Question

Hey pls Solve this question ..

Only for genuis and
MathAryabhatta ...

All other good users who can give the ans ....

Don't Spam ...

Answer 1

Hey friend , Harish here.

Here is your answer. (In Image)

NOTE :

1st we have multiplied and divided 'a' by √2 to make the numerator a perfect square.

Now, after getting the 'a' value, we have to find b which is it's reciprocal. And then it is rationalized by multiplying and dividing the denominator by it's conjugate.

Now, adding a and b , and finding their reciprocal gives us the answer.

______________________________

Hope my answer is helpful to you.

Answer 2

Given that:-

$a=\sqt{4-\sqrt{15}}$

We know that $4=\frac{8}{2}$

We also know that $\sqrt{15}=\frac{2*\sqrt{15}}{2}$

Now we can substitute these values into 'a' and complete the perfect square.

Remember perfect square should be of the form:$a^2+2ab+b^2$

$\implies a=\sqrt{\frac{8}{2}-\frac{2*\sqrt{15}}{2}$

Now note that:

$\frac{8}{2} \implies \frac{5+3}{2} \implies \frac{5}{2}+\frac{3}{2}$

This gives us:-

$a=\sqrt{\frac{5}{2}+\frac{3}{2}-\frac{2*\sqrt{15}}{2}$.............................(1)

We know $(a-b)^2=a^2+b^2-2ab$

Thus $(\frac{\sqrt{5}}{\sqrt{2}}-\frac{\sqrt{3}}{\sqrt{2}})^2$

$\implies \sqrt{\frac{5}{2}+\frac{3}{2}-\frac{2*\sqrt{15}}{2}}$

$a=\sqrt{(\frac{\sqrt{5}}{\sqrt{2}}-\frac{\sqrt{3}}{\sqrt{2}})^2}$[See(1)]

$\implies \frac{\sqrt{5}}{\sqrt{2}}-\frac{\sqrt{3}}{\sqrt{2}}$

$\implies \frac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}$..............................(2)

$b=\frac{1}{a}$

Hence b is the reciprocal of 'a'.

$\implies b=\frac{\sqrt{2}}{\sqrt{5}-\sqrt{3}}$

Now rationalize the denominators

$\implies b=\frac{\sqrt{2}}{\sqrt{5}-\sqrt{3}}*\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}$

$\implies b=\frac{\sqrt{2}(\sqrt{5}+\sqrt{3})}{(\sqrt{5})^2-(\sqrt{3})^2}$

$\implies \frac{\sqrt{2}(\sqrt{5}+\sqrt{3})}{2}$

We see that $2=\sqrt{2}*\sqrt{2}$

So $\sqrt{2}$ from the numerator gets cancelled.

$b=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}$.................................(3)

From (1) and (2) if we add both equations we will get:-

$a+b=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}$

$\implies \frac{\sqrt{5}+\sqrt{5}+\sqrt{3}-\sqrt{3}}{\sqrt{2}}$

$\implies \frac{2*\sqrt{5}}{\sqrt{2}}$

Again rationalize the denominators:-

$\implies \frac{2\sqrt{5}}{\sqrt{2}}*\frac{\sqrt{2}}{\sqrt{2}}$

$\implies \frac{2\sqrt{10}}{2}$

$\implies \sqrt{10}$................(4)

So $\frac{1}{a+b} \implies \frac{1}{\sqrt{10}}$ [See(4)]

Here is your answer:-

Option (1)-$\frac{1}{\sqrt{10}}$

Hope it helps :)