Question

If $y=\frac{x^4-x^2+1}{x^2+\sqrt{3}x+1}$ and $\frac{dy}{dx}=ax+b$ then the value of $a+b$ is equal to

• A. $\cot \frac{5\pi }{8}$
• B. $\cot \frac{5\pi }{12}$
• C. $\tan \frac{5\pi }{12}$
• D. $\tan \frac{5\pi }{8}$

Answer 1

The correct option is B $\cot \frac{5\pi }{12}$

$\frac{dy}{dx}=\frac{(4x^3-2x)(x^2+\sqrt{3}x+1)-(2x+\sqrt{3})(x^4-x^2+1)}{(x^2+\sqrt{3}x+1{)}^2}$

$=\frac{4x^5+4\sqrt{3}{x}^4+4x^3-2x^3-2\sqrt{3}{x}^2-2x-2x^5+2x^3-2x-\sqrt{3}{x}^4+\sqrt{3}{x}^2-\sqrt{3}}{(x^2+\sqrt{3}x+1{)}^2}$

$=\frac{2x^5+3\sqrt{3}{x}^4+4x^3-\sqrt{3}{x}^2-4x-\sqrt{3}}{(x^2+\sqrt{3}x+1{)}^2}$

which is equal to $(2x-\sqrt{3})$

so, $a=2$ and $b=-\sqrt{3}$

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

$cot75=cot(45+30)=\frac{cot45cot30-1}{cot45+cot30}$

$=\frac{\sqrt{3}-1}{1+\sqrt{3}}=\frac{(\sqrt{3}-1)(1-\sqrt{3})}{(1+\sqrt{3})(1-\sqrt{3})}$

$=\frac{-(3+1-2\sqrt{3})}{1-3}=2-\sqrt{3}$

Hence, option B is correct