Application Of Derivatives – JEE Main MathematicsPractice Questions & PYQs

The geometry of a tangent and a normal line relies on the fact that the normal at any point on a curve is perpendicular to the tangent at that same point, meaning their slopes are negative reciprocals of one another. By applying implicit differentiation to the curve $y^2 - 3x^2 + y + 10 = 0$, we find $2yy' - 6x + y' = 0$, which allows us to express the slope of the tangent $m$ at a point $P(\alpha, \beta)$ as $m = \frac{6\alpha}{2\beta + 1}$.

Since the normal line passes through the point $P(\alpha, \beta)$ and the point $(0, 3/2)$, its slope can also be determined by the coordinates as $\frac{\beta - 3/2}{\alpha - 0}$, which simplifies to $\frac{2\beta - 3}{2\alpha}$. Because this slope must equal the negative reciprocal of the tangent slope, we set $\frac{2\beta - 3}{2\alpha} = -\frac{2\beta + 1}{6\alpha}$. Simplifying this equation by cross-multiplying leads to $3(2\beta - 3) = -(2\beta + 1)$, or $6\beta - 9 = -2\beta - 1$, which solves to $8\beta = 8$, indicating that $\beta = 1$.

Substituting $\beta = 1$ back into the original curve equation $y^2 - 3x^2 + y + 10 = 0$ gives $1^2 - 3\alpha^2 + 1 + 10 = 0$, which results in $12 - 3\alpha^2 = 0$, or $\alpha^2 = 4$. Using these findings in the expression for $m$, we calculate $m = \frac{6\alpha}{2(1) + 1} = \frac{6\alpha}{3} = 2\alpha$. Given $\alpha^2 = 4$, $\alpha$ is $\pm 2$, meaning $m = \pm 4$. Therefore, the absolute value $|m|$ is 4.

Determining the nature of extrema involves finding the critical points where the derivative vanishes and examining sign changes in the neighborhood of these points. Applying the product rule to $f(x)= (3 x^{2}+a x-2-a ) e^{x}$ yields the derivative $f'(x) = e^x(3x^2 + (a+6)x - 2)$, and substituting $x=1$ as a critical point allows us to solve $3(1)^2 + (a+6)(1) - 2 = 0$ for $a = -7$. With $a$ determined, the derivative simplifies to $f'(x) = e^x(3x^2 - x - 2)$, which factors into $e^x(3x+2)(x-1)$, revealing critical points at $x=1$ and $x= - \frac{2}{3}$. Because $f'(x)$ changes from positive to negative as $x$ crosses $- \frac{2}{3}$ and from negative to positive as $x$ crosses $1$, the function exhibits a local maximum at $x= - \frac{2}{3}$ and a local minimum at $x=1$.

A function increases on intervals where its derivative is strictly positive. Expanding the original function gives $f(x) = 3x^{\frac{5}{3}} - 7x^{\frac{2}{3}}$, and applying the power rule allows us to calculate the derivative as $f'(x) = 5x^{\frac{2}{3}} - \frac{14}{3}x^{-\frac{1}{3}}$. To evaluate the sign of this derivative, we combine the terms over a common denominator to arrive at the simplified expression $f'(x) = \frac{15x - 14}{3x^{\frac{1}{3}}}$. The derivative is positive when the numerator and denominator share the same sign, which occurs when $x < 0$ (making both negative) and when $x > \frac{14}{15}$ (making both positive), confirming that the function is increasing on $(-\infty , 0) \cup \left(\frac{14}{15}, \infty \right)$.

To locate the point on the curve $y = x^{2} + 7x + 2$ closest to the line $y = 3x - 3$, one must find the point where the curve's tangent is parallel to the given line. Since the line $y = 3x - 3$ has a slope of $3$, we equate the derivative of the parabola, $\frac{dy}{dx} = 2x + 7$, to $3$. Solving $2x + 7 = 3$ gives $x = -2$, and substituting this back into the original parabola equation yields $y = (-2)^{2} + 7(-2) + 2 = -8$, confirming the contact point is $P(-2, -8)$.

The normal line at $P$ is perpendicular to the tangent, meaning its slope must be the negative reciprocal of $3$, which is $-\frac{1}{3}$. Using the point-slope form with the point $(-2, -8)$ and slope $-\frac{1}{3}$, we write the equation as $y - (-8) = -\frac{1}{3}(x - (-2))$. Multiplying through by $3$ results in $3y + 24 = -x - 2$, which simplifies to the final equation $x + 3y + 26 = 0$.

Rolle's theorem guarantees that for a function continuous on $[3, 4]$ and differentiable on $(3, 4)$, if $f(3) = f(4)$, there exists at least one point $c$ where $f'(c) = 0$. Equating the two boundary values $\log_e\left(\frac{9+\alpha}{21}\right) = \log_e\left(\frac{16+\alpha}{28}\right)$ leads to the proportion $\frac{9+\alpha}{21} = \frac{16+\alpha}{28}$, which simplifies to $\alpha = 12$ after cross-multiplication. With the function established as $f(x) = \log_e\left(\frac{x^{2+}12}{7x}\right)$, the derivative $f'(x) = \frac{2x}{x^{2+}12} - \frac{1}{x} = \frac{x^{2-}12}{x(x^{2+}12)}$ reveals that $f'(c) = 0$ when $c^2 = 12$.

Deriving $f'(x) = \frac{x^{2-}12}{x^{3+}12x}$ using the quotient rule allows us to find $f''(c)$ by evaluating the expression at $c^2 = 12$. Because the term involving $(x^{2-}12)$ in the quotient rule becomes zero at $c$, the expression simplifies significantly to $f''(c) = \frac{2c^2(c^{2+}12)}{c^2(c^{2+}12)^2} = \frac{2}{c^{2+}12}$. Substituting $c^2 = 12$ into this result yields $\frac{2}{12+12} = \frac{1}{12}$.

Because the parabola $y = x^2 - 1$ is symmetric about the y-axis, we can define the rectangle's width as $2t$ by placing its right vertex at an x-coordinate of $t$ and its left vertex at $-t$, where $0 < t < 1$. The height of the rectangle is the vertical distance between the x-axis and the parabola, which is $1 - t^2$ since the curve lies below the x-axis. Multiplying these dimensions yields the area function $A = 2t(1 - t^2)$, which simplifies to $A = 2t - 2t^3$.

Calculating the derivative $\frac{dA}{dt} = 2 - 6t^2$ and setting it equal to zero identifies the critical value $t = \frac{1}{\sqrt{3}}$ where the area is maximized. Substituting this value into the area formula gives $A = 2 \left( \frac{1}{\sqrt{3}} \right) \left( 1 - \frac{1}{3} \right)$, resulting in a maximum area of $\frac{4}{3\sqrt{3}}$.

Calculating the derivative of $f(x)$ yields $f'(x) = \frac{x - (1+x)\ln(1+x)}{x^{2}(1+x)}$. Assessing the sign of this derivative requires examining the numerator, so we define an auxiliary function $h(x) = x - (1+x)\ln(1+x)$. Differentiating $h(x)$ yields $h'(x) = -\ln(1+x)$, which is positive when $x \in (-1, 0)$ and negative when $x \in (0, \infty )$.

Because $h(0) = 0$ represents the maximum value of this auxiliary function, $h(x)$ is strictly less than zero for all other values in the domain $(-1, \infty )$. Given that the denominator $x^{2}(1+x)$ is always positive on the interval $(-1, \infty )$ excluding zero, the derivative $f'(x)$ is negative for all $x$. This confirms that $f(x)$ is a decreasing function throughout the interval $(-1, \infty )$.

The Mean Value Theorem ensures there exists a time $t$ where the instantaneous velocity matches the average velocity of the moving car over the interval $[t_1, t_2]$. Calculating the average speed gives $\frac{(at_2^2 + bt_2 + c) - (at_1^2 + bt_1 + c)}{t_2 - t_1}$, which simplifies to $a(t_1 + t_2) + b$ after factoring the difference of squares and dividing. Equating this expression to the derivative $f'(t) = 2at + b$ allows us to cancel common terms, resulting in $2at = a(t_1 + t_2)$ and leading to the solution $t = \frac{t_1 + t_2}{2}$.

A cubic polynomial that possesses local extrema at specific points must have a derivative that vanishes at those same points. Since $x=1$ and $x=2$ are the locations of the local maximum and minimum, the derivative $p'(x)$ must be a quadratic expression of the form $p'(x) = A(x-1)(x-2)$, where $A$ is a constant. Expanding this product yields $p'(x) = A(x^2 - 3x + 2)$, which can be integrated to determine the expression for $p(x)$ as $p(x) = A(\frac{x^3}{3} - \frac{3x^2}{2} + 2x) + C$, where $C$ represents the constant of integration.

Substituting the known values of the polynomial at these extrema allows us to construct a system of linear equations. At $x=1$, the condition $p(1) = 8$ leads to $8 = A(\frac{1}{3} - \frac{3}{2} + 2) + C$, which simplifies to $\frac{5A}{6} + C = 8$. Similarly, at $x=2$, the condition $p(2) = 4$ yields $4 = A(\frac{8}{3} - 6 + 4) + C$, which simplifies to $\frac{2A}{3} + C = 4$. Subtracting the second equation from the first isolates $A$ by showing that $\frac{A}{6} = 4$, which implies $A = 24$. Replacing $A$ with $24$ in the second equation results in $16 + C = 4$, revealing that $C = -12$. Because $p(0) = A(0) + C$, the value we are looking for is exactly $C$, confirming that $p(0) = -12$.

Maximizing the distance between a fixed point $Q(0,-4)$ and a point $P$ on the ellipse $4x^{2}+5y^{2}=20$ involves expressing the squared distance as a function of the ellipse's parametric coordinates. By converting the ellipse equation to $\frac{x^{2}}{5}+\frac{y^{2}}{4}=1$, we parameterize any point $P$ as $(\sqrt{5} \cos \theta, 2 \sin \theta)$ and calculate $PQ^{2} = (\sqrt{5} \cos \theta)^{2} + (2 \sin \theta + 4)^{2}$.

Expanding this expression yields $5 \cos^{2} \theta + 4 \sin^{2} \theta + 16 \sin \theta + 16$, which simplifies to $-\sin^{2} \theta + 16 \sin \theta + 21$ by substituting $\cos^{2} \theta = 1 - \sin^{2} \theta$. Since $\sin \theta$ ranges from -1 to 1, the expression $-\sin^{2} \theta + 16 \sin \theta + 21$ increases as $\sin \theta$ increases, reaching its maximum value when $\sin \theta = 1$. Consequently, the maximum squared distance is $-1 + 16 + 21 = 36$.

Substituting $\sin^{2}x = 1 - \cos^{2}x$ into the expression for $f(x)$ simplifies the function to $f(x) = \lambda \sin^{2}x + \sin^{3}x$. Differentiating this with respect to $x$ yields $f'(x) = 2\lambda \sin x \cos x + 3 \sin^{2}x \cos x$, which factors neatly into $f'(x) = \sin x \cos x (2\lambda + 3 \sin x)$. Given that $\cos x$ is never zero within the open interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, the critical points occur where $\sin x = 0$ or where $\sin x = -\frac{2\lambda}{3}$.

When $\lambda = 0$, the function becomes $f(x) = \sin^{3}x$, which is strictly increasing and thus lacks local extrema, necessitating that $\lambda \neq 0$. To ensure the function has exactly one maxima and one minima, the second critical point $\sin x = -\frac{2\lambda}{3}$ must exist as a distinct value within the valid range of $\sin x$, which is $(-1, 1)$. This requires the inequality $-1 < -\frac{2\lambda}{3} < 1$ to hold. Solving this inequality by multiplying across by $-\frac{3}{2}$ reveals that $\lambda$ must be in the range $\left(-\frac{3}{2}, \frac{3}{2}\right)$ while explicitly excluding zero.

The Lagrange Mean Value Theorem establishes a direct relationship between the derivative constraint and the function values at the endpoints of an interval, stating that for a function continuous on $[a, b]$ and differentiable on $(a, b)$, the inequality $\frac{f(b) - f(a)}{b - a} \leq K$ holds if $f'(x) \leq K$ for all $x \in (a, b)$. Given $f'(x) \leq 2$ and $f(-7) = -3$, we can determine upper bounds for $f(-1)$ and $f(0)$ by applying this theorem over the intervals $[-7, -1]$ and $[-7, 0]$.

For the interval $[-7, -1]$, the inequality becomes $\frac{f(-1) - (-3)}{-1 - (-7)} \leq 2$, which simplifies to $f(-1) + 3 \leq 12$, establishing $f(-1) \leq 9$. Similarly, applying the theorem to the interval $[-7, 0]$ yields $\frac{f(0) - (-3)}{0 - (-7)} \leq 2$, which simplifies to $f(0) + 3 \leq 14$, establishing $f(0) \leq 11$. Adding these two inequalities provides the combined bound $f(-1) + f(0) \leq 9 + 11 = 20$, and since there is no lower bound defined for the function, the expression lies in the interval $(-\infty , 20]$.

Differentiating the implicit function $x^4 e^y + 2 \sqrt{y+1} = 3$ with respect to $x$ using the product rule and chain rule yields $4x^3 e^y + x^4 e^y y' + \frac{y'}{\sqrt{y+1}} = 0$. Substituting the point $(1, 0)$ into this derivative leads to $4(1) + y' + y' = 0$, which simplifies to $4 + 2y' = 0$ and establishes a slope of $y' = -2$. The point-slope form $y - 0 = -2(x - 1)$ simplifies to the linear equation $2x + y = 2$. Among the choices provided, only the point $(-2, 6)$ satisfies this relationship because $2(-2) + 6 = 2$.

The intersection points of the curve $y=x^{2}-3x+2$ with the $x$-axis are determined by setting $y=0$, which factors into $(x-1)(x-2)=0$ to provide the coordinates $(1,0)$ and $(2,0)$. The rate of change of the curve is defined by the derivative $\frac{dy}{dx}=2x-3$, which evaluates to $-1$ at $x=1$ and $1$ at $x=2$.

The line $x+y=a$ can be rewritten as $y=-x+a$, identifying a slope of $-1$, which corresponds to the point $(1,0)$. Substituting these coordinates into the equation results in $1+0=a$, giving $a=1$. Similarly, the line $x-y=b$ rearranges to $y=x-b$, which has a slope of $1$ and corresponds to the point $(2,0)$. Substituting these values into the equation gives $2-0=b$, so $b=2$. Consequently, the ratio $\frac{a}{b}$ is $\frac{1}{2}$, which is equal to 0.50.

The volume $V$ of the ice layer surrounding the iron ball with a constant radius of 10 cm and a thickness $h$ is defined by $V = \frac{4}{3}\pi((10+h)^3 - 10^3)$. Differentiating this volume with respect to time $t$ provides the rate of change relationship
$\frac{dV}{dt} = 4\pi(10+h)^2 \frac{dh}{dt}$
Substituting the melting rate of $50\,\mathrm{cm}^3/\mathrm{min}$ and the thickness $h = 5\,\mathrm{cm}$ into this expression results in
$50 = 4\pi(10+5)^2 \frac{dh}{dt}$
Solving the simplified equation $50 = 900\pi \frac{dh}{dt}$ yields a rate of decrease for the thickness of $\frac{1}{18\pi}\,\mathrm{cm/min}$.

When $x$ is in the interval $[0, \pi/2]$, the absolute value simplifies to $|x| = x$, allowing us to write the function as $f(x) = x(\frac{\pi}{2} + x) = \frac{\pi}{2}x + x^2$. Conversely, for $x \in [-\pi/2, 0)$, the absolute value becomes $|x| = -x$, leading to $f(x) = x(\frac{\pi}{2} - x) = \frac{\pi}{2}x - x^2$. Calculating the derivatives for these defined ranges yields $f'(x) = \frac{\pi}{2} + 2x$ when $x > 0$ and $f'(x) = \frac{\pi}{2} - 2x$ when $x < 0$. Because the derivative expression $f'(x) = \frac{\pi}{2} - 2x$ possesses a negative slope, the function $f'$ decreases on the interval $(-\pi/2, 0)$, whereas the expression $f'(x) = \frac{\pi}{2} + 2x$ has a positive slope, causing $f'$ to increase on the interval $(0, \pi/2)$.

Parallel lines share the same slope, so the slope of the tangent at $(c, f(c))$ is identical to the slope of the line segment connecting $(1, 0)$ and $(e, e)$, which is $\frac{e - 0}{e - 1} = \frac{e}{e - 1}$. Differentiating the curve $f(x) = x \log_e x$ via the product rule yields $f'(x) = 1 + \log_e x$, so the tangent slope at $x = c$ is $1 + \log_e c$. Setting these expressions equal results in the equation $1 + \log_e c = \frac{e}{e - 1}$, which simplifies by subtracting 1 from both sides to $\log_e c = \frac{e - (e - 1)}{e - 1} = \frac{1}{e - 1}$. Raising $e$ to the power of both sides provides the value
$c = e^{\frac{1}{e - 1}}$

Lagrange's mean value theorem asserts that for a function continuous and differentiable on an interval $[a, b]$, there is at least one point $c$ where the instantaneous slope of the function equals the slope of the secant line passing through the endpoints. Applying this to $f(x) = x^3 - 4x^2 + 8x + 11$ on the interval $[0, 1]$, find the slope of the secant line by computing $f(0) = 11$ and $f(1) = 16$, which gives an average slope of $\frac{16 - 11}{1 - 0} = 5$. Differentiating the original function yields $f'(x) = 3x^2 - 8x + 8$, and equating this derivative to the secant slope results in $3c^2 - 8c + 8 = 5$, which simplifies to the quadratic equation $3c^2 - 8c + 3 = 0$. Applying the quadratic formula provides two potential values for $c$, but since $c$ must lie within the interval $[0, 1]$, the only acceptable solution is $\frac{4 - \sqrt{7}}{3}$.

Representing any point on the curve $y = \sqrt{x}$ as $(t^2, t)$ for $t > 0$ allows the squared distance from the point $(\frac{3}{2}, 0)$ to be expressed as $D^2 = (t^2 - \frac{3}{2})^2 + t^2$. Expanding this expression yields $D^2 = t^4 - 3t^2 + \frac{9}{4} + t^2$, which simplifies to $D^2 = t^4 - 2t^2 + \frac{9}{4}$. By completing the square, we obtain $D^2 = (t^2 - 1)^2 + \frac{5}{4}$, where the minimum value is clearly attained when the squared term vanishes. Therefore, the minimum distance simplifies to $\sqrt{\frac{5}{4}}$, resulting in $\frac{\sqrt{5}}{2}$.

When determining the point on a curve closest to a specific external point, the geometric condition requires the line segment connecting the two points to be perpendicular to the tangent of the curve at that location, effectively acting as the normal. For the helicopter following the path defined by $y = x^{3/2} + 7$, the slope of the tangent at any point is given by $\frac{dy}{dx} = \frac{3}{2}x^{1/2}$, which implies that the slope of the normal line is $-\frac{2}{3\sqrt{x}}$. By equating this normal slope to the slope of the line joining the current position $(x, x^{3/2} + 7)$ to the shooter at $(\frac{1}{2}, 7)$, we establish the relationship $\frac{x^{3/2}}{x - 1/2} = -\frac{2}{3\sqrt{x}}$.

Cross-multiplying results in $3x^2 = -2(x - 1/2)$, which simplifies to the quadratic equation $3x^2 + 2x - 1 = 0$. Factoring this expression gives $(3x - 1)(x + 1) = 0$, providing roots of $x = 1/3$ and $x = -1$; since the path restricts $x \ge 0$, we proceed with $x = 1/3$. Calculating the distance between $(\frac{1}{2}, 7)$ and $(\frac{1}{3}, (\frac{1}{3})^{3/2} + 7)$ using the distance formula yields $\sqrt{(\frac{1}{2} - \frac{1}{3})^2 + ((\frac{1}{3})^{3/2})^2}$. This calculation resolves to $\sqrt{\frac{1}{36} + \frac{1}{27}}$, which equals $\sqrt{\frac{3+4}{108}}$, resulting in the final value $\frac{1}{6} \sqrt{ \frac{7}{3} }$.