Mathematics Test - 9

\(\overrightarrow{a}=2 \hat{i}+\hat{j}-3 \hat{k}\) and \(\overrightarrow{b}=3 \hat{i}-2 \hat{j}-\hat{k}\)

\(\vec{a} \cdot \vec{b}=(2 \hat{i}+\hat{j}-3 \hat{k})(3 \hat{i}-2 \hat{j}-\hat{k}) \)

\(=6-2+3=7 \ldots .(1) \)

\(|\vec{a}|=\sqrt{2^{2}+1^{2}+(-3)^{2}}=\sqrt{14} \ldots(2) \)

\(|\vec{b}|=\sqrt{3^{2}+(-2)^{2}+(-1)^{2}}=\sqrt{14} \ldots .(3) \)

\(\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}| \cdot|\vec{b}|}\)

Put the values from (1), (2) and (3) in above equation:

\(=\frac{7}{\sqrt{14}. \sqrt{14}}=\frac{1}{2} \)

\(\cos \theta=\cos 60^{\circ} , \theta=60^{\circ}\)

Maximize:

\(\mathrm{Z}=(\mathrm{x}+\mathrm{y})\)

\(\Rightarrow \mathrm{x}-\mathrm{y} \leq-1, \mathrm{y}-\mathrm{x} \leq 0, \mathrm{x}, \mathrm{y} \leq 0\)

Put \(\mathrm{x}=0\) in \(\mathrm{x}-\mathrm{y}=-1\)

\(\mathrm{y}=1\)

Put \(\mathrm{x}=1\) in equation (i)

\(\mathrm{y}=2 \)

Put \(\mathrm{x}=-1 \)

\(-1-\mathrm{y}=-1 \)

\(\mathrm{y}=0 \)

\(\mathrm{y}-\mathrm{x}=0 \)

Put \(\mathrm{x}=0 \text { in equation (ii) } \)

\(\mathrm{y}=0 \)

Put \(\mathrm{x}=1 \)

\(\mathrm{y}=1 \)

Put \(\mathrm{x}=2 \)

\(\mathrm{y}=2 \)

Put \(\mathrm{x}=-1 \)

\(\mathrm{y}=-1\)

No feasible region, hence, no maximum value.

Given:

\(\frac{(\sin x+\cos x+1)\left(\sin x+\cos {x}-1\right)}{(\tan x+\cot x)\left(\sin ^{2} x \cos ^{2} x\right)}\)

\(=\frac{(\sin x+\cos x)^{2}-1}{\left(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right)\left(\sin ^{2} x \cos ^{2} x\right)}\)

\(=\frac{\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x-1}{\frac{\sin ^{2} x+\cos x}{\sin x \cdot \\cos ^{2} x} \times \sin ^{2} x \cos ^{2} x}\)

\(=\frac{2 \sin x \cos x}{\sin x \cos x}\)

\(=2\)

The general equation of sphere is given as:

\(x^2+y^2+z^2+2 u x+2 v y+2 w z+d=0\)........(1)

centre at \((- u ,- v ,- w )\) and

radius \(r=\sqrt{u^2+v^2+w^2-d}\)..........(2)

Given:

\(x^2+y^2+z^2-x+z-2=0\)

Compare given equation with equation (1)

\( x^2+y^2+z^2+2 u x+2 v y+2 w z+d=0 \)

\( u=-0.5, v=0, w=0.5, d=-2\)

Radius can be calculated from equation (2)

\(=\sqrt{\left(-\frac{1}{2}\right)^2+0^2+\left(\frac{1}{2}\right)^2+2}\)

\( =\sqrt{\frac{5}{2}}\)

Given,

\(\sin ^{4} x+2 \cos ^{4} x=\frac{2}{3}\)

\(\Rightarrow \sin ^{4} x+\cos ^{4} x+\cos ^{4} x=\frac{2}{3}\)

\(\Rightarrow 1-2 \sin ^{2} x \cos ^{2} x+\cos ^{4} x=\frac{2}{3}\)

\(\Rightarrow 1-2\left(1-\cos ^{2} x\right) \cos ^{2} x+\cos ^{4} x=\frac{2}{3}\)

\(\Rightarrow 1-2 \cos ^{2} x+2 \cos ^{4} x+\cos ^{4} x=\frac{2}{3}\)

\(\Rightarrow 9 \cos ^{4} x-6 \cos ^{2} x+1=0\)

\(\Rightarrow\left(3 \cos ^{2} x-1\right)^{2}=0\)

\(\Rightarrow \cos ^{2} x=\frac{1}{3}\)

\(\sec ^{2} x=3\)

\(\therefore\) The value of \(\sec ^{2} x\) is \(3\).

Equation of an ellipse is:

\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)

Given, \(x=3(\cos t+\sin t) \)

\(y=4(\cos t-\sin t) \)

\(\left(\frac{x}{3}\right)^{2}=1+\sin 2 t \ldots(1)\)

\(\left(\frac{y}{4}\right)^{2}=1-\sin 2 t \ldots(2)\)

Adding equation (1) and (2), we get

\(\frac{x^{2}}{9}+\frac{y^{2}}{16}=2\)

Thus, the given curve represents an ellipse.

When, a coin is tossed

Outcome \(=\mathrm{S}=\{\mathrm{H}, \mathrm{T}\}\)

When a die is rolled,

\(S=\{1,2,3,4,5,6\}\)

Now, a coin is tossed and a die is rolled, and we need only head on the coin

\(\text { So, } \mathrm{S}=\{(\mathrm{H}, 1),(\mathrm{H}, 2),(\mathrm{H}, 3),(\mathrm{H}, 4),(\mathrm{H}, 5),(\mathrm{H}, 6)\} \)

\(\mathrm{n}(\mathrm{S})=6\)

As we know,

\(i=\sqrt{-1} \)

\(i^{2}=-1 \)

\(i^{3}=-i \)

\(i^{4}=1 \)

\(i^{4 n}=1\)

Given,

\((-1 \sqrt{-1})^{4 n+5}\) where \(n\) is a positive integer

\(=(-1 \times i)^{4 n+5} \)

\(=(-i)^{4 n+5} \)

\(=(-i)^{4 n} \times(-i)^{5} \)

\(=(-i)^{4 n} \times(-i)^{4} \times(-i)^{1} \)

\(=1 \times 1 \times(-i) \)

\(=-i\)

\(I=\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin x+\cos x} d x\) .....(i)

Using identity, \(\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}=\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{a}+\mathrm{b}-\mathrm{x}) \mathrm{d} \mathrm{x}\)

\(I=\int_{0}^{\frac{\pi}{2}} \frac{\sin \left(\frac{\pi}{2}-\mathrm{x}\right)}{\sin \left(\frac{\pi}{2}-\mathrm{x}\right)+\cos \left(\frac{\pi}{2}-\mathrm{x}\right)} \mathrm{dx}\)

\(\Rightarrow I=\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\cos x+\sin x} d x\) ....(ii)

On adding eqn. (i) and (ii), we get

\(2 I=\int_{0}^{\frac{\pi}{2}} \frac{\sin x+\cos x}{\sin x+\cos x} d x\)

\(\Rightarrow 2 I=\int_{0}^{\frac{\pi}{2}} d x\)

\(\Rightarrow 2 I=\frac{\pi}{2}\)

\(I=\frac{\pi}{4}\)

As we know,

\(i=\sqrt{-1} \)

\(i^{2}=-1 \)

\(i^{4 n}=1\)

Given,

\(i^{1325} \)

\(=i^{(1324+1)} \)

\(=\left(i^{4}\right)^{331} \times i \)

\(=1 \times i \quad\left[\because i^{4 n}=1\right] \)

\(=i \)

\(\therefore i^{1325}=i\)