Hkcee 2010 Maths Paper 2 Solution Today

Critical points: ( x = 1 ) and ( x = -2 ) (undefined). Step 2: Test intervals: ( x < -2 ) → positive; between -2 and 1 → negative or zero; ( x > 1 ) → positive. Step 3: Include ( x = 1 ), exclude ( x = -2 ). Answer: ( -2 < x \le 1 ) → Option C.

( \sqrt(x-2)^2 + y^2 = 2\sqrtx^2+y^2 ) Square: ( x^2 -4x +4 + y^2 = 4x^2+4y^2 ) → ( -4x+4 = 3x^2+3y^2 ) → ( 3x^2+3y^2+4x-4 = 0 ) → Circle equation. hkcee 2010 maths paper 2 solution

: Start with option C and plug it into the equation. This "trial and error" approach can often be faster than solving complex equations from scratch. Critical points: ( x = 1 ) and ( x = -2 ) (undefined)

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