Mathematics Test - 11

∵ The equation of hyperbola is \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)

∵ Equation of hyperbola passes through (3,3)

\(\frac{1}{a^2}-\frac{1}{b^2}=\frac{1}{9} \ldots \text { (i) }\)

Equation of normal at point (3,3) is:

\(\frac{x-3}{\frac{1}{a^2} \cdot 3}=\frac{y-3}{-\frac{1}{b^2} \cdot 3}\)

∵ It passes through (9,0)

\(\frac{6}{\frac{1}{a^2}}=\frac{-3}{-\frac{1}{b^2}}\)

\(\frac{1}{b^2}=\frac{1}{2 a^2} \ldots \text { (ii) }\)

From equations (i) and (ii),

\(a^2=\frac{9}{2}, b^2=9\)

\(\because \text { Eccentricity }=\mathrm{e}, \text { then } \mathrm{e}^2=1+\frac{\mathrm{b}^2}{\mathrm{a}^2}=3\)

\(\therefore\left(\mathrm{a}^2, \mathrm{e}^2\right)=\left(\frac{9}{2}, 3\right)\)

Given that:

There are 20 cricket players, out of which 5 players can bowl.

We have to make a team of 11 players so to include 4 bowlers.

So, we select 4 bowlers out of 5 players and the remaining 7 players must be selected from 15 players, i.e.,

Total ways \(={ }^{5} {C}_{4} \times{ }^{15} {C}_{7}\)

Given,

|x – 1| ≤ 5, |x| ≥ 2

⇒ -(5 ≤ (x – 1) ≤ 5), (x ≤ -2 or x ≥ 2)

⇒ -(4 ≤ x ≤ 6), (x ≤ -2 or x ≥ 2)

Now, required solution is

x ∈ [-4, -2] ∪ [2, 6]

Given,

\(\sec ^{2} \theta+\tan ^{2} \theta=3\)...(i)

Subtracting 1 from both sides in equation (i) we get,

\(\sec ^{2} \theta+\tan ^{2} \theta-1=3-1\)\(\quad(\because\)\(\sec ^{2} \theta-1=\tan ^{2} \theta)\)

\(\Rightarrow \tan ^{2} \theta+\tan ^{2} \theta=2\)

\(\Rightarrow 2 \tan ^{2} \theta=2\)

\(\Rightarrow \tan ^{2} \theta=1\)

\(\Rightarrow \tan \theta=1\)

Now,

\(\cot \theta=\frac{1}{\tan \theta}\)

\(=1\)

\(L_1=\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}=\lambda\)

M(λ,1+2λ,2+3λ)

\(\overrightarrow{\mathrm{PM}}=(\lambda-1) \hat{\mathrm{i}}+(1+2 \lambda) \hat{\mathrm{j}}+(3 \lambda-5) \hat{\mathrm{k}}\)

\(\overrightarrow{\mathrm{PM}} \text { is perpendicular to line } \mathrm{L}_1\)

\( \overrightarrow{\mathrm{PM}} \cdot \overrightarrow{\mathrm{b}}=0 \quad(\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \\\)

\( \Rightarrow \lambda-1+4 \lambda+2+9 \lambda-15=0\)

14λ=14⇒λ=1

∴M=(1,3,5)

\( \vec{Q}=2 \vec{M}-\vec{P}[M \text { is midpoint of } \vec{P} \& \vec{Q}] \\\)

\( \vec{Q}=2 \hat{i}+6 \hat{j}+10 \hat{k}-\hat{i}-7 \hat{k} \\\)

\( \vec{Q}=\hat{i}+6 \hat{j}+3 \hat{k}\)

∴(α,β,γ)=(1,6,3)
Required line having direction cosine (l,m,n)
l2+m2+n2=1

\(\Rightarrow 1^2+\left(-\frac{1}{2}\right)^2+\left(-\frac{1}{\sqrt{2}}\right)^2=1\)

\(1^2=\frac{1}{4}\)

\(\therefore 1=\frac{1}{2}\) [Line make acute angle with x-axis]

Equation of line passing through (1,6,3) will be

\(\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}+6 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\mu\left(\frac{1}{2} \hat{\mathrm{i}}-\frac{1}{2} \hat{\mathrm{j}}-\frac{1}{\sqrt{2}} \hat{\mathrm{k}}\right)\)

We have to find the value of \(\sin ^{-1}\left(\frac{4}{5}\right)-\sin ^{-1}\left(\frac{3}{5}\right)\).

We know that: \(\sin ^{-1} {x}-\sin ^{-1} {y}=\sin ^{-1}\left({x} \sqrt{1-{y}^{2}}-{y} \sqrt{1-{x}^{2}}\right)\)

\(\therefore \sin ^{-1}\left(\frac{4}{5}\right)-\sin ^{-1}\left(\frac{3}{5}\right)=\sin ^{-1}\left(\left(\frac{4}{5} \sqrt{1-\left(\frac{3}{5}\right)^{2}}\right)-\left(\frac{3}{5} \sqrt{1-\left(\frac{4}{5}\right)^{2}}\right)\right)\)

\(=\sin ^{-1}\left(\left(\frac{4}{5} \sqrt{1-\frac{9}{25}}\right)-\left(\frac{3}{5} \sqrt{1-\frac{16}{25}}\right)\right)\)

\(=\sin ^{-1}\left(\left(\frac{4}{5} \sqrt{\frac{25-9}{25}}\right)-\left(\frac{3}{5} \sqrt{\frac{25-16}{25}}\right)\right)\)

\(=\sin ^{-1}\left(\left(\frac{4}{5} \sqrt{\frac{16}{25}}\right)-\left(\frac{3}{5} \sqrt{\frac{9}{25}}\right)\right)\)

\(=\sin ^{-1}\left(\left(\frac{4}{5} \sqrt{\left(\frac{4}{5}\right)^{2}}\right)-\left(\frac{3}{5} \sqrt{\left(\frac{3}{5}\right)^{2}}\right)\right)\)

\(=\sin ^{-1}\left(\left(\frac{4}{5} \times \frac{4}{5}\right)-\left(\frac{3}{5} \times \frac{3}{5}\right)\right)\)

\(=\sin ^{-1}\left(\left(\frac{16}{25}\right)-\left(\frac{9}{25}\right)\right)=\sin ^{-1}\left(\frac{7}{25}\right)\)

Given:

Numbers \(=\{12,15,16,14,18\}\)

Sum of numbers \(=12,15,16,14,18\)

\(\Rightarrow\) Sum \(=75\)

Mean = Sum of numbers / Number of observations

Variance \(=\left[\sum(\mathrm{x}-\overline{\mathrm{x}})\right] / \mathrm{n}\)

Standard deviation \(=\sqrt{\text { Variance }}\)

Where,

\(\mathrm x=\) number

\(\bar{\mathrm x}=\) mean of numbers

\(n=\) number of observations

Mean \(=75 / 5=15\)

Variance \(=\left[(12-15)^{2}+(15-15)^{2}+(16-15)^{2}+(14-15)^{2}+(18-15)^{2}\right] / 5\)

\(\Rightarrow\left(3^{2}+0^{2}+1^{2}+1^{2}+3^{2}\right) / 5\)

\(\Rightarrow 20 / 5\)

\(\Rightarrow 4\)

Standard deviation \(=\sqrt{4}\)

\(\therefore\) The standard deviation of the given set of numbers is 2.

Given,

\(A=\left|\begin{array}{ccc}a & b & c \\ b+c & c+a & a+b \\ a^{2} & b^{2} & c^{2}\end{array}\right|\)

Applying the column operations

\(\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}-\mathrm{C}_{2}\) and \(\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{3}\)

\(A=\left|\begin{array}{ccc}a-b & b-c & c \\ b-a & c-b & a+b \\ a^{2}-b^{2} & b^{2}-c^{2} & c^{2}\end{array}\right|\)

Taking common (a-b) and (b-c) From first and second column respectively.

\(\Rightarrow(a-b)(b-c)\left|\begin{array}{ccc}1 & 1 & c \\ -1 & -1 & a+b \\ a+b & b+c & c^{2}\end{array}\right|\)

Applying column operation \(C_{1} \rightarrow C_{1}-C_{2}\)

\(\Rightarrow(a-b)(b-c)\left|\begin{array}{ccc}0 & 1 & c \\ 0 & -1 & a+b \\ a-c & b+c & c^{2}\end{array}\right|\)

Now solving the determinant we get,

\(A=(a-b)(b-c)[(a-c) \times\{(a+b)-(-c)\}]\)

\(A=-(a-b)(b-c)(c-a)(a+b+c)\)