Question:

A scientist wants to find the root of the equation \(2x^3 + x^2 - 1 = 0\) lying in \((0,1)\). He applies the Secant method only once by taking two initial guesses, 0.5 and 0.7. The value of the root is approximately:

Updated On: Jul 20, 2024

0.17
0.52
0.65
0.75

Hide Solution

Verified By Collegedunia

The Correct Option is C

Solution and Explanation

The correct option is (C): 0.65

Was this answer helpful?

0

0

Top Questions on Differential Equations

If \( y = y(x) \) is the solution of the differential equation, \[ \sqrt{4 - x^2} \frac{dy}{dx} = \left( \left( \sin^{-1} \left( \frac{x}{2} \right) \right)^2 - y \right) \sin^{-1} \left( \frac{x}{2} \right), \] where \( -2 \leq x \leq 2 \), and \( y(2) = \frac{\pi^2 - 8}{4} \), then \( y^2(0) \) is equal to:
- JEE Main - 2025
- Mathematics
- Differential Equations
View Solution
If \( p \) and \( q \) are respectively the order and degree of the differential equation \( \frac{d}{dx} \left( \frac{dy}{dx} \right)^3 = 0 \), then \( (p - q) \) is:
- CBSE CLASS XII - 2025
- Mathematics
- Differential Equations
View Solution
If \( \frac{d}{dx}(y) = y \sin 2x \) and \( y(0) = 1 \), then what is the required solution?
- CUET (PG) - 2025
- Mechanical Engineering
- Differential Equations
View Solution
Let $ y = y(x) $ be the solution of the differential equation $ (x^2 + 1)y' - 2xy = (x^4 + 2x^2 + 1)\cos x, $ with the initial condition $ y(0) = 1 $. Then $ \int_{-3}^{3} y(x) \, dx \text{ is:} $
- JEE Main - 2025
- Mathematics
- Differential Equations
View Solution
Let $ y = y(x) $ be the solution curve of the differential equation $ x(x^2 + e^x) \, dy + \left( e^x(x - 2) y - x^3 \right) \, dx = 0, \quad x>0, $ passing through the point $ (1, 0) $. Then $ y(2) $ is equal to:
- JEE Main - 2025
- Mathematics
- Differential Equations
View Solution

View More Questions

Questions Asked in GATE TF exam

View More Questions