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Calculation:

Given ab + bc + ca = 0      ...(1)

6ab + 9bc + 8ca = 0          ...(2)

6 × eq(1) - eq(2)

⇒ 6(ab + bc + ca) - (6ab + 9bc + 8ca) = 0

⇒ - 3bc - 2ac = 0

⇒ 3b + 2a = 0           ...(3)

Similarly eliminating ab from eq(1) and eq(2), we get:

2a - c = 0           ...(4)

From eq(3) and eq(4)

2a = - 3b = c

⇒ \(\rm \frac{a}{-3/7}=\frac{b}{2/7}=\frac{c}{-6/7}\)

∴ Direction cosines of L is \(\left<-\frac{3}{7},\frac{2}{7},-\frac{6}{7}\right>\)

Concept:

Transpose of a Matrix:

The new matrix obtained by interchanging the rows and columns of the original matrix is called the transpose of the matrix.

It is denoted by A′ or AT

Calculation:

Given:

A = \(\begin{bmatrix} 4 & -3\ 1 & 0 \end{bmatrix}\)

Now find the transpose of a matrix A,

AT = \(\begin{bmatrix} 4 & 1\ -3 & 0 \end{bmatrix} \)

Now,

A + AT = \(\begin{bmatrix} 4 & -3\ 1 & 0 \end{bmatrix}+ \begin{bmatrix} 4 & 1\ -3 & 0 \end{bmatrix}\)

\(= \begin{bmatrix} 4+4 &-3+1 \ 1+(-3)&0+0 \end{bmatrix}\)

\(= \begin{bmatrix} 8 & -2\ -2 & 0 \end{bmatrix} \)

Given:

P(B) = 0.4, P(A ∪ B) = 0.8

Formula used:

A and B are independent events, P(A∩B) = P(A). P(B)

P(AUB) = P(A)+ P(B) – P(A)P(B)

Calculation:

0.8 = P(A) + 0.4 – 0.4 × P(A)

⇒ 0.4 = 0.6P(B)

⇒ P(B) = 0.4/0.6 = 4/6 = 2/3

A and B are independent then P(A) is 2/3.

∴ Option 2 is correct

Formula used:

\(\frac{d}{dx}e^x = e^x\)

\(\frac{d}{dx}sin\ x = cos\ x\)

Calculation:

Given,  y =esin x

⇒  y = esin x

⇒ \(\rm \frac{dy}{dx}\) = e^{sin x} \(\frac{d}{dx}sin\ x \)

⇒ \(\rm \frac{dy}{dx}\) = e^{sin x} (cos x)

∴ \(\rm \frac{dy}{dx}\) = cos x esin x

Formula used : 

1. log mn = n log m
2. Differentiation by part: \(\frac{d}{dx}(uv) = v\frac{du}{dx}+u\frac{dv}{dx}\)
3. \(\frac{d}{dx}x^n=nx^{n-1}\)
4. \(\frac{d}{dx}log\ x= \frac{1}{x}\)

Calculation :

Given that y = (xx)x,

Taking log on both sides

log y = log (xx)x

Using the formula (1)

⇒ log y = x log x^x

⇒ log y = x² log x

Diff. with respect to x 

\(\frac{d}{dx}\)(log y) = \(\frac{d}{dx}\)(x²log x)

Doing diff. by part as discussed in formula (2)

⇒ \(\frac{1}{y}\frac{dy}{dx} = x^2\frac{d}{dx}log\ x+ log \ x\frac{d}{dx}x^2\)

Since, \(\frac{d}{dx}x^n=nx^{n-1}\) & \(\frac{d}{dx}log\ x= \frac{1}{x}\)

⇒ \(\frac{1}{y}\frac{dy}{dx} = \frac{x^2}{x}+2x\ log \ x\)

⇒ \(\frac{1}{(x^x)^x}\frac{dy}{dx} = \frac{x^2}{x}+2x\ log \ x\)       (∵ y = (xx)x)

Put x = 1 in above equation

⇒ \(\frac{dy}{dx} \) = 1 + 2 × 1 log 1

⇒ \(\frac{dy}{dx} \) = 1          (∵ log 1 = 0)

∴ \(\frac{dy}{dx}\) at x = 1 is 1.

Concept:

Some useful formulas are:

∫cosx dx = sinx + c

∫ secx dx = ln(sec x + tan x) + c

cos2θ = 2cos²θ - 1

\(\rm 1\over cos θ \)= secθ 

Calculation:

Given integration is, \(\rm ∫{ {cos2x}\over {cosx} }dx\)

= \(\rm ∫{ {2cos^2x-1}\over {cosx} }dx\)

= \(\rm ∫({ {{2cos^2x}\over {cosx}}-{{1}\over {cosx} }})dx\)

= \(\rm ∫({ {{2cosx}}-{secx} })dx\)

= 2sinx - ln(sec x + tan x) + C, C = constant of integration

Calculation:

Given:

 \(\frac{dy}{dx}=1+tan(y-x)\)      ------(i)

let y - x = v 

on differentiating w.r.t x we get 

⇒ \(\frac{dy}{dx}-1=\frac{dv}{dx} \)

⇒ \(\frac{dy}{dx}=\frac{dv}{dx}+1 \)

equation (i) become

⇒ \(\frac{dv}{dx}+1=1+tan(v) \)

⇒ \(\frac{dv}{tanv}=dx \)

⇒ \(\frac{cosv}{sinv}dv=dx\)

On integrating both the side we get, 

⇒ log sin v = x + c

taking exponential both sides 

⇒ sin v = e^x + C

⇒ sin (y - x) = ex + C

Given:

The given integral is \(\int {\sqrt {{\rm{1 + }},{\rm{sec}},{\rm{x}}} ,{\rm{dx}}} \)

Formula Used:

1 + cos x = 2cos²x/2

1 - 2sin²x/2 = cos x 

\(\int \frac{1}{\sqrt{a^2 \ - \ x^2}}dx = sin^{-1} \left(\frac{x}{a} \right) + c\)

Calculation:

Let I = \(\int {\sqrt {{\rm{1 + }},{\rm{sec}},{\rm{x}}} ,{\rm{dx}}} \)

⇒ I = \(\int \sqrt{ 1 + \frac{1}{cosx} } dx\)

⇒ I = \(\int \sqrt{ \frac{1 \ + \ cosx}{cosx}}\)

⇒ I = \(\int \sqrt{\frac{2cos^2\frac{x}{2}}{1 - 2sin^2\frac{x}{2}}} dx \)

⇒ I = \(\int \frac{\sqrt2 cos \frac{x}{2}}{\sqrt{1 - 2sin^2 \frac{x}{2}}}dx\)

Let t = \(sin\frac{x}{2}\)

⇒ \(dt = \frac{1}{2} . cos\frac{x}{2} dx\)

⇒ \(2dt = cos\frac{x}{2} dx\)

⇒ I = \( 2\sqrt2 \int \frac{dt}{\sqrt{1 \ - \ 2t^2}}\)

⇒ I = \(2\sqrt2\times \int \frac{dt}{\sqrt{ 2 \left(\frac{1}{2} \ - \ t^2 \right)}}\)

⇒ I = \(2 \int \frac{dt}{\sqrt{ \left(\frac{1}{\sqrt2}\right)^2 \ -\ t^2}}\)

⇒ I = \(2 \times sin^{-1} \left( \frac{t}{\frac{1}{\sqrt2}} \right) + c\)

⇒ I = \(2 \times sin^{-1} (\sqrt{2}t) + c\)

⇒ I = \(2 \times sin^{-1} \left(\sqrt{2} sin \frac{x}{2} \right) + c\)

∴ The value of \(\int {\sqrt {{\rm{1 + }},{\rm{sec}},{\rm{x}}} ,{\rm{dx}}} \) equals to \(2 \times sin^{-1} \left(\sqrt{2} sin \frac{x}{2} \right) + c\)

Explanation:

Mean  = \(\frac{1+ 4+ 9 ... n^2}{n}\)

⇒ 130 = \(\frac{(n(n+1)(2n+1)}{6n}\)

⇒ 780 = 2n² + 3n + 1

⇒ 2n² + 3n – 779 =0

⇒ 2n² + 41n – 38n – 779 = 0

⇒ n(2n + 41) – 19(2n + 41) = 0 

⇒ (2n + 41) (n – 19) = 0

⇒ x = 19 ..... ( n = \(-\frac{41}{2}\) is not possible)

∴ Option (b) is correct.

CONCEPT**:**

In order to find the minor of an element of a determinant, we need to delete the row and column passing through the element aij , thus obtained is called the minor of aij and is usually denoted by Mij .

The co-factor of an element aij is given by (-1)i + j ⋅ Mij, where Mij is the minor of element aij and it is denoted by Cij.

Thus \({C_{ij}} = ;\left{ {\begin{array}{*{20}{c}} {{M_{ij}},;when;i + j;is;even}\ { - ;{M_{ij}},;when;i + j;is;odd} \end{array}} \right.\)

CALCULATION**:**

Given: \(A = \left[ {\begin{array}{*{20}{c}} { - 1}&2&3\ { - 2}&3&5\ 4&{ - 2}&1 \end{array}} \right]\)

Here, we have to find the cofactor matrix for the given matrix A

As we know that, the cofactor of an element aij is given by: \({C_{ij}} = ;\left{ {\begin{array}{*{20}{c}} {{M_{ij}},;when;i + j;is;even}\ { - ;{M_{ij}},;when;i + j;is;odd} \end{array}} \right.\)

⇒ \({C_{11}} = {\left( { - 1} \right)^2} \times \left| {\begin{array}{*{20}{c}} 3&5\ { - 2}&1 \end{array}} \right| = 13\)

⇒ \({C_{12}} = {\left( { - 1} \right)^3} \times \left| {\begin{array}{*{20}{c}} { - 2}&5\ 4&1 \end{array}} \right| = ;22\)

⇒ \({C_{13}} = {\left( { - 1} \right)^4} \times \left| {\begin{array}{*{20}{c}} { - 2}&3\ 4&{ - 2} \end{array}} \right| = ; - 8\)

Similarly, we can say that C₂₁ = - 8, C₂₂ = - 13 and C₂₃ = 6

Similarly, we can also say that, C₃₁ = 1, C₃₂ = - 1 and C₃₃ = 1

So, the required cofactor matrix is \(\left[ {\begin{array}{*{20}{c}} {13}&{22}&{ - 8}\ { - 8}&{ - 13}&6\ 1&{ - 1}&1 \end{array}} \right]\)

Hence, option B is the correct answer.