Question

If the tangent to the curve y = x³ – x² + x at the point (a, b) is also tangent to the curve y = 5x² + 2x – 25 at the point (2, –1), then |2a + 9b| is equal to _____ .

Answer 1

Correct Answer: 195

First, find the derivative of the first curve \(y = x^3 - x^2 + x\).
The derivative is: \( \frac{dy}{dx} = 3x^2 - 2x + 1 \).
At the point \((a, b)\), the slope of the tangent is \(3a^2 - 2a + 1\).
The second curve is \(y = 5x^2 + 2x - 25\).
The derivative of the second curve is: \(\frac{dy}{dx} = 10x + 2\).
At the point \((2, -1)\), the slope of the tangent is \(10(2) + 2 = 22\).
Set the slopes equal since both tangents are identical:
\(3a^2 - 2a + 1 = 22\)
Solve for \(a\):
\(3a^2 - 2a - 21 = 0\)
Using the quadratic formula: \(a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
Where \(a = 3\), \(b = -2\), and \(c = -21\).
\(a = \frac{2 \pm \sqrt{4 + 252}}{6}\)
\(a = \frac{2 \pm \sqrt{256}}{6} = \frac{2 \pm 16}{6}\)
Thus, \(a = 3\) or \(-\frac{7}{3}\).
Since we need to find the point \((a, b)\) on the first curve,
Substitute \(a = 3\) into \(y = x^3 - x^2 + x\):
\(b = 3^3 - 3^2 + 3 = 27 - 9 + 3 = 21\).

Next, calculate \(|2a + 9b|\).
\(|2(3) + 9(21)| = |6 + 189| = |195|\).

The absolute value is 195, fitting within the range of 195,195.

Answer 2

Slope of tangent to curve y = 5x² + 2x – 25,
\(m=\left(\frac{dy}{dx}\right)_{at(2,−1)}=22\)
Equation of tangent: y + 1 = 22(x – 2)
y = 22x – 45
Slope of tangent to y = x³ – x² + x at point (a, b) = 3a² – 2a + 1
3a² – 2a + 1 = 22
3a² – 2a – 21 = 0
\(∴a=3\) or \(a=−\frac{7}{3}\)
Also b = a³ – a² + a
Then \((a,b)=(3,21)\) or, \((−\frac{7}{3},–\frac{151}{9})\)
\((−\frac{7}{3},–\frac{151}{9})\) does not satisfy the equation of tangent
\(∴a=3,b=21\)
the, \(|2a+9b|=|2\times3+9\times21| =|195| = 195\)
So, the answer is 195.