
The value of $a$, \[\left( {a > 0} \right)\;\] for which the area bounded by the curves $y = \dfrac{x}{6} + \dfrac{1}{{{x^2}}}$, $y = 0$, $x = a$ and $x = 2a$ has the least value is
A. $2$
B. $\sqrt 2 $
C. ${\left( 2 \right)^{\dfrac{1}{3}}}$
D. $1$
Answer
162.9k+ views
Hint: In this question, we are given the equation of curve $y = \dfrac{x}{6} + \dfrac{1}{{{x^2}}}$ which lies in the $x - axis$ from $x = a$ and $x = 2a$. We have to find the value of $a$ at which the area will be least. Also, \[\left( {a > 0} \right)\;\] (given). So, calculate the area by integrating the equation of the curve from $a$ to $2a$. Then, differentiate the required answer with respect to $a$ and equate it to $0$. Solve it further.
Formula Used: Integration formula –
$\int {f\left( y \right)dx = \dfrac{{\int {f\left( y \right)dx} }}{{\left( {\dfrac{{dy}}{{dx}}} \right)}}} $
Differentiation formula –
$\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$
Complete step by step Solution:
Given that,
Equation of the curve is $y = \dfrac{x}{6} + \dfrac{1}{{{x^2}}}$ from $x = a$ to $x = 2a$
Therefore, the area bounded by the curve will be the integration of the equation of the curve from $x = a$ to $x = 2a$
Therefore, Area $ = \int\limits_a^{2a} {\dfrac{x}{6} + \dfrac{1}{{{x^2}}}} dx$
$ = \left[ {\dfrac{{{x^2}}}{{12}} - \dfrac{1}{x}} \right]_a^{2a}$
Now, resolving the limits
Thus, area $\left( A \right) = \dfrac{{{a^2}}}{4} + \dfrac{1}{{2a}}$
Differentiate the required area with respect to $a$,
We get,
$\dfrac{{dA}}{{da}} = \dfrac{{2a}}{4} + \dfrac{1}{{2{a^2}}}$
Also, written as $\dfrac{{dA}}{{da}} = \dfrac{a}{2} + \dfrac{1}{{2{a^2}}}$
Now, for minimum value
Put $\dfrac{{dA}}{{da}} = 0$
It implies that,
$\dfrac{a}{2} + \dfrac{1}{{2a}} = 0$
L.C.M of $2$ and $2a$ is $2a$
Therefore, $\dfrac{{{a^2} - 1}}{{2a}} = 0$
We get, $a = \pm 1$
According to the question, $a$ should be greater than $0$. So, $a = - 1$ will not exist.
Thus, at $a = 1$ the area of the given curve will be least.
Therefore, the correct option is (D).
Note:Different methods are used to determine the area under the curve, with the antiderivative method being the most prevalent. The area under the curve can be calculated by knowing the curve's equation, borders, and the axis surrounding the curve. There exist formulas for obtaining the areas of regular shapes such as squares, rectangles, quadrilaterals, polygons, and circles, but no formula for finding the area under a curve. The integration procedure aids in solving the equation and determining the required area. Antiderivative methods are highly useful for determining the areas of irregular planar surfaces.
Formula Used: Integration formula –
$\int {f\left( y \right)dx = \dfrac{{\int {f\left( y \right)dx} }}{{\left( {\dfrac{{dy}}{{dx}}} \right)}}} $
Differentiation formula –
$\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$
Complete step by step Solution:
Given that,
Equation of the curve is $y = \dfrac{x}{6} + \dfrac{1}{{{x^2}}}$ from $x = a$ to $x = 2a$
Therefore, the area bounded by the curve will be the integration of the equation of the curve from $x = a$ to $x = 2a$
Therefore, Area $ = \int\limits_a^{2a} {\dfrac{x}{6} + \dfrac{1}{{{x^2}}}} dx$
$ = \left[ {\dfrac{{{x^2}}}{{12}} - \dfrac{1}{x}} \right]_a^{2a}$
Now, resolving the limits
Thus, area $\left( A \right) = \dfrac{{{a^2}}}{4} + \dfrac{1}{{2a}}$
Differentiate the required area with respect to $a$,
We get,
$\dfrac{{dA}}{{da}} = \dfrac{{2a}}{4} + \dfrac{1}{{2{a^2}}}$
Also, written as $\dfrac{{dA}}{{da}} = \dfrac{a}{2} + \dfrac{1}{{2{a^2}}}$
Now, for minimum value
Put $\dfrac{{dA}}{{da}} = 0$
It implies that,
$\dfrac{a}{2} + \dfrac{1}{{2a}} = 0$
L.C.M of $2$ and $2a$ is $2a$
Therefore, $\dfrac{{{a^2} - 1}}{{2a}} = 0$
We get, $a = \pm 1$
According to the question, $a$ should be greater than $0$. So, $a = - 1$ will not exist.
Thus, at $a = 1$ the area of the given curve will be least.
Therefore, the correct option is (D).
Note:Different methods are used to determine the area under the curve, with the antiderivative method being the most prevalent. The area under the curve can be calculated by knowing the curve's equation, borders, and the axis surrounding the curve. There exist formulas for obtaining the areas of regular shapes such as squares, rectangles, quadrilaterals, polygons, and circles, but no formula for finding the area under a curve. The integration procedure aids in solving the equation and determining the required area. Antiderivative methods are highly useful for determining the areas of irregular planar surfaces.
Recently Updated Pages
JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

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

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

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

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

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

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

NEET 2025 – Every New Update You Need to Know

Verb Forms Guide: V1, V2, V3, V4, V5 Explained

NEET Total Marks 2025

1 Billion in Rupees
