Mathematics Test - 26

Given, \(f(x)=2 x, g(x)=x^{2}+2\)

then, \((f+g)(2)=f(2)+g(2)\)

\(=(2 \times 2)+\left(2^{2}+2\right)\)

\(=4+6\)

\(=10\)

Given,

The line passing through the points \((3,2)\) and \((1,-1)\)

Equation of a line passing through \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\) is:

\(\frac{y-y_{1}}{x-x_{1}}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

Therefore,

\(\frac{y-2}{x-3}=\frac{-1-2}{1-3} \)\(\quad\quad(\because \frac{y-2}{x-3}=\frac{-1-2}{1-3} )\)

\(3 x-2 y-5=0\)

\(y=\frac{3}{2} x-\frac{5}{2} \)

\(\Rightarrow \text { Slope }\left(m_{1}\right)=\frac{3}{2} \text { and } c_{1}=\frac{-5}{2}\)

Now for the slope of the perpendicular line \(\left(\mathrm{m}_{2}\right)\)

\(\mathrm{m}_{1} \times \mathrm{m}_{2}=-1 \)

\(\Rightarrow \frac{3}{2} \times \mathrm{m}_{2}=-1 \)

\(\Rightarrow \mathrm{m}_{2}=\frac{-2}{3}\)

\( \int_0^1 \frac{1}{\sqrt{3+x}+\sqrt{1+x}} d x=\int_0^1 \frac{\sqrt{3+x}-\sqrt{1+x}}{(3+x)-(1+x)} d x \\\)

\( \frac{1}{2}\left[\int_0^1 \sqrt{3+x} d x-\int_0^1(\sqrt{1+x}) d x\right] \\\)

\( \frac{1}{2}\left[2 \frac{(3+x)^{\frac{3}{2}}}{3}-\frac{2(1+x)^{\frac{3}{2}}}{3}\right]_0^1 \\\)

\( \frac{1}{2}\left[\frac{2}{3}(8-3 \sqrt{3})-\frac{2}{3}\left(2^{\frac{3}{2}}-1\right)\right] \\\)

\( \frac{1}{3}[8-3 \sqrt{3}-2 \sqrt{2}+1] \\\)

\( =3-\sqrt{3}-\frac{2}{3} \sqrt{2}=a+b \sqrt{2}+c \sqrt{3} \\\)

\( a=3, b=-\frac{2}{3}, c=-1 \\\)

\( 2 a+3 b-4 c=6-2+4=8\)

\(\begin{aligned} & \operatorname{Lt}_{n \rightarrow \infty} \sqrt{\frac{d}{n}}\left(\frac{\sqrt{a_1}-\sqrt{a_2}}{a_1-a_2}+\frac{\sqrt{a_2}-\sqrt{a_3}}{a_2-a_3}+\ldots . .+\frac{\sqrt{a_{n-1}}-\sqrt{a_n}}{a_{n-1}-a_n}\right) \\ & =\operatorname{Lt}_{n \rightarrow \infty} \sqrt{\frac{d}{n}}\left(\frac{\sqrt{a_1}-\sqrt{a_2}+\sqrt{a_2}+\sqrt{a_3}+\ldots . .+\sqrt{a_{n-1}}-\sqrt{a_n}}{-d}\right) \\ & =\operatorname{Lt}_{n \rightarrow \infty} \sqrt{\frac{d}{n}}\left(\frac{\sqrt{a_n}-\sqrt{a_1}}{d}\right) \\ & =\operatorname{Lt}_{n \rightarrow \infty} \frac{1}{\sqrt{n}}\left(\frac{\sqrt{a_1+(n-1) d}-\sqrt{a_1}}{\sqrt{d}}\right) \\ & =\operatorname{Lt}_{n \rightarrow \infty} \frac{1}{\sqrt{d}}\left(\sqrt{\frac{a_1}{n}+d-\frac{d}{n}}-\frac{\sqrt{a_1}}{n}\right) \\ & =1\end{aligned}\)

\(2 \tan ^2 \theta-5 \sec \theta-1=0 \\\)

\( \Rightarrow 2 \sec ^2 \theta-5 \sec \theta-3=0 \\\)

\( \Rightarrow(2 \sec \theta+1)(\sec \theta-3)=0 \\\)

\( \Rightarrow \sec \theta=-\frac{1}{2}, 3 \\\)

\( \Rightarrow \cos \theta=-2, \frac{1}{3} \\\)

\( \Rightarrow \cos \theta=\frac{1}{3}\)

For 7 solutions \(n=13\)

\(\text { So, } \sum_{\mathrm{k}=1}^{13} \frac{\mathrm{k}}{2^{\mathrm{k}}}=\mathrm{S} \text { (say) } \\\)

\( \mathrm{S}=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\ldots+\frac{13}{2^{13}} \\\)

\( \frac{1}{2} \mathrm{~S}=\frac{1}{2^2}+\frac{1}{2^3}+\ldots+\frac{12}{2^{13}}+\frac{13}{2^{14}} \\\)

\(\Rightarrow \frac{\mathrm{S}}{2}=\frac{1}{2} \cdot \frac{1-\frac{1}{2^{13}}}{1-\frac{1}{2}}-\frac{13}{2^{14}} \Rightarrow \mathrm{S}=2 \cdot\left(\frac{2^{13}-1}{2^{13}}\right)-\frac{13}{2^{13}}\)

We know mirror image of point \(\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)\) in the plane \(\mathrm{ax}+\mathrm{by}+\mathrm{cz}=\mathrm{d}\)

\(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=\frac{-2\left(a x_1+b y_1+c z_1-d\right)}{a^2+b^2+c^2}\)

Here given point \((2,4,7)\) and plane \(3 x-y+4 z=2\) then mirror image is

\(\frac{x-2}{3}=\frac{y-4}{-1}=\frac{z-7}{4}=\frac{-2(6-4+28-2)}{9+1+16} \\\)

\(\Rightarrow \frac{x-2}{3}=\frac{y-4}{-1}=\frac{z-7}{4}=-\frac{28}{13} \\\)

\(\therefore x=-\frac{58}{13}=a \\\)

\(y=\frac{80}{13}=b \\\)

\(z=-\frac{21}{13}=c \\\)

\(\therefore 2 a+b+2 c \\\)

\(=2\left(-\frac{58}{13}\right)+\frac{80}{13}+2\left(-\frac{21}{13}\right) \\\)

\(=\frac{-116+80-42}{13}=\frac{-78}{13}=-6\)

Given,

\(A=\left[\begin{array}{ccc}3 & 4 & 9 \\ 11 & 6 & 7 \\ 8 & 9 & 5\end{array}\right]\)

\(|2 \mathrm{~A}|=\mathrm{k}\)

\(\Rightarrow|\mathrm{A}|=3 \times(30-63)-4 \times(55-56)+9 \times(99-48)\)

\(\Rightarrow|\mathrm{A}|=-99+4+459=364\)

We know that,

If \(\mathrm{A}\) is a matrix of order \(\mathrm{n}\), then \(|\mathrm{k} \cdot \mathrm{A}|=\mathrm{k}^{n} \cdot|\mathrm{A}|\)

where \(\mathrm{k} \in \mathrm{R}\)

\(\Rightarrow|2 \mathrm{~A}|=2^{3} \cdot 364=2912\)

The optimal value of the objective function is attained at the points given by corner points of the feasible region.Any point in the feasible region that gives the optimal value (maximum or minimum) of the objective function is called an optimal solution.The optimal value of the objective function \(Z=a x+by\) may or may not exist, if the feasible region for a LPP is unbounded.

P - Q = The set of elements which belong to P but not to Q.

P U Q = The set which consists of all elements of P and all elements of Q, the common elements being taken only once.

P∩ Q = The set of all elements which are common to both Pand Q.

let us supposeP = {1, 2, 3, 4} and Q = {3, 4, 5, 6}

⇒ P - Q = {1, 2}

⇒ Q - P = {5, 6}

⇒ P U Q = {1, 2, 3, 4, 5, 6}

⇒ P∩ Q = {3, 4}

⇒(P - Q) ∪ (Q - P) ∪ (P ∩ Q) = {1, 2} U {5, 6} U {3, 4} = {1, 2, 3, 4, 5, 6} = P U Q