JAMIA 2026 — Mathematics PYQ

Step 1: Square both sides

Squaring both sides to simplify the expression:

$${\left(\sqrt{x+y}+\sqrt{x-y}\right)}^2={\left(\sqrt{2}\right)}^2$$

Using the identity $(a+b{)}^2=a^2+2ab+b^2$:

$$(x+y)+2\sqrt{(x+y)(x-y)}+(x-y)=2$$

Combine like terms ($+y$ and $-y$ cancel out):

$$2x+2\sqrt{x^2-y^2}=2$$

Divide the entire equation by 2:

$$x+\sqrt{x^2-y^2}=1$$

Step 2: Isolate the remaining radical and square again

Rearrange the equation to isolate the square root term on one side:

$$\sqrt{x^2-y^2}=1-x$$

Now, square both sides once more:

$${\left(\sqrt{x^2-y^2}\right)}^2=(1-x{)}^2$$

$$x^2-y^2=1-2x+x^2$$

Subtract $x^2$ from both sides:

$$-y^2=1-2x$$

Multiply by $-1$ to make it simpler:

$$y^2=2x-1$$

Step 3: Differentiate with respect to $x$

Now, perform implicit differentiation on both sides with respect to $x$:

$$\frac{d}{dx}(y^2)=\frac{d}{dx}(2x-1)$$

Using the chain rule on the left side:

$$2y\cdot \frac{dy}{dx}=2$$

Divide both sides by $2y$:

$$\frac{dy}{dx}=\frac{2}{2y}$$

$$\frac{dy}{dx}=\frac{1}{y}$$

Correct Answer:

(c) 1/y