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06 Jan 2023, 10:00
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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)!
Difficulty:
65%
(hard)
Question Stats:
60% (02:38) correct
40%
(02:06)
wrong
based on 279
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If \(f(x^2) = x^4–6x^2+5\), which of the following must be true?
A. \(f(x^2)=f(1+x^2)\)
B. \(f(x^2)=f(2-x^2)\)
C. \(f(x^2)=f(3-x^2)\)
D. \(f(x^2)=f(5-x^2)\)
E. \(f(x^2)=f(6-x^2)\)
A. \(f(x^2)=f(1+x^2)\)
B. \(f(x^2)=f(2-x^2)\)
C. \(f(x^2)=f(3-x^2)\)
D. \(f(x^2)=f(5-x^2)\)
E. \(f(x^2)=f(6-x^2)\)
14 Jan 2023, 11:34
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Given that \(f(x^2) = x^4–6x^2+5\) and we need to find which of the following must be true
\(f(x^2) = x^4–6x^2+5\) = \(x^4–x^2 - 5x^2+5\) = \(x^2(x^2–1) - 5(x^2–1)\) = \((x^2 – 1)*(x^2 - 5)\)
A. \(f(x^2)=f(1+x^2)\)
\(f(1+x^2)\) = \((1 + x^2 – 1)*(1 + x^2 - 5)\) = \(x^2*(x^2 - 4)\) ≠ \((x^2 – 1)*(x^2 - 5)\) => FALSE
B. \(f(x^2)=f(2-x^2)\)
\(f(2-x^2)\) = \((2-x^2 – 1)*(2-x^2 - 5)\) = \((1-x^2)*(-x^2 - 3)\) ≠ \((x^2 – 1)*(x^2 - 5)\) => FALSE
C. \(f(x^2)=f(3-x^2)\)
\(f(3-x^2)\) = \((3-x^2 – 1)*(3-x^2 - 5)\) = \((2-x^2)*(-x^2 - 2)\) ≠ \((x^2 – 1)*(x^2 - 5)\) => FALSE
D. \(f(x^2)=f(5-x^2)\)
\(f(5-x^2)\) = \((5-x^2 – 1)*(5-x^2 - 5)\) = \((4-x^2)*(-x^2)\) ≠ \((x^2 – 1)*(x^2 - 5)\) => FALSE
E. \(f(x^2)=f(6-x^2)\)
\(f(6-x^2)\) = \((6-x^2 – 1)*(6-x^2 - 5)\) = \((5-x^2)*(1-x^2)\) = \((x^2 – 1)*(x^2 - 5)\) => TRUE
So, Answer will be E
Hope it helps!
Watch the following video to learn the Basics of Functions and Custom Characters
\(f(x^2) = x^4–6x^2+5\) = \(x^4–x^2 - 5x^2+5\) = \(x^2(x^2–1) - 5(x^2–1)\) = \((x^2 – 1)*(x^2 - 5)\)
A. \(f(x^2)=f(1+x^2)\)
\(f(1+x^2)\) = \((1 + x^2 – 1)*(1 + x^2 - 5)\) = \(x^2*(x^2 - 4)\) ≠ \((x^2 – 1)*(x^2 - 5)\) => FALSE
B. \(f(x^2)=f(2-x^2)\)
\(f(2-x^2)\) = \((2-x^2 – 1)*(2-x^2 - 5)\) = \((1-x^2)*(-x^2 - 3)\) ≠ \((x^2 – 1)*(x^2 - 5)\) => FALSE
C. \(f(x^2)=f(3-x^2)\)
\(f(3-x^2)\) = \((3-x^2 – 1)*(3-x^2 - 5)\) = \((2-x^2)*(-x^2 - 2)\) ≠ \((x^2 – 1)*(x^2 - 5)\) => FALSE
D. \(f(x^2)=f(5-x^2)\)
\(f(5-x^2)\) = \((5-x^2 – 1)*(5-x^2 - 5)\) = \((4-x^2)*(-x^2)\) ≠ \((x^2 – 1)*(x^2 - 5)\) => FALSE
E. \(f(x^2)=f(6-x^2)\)
\(f(6-x^2)\) = \((6-x^2 – 1)*(6-x^2 - 5)\) = \((5-x^2)*(1-x^2)\) = \((x^2 – 1)*(x^2 - 5)\) => TRUE
So, Answer will be E
Hope it helps!
Watch the following video to learn the Basics of Functions and Custom Characters
14 Jan 2023, 11:58
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Bunuel
Assume \(x^2 = y\)
\(f(x^2) = x^4–6x^2+5\)
\(f(y) = y^2–6y+5\)
\(f(y) = (y-5)(y-1)\)
Note, that in the options we are still finding out an alternate expression for f(y) (which is nothing but \(f(x^2)\)). Hence, essentially the way to represent (y-5)(y-1) essentially changes. Option D and Option E looks quite promising. I may infact start with Option E in the exam, however we can test all the options now.
A. \(f(x^2)=f(1+x^2)\)
\(f(y) = (1+y-5)(1+y-1) = (y-4)(y)\) ⇒ Eliminate
B. \(f(x^2)=f(2-x^2)\)
\(f(y) = (2-y-5)(2-y-1) = (-y-3)(1-y) \)⇒ Eliminate
C. \(f(x^2)=f(3-x^2)\)
\(f(y) = (3-y-5)(3-y-1) = (-2-y)(2-y)\) ⇒ Eliminate
D. \(f(x^2)=f(5-x^2)\)
\(f(y) = (5-y-5)(5-y-1) = (-y)(4-y)\) ⇒ Eliminate
E. \(f(x^2)=f(6-x^2)\)
\(f(y) = (6-y-5)(6-y-1) = (1-y)(5-y) = -(y-1) * -(y-5) =(y-1)(y-5)\) ⇒ Answer
