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For $x\epsilon R$ , $f(x)=\mid \log2-\sin x\mid$ and $g(x)=f((f(x)), then$  

Answer:

Explanation:

We have $f(x)=\mid\log2-\sin x$  and $g(x)=f(f(x)),x\epsilon R$

   Note that , for $x\rightarrow0, \log2>\sin x$

 $f(x)=\log 2-\sin x$

 $\Rightarrow g(x)=\log 2-\sin(f(x))$

 Clearly g(x) is differentiable at x=0 as sin x is diffenerntiable

 Now    $g'(x)=-\cos (\log 2-\sin x)(-\cos x)$

               =$\cos x.\cos(\log2-\sin x)$

  $\Rightarrow g'(0)=1.\cos(\log 2)$

Let $p=\lim_{x \rightarrow 0}(1+\tan^{2}\sqrt{x})^{\frac{1}{2x}}$, then  $\log p$ is equal to 

Answer:

Explanation:

Given $p=\lim_{x \rightarrow 0}(1+\tan^{2}\sqrt{x})^{\frac{1}{2x}}$ ( 1^∞ form)

$e^{\lim_{x \rightarrow0}\frac{\tan^{2}\sqrt{x}}{2x}}=e^{\frac{1}{2}\lim_{x \rightarrow 0}\left(\frac{\tan\sqrt{x}}{\sqrt{x}}\right)^{2}}=e^{\frac{1}{2}}$

$\log p=\log e^{\frac{1}{2}}=\frac{1}{2}$

If the sum of first ten terms of the series $(1\frac{3}{5})^{2}+(2\frac{2}{5})^{2}+(3\frac{1}{5})^{2}+4^{2}+(4\frac{4}{5})^{2}+ ... is \frac{16}{5}m,$, then m is equal to

Answer:

Explanation:

Let S₁₀ be the sum of the first ten terms of the series, then we have 

$S_{10}= (1\frac{3}{5})^{2}+(2\frac{2}{5})^{2}+(3\frac{1}{5})^{2}+4^{2}+(4\frac{4}{5})^{2}+... to $ 10 terms

  $= (\frac{8}{5})^{2}+(\frac{12}{5})^{2}+(\frac{16}{5})^{2}+4^{2}+(\frac{24}{5})^{2}+... to $ 10 terms

= $\frac{1}{5^{2}}({8}^{2}+{12}^{2}+16^{2}+{20}^{2}+{24}^{2}... to $  10 terms

=$\frac{4^{2}}{5^{2}}({2}^{2}+{3}^{2}+4^{2}+{5}^{2}+... to $  to  10 terms)

= $ \frac{4^{2}}{5^{2}}({2}^{2}+{3}^{2}+4^{2}+{5}^{2}+... +11^{2})$

= $\frac{16}{25}(({1}^{2}+{2}^{2}+3^{2}+{4}^{2}+... +11^{2})-1^{2})$

=$ \frac{16}{25}(\frac{11.(11+1)(2.11+1)}{6}-1)$

 =$ \frac{16}{25}(506-1)=\frac{16}{25}\times 505$

$\Rightarrow \frac{16}{5}m=\frac{16}{25}\times 505=101$

If the 2nd, 5 th and 9th  terms  of a non-constant AP are in GP., then the common ratio of this GP is 

Answer:

Explanation:

Let a be the first term and d be the common difference. Then , we have a+d, a+4d, a+8d in GP

 i.e., (a+4d)² = (a+d)(a+8d)

$\Rightarrow a^{2}+16d^{2}+8ad= a^{2}+8ad+ad+8d^{2}$

$\Rightarrow 8d^{2}=a d       $

$\Rightarrow 8d=a $     [ d≠ 0]

Now, 

   common ratio

 $r=\frac{a+4d}{a+d}$

      $=\frac{8d+4d}{8d+d}$

              $=\frac{12d}{9d}$=$\frac{4}{3}$

If the number of terms in the expansion if $(1-\frac{2}{x}+\frac{4}{x^{2}}), x\neq 0, is 28$   . then the sum of the coefficients of all terms in this expansion is

Answer:

Explanation:

Clearly  number of terms in the expansion of $(1-\frac{2}{x}-\frac{4}{x^{2}})^{n}$  is  $\frac{(n+2((n+1)}{2} or ^{n+2}C_{2}$   { Assuming $\frac{1}{x}$ $\frac{1}{x^{2}}$ are distinct }

$\frac{(n+2)(n+1)}{2}=28$

$\Rightarrow n=0$

$\Rightarrow (n+2)(n+1)=56=(6+1)(6+2)$

= n=6

 Hence sum of coefficeints

    $(1-2+4)^{6}=3^{6}=729$

  NOTE:   As $\frac{1}{x}$ $\frac{1}{x^{2}}$ are function of same variables , therefore number of disimilar  terms will be 2n+1 . ie. odd, which is not possible . Hence it contaibns error

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