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If f(x) = ax2 + bx + c and f(-1) >= -4, f(1) and f(3) >= 5, then the least value of a is (a) 1/4(b) 1/8(c) 1/3(d) -1/3 Please explain the answer with proper steps.

Viraaj Sandip Raulgaonkar , 4 Years ago
Grade 11
anser 2 Answers
Srutarshi Tripathi

Last Activity: 4 Years ago

First of all, You have missed one important information. f(1)  0.
Now, we have
f(-1) > 4
or,a-b+c > 4………(i)
f(1)  0
⇒ a + b + c  0
or, -a-b-c > 0………(ii)(ii)
f(3) > 5
or, 9a+3b+c > 5……….(iii)
From  (i)+(ii) ⇒ -2b > -4……......(iv)
From (ii)+(iii)+(iv), 8a > 1 ⇒ a > 1/8 (ANSWER is B)

Srutarshi Tripathi

Last Activity: 4 Years ago

First of all, You have missed one important information. f(1)  0.
Now, we have
f(-1) > 4
or,a-b+c > 4………(i)
f(1)  0
⇒ a + b + c  0
or, -a-b-c > 0………(ii)(ii)
f(3) > 5
or, 9a+3b+c > 5……….(iii)
From  (i)+(ii) ⇒ -2b > -4……......(iv)
From (ii)+(iii)+(iv), 8a > 1 ⇒ a > 1/8 (ANSWER is B)

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