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 _____ .

Updated On: Dec 29, 2025
Hide Solution
collegedunia
Verified By Collegedunia

Correct Answer: 195

Approach Solution - 1

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.
Was this answer helpful?
3
3
Hide Solution
collegedunia
Verified By Collegedunia

Approach Solution -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.

Was this answer helpful?
0
0

Concepts Used:

Tangents and Normals

  • A tangent at a degree on the curve could be a straight line that touches the curve at that time and whose slope is up to the derivative of the curve at that point. From the definition, you'll be able to deduce the way to realize the equation of the tangent to the curve at any point.
  • Given a function y = f(x), the equation of the tangent for this curve at x = x0 
  • Slope of tangent (at x=x0) m=dy/dx||x=x0
  • A normal at a degree on the curve is a line that intersects the curve at that time and is perpendicular to the tangent at that point. If its slope is given by n, and also the slope of the tangent at that point or the value of the derivative at that point is given by m. then we got 

m×n = -1

  • The normal to a given curve y = f(x) at a point x = x0
  • The slope ‘n’ of the normal: As the normal is perpendicular to the tangent, we have: n=-1/m

Diagram Explaining Tangents and Normal: