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Bunuel
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If f(x) = ax^4 + bx^2 + 1, where a and b are constants 31 Aug 2026, 19:00
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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15% (low)

Question Stats:

91% (01:07) correct 9% (01:29) wrong based on 65 sessions

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If \(f(x)=ax^4+bx^2+1\), where \(a\) and \(b\) are constants, and \(f(3)=9\), what is the value of \(f(-3)\) ?

A. -9
B. -3
C. 0
D. 3
E. 9


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E
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Jyoti3380
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If f(x) = ax^4 + bx^2 + 1, where a and b are constants 31 Aug 2026, 20:36
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Getting answer as "3" when solving this. Pls tell if this is correct.
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icandoit4u
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If f(x) = ax^4 + bx^2 + 1, where a and b are constants 01 Sep 2026, 05:46
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solution : E
putting 3 as x ; f(3)=9
a(3)^4+b(3)^2+1 = 9
81 a +9b = 8 (i)
solving for f(-3)
a(-3)^4+b(-3)^2+1 = 81 a +9b +1 (ii)
putting value of 81 a + 9b = 8 in eq (ii)
8+1 = 9 hence ans is option e = 9

Bunuel
If \(f(x)=ax^4+bx^2+1\), where \(a\) and \(b\) are constants, and \(f(3)=9\), what is the value of \(f(-3)\) ?

A. -9
B. -3
C. 0
D. 3
E. 9


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Gaurav015
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If f(x) = ax^4 + bx^2 + 1, where a and b are constants 01 Sep 2026, 06:24
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No, the correct answer choice is E. If you look closely, by substituting x = 3 and x = -3, you will get the same equation. So f(3)= f(-3)= 9.
Jyoti3380
Getting answer as "3" when solving this. Pls tell if this is correct.
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If f(x) = ax^4 + bx^2 + 1, where a and b are constants 01 Sep 2026, 10:47
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Answer: Option E (9)

STIMULUS SETUP:
- Function: f(x) = a*x^4 + b*x^2 + 1
- Given: f(3) = 9
- Target: Find f(-3)

CONCEPT (EVEN FUNCTIONS):
Notice that all exponents of x are even (4 and 2).
f(-x) = a*(-x)^4 + b*(-x)^2 + 1 = a*x^4 + b*x^2 + 1 = f(x).
Therefore, f(x) is an even function, which means f(-x) = f(x) for any x.

CALCULATION:
f(-3) = f(3) = 9


Hope this helps
Jyoti3380
Getting answer as "3" when solving this. Pls tell if this is correct.
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Jyoti3380
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If f(x) = ax^4 + bx^2 + 1, where a and b are constants 01 Sep 2026, 12:36
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Yes, 9 is correct. It was an error at my end, instead of equating it to against 9, I considered this as 3, silly mistake : (. Now I have got this as well if the powers are all even, answer will be equal to given equation. Thanks for clarifying.
Gaurav015
No, the correct answer choice is E. If you look closely, by substituting x = 3 and x = -3, you will get the same equation. So f(3)= f(-3)= 9.

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If f(x) = ax^4 + bx^2 + 1, where a and b are constants 01 Sep 2026, 12:37
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Hey, it was a stupid error at my end. Now I have got correct answer, thanks for clarification below. This helps.
counterclash1
Answer: Option E (9)

STIMULUS SETUP:
- Function: f(x) = a*x^4 + b*x^2 + 1
- Given: f(3) = 9
- Target: Find f(-3)

CONCEPT (EVEN FUNCTIONS):
Notice that all exponents of x are even (4 and 2).
f(-x) = a*(-x)^4 + b*(-x)^2 + 1 = a*x^4 + b*x^2 + 1 = f(x).
Therefore, f(x) is an even function, which means f(-x) = f(x) for any x.

CALCULATION:
f(-3) = f(3) = 9


Hope this helps

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If f(x) = ax^4 + bx^2 + 1, where a and b are constants 02 Sep 2026, 01:45
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\(f(3)=9\) means: \(a*(3)^4+b*(3)^2+1=9\) ⭢ \(81a+9b+1=9\)

Since the exponents are even numbers, we obtain the same result by substituting \(-3\) as we do by substituting \(3\): \(f(-3)= 81a+9b+1=9\)


Answer: E
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