Mathematics Test-37

Given:

\(I=\int_{\frac{\pi }{ 4}}^{\frac{3 \pi }{ 4}} \frac{d x}{1+\cos x}\)

Multiply and divide by (1 - cos x) in numrator and denominator.

\(=\int_{\frac{\pi}{4}}^{\frac{3 x}{4}} \frac{1-\cos x}{(1+\operatorname{cos} x)(1-\cos x)} d x\)

\(=\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{1-\cos x}{\sin ^{2} x} d x\)

\(=\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}}\left(\operatorname{cosec}^{2} x-\operatorname{cosec} x \cdot \cot x\right) d x\)

\(=[-\cot x+\operatorname{cosec} x]_{\frac{\pi }{ 4}}^{\frac{3 \pi }{ 4}}\)

\(=-\cot \frac{3 \pi}{4}+\cot \frac{\pi}{4}+\operatorname{cosec} \frac{3 \pi}{4}-\operatorname{cosec} \frac{\pi}{4}\)

\(=1+1+\sqrt{2}-\sqrt{2}\)

\(=2\)

Position vector of \(A\) is \(\hat{i}+\hat{j}+\hat{k}\), gives coordinates as \((1,1,1)\).

Position vector of \(B\) is \(4 \hat{i}+\hat{j}+\hat{k}\), gives coordinates as \((4,1,1)\).

Position vector of \(C\) is \(4 \hat{i}+5 \hat{j}+\hat{k}\), gives coordinates as \((4,5,1)\).

\(\overrightarrow{A B}=\vec{B}-\vec{A}=3 \hat{i}\)

\(\overrightarrow{B C}=\vec{C}-\vec{B}=4 \hat{j}\)

\(\overrightarrow{A C}=\vec{C}-\vec{A}=3 \hat{i}+4 \hat{j}\)

Here, we can see that \(\triangle A B C\) is a right-angled triangle with a right angle at \(B\).

Hence, the circumcenter will be the midpoint of \(AC\).

Midpoint of \(AC = D\)

\(=\left(\frac{1+4}{2}, \frac{1+5}{2}, \frac{1+1}{2}\right)=\left(\frac{5}{2}, 3,1\right)\)

So, the position vector of \(D\) will be\(\frac{1}{2}(5 \hat{i}+6 \hat{j}+2 \hat{k})\)which is a circumcenter.

Number of ways 10 people can be seated in a row = \({ }^{10} {C}_{5}\)= 252

Number of ways 5 people can be seated in one row = 5!

Number of ways 5 people can be seated in the second row = 5!

Number of ways the people of the rows can be arranged = 5! × 5!

Number of ways 2 food items can be arranged among 2 rows = 2!

= 2

Total number of ways the food can be served to these people = 252 × 5! × 5! × 2

= 7257600

∴ The people can be served in 7257600 ways.

Given:

The radius of the circle \(=\mathrm{r}\)

As we know,

\(A B\) is perpendicular bisector of \(O C\)

\(\mathrm{OA}=\mathrm{r}\) (OA radius of the circle)

\(\mathrm{OG}=\frac{1}{2} \times \mathrm{OC}=\frac{1}{2} \times \mathrm{r}\)

In, \(\triangle \mathrm{OAG}\)

\(\mathrm{AG} \perp \mathrm{OC}\)

By Pythagoras theorem,

\((\mathrm{H})^2=(\mathrm{P})^2+(\mathrm{B})^2\)

\(\Rightarrow(\mathrm{GA})^2=(\mathrm{r})^2-\left(\frac{\mathrm{r}}{2}\right)^2\)

\(\Rightarrow(\mathrm{GA})^2=\frac{3}{4} \times \mathrm{r}^2\)

\(\Rightarrow G A=\frac{\sqrt{3}}{2} \times r\)

\(A B=2 \times G A\)

\(\Rightarrow A B=2 \times \frac{\sqrt{3}}{2} \times r\)

\(\Rightarrow A B=\sqrt{3 r}\)

So, The length of common chord of the circle is \(\sqrt{3} r\).

As we know,

The standard form of a linear equation of the first order is given by \(\frac{d y}{d x}+P y=Q\) where \(P, Q\) are arbitrary function of \(x\).

The integrating factor of the linear equation is given by: \(I . F .=e^{\int p d x}\)

The solution of the linear equation is given by: \(y(I . F.)=\int Q(I . F.) d x+\)\(c \)

\(y d x=(y-x) d y\)

\(y \frac{d x}{d y}=y-x\)

\(\frac{d x}{d y}+\frac{x}{y}=1\)

It is form of\(\frac{d x}{d y}+P x=Q\).

\(I. F. =e^{\int p d y}\)

\(I. F .=e^{l n y}=y\)

The solution of the linear equation is given by,

\(x(I . F)=.\int Q(I . F) d y+c\)

\(x(y)=\int 1(y) d y+c\)

\(x y=\frac{y^2}{2}+c\)

\(x=\frac{y}{2}+\frac{c}{y}\)

\(\lim _{x \rightarrow 1} \frac{x+x^{2}+x^{3}-3}{x-1}\) Form \(\left(\frac{0}{0}\right)\)

Apply L-Hospital Rule,

\(=\lim _{x \rightarrow 1} \frac{\frac{d\left(x+x^{2}+x^{3}-3\right)}{d x}}{\frac{d(x-1)}{d x}}\)

\(=\lim _{x \rightarrow 1} \frac{1+2 x+3 x^{2}-0}{1-0}\)

\(=\frac{1+2 \times 1+3 \times 1^{2}-0}{1-0}\)

\(=6\)

Formula used:

\((a+b)^{2}=a^{2}+b^{2}+2 a b\)

\(\sin \alpha=\frac 1 { \operatorname{cosec} \alpha}\)

\(\cos \alpha=\frac 1 { \sec \alpha}\)

\(\sin ^{2} \alpha+\cos ^{2} \alpha=1\)

\(\operatorname{cosec}^{2} \alpha=1+\cot ^{2} \alpha\)

\(\sec ^{2} \alpha=1+\tan ^{2} \alpha\)

\((\sin \alpha+\operatorname{cosec} \alpha)^{2}+(\cos \alpha+\sec \alpha)^{2}\)

\(\Rightarrow \sin ^{2} \alpha+\operatorname{cosec}^{2} \alpha+2 \sin \alpha \cdot \operatorname{cosec} \alpha+\cos ^{2} \alpha+\sec ^{2} \alpha+2 \cos \alpha \cdot \sec \alpha \quad[\because(a+b)^{2}=a^{2}+b^{2}+2 a b]\)

\(\Rightarrow \sin ^{2} \alpha+\operatorname{cosec}^{2} \alpha+2+\cos ^{2} \alpha+\sec ^{2} \alpha+2 \quad\left[(\because \sin \alpha=\frac 1{ \operatorname{cosec} \alpha}\right.\) and \(\left.\cos \alpha=\frac 1{\sec \alpha}\right]\)

\(\Rightarrow\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)+1+\cot ^{2} \alpha+1+\tan ^{2} \alpha+4 \quad\left[\because \sin ^{2} \alpha+\cos ^{2} \alpha=1 ; \operatorname{cosec}^{2} \alpha\right.=1+\cot ^{2} \alpha\) and \(\left.\sec ^{2} \alpha=1+\tan ^{2} \alpha\right]\)

\(\Rightarrow \tan ^{2} \alpha+\cot ^{2} \alpha+7\)

We have \(\left[\begin{array}{ccc}1 & 2 & 3 \\ 4 & 5 & 6 \\ 5 & 8 & 9\end{array}\right]\)

Eliminate row and column containing 4 we get:

\(\left[\begin{array}{ll}2 & 3 \\ 8 & 9\end{array}\right]\)

Now, determinant \(=2 \times 9-(8 \times 3)\)

\(=-6\)

Now from sign convention of \(3 \times 3\), matrix

For element 4, sign is -ve: \(\quad \quad\quad\left(\because\left[\begin{array}{lll}+ & - & + \\ - & + & - \\ + & - & +\end{array}\right]\right)\)

\(\therefore\) Cofactor of element 4 is 6.

Given : \(A =(0,7,10), B =(-1,6,6)\) and \(C =(-4,9,6)\)

Position vector(P.V.) of \(A B\) is \((-1,-1,-4)\)

\(\Rightarrow|A B|=\sqrt{(-1)^2+(-1)^2+(-4)^2}=\sqrt{18}\)

P.V. of \(BC\) is \((-3,3,0)\)

\(\Rightarrow| BC |=\sqrt{(-3)^2+3^2+0}=\sqrt{18}\)

P.V. of \(AC\) is \((-4,2,-4)\)

\(\Rightarrow| AC |=\sqrt{(-4)^2+2^2+(-4)^2}=6\)

Now, \(D\) is the midpoint of \(AC\)

\(\Rightarrow D =\left(\frac{0-4}{2}, \frac{7+9}{2}, \frac{10+6}{2}\right)=(-2,8,8)\)

\(\Rightarrow \overrightarrow{ BD }=-\hat{i}+2 \hat{ j }+2 \hat{ k }\)