Application Of Derivatives – JEE Main MathematicsPractice Questions & PYQs

Integrating a differential equation allows us to find the family of curves satisfying a specific slope, while a point through which the curve passes determines the unique constant of integration. The circle equation $x^2 + y^2 - 2x - 2y = 0$ simplifies to $(x-1)^2 + (y-1)^2 = 2$, identifying the center of the circle as $(1, 1)$.

Separating variables in the differential equation $\frac{dy}{dx} = \frac{2y}{x^2}$ gives $\int \frac{1}{y} dy = \int \frac{2}{x^2} dx$, which integrates to $\log_{e}|y| = -\frac{2}{x} + C$. Substituting the center point $(1, 1)$ into this result reveals that $0 = -2 + C$, leading to $C = 2$. The final equation becomes $\log_{e}|y| = -\frac{2}{x} + 2$, which, upon multiplying both sides by $x$, results in $x \log_{e}|y| = -2 + 2x$ or $x \log_{e}|y| = 2(x-1)$.

Finding local extrema requires locating critical points where the function's rate of change is zero and then determining the behavior of the curve as it passes through those points. By differentiating the function, we find:
$f'(x)=36 x^{3}+36 x^{2}-72 x=36 x(x-1)(x+2)$
The roots of this derivative, $x = 0$, $x = 1$, and $x = -2$, serve as the potential locations for local maxima and minima.

Determining the nature of these points involves assessing how $f'(x)$ changes sign across the intervals defined by these roots. A local minimum occurs when the function transitions from decreasing to increasing, whereas a local maximum occurs when it switches from increasing to decreasing. Observing these sign changes across the number line reveals that $x = -2$ and $x = 1$ correspond to local minima, while $x = 0$ corresponds to a local maximum. Therefore, the set of local minimum points $S_{1}$ is $\{-2, 1\}$ and the set of local maximum points $S_{2}$ is $\{0\}$.

Parallel lines must have identical slopes, meaning the derivative of the curve at a specific point $x$ must equal the slope of the line segment connecting the points $(1, -2)$ and $(-1, 0)$. The slope of this segment is $\frac{-2 - 0}{1 - (-1)}$, which simplifies to $-1$. Because the derivative of the function is $f'(x) = 3x^2 - 2x - 2$, we set $3x^2 - 2x - 2 = -1$ to find the points where the tangent is parallel to this segment. Solving $3x^2 - 2x - 1 = 0$ leads to the factored form $(3x + 1)(x - 1) = 0$, identifying the values $x = -\frac{1}{3}$ and $x = 1$ as the set $S$.

Two lines are parallel when their slopes are identical, so the tangent to a curve must share the same slope as the line it is parallel to. Rearranging the line equation $2y = 4x + 1$ into slope-intercept form gives $y = 2x + \frac{1}{2}$, revealing a slope of $2$. By finding the derivative of the curve $y = x^2 - 5x + 5$, we obtain the slope function $\frac{dy}{dx} = 2x - 5$. Equating this slope function to the required value of $2$ leads to $2x - 5 = 2$, which gives $x = \frac{7}{2}$.

With the x-coordinate determined, we calculate the corresponding y-coordinate on the curve: $y = \left(\frac{7}{2}\right)^2 - 5\left(\frac{7}{2}\right) + 5 = \frac{49}{4} - \frac{35}{2} + 5 = -\frac{1}{4}$. Now that we have the point of tangency, $\left(\frac{7}{2}, -\frac{1}{4}\right)$, and the slope of $2$, the equation of the tangent line follows the point-slope form $y - \left(-\frac{1}{4}\right) = 2\left(x - \frac{7}{2}\right)$. Simplifying this yields $y + \frac{1}{4} = 2x - 7$, which rearranges to $2x - y - \frac{29}{4} = 0$. Testing the coordinates from the provided options, substituting $x = \frac{1}{8}$ and $y = -7$ into the expression $2x - y - \frac{29}{4}$ results in $2\left(\frac{1}{8}\right) - (-7) - \frac{29}{4} = \frac{1}{4} + 7 - \frac{29}{4} = \frac{1 + 28 - 29}{4} = 0$, confirming this point lies on the tangent.

A polynomial of degree four with local extrema at $x = -1, 0, 1$ must have a derivative that vanishes at these points, expressed as $f'(x) = \lambda x(x^2 - 1) = \lambda(x^3 - x)$ for some non-zero constant $\lambda$. Integrating this derivative identifies the general form of the function as:
$f(x) = \lambda \left( \frac{x^4}{4} - \frac{x^2}{2} \right) + C$

Equating $f(x)$ to $f(0)$, which is simply the constant $C$, leads to $\lambda \left( \frac{x^4}{4} - \frac{x^2}{2} \right) + C = C$. Dividing by the non-zero constant $\lambda$ yields the simplified equation $\frac{x^2}{2} \left( \frac{x^2}{2} - 1 \right) = 0$, which provides roots at $x = 0$ and $x = \pm\sqrt{2}$. This confirms the set contains exactly one rational number, $0$, and two irrational numbers, $\sqrt{2}$ and $-\sqrt{2}$.

Determining the maximum value of a function over a restricted domain relies on first identifying the valid interval for the input variable and subsequently evaluating the function's behavior within that range. The inequality $x^2 + 30 \le 11x$ is equivalent to $x^2 - 11x + 30 \le 0$, which factors into $(x - 5)(x - 6) \le 0$, confining the variable $x$ to the closed interval $[5, 6]$.

To find the maximum within this interval, examine the derivative $f'(x) = 9x^2 - 36x + 27$, which simplifies to $9(x - 1)(x - 3)$. For any $x$ in the interval $[5, 6]$, both $(x - 1)$ and $(x - 3)$ are positive, making $f'(x)$ positive across the entire domain $S$. Since the derivative is positive, the function $f(x)$ is strictly increasing, ensuring the largest value occurs at the rightmost endpoint, $x = 6$. Evaluating the function at this point yields $3(6)^3 - 18(6)^2 + 27(6) - 40$, which calculates to $648 - 648 + 162 - 40$, resulting in a maximum value of 122.

The slope of a tangent line at any point on a curve corresponds to the derivative of the function at that location, while two lines are perpendicular if the product of their slopes equals $-1$. First, rewrite the given line equation $-x + y + 4 = 0$ in slope-intercept form to find its slope of $1$. Because the tangent is perpendicular to this line, the slope of the tangent must be the negative reciprocal, which is $-1$.

Differentiating the curve equation $y = x^3 + ax - b$ gives the slope function $\frac{dy}{dx} = 3x^2 + a$. Substituting the x-coordinate of the point $(1, -5)$ into this derivative gives the equation $3(1)^2 + a = -1$, which simplifies to $3 + a = -1$ and confirms that $a = -4$. With $a$ determined, substitute the point $(1, -5)$ and the value of $a$ into the original curve equation:
$-5 = 1^3 + (-4)(1) - b$
Solving this yields $-5 = -3 - b$, which means $b = 2$. Consequently, the equation for the curve is $y = x^3 - 4x - 2$. By testing the provided options, the point $(2, -2)$ satisfies the equation since $2^3 - 4(2) - 2 = 8 - 8 - 2 = -2$.

The volume of a right circular cone is defined by the formula $V = \frac{1}{3} \pi r^2 h$. Because the slant height is fixed at 3, the radius squared and the height are related by the Pythagorean theorem, such that $r^2 = 9 - h^2$. Substituting this relationship into the volume formula gives the function $V = \frac{1}{3} \pi (9h - h^3)$. To identify the height that maximizes this volume, calculate the derivative with respect to $h$, set the resulting expression to zero, and solve: $\frac{dV}{dh} = \frac{1}{3} \pi (9 - 3h^2) = 0$, which leads to $h = \sqrt{3}$. Substituting this value of $h$ back into the volume formula yields $V = \frac{1}{3} \pi (9\sqrt{3} - 3\sqrt{3}) = 2\sqrt{3} \pi$.

This scenario is best understood through the Pythagorean theorem, which links the distance of the ladder's base from the wall $x$ and its top from the floor $y$ to the constant ladder length $L$ via $x^2 + y^2 = L^2$. Differentiating this expression with respect to time $t$ yields $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$, which demonstrates that the rate at which the base slides away from the wall is fundamentally tied to the rate at which the top slides down.

With the ladder length set at 2 meters, the geometric relationship becomes $x^2 + y^2 = 4$. When the top is 1 meter above the ground, $y=1$, which implies $x = \sqrt{3}$ meters. Substituting the known values $x=\sqrt{3}$, $y=1$, and the downward velocity $\frac{dy}{dt} = -25$ cm/sec into the differentiated rate equation $x \frac{dx}{dt} + y \frac{dy}{dt} = 0$ results in $\sqrt{3} \frac{dx}{dt} + 1(-25) = 0$. Solving this for $\frac{dx}{dt}$ gives the rate as $\frac{25}{\sqrt{3}}$ cm/sec.

The geometry of a right circular cone establishes a fixed relationship between its radius and height based on the semi-vertical angle. Since the tangent of this angle is $\frac{1}{2}$, the ratio of the radius to the height remains constant at $r = \frac{h}{2}$. By substituting this ratio into the standard volume formula $V = \frac{1}{3}\pi r^2 h$, the volume is expressed solely as a function of the height, resulting in $V = \frac{\pi}{12} h^3$.

Differentiating this relationship with respect to time leads to $\frac{dV}{dt} = \frac{\pi}{4} h^2 \frac{dh}{dt}$. Inserting the known rate of water inflow of 5 m$^3$/min and the depth of 10 m gives the equation $\frac{\pi}{4} (10)^2 \frac{dh}{dt} = 5$. Simplifying this, $25\pi \frac{dh}{dt} = 5$, which results in the rate of change in height being $\frac{1}{5\pi}$ m/min.

The slope of the tangent line to the curve $y = xe^{x^2}$ at any point is determined by the derivative of the function, calculated using the product rule as $y' = e^{x^2} + x(e^{x^2} \cdot 2x) = e^{x^2}(1 + 2x^2)$. Substituting $x = 1$ into this derivative yields a slope of $3e$ at the point $(1, e)$. Using the point-slope form $y - y_1 = m(x - x_1)$, the equation of the tangent line becomes $y - e = 3e(x - 1)$, which simplifies to $y = 3ex - 2e$. The point $\left(\frac{4}{3}, 2e\right)$ lies on this line because substituting $x = \frac{4}{3}$ into the tangent equation results in $y = 3e(\frac{4}{3}) - 2e = 4e - 2e = 2e$.

The slope of a tangent line to the curve $y = \frac{x}{x^{2}-3}$ at any point $(\alpha, \beta)$ is determined by the derivative of the function, which must be equal to the slope of the line $2x + 6y - 11 = 0$. First, applying the quotient rule to differentiate $y$, we find the expression for the derivative:

$\frac{dy}{dx} = \frac{1(x^2 - 3) - x(2x)}{(x^2 - 3)^2} = \frac{-(x^2 + 3)}{(x^2 - 3)^2}$

The given line $2x + 6y - 11 = 0$ can be rewritten as $y = -\frac{1}{3}x + \frac{11}{6}$, which reveals a slope of $-\frac{1}{3}$. Setting the derivative at $(\alpha, \beta)$ equal to this slope gives the following equation:

$\frac{-(\alpha^2 + 3)}{(\alpha^2 - 3)^2} = -\frac{1}{3}$

Multiplying both sides by $-3(\alpha^2 - 3)^2$ leads to $3(\alpha^2 + 3) = (\alpha^2 - 3)^2$, which simplifies to $3\alpha^2 + 9 = \alpha^4 - 6\alpha^2 + 9$. This further reduces to $\alpha^4 - 9\alpha^2 = 0$, implying $\alpha^2(\alpha^2 - 9) = 0$. Since the point cannot be $(0,0)$, we have $\alpha^2 = 9$, meaning $\alpha = 3$ or $\alpha = -3$. For $\alpha = 3$, the corresponding coordinate is $\beta = \frac{3}{3^2 - 3} = \frac{1}{2}$. Evaluating the expression $2\alpha + 6\beta$ using these values yields $2(3) + 6(1/2) = 6 + 3 = 9$.

The function $f(x) = x \sqrt{kx - x^2}$ is increasing on the interval $[0, 3]$ only if its derivative remains non-negative throughout this domain. A useful approach involves examining the squared function to simplify the derivative calculations:
$g(x) = f(x)^2 = kx^3 - x^4$
The derivative of this function is:
$g'(x) = 3kx^2 - 4x^3 = x^2(3k - 4x)$
For $g'(x) \ge 0$ to hold for all $x \in [0, 3]$, the term $(3k - 4x)$ must be non-negative. Because $4x$ reaches its maximum value at $x=3$, the constraint $3k - 12 \ge 0$ must be satisfied, which identifies the minimum value as $m = 4$. With $k=4$, the function $f(x) = x \sqrt{4x - x^2}$ is monotonically increasing on the interval, ensuring the maximum value $M$ occurs at the upper boundary $x=3$. Calculating the output at this point gives:
$M = f(3) = 3 \sqrt{4(3) - 3^2} = 3 \sqrt{12 - 9} = 3 \sqrt{3}$
This results in the ordered pair $(m, M) = (4, 3 \sqrt{3})$.

Related rates problems apply the chain rule to connect the time-dependent changes of volume and physical dimensions. Since the iron ball has a fixed radius of 10 cm and is covered by an ice layer of thickness $x$, the total radius of the sphere is $10 + x$. The total volume $V$ depends on this total radius, so we differentiate the volume formula with respect to time $t$ to determine how the thickness changes as the volume decreases.

$V = \frac{4}{3}\pi (10+x)^3$
$\frac{dV}{dt} = 4\pi (10+x)^2 \frac{dx}{dt}$

Given that the ice melts at a rate of 50 cm$^3$/min at the instant the thickness $x$ is 5 cm, we substitute these values into the differentiated equation:

$50 = 4\pi (10+5)^2 \frac{dx}{dt}$
$50 = 4\pi (15)^2 \frac{dx}{dt}$
$50 = 4\pi (225) \frac{dx}{dt}$
$50 = 900\pi \frac{dx}{dt}$

Dividing both sides by $900\pi$ yields the rate at which the thickness decreases:

$\frac{dx}{dt} = \frac{50}{900\pi} = \frac{1}{18\pi}$

The monotonicity of a function is determined by examining the sign of its first derivative. Differentiating $f(x) = \frac{x}{\sqrt{a^{2+}x^2}} - \frac{d-x}{\sqrt{b^{2+}(d-x)^2}}$ yields the expression $f'(x) = \frac{a^2}{(a^{2+}x^2)^{3/2}} + \frac{b^2}{(b^{2+}(d-x)^2)^{3/2}}$. Because $a^2$ and $b^2$ are positive and the denominators are powers of positive squared terms, $f'(x)$ remains strictly greater than zero for all $x \in R$. This positive derivative confirms that the function is increasing.

Maximizing the volume of a cylinder inside a sphere relies on expressing the cylinder's volume as a function of a single variable, its height. Within a sphere of radius 3, the Pythagorean relationship between the sphere's radius, half the cylinder's height, and the cylinder's base radius gives $r^2 + (\frac{h}{2})^2 = 3^2$, which simplifies to $r^2 = 9 - \frac{h^2}{4}$. Substituting this expression into the cylinder volume formula $V = \pi r^2 h$ results in $V = \pi(9 - \frac{h^2}{4})h$. Setting the first derivative $\frac{dV}{dh} = \pi(9 - \frac{3h^2}{4})$ equal to zero leads to the equation $9 = \frac{3h^2}{4}$, which gives $h^2 = 12$. Solving for height yields $h = 2\sqrt{3}$, and the negative second derivative at this point confirms that this dimension provides the maximum possible volume.

When the second derivative of a function is positive, its rate of change-the first derivative-is strictly increasing throughout the domain. To determine the behavior of $\phi(x)=f(x)+f(2-x)$, we examine its derivative, $\phi'(x)=f'(x)-f'(2-x)$, which depends entirely on the relative magnitude of $f'(x)$ and $f'(2-x)$.

In the interval $(0, 1)$, any chosen $x$ is strictly less than its symmetric counterpart $2-x$. Because $f'(x)$ is an increasing function, it follows that $f'(x) < f'(2-x)$, resulting in a negative derivative $\phi'(x)$ and indicating that the function is decreasing. Conversely, throughout the interval $(1, 2)$, the value of $x$ is always greater than $2-x$, which implies $f'(x) > f'(2-x)$. This makes the difference $\phi'(x)$ positive, causing the function to increase across this range.

The derivative of a function evaluated at a turning point where it transitions from increasing to decreasing is always zero. Given $f(x) = x^3 - 3(a - 2)x^2 + 3ax + 7$, its derivative is $f'(x) = 3x^2 - 6(a - 2)x + 3a$. Applying the condition at $x = 1$, we set $f'(1) = 3(1)^2 - 6(a - 2)(1) + 3a = 0$. Simplifying this arithmetic gives $3 - 6a + 12 + 3a = 0$, or $15 = 3a$, which clearly establishes that $a = 5$.

With $a$ identified as 5, the function becomes $f(x) = x^3 - 3(3)x^2 + 3(5)x + 7$, which simplifies to $f(x) = x^3 - 9x^2 + 15x + 7$. We are tasked with finding the root of $\frac{f(x)-14}{(x-1)^2} = 0$ for $x \neq 1$, which is equivalent to solving $f(x) - 14 = 0$. Subtracting 14 from the polynomial results in $x^3 - 9x^2 + 15x - 7 = 0$. Dividing this cubic equation by $(x-1)^2$ yields the factor $(x - 7)$, confirming that the only valid root for the equation is $x = 7$.

The volume of a cone inscribed in a sphere of radius $R$ is expressed as $V = \frac{1}{3}\pi r^2 h$, where $r$ is the base radius and $h$ is the height. Expressing $r$ and $h$ in terms of $R$ and the distance $x$ from the sphere's center to the base, we have $h = R + x$ and $r^2 = R^2 - x^2$, leading to $V = \frac{1}{3}\pi(R^2 - x^2)(R + x)$. Maximizing this volume by setting the derivative with respect to $x$ to zero, $\frac{dV}{dx} = \frac{1}{3}\pi(R^2 + 2Rx - 3x^2) = 0$, results in $x = \frac{R}{3}$. Substituting $R = 3$ cm gives a height $h = 3 + 1 = 4$ cm and a radius $r = \sqrt{3^2 - 1^2} = 2\sqrt{2}$ cm. The slant height is calculated as $l = \sqrt{r^2 + h^2} = \sqrt{8 + 16} = 2\sqrt{6}$ cm. The curved surface area of the cone, defined by the formula $\pi r l$, is $\pi(2\sqrt{2})(2\sqrt{6}) = 4\pi\sqrt{12} = 8\sqrt{3}\pi \text{ cm}^2$.

By introducing the substitution $t = x - \frac{1}{x}$, the expression $f(x)$ can be rewritten as $t^2 + 2$ because expanding $t^2$ yields $x^2 + \frac{1}{x^2} - 2$. Substituting this into the definition of $h(x)$ transforms the function into $h(t) = \frac{t^{2+}2}{t}$, which simplifies to $h(t) = t + \frac{2}{t}$. Applying the arithmetic mean-geometric mean inequality to this expression for positive values of $t$ provides $t + \frac{2}{t} \geq 2\sqrt{t \cdot \frac{2}{t}} = 2\sqrt{2}$. Consequently, the local minimum value of the function is $2\sqrt{2}$.