If 0 and 1 are th

∵ 0 and 1 are the zeroes of the polynomial f(x) = 2x³-3x²+ ax + b

∴ f(0) = 0 and f(1) = 1

Now,

f(x) = 2x³-3x²+ ax + b

⇒ f(0) = 2(0)³ – 3(0)²+ a(0) + 0

⇒ 2 × 0– 3 × 0 + a × 0 + b = 0

⇒ 0– 0 + 0 + b = 0

⇒ b = 0

And,

⇒ f(1) = 2(1)³ – 3(1)²+ a(1) +1

⇒ 2 × 1– 3 × 1 + a × 1 + b = 0

⇒ 2– 3 + a + b = 0

⇒ 2– 3 + a + 0 = 0 [∵ b = 0]

⇒ -1 + a = 0

⇒ a = 1