Mathematics Test - 28

Let \(\mathrm{P}(\mathrm{n}): 3^{\mathrm{n}}>2^{\mathrm{n}}\)

Step (i) for \(\mathrm{n}=1\)

\(\mathrm{P}(1): 3>2\)

which is true for \(\mathrm{n}=1\)

Step (ii)

Let it is true for \(\mathrm{n}=\mathrm{k}\)

So, \(3^{\mathrm{k}}>2^{\mathrm{k}}\)

Step (iii)

We have to prove that it is true for \(\mathrm{n}=\mathrm{k}+1\) using step (ii)

Now \(3^{\mathrm{k}+1}=3^{\mathrm{k}} \times 3\)

\(3^{\mathrm{k}+1}>2^{\mathrm{k}} \times 3\) using step (ii)

\(3^{\mathrm{k}+1}>2^{\mathrm{k}}(2+1)\)

\(3^{\mathrm{k}+1}>2^{\mathrm{k}+1}+2^{\mathrm{k}}\)

Therefore, \(3^{\mathrm{k}+1}>2^{\mathrm{k}+1}\)

So it is true for \(\mathrm{n}=\mathrm{k}+1\)

Therefore, it is true for all \(\mathrm{n} \in \mathrm{N}\)

Scalar product of \(\vec{a}=-2 \hat{i}+3 \hat{j}+11 \hat{k}\) and \(\vec{ b }=5 \hat{ i }+7 \hat{ j }+-4 \hat{ k }\) is given by,

\(\vec{a} \cdot \vec{b}=(-2 \hat{i}+3 \hat{j}+11 \hat{k}) \cdot(5 \hat{i}+7 \hat{j}-4 \hat{k}) \)

\(=-2 \cdot 5+3 \cdot(7)+11.(-4) \)

\(=-10+21-44[\hat{i}.\hat{i}=1, \hat{j} \cdot \hat{j}=1 , \hat{k} \cdot \hat{k}=1] \)

\(=-54+21 \)

\(=-33\)

Given that

\(\mathrm{P}=\frac{1}{4} \Rightarrow \mathrm{q}=1-\frac{1}{4}=\frac{3}{4} \)

\(\text { and } \mathrm{P}(\mathrm{x} \geq 1) \geq \frac{9}{10} \)

\(\Rightarrow 1-\mathrm{P}(\mathrm{x}=0) \geq \frac{9}{10} \)

\(\Rightarrow 1-{ }^{\mathrm{n}} \mathrm{C}_0\left(\frac{1}{4}\right)^0\left(\frac{3}{4}\right)^{\mathrm{n}} \geq \frac{9}{10} \)

\(\Rightarrow 1-\frac{9}{10} \geq\left(\frac{3}{4}\right)^{\mathrm{n}} \)

\( \Rightarrow\left(\frac{3}{4}\right)^{\mathrm{n}} \leq\left(\frac{1}{10}\right)\)

Taking log at the base \(3 / 4\), on both sides, we get

\( n \log _{3 / 4}\left(\frac{3}{4}\right) \leq \log _{3 / 4}\left(\frac{1}{10}\right) \)

\(\Rightarrow \mathrm{n} \geq-\log _{3 / 4} 10=\frac{-\log _{10} 10}{\log _{10}\left(\frac{3}{4}\right)}=\frac{-1}{\log _{10} 3-\log _{10} 4} \)

\(\Rightarrow \mathrm{n} \geq \frac{1}{\log _{10} 4-\log _{10} 3}\)

Given a parallelogram \({OABC}\) such that \(\overrightarrow{{OC}}=\overrightarrow{{a}}\) and \(\overrightarrow{{AB}}=\overrightarrow{{b}}\). Then,

\(\overrightarrow{{OB}}+\overrightarrow{{BC}}=\overrightarrow{{OC}}\)

\(\Rightarrow\overrightarrow{OB}=\overrightarrow{OC}-\overrightarrow{BC}\)

\(\Rightarrow \overrightarrow{{OB}}=\overrightarrow{{OC}}-\overrightarrow{{OA}}\quad[\because \overrightarrow{{BC}}=\overrightarrow{{OA}}] \)

\(\Rightarrow \overrightarrow{{OB}}=\overrightarrow{{a}}-\overrightarrow{{OA}} \)...(1)

Therefore,

\(\overrightarrow{{OA}}+\overrightarrow{{AB}}=\overrightarrow{{OB}}\)

\(\Rightarrow \overrightarrow{{OA}}+\overrightarrow{{b}}=\overrightarrow{{a}}-\overrightarrow{{OA}}\quad[\) Using (1) \(]\)

\(\Rightarrow 2\overrightarrow{{OA}}=\overrightarrow{{a}}-\overrightarrow{{b}}\)

\(\Rightarrow \overrightarrow{{OA}}=\frac{1}{2}(\overrightarrow{{a}}-\overrightarrow{{b}})\)

Given quadratic equation is

\( x(x+1)+(x+1)(x+2)+\ldots+(x+\overline{ n -1})(x+ n )=10 n\)

\( \Rightarrow\left(x^2+x^2+\ldots+x^2\right)+[(1+3+5)+\ldots+(2 n-1)] x+[1.2+2.3+\ldots+(n-1) n]=10 n \)

\( \Rightarrow n x^2+n^2 x+\frac{n\left(n^2-1\right)}{3}-10 n=0 \)

\( \Rightarrow 3 x^2+3 n x+n^2-31=0 \)

\( \text { Let } a \text { and } \beta \text { be the roots. } \)

\( \text { Since, } a \text { and } \beta \text { are consecutive: } \)

\( \therefore|a-\beta|=1 \Rightarrow(\alpha-\beta)^2=1 \)

\( \text { Again, }(a-\beta)^2=(a+\beta)^2-4 a \beta \)

\( \Rightarrow 1=\left(\frac{-3 n}{3}\right)^2-4\left(\frac{n^2-31}{3}\right) \)

\( \Rightarrow 1=n^2-\frac{4}{3}\left(n^2-31\right) \Rightarrow 3=3 n^2-4 n^2+124 \)

\( \Rightarrow n^2=121 \Rightarrow n=\pm 11 \)

\( \therefore n=11[\because n>0]\)

Given,

\(\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=\pi\)

\(\cos ^{-1}(x)+\cos ^{-1}(y)=\pi-\cos ^{-1}(z)\)

\(\cos ^{-1}\left(x y-\sqrt{1-x^2} \sqrt{1-y^2}\right)=\pi-\cos ^{-1}(z)\)

Taking cos on both the sides, we get

\(x y-\sqrt{1-x^2} \sqrt{1-y^2}=-z \)

\(x y+z=\sqrt{1-x^2} \sqrt{1-y^2}\)

Squaring on both the sides, we get

\(x^2 y^2+z^2+2 x y z=1-y^2-x^2+x^2 y^2 \)

\(x^2+y^2+z^2+2 x y z=1\)

Given, \(z_{1}=2-i\) and \(z_{2}=1+{i}\)

\(\therefore\left|\frac{z_{1}+z_{2}+1}{z_{1}-z_{2}+1}\right|=\left|\frac{(2-i)+(1+i)+1}{(2-i)-(1+i)+1}\right|\)

\(=\left|\frac{4}{2-2 i}\right|=\left|\frac{4}{2(1-i)}\right|\)

\(=\left|\frac{2}{1-i} \times \frac{1+i}{1+i}\right|=\left|\frac{2(1+i)}{1^{2}-i^{2}}\right|\)

\(=\left|\frac{2(1+i) }{1+1}\right| \quad\quad \left[\because i^{2}=-1\right]\)

\(=\left|\frac{2(1+i)}{2}\right|\)

\(=|1+i|=\sqrt{1^{2}+1^{2}}=\sqrt{2}\)

\(\begin{aligned} & \text { Given, } x=\frac{1}{2}, y=\frac{-1}{4} \Rightarrow x y=\frac{-1}{8} \\ & y \cdot \frac{1 \cdot(-2 x)}{2 \sqrt{1-x^2}}+y^{\prime} \sqrt{1-x^2} \\ & =-\left\{1 \cdot \sqrt{1-y^2}+\frac{x \cdot(-2 y)}{2 \sqrt{1-y^2}} y^{\prime}\right\} \\ & \Rightarrow-\frac{x y}{\sqrt{1-x^2}}+y^{\prime} \sqrt{1-x^2}=-\sqrt{1-y^2}+\frac{x y \cdot y^{\prime}}{\sqrt{1-y^2}} \\ & \Rightarrow y^{\prime}\left(\sqrt{1-x^2}-\frac{x y}{\sqrt{1-y^2}}\right)=\frac{x y}{\sqrt{1-x^2}}-\sqrt{1-y^2} \\ & \Rightarrow \mathrm{y}^{\prime}\left(\frac{\sqrt{3}}{2}+\frac{1}{8 \cdot \frac{\sqrt{15}}{4}}\right)=\frac{-1}{8 \cdot \sqrt{\frac{3}{2}}}-\frac{\sqrt{15}}{4} \\ & \Rightarrow \mathrm{y}^{\prime}\left(\frac{\sqrt{45}+1}{2 \sqrt{15}}\right)=-\frac{(1+\sqrt{45})}{4 \sqrt{3}} \\ & \therefore \mathrm{y}^{\prime}=-\frac{\sqrt{5}}{2} \\ & \end{aligned}\)