Application Of Derivatives – JEE Main MathematicsPractice Questions & PYQs

Two curves intersect at right angles when the product of their slopes at the point of intersection is $-1$. Differentiating the equation $y^2=6x$ gives $\frac{dy}{dx}=\frac{3}{y}$, while differentiating $9x^{2+}by^2=16$ gives $\frac{dy}{dx}=-\frac{9x}{by}$. Equating the product of these slopes to $-1$ results in $\frac{27x}{by^2}=1$, and substituting $y^2=6x$ from the first curve into this expression allows for the simplification $\frac{27x}{b(6x)}=1$. Solving this equation gives $\frac{27}{6b}=1$, which confirms that $b=\frac{9}{2}$.

The absolute extrema of a continuous function on a closed interval occur at critical points, where the first derivative is zero, or at the interval endpoints. The derivative $f'(x) = 6x^2 - 18x + 12 = 6(x-1)(x-2)$ identifies the critical points $x=1$ and $x=2$, both of which reside within the interval $[0, 3]$. Evaluating $f(x)$ at these points and the boundaries yields $f(0)=5$, $f(1)=10$, $f(2)=9$, and $f(3)=14$. Consequently, the absolute maximum value $M$ is $14$ and the absolute minimum value $m$ is $5$, which results in $M-m = 9$.

The equation of the normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at point $(x_1, y_1)$ is $\frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2$, which implies the slope of the normal is $m = \frac{a^2y_1}{b^2x_1}$. For the ellipse $x^2 + 3y^2 = 9$ where $a^2 = 9$ and $b^2 = 3$, the normal slopes at $P_1(3\cos\theta, \sqrt{3}\sin\theta)$ and $P_2(-3\sin\theta, \sqrt{3}\cos\theta)$ are $m_1 = \frac{9(\sqrt{3}\sin\theta)}{3(3\cos\theta)} = \sqrt{3}\tan\theta$ and $m_2 = \frac{9(\sqrt{3}\cos\theta)}{3(-3\sin\theta)} = -\sqrt{3}\cot\theta$.

The angle $\beta$ between these normals satisfies the relation $\tan\beta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$, which yields $\tan\beta = \left| \frac{\sqrt{3}\tan\theta + \sqrt{3}\cot\theta}{1 + (\sqrt{3}\tan\theta)(-\sqrt{3}\cot\theta)} \right|$. Simplifying the denominator gives $1 - 3 = -2$, and the numerator becomes $\sqrt{3}(\frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta}) = \frac{2\sqrt{3}}{\sin 2\theta}$. Thus, $\tan\beta = \left| \frac{2\sqrt{3}}{\sin 2\theta} \cdot \frac{1}{-2} \right| = \frac{\sqrt{3}}{\sin 2\theta}$, so $\cot\beta = \frac{\sin 2\theta}{\sqrt{3}}$. Substituting this into the required expression results in $\frac{2\cot\beta}{\sin 2\theta} = \frac{2(\frac{\sin 2\theta}{\sqrt{3}})}{\sin 2\theta} = \frac{2}{\sqrt{3}}$.

A function $f(x)$ is strictly increasing on an interval if its first derivative $f'(x) > 0$ for all values of $x$ within that interval. Differentiating $f(x) = x^3 - 3x^2 + 5x + 7$ with respect to $x$ gives $f'(x) = 3x^2 - 6x + 5$. To determine the sign of this quadratic expression, calculate the discriminant $D = b^2 - 4ac$, where $a=3$, $b=-6$, and $c=5$:

$D = (-6)^2 - 4(3)(5) = 36 - 60 = -24$

Since the leading coefficient is positive and the discriminant is negative, the quadratic expression $3x^2 - 6x + 5$ is always positive for all real $x$. Therefore, $f'(x) > 0$ for all $x \in \mathbb{R}$, which implies the function is strictly increasing throughout its domain.

The equation of the tangent to the curve $y=f(x)$ at point $P(x, y)$ is $Y - y = \frac{dy}{dx}(X - x)$. Setting $Y=0$ gives the x-intercept $A = \left(x - y \frac{dx}{dy}, 0\right)$, and setting $X=0$ gives the y-intercept $B = \left(0, y - x \frac{dy}{dx}\right)$, where $\frac{dy}{dx}$ represents the derivative of the function. Since the point $P$ divides the segment $AB$ in the ratio $1:3$, the section formula states $P = \frac{3A + 1B}{4}$. This provides the coordinate equation $x = \frac{3}{4} \left(x - y \frac{dx}{dy}\right)$, which simplifies to $4x = 3x - 3y \frac{dx}{dy}$, or $x = -3y \frac{dx}{dy}$.

Rearranging this differential equation to $\frac{dy}{dx} = -\frac{3y}{x}$ and separating the variables yields $\int \frac{dy}{y} = -3 \int \frac{dx}{x}$. Integrating both sides results in $\ln |y| = -3 \ln |x| + C$, which simplifies to $y = \frac{k}{x^3}$ for some constant $k$. Applying the condition $f(1)=1$ gives $1 = \frac{k}{1^3}$, so $k=1$ and the curve is defined by $f(x) = \frac{1}{x^3}$. Evaluating this function at $x=2$ results in $f(2) = \frac{1}{2^3} = \frac{1}{8}$, confirming that the point $\left(2, \frac{1}{8}\right)$ lies on the curve.

Determining the normal to a curve requires identifying the point of tangency and the slope of the curve at that point. By setting $x = 0$ in the equation $y(x - 2)(x - 3) = x + 6$, we find the curve's y-intercept at the point $(0, 1)$. Expressing $y$ as $\frac{x + 6}{x^2 - 5x + 6}$ and applying the quotient rule to find the derivative gives a slope of $1$ at $x = 0$. Because the normal line is perpendicular to the tangent, its slope is the negative reciprocal, $-1$, which leads to the normal equation $y - 1 = -1(x - 0)$, or $x + y = 1$. The point $\left(\frac{1}{2}, \frac{1}{2}\right)$ lies on this line, satisfying the linear relationship.

The slope of a tangent line at a specific point on a curve is defined by the derivative of the function, which dictates the rate of change at that exact coordinate. Beginning with the implicit equation $x^2y^2 - 2x = 4 - 4y$, differentiate both sides with respect to $x$ to obtain $2xy^2 + 2x^2y \frac{dy}{dx} - 2 = -4 \frac{dy}{dx}$. Rearranging this expression to isolate the slope term results in $\frac{dy}{dx}(2x^2y + 4) = 2 - 2xy^2$. Substituting the coordinates $(2, -2)$ into this equation, the slope becomes $\frac{dy}{dx} = \frac{2 - 2(2)(-2)^2}{2(2^2)(-2) + 4}$, which simplifies to $\frac{-14}{-12}$ or $\frac{7}{6}$.

With the slope determined, the equation of the tangent line follows the point-slope form $y - (-2) = \frac{7}{6}(x - 2)$, which simplifies to the linear relation $7x - 6y = 26$. To identify which point does not lie on this line, evaluate each option by substituting the coordinates into this equation. The points $\left(4, \frac{1}{3}\right)$, $(8, 5)$, and $(-4, -9)$ all satisfy the equation, yielding a result of 26. However, testing the coordinates $(-2, -7)$ results in $7(-2) - 6(-7) = -14 + 42 = 28$, which does not equal 26, indicating that the tangent line does not pass through this point.

The perimeter of a circular sector is defined by two radii and the arc length, leading to the relationship $2r + r\theta = 20$ for a total wire length of $20$. By rearranging this equation, we find that the central angle $\theta$ is equal to $\frac{20 - 2r}{r}$. Substituting this expression into the standard formula for the area of a sector, $A = \frac{1}{2} r^2 \theta$, allows us to express the area as a function of the radius alone, specifically $A = \frac{1}{2} r^2 \left( \frac{20 - 2r}{r} \right)$.

This expression simplifies to the quadratic equation $A = 10r - r^2$. To find the maximum area, we calculate the derivative with respect to the radius, $\frac{dA}{dr} = 10 - 2r$, and set it to zero to identify the critical point. Solving this equation yields an optimal radius of $r = 5$. Substituting this value back into the area equation gives $10(5) - (5)^2$, which results in a maximum area of $25$ square meters.

A function is strictly increasing on an interval whenever its first derivative maintains a positive value throughout that range. For the function $f(x) = \sin^4 x + \cos^4 x$, the derivative is calculated as $f'(x) = 4\sin^3 x \cos x - 4\cos^3 x \sin x$, which simplifies to $f'(x) = 4\sin x \cos x (\sin^2 x - \cos^2 x)$ by factoring. Utilizing the double-angle identities $\sin 2x = 2\sin x \cos x$ and $\cos 2x = \cos^2 x - \sin^2 x$, this expression transforms into $f'(x) = -2\sin 2x \cos 2x$, or more compactly, $f'(x) = -\sin 4x$. Because an increasing function requires $f'(x) > 0$, we require $-\sin 4x > 0$, which is equivalent to $\sin 4x < 0$. The sine function is negative when its argument lies between $\pi$ and $2\pi$, so solving $\pi < 4x < 2\pi$ yields the interval $\frac{\pi}{4} < x < \frac{\pi}{2}$.

The slope of the tangent at any point on a parametric curve is determined by calculating the ratio of the derivatives of the coordinates with respect to the parameter. For the curve defined by $x=4t^{2+}3$ and $y=8t^{3-}1$, the derivative $\frac{dx}{dt}$ is $8t$ and the derivative $\frac{dy}{dt}$ is $24t^2$, which yields a slope of $m = \frac{24t^2}{8t} = 3t$. The equation of the tangent line at parameter $t$ is $y - (8t^3 - 1) = 3t(x - (4t^2 + 3))$.

If this line intersects the curve again at a point $Q$ corresponding to a parameter $t_1$, its coordinates $x = 4t_1^2 + 3$ and $y = 8t_1^3 - 1$ must satisfy the tangent equation. Substituting these into the equation yields $8t_1^3 - 8t^3 = 3t(4t_1^2 - 4t^2)$. Dividing by common factors simplifies the expression to $2(t_1^2 + t_1t + t^2) = 3t(t_1 + t)$, which rearranges to $2t_1^2 - t_1t - t^2 = 0$. Factoring this quadratic gives $(t_1 - t)(2t_1 + t) = 0$, and since $t_1$ corresponds to a distinct point $Q$, we identify $t_1 = -\frac{t}{2}$. Evaluating the original parametric equations at this parameter results in the coordinates $x = 4(-\frac{t}{2})^2 + 3 = t^2 + 3$ and $y = 8(-\frac{t}{2})^3 - 1 = -t^3 - 1$.

To find the minimum distance from the origin $(0, 0)$ to the curve $y = x^2 - 4$, one must minimize the squared distance function $f(x) = x^2 + (x^2 - 4)^2$ for an arbitrary point $(x, x^2 - 4)$ on the parabola. Differentiating $f(x)$ with respect to $x$ yields $f'(x) = 2x + 2(x^2 - 4)(2x)$, which simplifies to $4x^3 - 14x$. Setting this derivative to zero results in the critical points $x = 0$ and $x^2 = \frac{7}{2}$.

Evaluating the distance at these critical points demonstrates that $x^2 = \frac{7}{2}$ produces the minimum value. Substituting $x^2 = \frac{7}{2}$ back into the expression for the distance gives:

$\sqrt{\frac{7}{2} + \left( \frac{7}{2} - 4 \right)^2} = \sqrt{\frac{7}{2} + \left( -\frac{1}{2} \right)^2} = \sqrt{\frac{14}{4} + \frac{1}{4}} = \frac{\sqrt{15}}{2}$

The expression inside the inverse tangent simplifies using trigonometric half-angle identities, where $1+\sin x$ corresponds to $(\cos(x/2) + \sin(x/2))^2$ and $1-\sin x$ to $(\cos(x/2) - \sin(x/2))^2$. Taking the square root of their quotient produces $\tan(\pi/4 + x/2)$, which reduces the function to $f(x) = \pi/4 + x/2$ for the given domain.

Differentiating this linear expression results in $f'(x) = 1/2$, the slope of the tangent line. The slope of the normal line is the negative reciprocal, which is $-2$. At $x = \pi/6$, the function value is $f(\pi/6) = \pi/3$, identifying the point of tangency as $(\pi/6, \pi/3)$.

The equation of the normal line is expressed as $y - \pi/3 = -2(x - \pi/6)$. Inserting the coordinates $(0, 2\pi/3)$ into this equation yields $2\pi/3 - \pi/3 = -2(0 - \pi/6)$, which simplifies to $\pi/3 = \pi/3$. This equality confirms that the normal line passes through the point $(0, 2\pi/3)$.

Minimizing the total area generated by cutting a wire of length 2 into two segments involves expressing the combined area as a function of the square's side length $x$. Because the wire has a total length of 2, the square consumes $4x$ of that length, leaving $2 - 4x$ to form the circumference of a circle, $2\pi r$. This relationship establishes the radius as $r = \frac{1-2x}{\pi}$. The sum of the areas, given by $x^2 + \pi r^2$, becomes the function $f(x) = x^2 + \frac{(1-2x)^2}{\pi}$.

Setting the derivative $f'(x) = 2x - \frac{4(1-2x)}{\pi}$ equal to zero identifies the point where the total area is minimal. Rearranging this equation yields $2x = \frac{4(1-2x)}{\pi}$. Since the radius is defined as $r = \frac{1-2x}{\pi}$, the right side of the equation simplifies to $4r$, resulting in the expression $2x = 4r$. Dividing both sides by 2 confirms the relationship $x = 2r$, which is the condition required for the sum of the areas of the square and the circle to be at its minimum.

The existence of a finite limit as $x$ approaches zero implies that the terms of $f(x)$ must be at least of the second degree. Given that \lim_{x\to 0}\[1+\frac{f(x)}{x^2}]$=3$, the limit of$\frac{f(x)}{x^2}$must be equal to 2. For a polynomial of degree four, this necessitates the form$f(x) = ax^4 + bx^3 + 2x^2$, because any constant or linear term would result in an undefined limit as$x$ approaches zero.

Since the function possesses extreme values at $x = 1$ and $x = 2$, its derivative must vanish at these points. Computing the derivative gives $f'(x) = 4ax^3 + 3bx^2 + 4x$. Substituting the conditions $f'(1) = 0$ and $f'(2) = 0$ results in the system of equations $4a + 3b + 4 = 0$ and $32a + 12b + 8 = 0$. Simplifying the second equation yields $8a + 3b = -2$. Subtracting the first equation from this simplified result gives $4a = 2$, which implies $a = \frac{1}{2}$, subsequently leading to $b = -2$. Substituting these values back into the expression for $f(x)$ results in $\frac{1}{2}x^4 - 2x^3 + 2x^2$. Evaluating this polynomial at $x = 2$ provides $8 - 16 + 8$, which equals 0.

The equation $x^2 + 2xy - 3y^2 = 0$ represents a pair of intersecting lines that can be factored into $x - y = 0$ and $x + 3y = 0$. Since the point $(1, 1)$ lies on the component line $x - y = 0$, we find the slope of the tangent at this point to be 1 via implicit differentiation, which implies the normal line has a slope of $-1$ and follows the equation $x + y = 2$. To determine where this normal meets the curve again, we solve for the intersection with the other component, $x + 3y = 0$, by treating the two equations as a system. The resulting intersection point is $(3, -1)$, which, having a positive x-coordinate and a negative y-coordinate, is located in the fourth quadrant.

To establish a relationship between the derivatives of two functions $f$ and $g$, we construct an auxiliary function $h(x) = f(x) - 2g(x)$ that satisfies the conditions of Rolle's Theorem. Evaluating this expression at the boundaries, we find that $h(0) = f(0) - 2g(0) = 2 - 2(0) = 2$ and $h(1) = f(1) - 2g(1) = 6 - 2(2) = 2$. Since $h(0) = h(1)$, Rolle's Theorem guarantees the existence of a point $c \in \left]0,1\right[$ such that $h'(c) = 0$. Differentiating the auxiliary function gives $h'(x) = f'(x) - 2g'(x)$, and substituting $c$ yields $f'(c) - 2g'(c) = 0$, which confirms $f'(c) = 2g'(c)$.

Extreme points occur where the first derivative of a function equals zero, signifying stationary points on the graph. Finding the derivative of $f(x) = \alpha \log |x| + \beta x^2 + x$ leads to $f'(x) = \frac{\alpha}{x} + 2\beta x + 1$, which equates to $2\beta x^2 + x + \alpha = 0$ when set to zero.

Since -1 and 2 are the roots of this quadratic equation, the sum of roots $-\frac{1}{2\beta}$ must equal 1, implying $2\beta = -1$ or $\beta = -\frac{1}{2}$. The product of roots $\frac{\alpha}{2\beta}$ must equal -2, which means $\frac{\alpha}{-1} = -2$, confirming $\alpha = 2$.

The slope of a tangent line to a curve defined by an integral is obtained by differentiating the function, which, according to the Fundamental Theorem of Calculus, simplifies the derivative of $y=\int_0^x |t|dt$ to $dy/dx=|x|$. Given that the tangent is parallel to the line $y=2x$, its slope must equal 2, requiring $|x|=2$ and identifying the points of tangency at $x=2$ and $x=-2$. Evaluating the integral at these specific $x$ values confirms the corresponding vertical positions are $y=2$ and $y=-2$.

Constructing the tangent lines using the point-slope formula yields the equations $y-2=2(x-2)$ and $y+2=2(x+2)$, which simplify to $y=2x-2$ and $y=2x+2$. Setting $y=0$ in these linear equations reveals that the intercepts on the x-axis are $x=1$ and $x=-1$.

The polynomial $f(x) = 2x^3 + 3x + k$ has a derivative of $f'(x) = 6x^2 + 3$. Since any real number squared results in a non-negative value, the derivative $6x^2 + 3$ remains strictly greater than zero for all inputs. A function with a strictly positive derivative is always increasing, which means it can intersect the x-axis at most one time. Therefore, it is impossible for the equation $2x^3 + 3x + k = 0$ to have two distinct real roots in the interval $[0, 1]$, and no such value of $k$ exists.

The rate at which the radius of a spherical balloon changes is governed by the volume formula $V = \frac{4}{3} \pi r^3$. By applying the chain rule to differentiate this relationship with respect to time, the connection between the changing volume and the radius is defined as $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$.

To find the specific rate after 49 minutes, we first determine the amount of gas remaining in the balloon. Given the escape rate of $72 \pi$ cubic meters per minute, the balloon loses $49 \times 72 \pi = 3528 \pi$ cubic meters of gas over that duration. Subtracting this from the initial $4500 \pi$ cubic meters leaves a current volume of $972 \pi$ cubic meters. Equating this to the volume formula $\frac{4}{3} \pi r^3 = 972 \pi$ results in $r^3 = 729$, which confirms the radius is 9 meters at that moment.

Finally, substituting the volume leakage rate of $72 \pi$ and the radius of 9 meters into the derivative equation $72 \pi = 4 \pi r^2 \frac{dr}{dt}$ leads to $72 \pi = 4 \pi (9)^2 \frac{dr}{dt}$. Simplifying this expression gives $72 \pi = 324 \pi \frac{dr}{dt}$, which shows that the radius decreases at a rate of $\frac{2}{9}$ meters per minute.