Question

If the tangents to the curve $\sqrt{x}+\sqrt{y}=\sqrt{a}$ at any point on it cuts the axes OX and OY at P and Q respectively, then OP + OQ is

• A. $\frac{a}{2}$
• B. $a$
• C. $2a$
• D. $4a$

Answer 1

The correct option is B $a$

$5f\left(x\right)+3f\left(\frac{1}{x}\right)=x+2$ . . . . (i)
Replacing x by $\frac{1}{x}$
$\therefore 5f\left(\frac{1}{x}\right)+3f\left(x\right)=\frac{1}{x}+2$ . . . . (ii)
From Eq. (i),
$25f\left(x\right)+15f\left(\frac{1}{x}\right)=5x+10$ . . . . (iii)
and from Eq. (ii)
$9f\left(x\right)+15f\left(\frac{1}{x}\right)=\frac{3}{x}+6$ . . . . (iv)
Subtracting Eq. (iv) from (iii), we get
$\therefore 16f\left(x\right)=5x-\frac{3}{x}+4\therefore xf\left(x\right)=\frac{5x^2-3+4x}{16}=y\therefore \frac{dy}{dx}=\frac{10x+4}{16}\therefore \frac{dy}{dx}{|}_{(2,2)}=\frac{10+4}{16}=\frac{7}{8}$