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Question:

Consider the polynomials f(x) = ax² + bx + c, where a > 0, b, c are real, g(x) = -2x. If f(x) cuts the x-axis at (-2, 0) and g(x) passes through (a, b), then the minimum value of f(x) + 9a + 1 is

Explanation:

g(x) = -2x passes through (a, b)
⇒ b = -2a ...(1)

Now, f(x) = ax² + bx + c
⇒ f(x) = ax² - 2ax + c [from (1)]

f(x) passes through (2, 0)
⇒ a(-2)² - 2a(-2) + c = 0
⇒ 8a + c = 0
⇒ c = -8a

∴ f(x) = ax² - 2ax - 8a

Now, we have to find the least value of f(x) + 9a + 1
= ax² - 2ax - 8a + 9a + 1
= ax² - 2ax + a + 1
= a(x² - 2x + 1) + 1
= a(x - 1)² + 1

Least value of this expression will be 1, when x = 1.

Hence, option (b).

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Doubts

Volumemute said (2023-03-28 15:37:29)

i think the answer should be 1 since the end equation we get is smthng like a(x-1)^2 + 1......when x=1, the value of fx+9a+1 would then be 1.....pls help

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Solution updated.

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