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A function, f(x) = 3x^2 - tx + 5 is tangent to x-axis.
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07 Sep 2021, 09:41
07 Sep 2021, 09:41
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53% (01:39) correct
46% (02:09) wrong based on 30 sessions
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A function, \(f(x) = 3x^2 - tx + 5\) is tangent to x-axis.
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Quantity A |
Quantity B |
\(t\) |
\(2\sqrt{15}\) |
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
A function, f(x) = 3x^2 - tx + 5 is tangent to x-axis.
[#permalink]
Updated on: 09 Sep 2021, 04:10
Updated on: 09 Sep 2021, 04:10
1
1
Given : \(f(x) = 3x^2 - tx + 5\) is tangent to x-axis
so two observation from the above statement
1. at the point of tangent intersection at x, y=0
2. and \(f'(x) = 0\)
so first take the derivative and equate to 0 to find the point of minimum
\( f' = 6x - t = 0 => x = t/6\) when x = t/6, the value for f(x) is minimum that's 0
so substitute \(x =t/6 \) in f(x) and this will give \(t = +/-2\sqrt{15}\)
Answer is D
Thanks to gernico for correcting.
so two observation from the above statement
1. at the point of tangent intersection at x, y=0
2. and \(f'(x) = 0\)
so first take the derivative and equate to 0 to find the point of minimum
\( f' = 6x - t = 0 => x = t/6\) when x = t/6, the value for f(x) is minimum that's 0
so substitute \(x =t/6 \) in f(x) and this will give \(t = +/-2\sqrt{15}\)
Answer is D
Thanks to gernico for correcting.
A function, f(x) = 3x^2 - tx + 5 is tangent to x-axis.
[#permalink]
08 Sep 2021, 19:02
08 Sep 2021, 19:02
2
eskay1981 wrote:
Given : \(f(x) = 3x^2 - tx + 5\) is tangent to x-axis
so two observation from the above statement
1. at the point of tangent intersection at x, y=0
2. and \(f'(x) = 0\)
so first take the derivative and equate to 0 to find the point of minimum
\( f' = 6x - t = 0 => x = t/6\) when x = t/6, the value for f(x) is minimum that's 0
so substitute \(x =t/6 \) in f(x) and this will give \(t = 2\sqrt{15}\)
Answer is C
so two observation from the above statement
1. at the point of tangent intersection at x, y=0
2. and \(f'(x) = 0\)
so first take the derivative and equate to 0 to find the point of minimum
\( f' = 6x - t = 0 => x = t/6\) when x = t/6, the value for f(x) is minimum that's 0
so substitute \(x =t/6 \) in f(x) and this will give \(t = 2\sqrt{15}\)
Answer is C
When subbing in \(\frac{t}{6}\) for \(x\), don't forget that when you get to:
\(t^2 = 60\)
\(t = 2\sqrt{15}\) OR \(-2\sqrt{15}\)
Which is why the answer is D.
