Q73 / 155 · WBJEE 2026 · Mathematics
Consider the function $y = f(x)$ defined implicitly by the equation $y^3 - 3y + x = 0$ on the interval $(-\infty, -2) \cup (2, \infty)$. The area of the region bounded by the curve $y = f(x)$, the x-axis and the lines $x = a, x = b$, where $-\infty < a < b < -2$ is
- A$\displaystyle\int_{a}^{b} \dfrac{x\,dx}{3((f(x))^2 - 1)} - bf(b) + af(a)$
- Correct answerB$\displaystyle\int_{a}^{b} \dfrac{x\,dx}{3((f(x))^2 - 1)} + bf(b) - af(a)$
- C$-\displaystyle\int_{a}^{b} \dfrac{x\,dx}{3((f(x))^2 - 1)} - bf(b) + af(a)$
- D$-\displaystyle\int_{a}^{b} \dfrac{x\,dx}{3((f(x))^2 - 1)} + bf(b) - af(a)$
Solution
Given the implicit equation $y^3 - 3y + x = 0$, we can write $x = 3y - y^3$.
For $x \in (-\infty, -2)$, we have $3y - y^3 < -2$, which implies $y^3 - 3y - 2 > 0$. Factoring this gives $(y+1)^2(y-2) > 0$, which means $y > 2$.
Since $y = f(x) > 2$ for all $x \in (-\infty, -2)$, the function $f(x)$ is strictly positive in the interval $[a, b]$. Thus, the area of the bounded region is given by:
$A = \displaystyle\int_{a}^{b} f(x) \, dx$
Using integration by parts, taking $u = f(x)$ and $dv = dx$:
$A = [x f(x)]_{a}^{b} - \displaystyle\int_{a}^{b} x f'(x) \, dx$
$A = b f(b) - a f(a) - \displaystyle\int_{a}^{b} x f'(x) \, dx$
To find $f'(x)$, we differentiate the given equation $y^3 - 3y + x = 0$ with respect to $x$:
$3y^2 \dfrac{dy}{dx} - 3 \dfrac{dy}{dx} + 1 = 0$
$\dfrac{dy}{dx} (3y^2 - 3) = -1$
$f'(x) = \dfrac{-1}{3(y^2 - 1)} = \dfrac{-1}{3((f(x))^2 - 1)}$
Substituting $f'(x)$ back into the area expression:
$A = b f(b) - a f(a) - \displaystyle\int_{a}^{b} x \left( \dfrac{-1}{3((f(x))^2 - 1)} \right) \, dx$
$A = \displaystyle\int_{a}^{b} \dfrac{x \, dx}{3((f(x))^2 - 1)} + b f(b) - a f(a)$
Answer: $\displaystyle\int_{a}^{b} \dfrac{x\,dx}{3((f(x))^2 - 1)} + bf(b) - af(a)$
For $x \in (-\infty, -2)$, we have $3y - y^3 < -2$, which implies $y^3 - 3y - 2 > 0$. Factoring this gives $(y+1)^2(y-2) > 0$, which means $y > 2$.
Since $y = f(x) > 2$ for all $x \in (-\infty, -2)$, the function $f(x)$ is strictly positive in the interval $[a, b]$. Thus, the area of the bounded region is given by:
$A = \displaystyle\int_{a}^{b} f(x) \, dx$
Using integration by parts, taking $u = f(x)$ and $dv = dx$:
$A = [x f(x)]_{a}^{b} - \displaystyle\int_{a}^{b} x f'(x) \, dx$
$A = b f(b) - a f(a) - \displaystyle\int_{a}^{b} x f'(x) \, dx$
To find $f'(x)$, we differentiate the given equation $y^3 - 3y + x = 0$ with respect to $x$:
$3y^2 \dfrac{dy}{dx} - 3 \dfrac{dy}{dx} + 1 = 0$
$\dfrac{dy}{dx} (3y^2 - 3) = -1$
$f'(x) = \dfrac{-1}{3(y^2 - 1)} = \dfrac{-1}{3((f(x))^2 - 1)}$
Substituting $f'(x)$ back into the area expression:
$A = b f(b) - a f(a) - \displaystyle\int_{a}^{b} x \left( \dfrac{-1}{3((f(x))^2 - 1)} \right) \, dx$
$A = \displaystyle\int_{a}^{b} \dfrac{x \, dx}{3((f(x))^2 - 1)} + b f(b) - a f(a)$
Answer: $\displaystyle\int_{a}^{b} \dfrac{x\,dx}{3((f(x))^2 - 1)} + bf(b) - af(a)$
Hence, the correct answer is option: 'B'.