Consider the functions defined implicitly by the equation y³ - 3y + x = 0 on various intervals in th

Q: Wait, why can't we just solve for $y$ explicitly in $y^3 - 3y + x = 0$?

That is a great question! While it is tempting to try and solve for $y$ using the cubic formula, it would lead to an incredibly messy expression involving complex roots and radicals. In JEE Advanced, we often deal with implicit functions precisely because the algebraic form is elegant, even if the explicit form is not. Using implicit differentiation allows us to work with the function's properties directly without needing to 'solve' for $y$.

Q: Why is $g(x)$ an odd function?

Look at the original equation: $y^3 - 3y + x = 0$. If you replace $x$ with $-x$ and $y$ with $-y$, you get $(-y)^3 - 3(-y) + (-x) = -y^3 + 3y - x = -(y^3 - 3y + x) = 0$. This means that if $(x, y)$ is a solution, then $(-x, -y)$ is also a solution. Since $g(x)$ is the unique branch passing through the origin, this symmetry forces $g(-x) = -g(x)$, making it an odd function.

Q: How do I know when to use Integration by Parts for area?

Integration by Parts is your best friend when you have an integral of a function that is hard to integrate directly, but whose derivative is known or simpler. Here, we have $\int y dx$. We know $y$ is a function of $x$, but we don't have $y(x)$ explicitly. However, we know $y'$ from implicit differentiation. By using $\int u dv = uv - \int v du$ with $u=y$ and $dv=dx$, we transform the integral into one involving $y'$, which we already calculated. It's a classic technique to turn an impossible integral into a solvable one!

Question

Passage:
Consider the functions defined implicitly by the equation $y^3 - 3y + x = 0$ on various intervals in the real line. If $x \in (-\infty, -2) \cup (2, \infty)$, the equation implicitly defines a unique real valued differentiable function $y = f(x)$. If $x \in (-2, 2)$, the equation implicitly defines a unique real valued differentiable function $y = g(x)$ satisfying $g(0) = 0$.

Question 1:
If $f(-10\sqrt{2}) = 2\sqrt{2}$, then $f''(-10\sqrt{2}) =$
(A) $\frac{4\sqrt{2}}{7^3 3^2}$
(B) $-\frac{4\sqrt{2}}{7^3 3^2}$
(C) $\frac{4\sqrt{2}}{7^3 3}$
(D) $-\frac{4\sqrt{2}}{7^3 3}$

Question 2:
The area of the region bounded by the curve $y = f(x)$, the x-axis, and the lines $x = a$ and $x = b$, where $-\infty < a < b < -2$, is
(A) $\int_a^b \frac{x}{3(f(x)^2-1)} dx + bf(b) - af(a)$
(B) $-\int_a^b \frac{x}{3(f(x)^2-1)} dx + bf(b) - af(a)$
(C) $\int_a^b \frac{x}{3(f(x)^2-1)} dx - bf(b) + af(a)$
(D) $-\int_a^b \frac{x}{3(f(x)^2-1)} dx - bf(b) + af(a)$

Question 3:
$\int_{-1}^1 g'(x) dx =$
(A) $2g(-1)$
(B) 0
(C) $-2g(1)$
(D) $2g(1)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The equation $y^3 - 3y + x = 0$ defines an implicit curve in the Cartesian plane. While one might be tempted to solve for $y$ explicitly, the most efficient approach is to work directly with the implicit relationship.

Resisting the urge to isolate $y$ allows us to maintain the geometric integrity of the curve and simplifies the subsequent calculus operations.

### Visualizing the Branches

Consider the rearranged form $x = 3y - y^3$. This curve intersects the $x$-axis at $x=0$ (where $y=0, \pm\sqrt{3}$) and exhibits vertical tangents at $x = 2$ and $x = -2$.

These vertical tangents act as the boundaries for the function's domain. The outer branches define $f(x)$ for $x > 2$ or $x < -2$, while the middle branch defines $g(x)$ within the interval $(-2, 2)$.

### The Power of Implicit Differentiation

To find the slope, we differentiate $y^3 - 3y + x = 0$ with respect to $x$. Applying the chain rule, we obtain:

$$3y^2 y' - 3y' + 1 = 0$$

Factoring out $y'$, we arrive at the expression for the first derivative:

$$y' = \frac{1}{3(1 - y^2)}$$

Differentiating $y'$ with respect to $x$ to find the second derivative $y''$ yields:

$$y'' = \frac{2y}{9(1 - y^2)^3}$$

When evaluating these at specific points, such as $x = -10\sqrt{2}$, the algebraic complexity collapses into a manageable form.

### The Area Challenge

To compute the area bounded by $f(x)$, denoted by $\int_a^b f(x) dx$, we employ **Integration by Parts**. Setting $u = y$ and $dv = dx$, we have $du = y' dx$ and $v = x$.

Using the formula $\int u dv = uv - \int v du$, the integral becomes:

$$\int y dx = xy - \int x y' dx$$

Substituting our expression for $y'$, the integral transforms into:

$$\int_a^b \frac{x}{3(1 - y^2)} dx$$

By carefully adjusting the signs and limits, this expression aligns with the standard solutions for such implicit area problems.

### The Symmetry Secret

For the function $g(x)$ on the interval $(-2, 2)$, we evaluate $\int_{-1}^1 g'(x) dx$. By the **Fundamental Theorem of Calculus**, this is equivalent to $g(1) - g(-1)$.

The equation $y^3 - 3y + x = 0$ is invariant under the transformation $(x, y) 	o (-x, -y)$. This confirms that $g(x)$ is an **odd function**, implying $g(-1) = -g(1)$.

Substituting this into our expression, we find:

$$g(1) - (-g(1)) = 2g(1)$$

Symmetry serves as a powerful tool, reducing a seemingly impossible evaluation into a direct consequence of the system's fundamental properties.

### Final Thoughts

Implicit functions are simply curves waiting to be decoded. By utilizing implicit differentiation, integration by parts, and symmetry, you can uncover the underlying structure of any such equation.

**Mastering these techniques ensures that no complex implicit relation can intimidate you.** Keep practicing, and you will find that even the most abstract curves yield to logical analysis.

Animated Solution for Mathematics - Definite Integration: Comprehension Passage

If $f (- 102) = 22$ , then $f^{''} (- 102) =$

Select Answer:

The area of the region bounded by the curve $y = f (x)$ , the x-axis, and the lines $x = a$ and $x = b$ , where $- \infty < a < b < - 2$ , is

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$\int_{- 1}^{1} g^{'} (x) d x =$

Select Answer:

Visualized Solution

Visualizing the Implicit Equation

Finding the First Derivative $y^{'}$

Calculating the Second Derivative $y^{''}$

Evaluating $f^{''} (- 102)$

Area Calculation Setup

Integration by Parts for Area

Finalizing the Area Expression

Analyzing the Function $g (x)$

Evaluating the Integral of $g^{'} (x)$

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

Visualizing the Branches

The Power of Implicit Differentiation

The Area Challenge

The Symmetry Secret

Final Thoughts

Similar Questions

Let $f (x)$ be a continuous function given by $f (x) = {2 x, x^{2} + a x + b, ∣ x ∣ \leq 1 ∣ x ∣ > 1$ . Find the area of the region in the third quadrant bounded by the curves $x = - 2 y^{2}$ and $y = f (x)$ lying on the left of the line $8 x + 1 = 0$ .

Let the functions $f : R \to R$ and $g : R \to R$ be defined by $f (x) = e^{x - 1} - e^{- ∣ x - 1∣}$ and $g (x) = \frac{1}{2} (e^{x - 1} + e^{1 - x})$ . Then the area of the region in the first quadrant bounded by the curves $y = f (x)$ , $y = g (x)$ and $x = 0$ is

Let $f (x)$ be a non-negative continuous function such that the area bounded by the curve $y = f (x)$ , x-axis and the ordinates $x = π /4$ and $x = β > π /4$ is $[β sin β + \frac{π}{4} cos β + 2 β]$ . Then $f (π /2)$ is

The area of the region bounded by the curves $y = ∣ x - 2∣, x = 1, x = 3$ and the x-axis is

The area of the bounded region enclosed by the curve $y = 3 - x - \frac{1}{2} - ∣ x + 1∣$ and the x-axis is

Find all maxima and minima of the function $y = x (x - 1)^{2}, 0 \leq x \leq 2$ . Also determine the area bounded by the curve $y = x (x - 1)^{2}$ , the $y$ -axis and the line $y = 2$ .

The area of the region bounded by the parabola $(y - 2)^{2} = (x - 1)$ , the tangent to it at the point whose ordinate is 3 and the $x$ -axis is :

Animated Solution for Mathematics - Definite Integration: Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(−102​)=22​, then f′′(−102​)=

Select Answer:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area of the region bounded by the curve y=f(x), the x-axis, and the lines x=a and x=b, where −∞<a<b<−2, is

Select Answer:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∫−11​g′(x)dx=

Select Answer:

Visualized Solution

Visualizing the Implicit Equation

Finding the First Derivative .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y′

Calculating the Second Derivative .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y′′

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′′(−102​)

Area Calculation Setup

Integration by Parts for Area

Finalizing the Area Expression

Analyzing the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

Evaluating the Integral of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g′(x)

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

Visualizing the Branches

The Power of Implicit Differentiation

The Area Challenge

The Symmetry Secret

Final Thoughts

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area of the region bounded by the curves y=∣x−2∣,x=1,x=3 and the x-axis is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area of the bounded region enclosed by the curve y=3−​x−21​​−∣x+1∣ and the x-axis is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find all maxima and minima of the function y=x(x−1)2,0≤x≤2. Also determine the area bounded by the curve y=x(x−1)2, the y-axis and the line y=2.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area of the region bounded by the parabola (y−2)2=(x−1), the tangent to it at the point whose ordinate is 3 and the x-axis is :

If $f (- 102) = 22$ , then $f^{''} (- 102) =$

The area of the region bounded by the curve $y = f (x)$ , the x-axis, and the lines $x = a$ and $x = b$ , where $- \infty < a < b < - 2$ , is

$\int_{- 1}^{1} g^{'} (x) d x =$

Finding the First Derivative $y^{'}$

Calculating the Second Derivative $y^{''}$

Evaluating $f^{''} (- 102)$

Analyzing the Function $g (x)$

Evaluating the Integral of $g^{'} (x)$

The area of the region bounded by the curves $y = ∣ x - 2∣, x = 1, x = 3$ and the x-axis is

The area of the bounded region enclosed by the curve $y = 3 - x - \frac{1}{2} - ∣ x + 1∣$ and the x-axis is

Find all maxima and minima of the function $y = x (x - 1)^{2}, 0 \leq x \leq 2$ . Also determine the area bounded by the curve $y = x (x - 1)^{2}$ , the $y$ -axis and the line $y = 2$ .

The area of the region bounded by the parabola $(y - 2)^{2} = (x - 1)$ , the tangent to it at the point whose ordinate is 3 and the $x$ -axis is :