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$\lim_{x \rightarrow 0}\frac{(1-\cos 2x)(3+\cos x)}{x \tan 4x}$  is equal to  

Answer:

Explanation:

 We have

  $\lim_{x \rightarrow 0}\frac{(1-\cos 2x)(3+\cos x)}{x \tan4x}$

   =  $\lim_{x \rightarrow 0}\frac{(2\sin ^{2}x(3+\cos x))}{x \times \frac{\tan 4x}{4x}\times 4x}$

   =   $\lim_{x \rightarrow 0}\frac{2\sin ^{2}x}{x^{2}}\times\lim_{x \rightarrow 0}\frac{(3+\cos x)}{4} \times\frac{1}{\lim_{x \rightarrow 0}\frac{\tan 4x}{4x}}$

 $=2\times \frac{4}{4}\times1$      $\left( \therefore \lim_{\theta \rightarrow 0}\frac{\sin \theta}{\theta}=1 and \lim_{\theta \rightarrow 0}\frac{\tan \theta}{\theta}=1\right)$

=2

 The sum of first 9 terms of the series $\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\frac{1^{3}+2^{3}+3^{3}}{1+3+5}+...is$

Answer:

Explanation:

Central Idea. Write  the nth term of the given  series and simplify it to get its lowest forth. Then , apply , S_n = ∑ Tπ

Given series is 

  $\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\frac{1^{3}+2^{3}+3^{3}}{1+3+5}+...\infty$

  Let T_n  be the nth term of the given series

   $\therefore$    $\frac{1^{3}+2^{3}+3^{3}+....+n^{3}}{1+3+5+.....+n terms}$

    =    $\frac{\left\{\frac{n(n+1)}{2}\right\}^{2}}{n^{2}}$

     = $\frac{(n+1)^{2}}{4}$

$S_{9}=\sum_{n=1}^{9}\frac{(n+1)^{2}}{4}=\frac{1}{4}[(2^{2}+3^{2}+.....+10^{2})$

                                                                     +  $1^{2}-1^{2}]$

  $=\frac{1}{4}\left[\frac{10(10+1)(20+1)}{6}-1\right]=\frac{384}{4}=96$

If m is the AM of two distinct real numbers l and n (l,n>1) and G₁, G₂ and G₃  are three geometric means between  l and n , then $G_{1}^{4}+2G_{2}^{4}+G^{4}_{3}$  equals 

Answer:

Explanation:

Given, m is the AM of l and n 

$\therefore$    l+n=2m          .......(i)

 and G₁, G₂, G₃  are geometric means between l

 and n

$\therefore$ l, G₁, G₂, G_3, n are in GP.

 Let r be the common ration of this GP

    $\therefore$        $G_{1}=lr^{}$

                              $G_{2}=lr^{2}$

                               $G_{3}=lr^{3}$

                                $n=lr^{4}$

$\Rightarrow$     $r=\left(\frac{n}{l}\right)^{1/4}$

Now,  $G_{1}^{4}+2G_{2}^{4}+G^{4}_{3}= (lr)^{4}+2(lr^{2})^{4}+(lr^{3})^{4}$

            =  $l^{4}\times r^{4 }(1+2r^{4}+r^{8})$

          =  $l^{4}\times r^{4 }(r^{4}+1)^{2}$

      $=l^{4}\times\frac{n}{l}\left( \frac{n+l}{l}\right)^{2}$

          $=ln\times4m^{2}=4lm^{2}n$

The sum of coefficients  of integral powers of x in the binomial  expansion of  $(1-2\sqrt{x})^{50}$  is 

Answer:

Explanation:

 Let   $T_{r+1}$ be the general term in the expansion  of  $(1-2\sqrt{x})^{50}$

  $\therefore$  $T_{r+1}$  =   $^{50}C_{r}(1)^{50-r}(-2x^{1/2})^{r}$

                                            = $^{50}C_{r}2^{r}x^{1/2}(-1)^{r}$

 For the integral power of x, r should be  even integer.

  $\therefore$   Sum of coefficients=   $\sum_{r=0}^{25}$    $^{50}C_{2r}(2)^{2r}$

              =  $\frac{1}{2}[(1+2)^{50}+(1-2)^{50}]$

            =  $\frac{1}{2}[3^{50}+1]$

Alter:

We have ,

 $(1-2\sqrt{x})^{50}=C_{0}-C_{1}2\sqrt{x}+C_{2}(2\sqrt{x})^{2}+...+C_{50}(2\sqrt{x})^{50}......(i)$

  $(1+2\sqrt{x})^{50}=C_{0}+C_{1}2\sqrt{x}+C_{2}(2\sqrt{x})^{2}+...+C_{50}(2\sqrt{x})^{50}......(ii)$

 On adding Eqs.(i) and (ii) , we get

   $(1-2\sqrt{x})^{50}+(1+2\sqrt{x})^{50}=2[C_{0}+C_{2}(2\sqrt{x})^{2}+...+C_{50}(2\sqrt{x})^{50}]$

$\Rightarrow   \frac{(1-2\sqrt{x)}^{50}+(1+2\sqrt{x})^{50}}{2}=C_{0}$

                                                    $+C_{2}(2\sqrt{x})^{2}+......+C_{50}(2\sqrt{x})^{50}$

   On putting  x=1, we get

  $\frac{(1-2\sqrt{1})^{50}+(1+2\sqrt{1})^{50}}{2}=C_{0}+C_{2}(2)^{2}+...+C_{50}(2)^{50}$

    $\Rightarrow \frac{(-1)^{50}+(3)^{50}}{2}=C_{0}+C_{2}(2)^{2}+......+C_{50}(2)^{50}$

$\Rightarrow \frac{1+3^{50}}{2}=C_{0}+C_{2}(2)^{2}+....+C_{50}(2)^{50}$

 The number of integers greater than 6000 that can be formed. using the digits 3,5,6,7 and 8 without repetition , is

Answer:

Explanation:

The integer greater than  6000 may be of 4 digits or 5 digit. So here two cases arise.

 Case I.

      When number is of 4 digit. Four-digit number can start from 6,7 or 8

 Thus, total number of 4 digit number. which are greater than 6000 =   $3\times4\times3\times2=72$

 Cae II. When number is of 5 digit

  Total number of five digit number which are greater than 6000=5!= 120

     $\therefore$  Total number of integers = 72+120=192

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