TS EAMCET 2018 :: Mathematics :: General questions

• Mathematics - General questions

If a,b,c are  distinct  real numbers are P,Q, R are three points  whose position  vectors are  respectively 

$a\hat{i}+b\hat{j}+c\hat{k},b\hat{i}+c\hat{j}+a\hat{k}$ and $c\hat{i}+a\hat{j}+b\hat{k}$  , then $\angle QPR$=

Answer:

Explanation:

position vector of P= $a\hat{i}+b\hat{j}+c\hat{k}$

Position vector of Q= $b\hat{i}+c\hat{j}+a\hat{k}$

Position vector R=$c\hat{i}+a\hat{j}+b\hat{k}$

Now, PQ= PV of Q-PV of P

 =($b\hat{i}+c\hat{j}+a\hat{k}$)-($a\hat{i}+b\hat{j}+c\hat{k}$)

=$(b-a)\hat{i}+(c-b)\hat{j}+(a-c)\hat{k}$

 Similarly, PR= $(c-a)\hat{i}+(a-b)\hat{j}+(b-c)\hat{k}$

Now, PQ.PR=[$(b-a)\hat{i}+(c-b)\hat{j}+(a-c)\hat{k}$].[$(c-a)\hat{i}+(a-b)\hat{j}+(b-c)\hat{k}$]

    =(b-a)(c-a)+(c-b)(a-b)+(a-c)(b-c)

=$a^{2}+b^{2}+c^{2}-ab-bc-ca$

 |PQ|= $\sqrt{(b-a)^{2}+(c-b)^{2}+(a-c)^{2}}$

|PR|= $\sqrt{(c-a)^{2}+(a-b)^{2}+(b-c)^{2}}$

Now, $\cos \theta $=$\frac{PQ.PR}{|PQ|.|PR|}$

 = $\frac{a^{2}+b^{2}+c^{2}-ab-bc-ca}{2a^{2}+2b^{2}+2c^{2}-2ab-2bc-2ca}$

 = $\frac{a^{2}+b^{2}+c^{2}-ab-bc-ca}{2(a^{2}+b^{2}+c^{2}-ab-bc-ca)}$

$\cos \theta =\frac{1}{2}$

 $\therefore$  $\theta$= $\frac{\pi}{3}$

 Here, $\angle QPR$=$\frac{\pi}{3}$

$\cos \left(\frac{\pi}{7}\right)\cos \left(\frac{2\pi}{7}\right)\cos \left(\frac{4\pi}{7}\right)=$

Answer:

Explanation:

We have

$\cos \left(\frac{\pi}{7}\right)\cos \left(\frac{2\pi}{7}\right)\cos \left(\frac{4\pi}{7}\right)$

 Let A =$\frac{\pi}{7}\Rightarrow \pi=7A$

  = $\cos A \cos 2A \cos 4A= \cos A \cos 2A \cos 2^{2} A$

 Now, we will use the formula

  $\cos A \cos 2A \cos 2^{2} A.......\cos ^{n-1} A= \frac{\sin  2^{n} A}{2^{n} \sin A}$

$\therefore$    $\cos A \cos2A\cos 2^{2}A=\frac{\sin  2^{3} A}{2^{3} \sin A}$

=$\frac{ \sin 8 A}{8 \sin A}= \frac{ \sin(7A+A)}{8 \sin A}= \frac{\sin(\pi+A)}{8 \sin A}$

                                                                      [$\because 7A=\pi]$

 =$\frac{-\sin A}{8 \sin A}$            $[\because \sin(\pi+\theta)=-\sin \theta)$

   = $-\frac{1}{8}$

If   $x=\frac{2.5}{3.6}-\frac{2.5.8}{3.6.9}\left(\frac{2}{5}\right)+\frac{2.5.8.11}{3.6.9.12}\left(\frac{2}{5}\right)^{2}-... \infty$  then

$7^{2}(12x+55)^{3}$=

Answer:

Explanation:

$x=\frac{2.5}{3.6}-\frac{2.5.8}{3.6.9}\left(\frac{2}{5}\right)+\frac{2.5.8.11}{3.6.9.12}\left(\frac{2}{5}\right)^{2}-... \infty$

$=\frac{2.5}{3^{2}.2!}-\frac{2.5.8}{3^{3}.3!}\left(\frac{2}{5}\right)+\frac{2.5.8.11}{3^{4}.4!}\left(\frac{2}{5}\right)^{2}$

mulitply with $(\frac{2}{5})^{2}$

$\frac{4}{25}x=\frac{2.5}{3^{2}.2!}\left(\frac{2}{5}\right)^{2}-\frac{2.5.8}{3^{3}.3!}\left(\frac{2}{5}\right)^{3}+......\infty$

$\frac{4}{25}x-\frac{2.5}{3^{}.1!}\left(\frac{2}{5}\right)^{}+1=1-\frac{2}{3^{}.1!}\left(\frac{2}{5}\right)^{}+\frac{2.5}{3^{2}.2!}\left(\frac{2}{5}\right)^{2}-\frac{2.5.8}{3^{3}.3!}\left(\frac{2}{5}\right)^{3}+......$

$\frac{4x}{25}-\frac{4}{15}+1=\left(1+\frac{2}{5}\right)^{-2/3}$

$\Rightarrow$    $\frac{4x}{25}+\frac{11}{15}=\frac{1}{5}\left(\frac{7}{5}\right)^{2/3}\Rightarrow\left(\frac{4x}{5}+\frac{11}{3}\right)=\left(\frac{7}{5}\right)^{-2/3}$

$\Rightarrow$    $\frac{1}{5}\left(\frac{12x+55}{15}\right)=\left(\frac{7}{5}\right)^{-2/3}\Rightarrow\frac{12x+55}{75}=\left(\frac{5}{7}\right)^{2/3}$

cube both sides,

$\Rightarrow \frac{(12x+55)^{3}}{(79)^{3}}=\frac{5^{2}}{7^{2}}\Rightarrow 7^{2}(12x+55)^{3}=75^{3}.5^{2}$

$=3^{3}.5^{6}.5^{2}=3^{3}.5^{8}$

The number of ways  in which four letters can be put in four addressed envelops so that no letter goes  into envelope  meant for it is

Answer:

Explanation:

Let n=4 be number of envelopes in which number letters goes into envelope meants for it

Then, the number of ways =4!= $\sum_{k=1}^{4}{^{4}}{C}_{k}$

[ $\because $ required number of ways = n!= $\sum_{k=1}^{n}{^{n}}{C}_{k}$

   =$24-[^{4}C_{1}+^{4}C_{2}+^{4}C_{3}+^{4}C_{4}]$

  =24-[4+6+4+1]=24-15=9

  Hence , required number of ways =9

If $\alpha_{1},\alpha_{2},.......,\alpha_{n}$  are the roots of $x^{n}+px+q=0$,  then ($\alpha_{n}-\alpha_{1})(\alpha_{n}-\alpha_{2})...............(\alpha_{n}-\alpha_{n-1})=$

Answer:

Explanation:

We have,

  $x^{n}+px+q=0$,

 If   $\alpha_{1},\alpha_{2},.......,\alpha_{n}$    are roots of given equation so,

$x^{n}+px+q=(x-\alpha_{1})(x-\alpha_{2} )(x-\alpha_{3}).......(x-\alpha_{n})$

$\Rightarrow$    $\frac{x^{n}+px+q}{x-\alpha_{n}}=$  $(x-\alpha_{1})(x-\alpha_{2} )(x-\alpha_{3}).......(x-\alpha_{n-1})$

$\therefore$  $\lim_{x \rightarrow \alpha_{n}}\frac{x^{n}+px+q}{x-\alpha_{n}}=\lim_{x \rightarrow \alpha_{n}}$ $(x-\alpha_{1})(x-\alpha_{2}) (x-\alpha_{3}).......(x-\alpha_{n-1})$

$\lim_{x \rightarrow \alpha_{n}}\frac{nx^{n-1}+P}{1}=(\alpha_{n}-\alpha_{1})(\alpha_{n}-\alpha_{2})......(\alpha_{n}-\alpha_{n-1})$

 $\Rightarrow$      $n \alpha_{n}^{n-1}+p=(\alpha_{n}-\alpha_{1})(\alpha_{n}-\alpha_{2})......(\alpha_{n}-\alpha_{n-1})$

• Mathematics - General questions