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Updated on: 16 Sep 2019, 20:50
Originally posted by DisciplinedPrep on 16 Sep 2019, 14:44.
Last edited by Bunuel on 16 Sep 2019, 20:50, edited 1 time in total.
Last edited by Bunuel on 16 Sep 2019, 20:50, edited 1 time in total.
Renamed the topic and edited the question
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
60% (02:39) correct
40%
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If \(f(x^2 – 1) = x^4 – 7x^2 + k_1\) and \(f(x^3 – 2) = x^6 – 9x^3 + k_2\), find the value of \((k_2 – k_1)\).
A. 6
B. 7
C. 8
D. 9
E. 12
A. 6
B. 7
C. 8
D. 9
E. 12
16 Sep 2019, 19:55
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DisciplinedPrep
Get \(f(x^2-1)=f(x^3-2)\)
1) When we take x = 0 in \(f(x^2-1)\), it becomes f(0-1)=f(-1)
f(\(x^2\) – 1) = \(x^4\) – 7\(x^2\) + k1 .....f(-1)=0-0+k1=k1..(a)
2) When we take x = 1 in \(f(x^3-2)\), it becomes f(1-2)=f(-1)
f(\(x^3\) – 2) = \(x^6\) – 9\(x^3\) + k2 .....f(-1)=1-9+k2=k2-8...(b)
From (a) and (b), we get f(-1)=k1=k2-8.......k2-k1=8
C
General Discussion
17 Sep 2019, 01:10
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f(\(x^2\)-1) = \(x^4\)-7\(x^2\)+k1
f(\(x^2\)-1) = \(x^4\)-2\(x^2\)-\(5x^2\)+1-1+5-5+k1
f(\(x^2\)-1) = \(x^4\)+1-2\(x^2\)-5(x^2-1)-6+k1 = \((x^2-1)^2\)-5(\(x^2\)-1)-6+k1
f(x) = \(x^2\)-5x-6+k1
now,
f(\(x^3\)-2) = \(x^6\)-9\(x^3\)+k2
f(\(x^3\)-2) = \(x^6\)+4-\(4x^3\)-5(\(x^3\)-2)-14+k2 = \((x^3-2)^2\)-5(\(x^3\)-2)-14+k2
f(x) = \(x^2\)-5x-14+k2
since at x = t f(x) has same value
\(t^2\)-5t-6+k1 = \(t^2\)-5t-14+k2
k2-k1 = 14-6 = 8
C correct answer
f(\(x^2\)-1) = \(x^4\)-2\(x^2\)-\(5x^2\)+1-1+5-5+k1
f(\(x^2\)-1) = \(x^4\)+1-2\(x^2\)-5(x^2-1)-6+k1 = \((x^2-1)^2\)-5(\(x^2\)-1)-6+k1
f(x) = \(x^2\)-5x-6+k1
now,
f(\(x^3\)-2) = \(x^6\)-9\(x^3\)+k2
f(\(x^3\)-2) = \(x^6\)+4-\(4x^3\)-5(\(x^3\)-2)-14+k2 = \((x^3-2)^2\)-5(\(x^3\)-2)-14+k2
f(x) = \(x^2\)-5x-14+k2
since at x = t f(x) has same value
\(t^2\)-5t-6+k1 = \(t^2\)-5t-14+k2
k2-k1 = 14-6 = 8
C correct answer
