ± 1

Hence there are four real roots of the equation.

Concept:

Let us consider the standard form of a quadratic equation, ax² + bx + c =0

Where a, b and c are constants with a ≠ 0

Let α and β be the two roots of the above quadratic equation. 

The sum of the roots of a quadratic equation are:

${\rm{\alpha }} + {\rm{\beta }} = - \frac{{\rm{b}}}{{\rm{a}}} = - \frac{{{\rm{coefficient;of;x}}}}{{{\rm{coefficient;of;}}{{\rm{x}}^2}}}$

The product of the roots is given by:

${\rm{\alpha \beta }} = \frac{{\rm{c}}}{{\rm{a}}} = \frac{{{\rm{constant;term}}}}{{{\rm{coefficient;of;}}{{\rm{x}}^2}}}$

Calculation:

We have, ax² + bx + c =0

Let α and β be the two roots of the above quadratic equation.

The sum of the roots of a quadratic equation are: ${\rm{\alpha }} + {\rm{\beta }} = - \frac{{\rm{b}}}{{\rm{a}}}$

The product of the roots is given by: ${\rm{\alpha \beta }} = \frac{{\rm{c}}}{{\rm{a}}}$

Given: Sum of the roots of the equation is equal to the sum of their squares.

∴ α + β = α² + β²

⇒ α + β = (α + β) ² - 2αβ

$\Rightarrow - \frac{{\rm{b}}}{{\rm{a}}} = ;{\left( { - \frac{{\rm{b}}}{{\rm{a}}}} \right)^2} - 2; \times \frac{{\rm{c}}}{{\rm{a}}}$

$\Rightarrow - \frac{{\rm{b}}}{{\rm{a}}} = \frac{{{{\rm{b}}^2}}}{{{{\rm{a}}^2}}} - \frac{{2{\rm{c}}}}{{\rm{a}}}$

$\Rightarrow - \frac{{\rm{b}}}{{\rm{a}}} = \frac{{{{\rm{b}}^2} - 2{\rm{ac}}}}{{{{\rm{a}}^2}}}$

⇒ -ab = b² -2ac

∴ b² + ab = 2ac

Concept:

• If \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} {{{\rm{a}}{11}}}&{{{\rm{a}}{12}}}\ {{{\rm{a}}{21}}}&{{{\rm{a}}{22}}} \end{array}} \right]\) then determinant of A is given by: |A| = (a­₁₁ × a₂₂) - (a₁₂ - a₂₁).
• cos (x + y) = cos x cos y - sin x sin y
• cos (x - y) = cos x cos y + sin x sin y

Calculation:

Given: α, β are complementary angles.

⇒ α + β = 90°

Let \({\rm{A}} = {\rm{;}}\left| {\begin{array}{*{20}{c}} {{{\cos }^2}\frac{{\rm{\alpha }}}{2}}&{{{\sin }^2}\frac{{\rm{\alpha }}}{2}}\ {{{\sin }^2}\frac{{\rm{\beta }}}{2}}&{{{\cos }^2}\frac{{\rm{\beta }}}{2}} \end{array}} \right|\)

Determinant of A = |A| = ${\cos ^2}\frac{{\rm{\alpha }}}{2}{\cos ^2}\frac{{\rm{\beta }}}{2} - {\sin ^2}\frac{{\rm{\alpha }}}{2}{\sin ^2}\frac{{\rm{\beta }}}{2}$

$\Rightarrow \left| {\rm{A}} \right| = \left( {\cos \frac{{\rm{\alpha }}}{2}\cos \frac{{\rm{\beta }}}{2} - \sin \frac{{\rm{\alpha }}}{2}\sin \frac{{\rm{\beta }}}{2}} \right)\left( {\cos \frac{{\rm{\alpha }}}{2}\cos \frac{{\rm{\beta }}}{2} + \sin \frac{{\rm{\alpha }}}{2}\sin \frac{{\rm{\beta }}}{2}} \right)$               (∵ a² - b² = (a - b) (a + b))

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

$\Rightarrow \left| {\rm{A}} \right| = \cos \left( {\frac{{90^\circ }}{2}} \right)\cos \left( {\frac{{{\rm{\alpha }} - {\rm{;\beta }}}}{2}} \right) = \cos 45^\circ \cos \left( {\frac{{{\rm{\alpha }} - {\rm{;\beta }}}}{2}} \right) = \frac{1}{{\sqrt 2 }}\cos \left( {\frac{{{\rm{\alpha }} - {\rm{\beta }}}}{2}} \right)$

∴ Statement 1^st is correct.

We know that, Maximum value of cos θ is one.

So, Maximum value of Determinant of A = |A| = (1/√2) × 1 = 1/√2

∴ Statement 2^nd is correct.

Calculation:

Given: \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} 1&0&{ - 2}\ 2&{ - 3}&4 \end{array}} \right]\)

We have, 2X + 3A = 0

⇒ 2X = -3A

∴ X = (-3/2) × A

Now,

\({\rm{x}} = \frac{{ - 3}}{2}\left[ {\begin{array}{*{20}{c}} 1&0&{ - 2}\ 2&{ - 3}&4 \end{array}} \right]\)

\(\Rightarrow {\rm{x}} = \left[ {\begin{array}{*{20}{c}} {1 \times \frac{{ - 3}}{2}}&{0 \times \frac{{ - 3}}{2}}&{ - 2 \times \frac{{ - 3}}{2}}\ {2 \times \frac{{ - 3}}{2}}&{ - 3 \times \frac{{ - 3}}{2}}&{4 \times \frac{{ - 3}}{2}} \end{array}} \right]\)

\(\therefore x = \left[ {\begin{array}{*{20}{c}} { - \frac{3}{2}}&0&3\ { - 3}&{\frac{9}{2}}&{ - 6} \end{array}} \right]\)

Concept:

Let z = x + iy be a complex number.

• Modulus of z = $\left| {\rm{z}} \right| = {\rm{;}}\sqrt {{{\rm{x}}^2} + {{\rm{y}}^2}} = {\rm{;}}\sqrt {{\rm{Re}}{{\left( {\rm{z}} \right)}^2} + {\rm{;Im;}}{{\left( {\rm{z}} \right)}^2}}$

Calculation:

Let z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂

Given: |z₁| = |z₂|

⇒ $\sqrt{x_1^2+y_1^2}=\sqrt{x_2^2+y_2^2}$

It is true for many values of x₁, x₂ and y₁, y₂.

So x₁ must not equal to x₂, and y₁ must not equal to y₂.

∴ z₁ ≠ z₂

All statements are false.

Verification:

Let z₁ = 1 + i and z₂ = -1 - i

$|z_1|=\sqrt{1^2+1^2}=\sqrt2$

$|z_2|=\sqrt{(-1)^2+(-1)^2}=\sqrt2$

Here we get |z₁| = |z₂|

But z₁ ≠ z₂, Real part of z₁ ≠ Real part of z₂ and Imaginary part of z₁ ≠ Imaginary part of z₂

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:

We have, single digit natural numbers = {1, 2, 3,..., 9}

We have to choose two numbers out of these numbers.

⇒ n (S) = ⁹C₂ = 36

For favourable outcomes:

The sum of two different numbers which are prime 2, 3, 5, 7, 11, 13, 17, 19

(1, 2), (1, 4), (1, 6), (2, 3), (2, 5), (3, 2), (4, 1), {6, 1), (9, 2), (7, 4), (4, 7), (7, 6), (6, 7), (8, 9)

⇒ n (E) = 14

Probability of an event happening = $\frac{{{\rm{Number;of;ways;it;can;happen}}}}{{{\rm{Total;number;of;outcomes}}}} = \frac{{{\rm{n;}}\left( {\rm{E}} \right)}}{{{\rm{n;}}\left( {\rm{S}} \right)}} = \frac{{14}}{{36}} = \frac{7}{{18}}$

Concept:

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

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

Note: If b_xy and b_yx are negative then the coefficient of correlation would be negative and if If b_xy and b_yx are positive then the coefficient of correlation would be positive.

Calculation:

Given:

b_xy = -0.64 and b_yx = -0.36

We know that,

\(\begin{array}{l} {\rm{Coefficient;of;correlation;}} = {\rm{;r;}} = {\rm{;}}\sqrt {{{\rm{b}}{{\rm{yx}}}} \times {{\rm{b}}{{\rm{xy}}}}} \ {\rm{r;}} = {\rm{;}}\sqrt { - 0.64 \times - 0.36} = ; \pm 0.48 \end{array}\)

Here b_xy and b_yx are negative then the coefficient of correlation would be negative

∴ r = -0.48

Concept:

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

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

Calculation:

Given: n = 5

P = 1/2

So, q = 1 - p = 1- 1/2 = 1/2

Now,

P (tail appear odd number of times) = P (1) + P (3) + P (5)

$= {{\rm{;}}^5}{{\rm{C}}_1}{\rm{;}}{\left( {\frac{1}{2}} \right)^1}{\left( {\frac{1}{2}} \right)^{5 - 1}} + {{\rm{;}}^5}{{\rm{C}}_3}{\rm{;}}{\left( {\frac{1}{2}} \right)^3}{\left( {\frac{1}{2}} \right)^{5 - 3}} + {{\rm{;}}^5}{{\rm{C}}_5}{\rm{;}}{\left( {\frac{1}{2}} \right)^5}{\left( {\frac{1}{2}} \right)^{5 - 5}}$

$= \frac{5}{{32}} + \frac{{10}}{{32}} + \frac{1}{{32}} = \frac{{16}}{{32}} = \frac{1}{2}$

Concept:

Let A₁, A₂, …. , A_n be n mutually exclusive and exhaustive events of the sample space S and A is event which can occur with any of the events then

• \({\rm{P}}\left( {\frac{{{{\rm{A}}{\rm{i}}}}}{{\rm{A}}}} \right) = \frac{{{\rm{P}}\left( {{{\rm{A}}{\rm{i}}}} \right){\rm{P}}\left( {\frac{{\rm{A}}}{{{{\rm{A}}{\rm{i}}}}}} \right)}}{{\mathop \sum \nolimits{{\rm{i}} = 1}^{\rm{n}} {\rm{P}}\left( {{{\rm{A}}{\rm{i}}}} \right){\rm{P}}\left( {\frac{{\rm{A}}}{{{{\rm{A}}{\rm{i}}}}}} \right)}}\)

Calculation:

Let E be the event that person meets with accident.

Let E₁ be the scooter driver, E₂ be the car driver and E₃ be the truck driver

P (E₁) = 2000/12000 = 1/6

P (E₂) = 4000/12000 = 1/3

P (E₃) = 6000/12000 = 1/2

Also,

P (E|E₁) = Probability the event that accident involving a scooter driver = 0.01

P (E|E₂) = Probability the event that accident involving a car driver = 0.03

P (E|E₃) = Probability the event that accident involving a truck driver = 0.15

By using Bayes’ theorem,

The probability that the person is a scooter driver is

${\rm{P}}\left( {{\rm{E}}1{\rm{|E}}} \right) = \frac{{{\rm{P}}\left( {{\rm{E}}1} \right){\rm{P}}\left( {{\rm{E|E}}1} \right)}}{{{\rm{P}}\left( {{\rm{E}}1} \right){\rm{P}}\left( {{\rm{E|E}}1} \right) + {\rm{;P}}\left( {{\rm{E}}2} \right){\rm{P}}\left( {{\rm{E|E}}2} \right) + ; + {\rm{;P}}\left( {{\rm{E}}3} \right){\rm{P}}({\rm{E}}|{\rm{E}}3)}}$

$\Rightarrow {\rm{P}}\left( {{\rm{E}}1{\rm{|E}}} \right) = \frac{{\frac{1}{6}{\rm{;}} \times {\rm{;}}0.01}}{{\frac{1}{6} \times {\rm{;}}0.01 + \frac{1}{3}{\rm{;}} \times {\rm{;}}0.03{\rm{;}} + {\rm{;}}\frac{1}{2}{\rm{;}} \times {\rm{;}}0.15{\rm{;}}}} = \frac{1}{{52}}$

Concept:

Complement of an event:

• The complement of an event is the subset of outcomes in the sample space that are not in the event. A complement is itself an event.
• The probability of the complement of an event is one minus the probability of the event.

P (A’) or P (A^c) or ${\rm{P}}\left( {{\rm{\bar A}}} \right)$ = 1 - P (A)

Where,

P (A) be probability of Event A and P (A’) or P (A^c) or ${\rm{P}}\left( {{\rm{\bar A}}} \right)$ be probability of the complement of Event A

Calculation:

Given:

Probability of solving problem by A = P (A) =1/2

Probability of solving problem by B = P (B) =1/3

Probability of solving problem by C = P (C) =1/4

P (problem will not solved by A) = 1 - (1/2) = 1/2

P (problem will not solved by B) = 1 - (1/3) = 2/3

P (problem will not solved by C) = 1 - (1/4) = 3/4

P (Problem will be solved) = 1 - P (problem will not solved by A, B and C)

$= 1 - \left( {\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4}} \right) = 1 - \frac{1}{4} = \frac{3}{4}$

Concept:

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

Calculation:

Given:

A bag contains 4 white and 2 black balls and another bag contains 3 white and 5 black balls.

We have to find probability that one ball is white and one ball is black.

Possible cases:

Let E₁ = Black ball from bag 1 and white ball from bag 2

And E₂ = White ball from bag 1 and black ball from bag 2.

P (one ball is white and one ball is black) = P (E₁) + P (E₂)

$= \frac{2}{6} \times \frac{3}{8} + \frac{4}{6} \times \frac{5}{8} = \frac{{26}}{{48}} = \frac{{13}}{{24}}$

Calculation:

Given: x² - 3x + 2 ≥ 0

⇒ x² - x - 2x + 2 ≥ 0

⇒ x (x - 1) - 2 (x - 1) ≥ 0

⇒ (x - 2) (x -1) ≥ 0

⇒ x ∈ (-∞, 1] ∪ [2, ∞)

Here x ∈ [0, 5]

Total number of solutions, n (S) = 5 which are [0, 1], [1, 2], [2,  3], [3, 4], [4, 5]

and favourable events are 4 which are [0, 1], [2,  3], [3, 4], [4, 5]

⇒ n (E) = 4

∴ Probability of an event happening =  $\frac{{{\rm{Number;of;ways;it;can;happen}}}}{{{\rm{Total;number;of;outcomes}}}} = ;\frac{{{\rm{n;}}\left( {\rm{E}} \right)}}{{{\rm{n;}}\left( {\rm{S}} \right)}} = \frac{4}{5}$

Calculation:

Total digit = 6 {0, 1, 2, 3, 4 and 5}

We have to formed 3-digit even numbers:

Case-1: When 0 is at unit place, then

Number of choices for the units place = 1

(0 is at unit place)

Number of choices for the ten's place = 5

(Any digit except the one that is already used as the 1st digit)

Number of choices for the hundredth's place = 4

(Any digit except the ones that are already used as the 1st and 2nd digit)

No. of such three digit even number = 1 × 5 × 4 = 20

Case-2: When 0 is at ten's place, then

For even number unit place must be filled with 2 or 4

Number of choices for the units place = 2

Number of choices for the ten's place = 1

Number of choices for the hundredth place = 4

No. of such three digit even number = 2 × 1 × 4 = 8

Case-3: When no digit in the 3-digit even number is zero.

For even number unit place must be filled with 2 or 4

Number of choices for the units place = 2

Number of choices for the ten's place = 4

Number of choices for the hundredth place = 3

No. of such three digit even number = 2 × 4 × 3 = 24

Total number of possible three digit even number = 8 + 24 + 20 = 52

Concept:

Logarithm properties

• Product rule: The log of a product equals the sum of two logs.

​${\log _{\rm{a}}}\left( {{\rm{mn}}} \right) = {\rm{;}}{\log _{\rm{a}}}{\rm{m}} + {\rm{;}}{\log _{\rm{a}}}{\rm{n}}$

• Quotient rule: The log of a quotient equals the difference of two logs.

${\log _{\rm{a}}}\frac{{\rm{m}}}{{\rm{n}}} = {\rm{;}}{\log _{\rm{a}}}{\rm{m}} - {\rm{;}}{\log _{\rm{a}}}{\rm{n}}$

• Power rule: In the log of a power the exponent becomes a coefficient.

​${\log _{\rm{a}}}{{\rm{m}}^{\rm{n}}} = {\rm{n}}{\log _{\rm{a}}}{\rm{m}}$

• If log_x y = a than y = x^a

Calculation:

Given: log₈ m + log₈ $\frac{1}{6} = \frac{2}{3}$

$\Rightarrow {\rm{;}}{\log _8}\left( {{\rm{m}} \times {\rm{}}\frac{1}{6}} \right) = {\rm{;}}\frac{2}{3}$  

$\Rightarrow {\rm{;}}\frac{{\rm{m}}}{6} = {\rm{;}}{8^{\frac{2}{3}}} = {\rm{;}}{\left( {{2^3}} \right)^{\frac{2}{3}}} = {\rm{;}}{2^2}$ 

∴ m = 4 × 6 = 24

Concept:

• The intercept form of the line ⇔ ${\rm{;}}\frac{{\rm{x}}}{{\rm{a}}} + \frac{{\rm{y}}}{{\rm{b}}} = 1$ , Where a is the x- intercept and b is the y- intercept
• Area of parallelogram = $\frac{1}{2}{\rm{}} \times {\rm{;Product;of;diagonals}}$

Calculation:

ax + by + c = 0

⇒ ax + by = -c

$\Rightarrow \frac{{{\rm{ax}}}}{{ - {\rm{c}}}} + {\rm{;}}\frac{{{\rm{by}}}}{{ - {\rm{c}}}} = 1 \Rightarrow \frac{{\rm{x}}}{{\left( {\frac{{ - {\rm{c}}}}{{\rm{a}}}} \right)}} + {\rm{;}}\frac{{\rm{y}}}{{\left( {\frac{{ - {\rm{c}}}}{{\rm{b}}}} \right)}} = 1$ 

ax – by + c = 0

⇒ ax – by = -c

$\Rightarrow \frac{{{\rm{ax}}}}{{ - {\rm{c}}}} - {\rm{;}}\frac{{{\rm{by}}}}{{ - {\rm{c}}}} = 1 \Rightarrow \frac{{\rm{x}}}{{\left( {\frac{{ - {\rm{c}}}}{{\rm{a}}}} \right)}} + {\rm{;}}\frac{{\rm{y}}}{{\left( {\frac{{\rm{c}}}{{\rm{b}}}} \right)}} = 1$ 

ax + by – c = 0

⇒ ax + by = c

$\Rightarrow \frac{{{\rm{ax}}}}{{\rm{c}}} + {\rm{;}}\frac{{{\rm{by}}}}{{\rm{c}}} = 1 \Rightarrow \frac{{\rm{x}}}{{\left( {\frac{{\rm{c}}}{{\rm{a}}}} \right)}} + {\rm{;}}\frac{{\rm{y}}}{{\left( {\frac{{\rm{c}}}{{\rm{b}}}} \right)}} = 1$ 

ax – by – c = 0

⇒ ax – by = c

$\Rightarrow \frac{{{\rm{ax}}}}{{\rm{c}}} - {\rm{;}}\frac{{{\rm{by}}}}{{\rm{c}}} = 1 \Rightarrow \frac{{\rm{x}}}{{\left( {\frac{{\rm{c}}}{{\rm{a}}}} \right)}} + {\rm{;}}\frac{{\rm{y}}}{{\left( {\frac{{ - {\rm{c}}}}{{\rm{b}}}} \right)}} = 1$ 

The figure formed by the lines ax + by + c = 0, ax – by + c = 0, ax + by – c = 0 and ax – by – c = 0 is shown below

Clearly, ABCD is a parallelogram and its area = $\frac{1}{2}{\rm{}} \times {\rm{Product;of;diagonals}}$

$= \frac{1}{2}{\rm{}} \times {\rm{}}\frac{{2{\rm{c}}}}{{\rm{a}}}{\rm{}} \times {\rm{}}\frac{{2{\rm{c}}}}{{\rm{b}}} = {\rm{}}\frac{{2{{\rm{c}}^2}}}{{{\rm{ab}}}}$

Concept:

• Equation of a line perpendicular to a given line ax + by + c = 0 is bx - ay + λ = 0, where λ is a constant.
• The intercept form of the line ⇔ ${\rm{;}}\frac{{\rm{x}}}{{\rm{a}}} + \frac{{\rm{y}}}{{\rm{b}}} = 1$ , Where a is the x- intercept and b is the y- intercept.

Calculation:

Given: Area of triangle is 5 square units

Given equation of line is 5x – y = 0

Equation of a line perpendicular to a given line is

⇒ x + 5y = λ      …. (1)

$\Rightarrow \frac{{\rm{x}}}{{\rm{\lambda }}} + {\rm{;}}\frac{{5{\rm{y}}}}{{\rm{\lambda }}} = 1 \Rightarrow \frac{{\rm{x}}}{{\rm{\lambda }}} + {\rm{;}}\frac{{\rm{y}}}{{\left( {\frac{{\rm{\lambda }}}{5}} \right)}} = 1$

Area of triangle = 5

$\Rightarrow \frac{1}{2} \times {\rm{\lambda }} \times \frac{{\rm{\lambda }}}{5} = 5$

⇒ λ² = 50

∴ λ = ± 5√2

Put the value of λ in equation 1^st

So, x + 5y = ± 5√2

⇒ x + 5y ± 5√2 = 1

Concept:

Ellipse: The locus of a point which moves such that its distance from a fixed point and a fixed straight line is always less than one (Eccentricity = e < 1) ⇔ $\frac{{{\rm{;}}{{\bf{x}}^2}}}{{{{\bf{a}}^2}}} + \frac{{{{\bf{y}}^2}}}{{{{\bf{b}}^2}}} = 1$ (a > b)

The sum of the focal distance of any point on an ellipse is constant and equal to the length of the major axis of the ellipse. ⇔ PS + PS' = 2a = Major axis

Focus: (± ae, 0)

Eccentricity: ${\rm{e}} = {\rm{;}}\sqrt {1 - {\rm{;}}\frac{{{{\rm{b}}^2}}}{{{{\rm{a}}^2}}}}$

Calculation:

We have $\frac{{{{\rm{x}}^2}}}{{25}} + \frac{{{{\rm{y}}^2}}}{9} = 1$

Here, a² = 25, b² = 9 ⇒ a = 5 and b = 3

Eccentricity:

${\rm{e}} = {\rm{;}}\sqrt {1 - {\rm{;}}\frac{{{{\rm{b}}^2}}}{{{{\rm{a}}^2}}}} = {\rm{;}}\sqrt {1 - {\rm{;}}\frac{9}{{25}}} = {\rm{;}}\sqrt {\frac{{16}}{{25}}} = {\rm{;}}\frac{4}{5}$ 

Focus = (± ae, 0) = (±4, 0)

PS + PS' = 2a = Major axis

⇒ r + s = 2 × 5 = 10 unit

Concept:

Perpendicular Distance of a Point from a Line:

Let us consider a plane given by the Cartesian equation, Ax + By + C = 0

And a point whose coordinate is, (x₁, y₁)

Now, distance = $\left| {\frac{{{\rm{A}}{{\rm{x}}_1}{\rm{;}} + {\rm{;B}}{{\rm{y}}_1}{\rm{;}} + {\rm{;c}}}}{{\sqrt {{{\rm{A}}^2}{\rm{;}} + {\rm{;;}}{{\rm{B}}^2}{\rm{;}}} }}} \right|$

Calculation:

Given a straight line x = y + 2 touches the circle 4(x² + y²) = r²

x = y + 2 ⇒ x – y – 2 = 0

⇒ x² + y² = r²/4

Centre = (0, 0)

Radius = r/2 

From above figure,

CP = Radius

$\Rightarrow {\rm{;}}\left| {\frac{{0 - 0 - 2}}{{\sqrt {{{\left( 1 \right)}^2} + {{\left( { - 1} \right)}^2}} }}} \right| = {\rm{;}}\frac{{\rm{r}}}{2}$

⇒ r/2 = √2

∴ r = 2√2

Concept:

• A set of lines or curves are said to be concurrent if they all intersect at the same point.

Calculation:

Given lines are 4x + 4y = 1, 8x – 3y = 2 and y = 0

Solving 1^st and 3^rd line of equation,

y = 0

4x + 4y = 1, Put y = 0

∴ x = 1/4

Point of intersection = (1/4, 0)

Now solving the 2^nd and 3^rd equations,

y = 0

8x – 3y = 2, Put y = 0

∴ x = 1/4

Point of intersection = (1/4, 0)

Since, the point of intersection of lines are same, we can say that the given lines are concurrent.

Concept:

Circumcentre: circumcenter is the point where the perpendicular bisectors of a triangle intersect.

The circumcenter of a right triangle lies exactly at the midpoint of the hypotenuse (longest side).

Calculation:

Given:

3x + 4y – 24 = 0

⇒ 3x + 4y = 24

$\Rightarrow \frac{{3{\rm{x}}}}{{24}} + {\rm{;}}\frac{{4{\rm{y}}}}{{24}} = 1 \Rightarrow \frac{{\rm{x}}}{8} + {\rm{;}}\frac{{\rm{y}}}{6} = 1$

Its x-intercept A = (8, 0) and y-intercept B = (0, 6)

Since, AOB is a right angled triangle. So, the mid-point of hypotenuse is the circumcentre.

Hence, circumcentre = $\left( {\frac{{8; + ;0}}{2},{\rm{;}}\frac{{0; + ;6}}{2}} \right)$

∴ circumcentre = (4, 3)

Concept:

Hyperbola: The locus of a point which moves such that its distance from a fixed point is greater than its distance from a fixed straight line. (Eccentricity = e > 1)

Calculation:

Given:

16x² – 9y² = 1

$\Rightarrow \frac{{{\rm{;}}{{\rm{x}}^2}}}{{\frac{1}{{16}}}} - \frac{{{{\rm{y}}^2}}}{{\frac{1}{9}}} = 1$

Compare with $\frac{{{\rm{;}}{{\rm{x}}^2}}}{{{{\rm{a}}^2}}} - \frac{{{{\rm{y}}^2}}}{{{{\rm{b}}^2}}} = 1$

∴ a² = 1/16 and b² = 1/9

Eccentricity = $\sqrt {1 + {\rm{;}}\frac{{{{\rm{b}}^2}}}{{{{\rm{a}}^2}}}} = ;\sqrt {1 + ;\frac{{\left( {\frac{1}{9}} \right)}}{{\left( {\frac{1}{{16}}} \right)}}} = ;\sqrt {1 + ;\frac{{16}}{9}} = ;\sqrt {\frac{{25}}{9}} = ;\frac{5}{3}$