
The real number k for which the equation $2{x^3} + 3x + k = 0$, has two distinct real roots in [0, 1]
(A). Lies between 2 and 3
(B). Lies between 1 and 0
(C). Does not exist
(D). Lies between 1 and 2
Answer
150k+ views
Hint: Start by taking the given equation as function of x or f(x) and differentiate with respect to x . Check whether f’(x) obtained is increasing or decreasing , if it is increasing then it will have at most 1 real root.
Complete step-by-step answer:
Given, $2{x^3} + 3x + k = 0$
Let $f(x) = 2{x^3} + 3x + k$
Differentiating with respect to x , we get
Here we will use the formula $\dfrac{{d({x^n})}}{{dx}} = n{x^{n - 1}}$
$
f'(x) = 6{x^2} + 3 \\
f'(x) > 0 \\
$
As for any value of x, f’(x) can never be negative because of the square term involved.
f’(x) is a strictly increasing function and has at most 1 real root .
And we know , if a polynomial of odd degree, in this case it is 3, has exactly 1 real root.
So, f(x) = has exactly one real root.
We see that the results found do not satisfy the conditions.
Therefore, k does not exist.
So , option C is the correct answer.
Note: Students must know the principle of differentiation , nature of function , graph plotting etc in order to solve such similar problems. Questions can also be asked in such a manner which would demand the application of Lagrange’s mean value theorem(LMVT) , Intermediate value theorem(IVT) , Rolle’s theorem, and are recommended to be practised very well as they make the approach to the solution very easy meanwhile giving valuable information about the function too.
Complete step-by-step answer:
Given, $2{x^3} + 3x + k = 0$
Let $f(x) = 2{x^3} + 3x + k$
Differentiating with respect to x , we get
Here we will use the formula $\dfrac{{d({x^n})}}{{dx}} = n{x^{n - 1}}$
$
f'(x) = 6{x^2} + 3 \\
f'(x) > 0 \\
$
As for any value of x, f’(x) can never be negative because of the square term involved.
f’(x) is a strictly increasing function and has at most 1 real root .
And we know , if a polynomial of odd degree, in this case it is 3, has exactly 1 real root.
So, f(x) = has exactly one real root.
We see that the results found do not satisfy the conditions.
Therefore, k does not exist.
So , option C is the correct answer.
Note: Students must know the principle of differentiation , nature of function , graph plotting etc in order to solve such similar problems. Questions can also be asked in such a manner which would demand the application of Lagrange’s mean value theorem(LMVT) , Intermediate value theorem(IVT) , Rolle’s theorem, and are recommended to be practised very well as they make the approach to the solution very easy meanwhile giving valuable information about the function too.
Recently Updated Pages
JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

Difference Between Area and Volume

Difference Between Mutually Exclusive and Independent Events

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Degree of Dissociation and Its Formula With Solved Example for JEE

Electrical Field of Charged Spherical Shell - JEE

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Ideal and Non-Ideal Solutions Raoult's Law - JEE

JEE Advanced 2025 Notes

JEE Main Chemistry Question Paper with Answer Keys and Solutions
