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If all the words (with or without meaning ) having five letters formed using the letters of the word "SMALL and arranged as ina dictionary , then the position of the word  "SMALL" is

Answer:

Explanation:

Clearly , number of words start with $A=\frac{4!}{2!}=12$

  Number of words start with =4!=24

Number of words start with= $\frac{4!}{2!}=12$

 Numer of start with = $\frac{3!}{2!}=3$

 Numer of words stsrt with  = 3!=6

Note that  next word will be "SMALL"

  Hence , the position of word  'SMALL" is 58th

The system of linear equation $x+\lambda y-z=\alpha,  hx-y-z=\alpha, x+y-\lambda z=0$

has a non-trivial solution for 

Answer:

Explanation:

Given a system of linear equation is $x+\lambda y-z=\alpha,  hx-y-z=\alpha, x+y-\lambda z=0$

  Note that given system will have a non trivial solution only if determined  of coefficient matris is zero, i.e,

$\Rightarrow 1(\lambda+1)-\lambda (-\lambda^{2}+1)-1(\lambda+1)=0$

$\Rightarrow \lambda+1+\lambda^{2}-\lambda -\lambda-1=0$

$\Rightarrow \lambda+1+\lambda^{2}-\lambda -\lambda-1=0$

$\Rightarrow \lambda^{3}-\lambda=0=\lambda(\lambda^{2}-1)=0$

$\Rightarrow \lambda=0 or \lambda=\pm 1$

Hence , given system of linear equation has a non trivial solution for exactly three values of λ

A value of θ for which $\frac{2+3i\sin\theta}{1-2i\sin\theta}$ is purely imaginary , is

Answer:

Explanation:

Let $z= \frac{2+3i\sin\theta}{1-2i\sin\theta}$

is purely

 imaaginary Then , we have Re(z)=0

 NOw conside $z= \frac{2+3i\sin\theta}{1-2i\sin\theta}$

   $z= \frac{(2+3i\sin\theta)(1+2i\sin\theta)}{(1-2i\sin\theta)(1+2i\sin\theta)}$

 $z= \frac{(2+4i\sin\theta+3i\sin\theta+6i^{2}\sin^{2}\theta)}{(1^{2}-(2i\sin\theta)^{2}}$

   $z= \frac{(2+7i\sin\theta-6\sin^{2}\theta)}{(1+4\sin^{2}\theta)}$

  $z= \frac{(2-6\sin^{2}\theta)}{(1+4\sin^{2}\theta)}+i\frac{7\sin\theta}{(1+4\sin^{2}\theta)}$

 Re(z)=0

   $\frac{(2-6\sin^{2}\theta)}{(1+4\sin^{2}\theta)}=0$

   $2=6\sin^{2}\theta$

  $\Rightarrow sin^{2}\theta=\frac{1}{\sqrt{3}}$

$\Rightarrow sin^{2}\theta=\pm\frac{1}{\sqrt{3}}$

$\Rightarrow \theta= \sin^{-}(\pm \frac{1}{\sqrt{3}})=\pm sin^{-}(\frac{1}{\sqrt{3}})$

If $f(x0 +2f(\frac{1}{x})=3x, x\neq  0$ and  $S= [x\epsilon R:f(x)=f(-x)]:$ then S 

Answer:

Explanation:

We have , $f(x)+2f(\frac{1}{x})=3x,x\neq 0.....(i)$

On replaceing x by 1/x in the above eq (i)

we get  $f(\frac{1}{x})+2f(x)=\frac{3}{x}$

  $\Rightarrow 2f(x)+2f(\frac{1}{x})=\frac{3}{x}$   .......(ii)

On multiplying Eq(ii)  by 2 and substracting Eq(i) from Wq(ii), we get

$4f(x)+2f(\frac{1}{x})=\frac{6}{x}$

 $\frac{f(x)+2f(\frac{1}{x})=3x}{3f(x)=\frac{6}{x}-3x}$

$\Rightarrow   f(x)=\frac{2}{x}-x$

now consider, f(x)=f(-x)

$\Rightarrow   \frac{2}{x}-4=-\frac{2}{x}+x$

$\Rightarrow   \frac{4}{x}=2x$

$\Rightarrow  2x^{2}=4\Rightarrow x^{2}=2$

$\Rightarrow  x=\pm \sqrt{2}$

 Hence, S contains exactly two elements

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