IPMAT 2021 - Quantitative Aptitude Question 20
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IPMAT 2021 - Quantitative Aptitude
f(x) = ax2 + bx + c
g(x) = – 2x
f(x) cuts x axis at (– 2 ,0) hence (– 2, 0) is a point of b(x)
f(– 2) = 4a – 2b + c = 0 ______ (1)
g(x) passes through (a, b)
g(a) = – 2a = b _________ (2)
By putting value of b in (1)
4a – 2 (– 2a) + c = 0
4a + 4a + c = 0
c = – 8a
f(x) + 9a + 1 = ax2 – 2ax – 8a + 9a + 1
= ax2 – 2ax + a + 1
= a (x – 1)2 + 1
Method 1:
Minimum of (x – 1)2 = 0
∴ minimum value of above expression = 1
Alternative Method:
f(x) = ax2 – 2ax + a + 1
f’(x) = 2ax – 2a
for minimum value
f’(x) = 0
2ax – 2a = 0
x = 1
f” (x) > 0
f” (x) = 2a > 0 (as given a > 0)
So minimum value of f(x) will be at x = 1
f(1) = a – 2a + a + 1
= 1