Application Of Derivatives – JEE Main MathematicsPractice Questions & PYQs

To maximize the volume of the box, we define the side of the cut-out square as $x$. The resulting open box has a height of $x$ and a square base with a side length of $30 - 2x$, leading to the volume formula $V(x) = x(30 - 2x)^2$. Differentiating this expression with respect to $x$ and setting the derivative to zero yields $12x^2 - 240x + 900 = 0$, which factors into $(x - 5)(x - 15) = 0$. Because $x = 15$ would make the side length zero, we determine that the volume is maximized when $x = 5$. The surface area of the box is the sum of the base area and the four side flaps, calculated using the formula $A = (30 - 2x)^2 + 4x(30 - 2x)$. Substituting $x = 5$ into this expression provides $20^2 + 4(5)(20)$, which equals $400 + 400$, resulting in a total surface area of 800 cm².

The quadratic expression $x^{2-}5x+6$ factors into $(x-2)(x-3)$, which allows us to define the function piecewise by identifying the intervals where the expression inside the modulus is positive or negative:

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The first branch $x^{2-}8x+8$ is monotonic on $[-1, 2]$ because its vertex at $x=4$ is outside the interval, yielding values of $f(-1)=17$ and $f(2)=-4$. Similarly, the second branch $-x^{2+}2x-4$ is monotonic on $[2, 3]$ because its vertex at $x=1$ is outside the range, producing endpoints at $f(2)=-4$ and $f(3)=-7$. Comparing these calculated boundaries, the absolute maximum is 17 and the absolute minimum is -7, resulting in a total sum of 10.

The perimeter of the triangle is given by $AB + BC + AC$, and since the length of the side $AB$ is fixed, maximizing and minimizing the perimeter depends entirely on the sum $BC + AC$, where $C(t,4)$ moves along the line $y=4$. To minimize this sum, we use the principle of reflection, mirroring the point $B(0,0)$ across the line $y=4$ to obtain the point $B_1(0,8)$. The shortest distance $BC + AC$ is equal to the straight-line distance between $A(2,1)$ and $B_1(0,8)$, which occurs when $C$ lies on the segment $AB_1$. The line passing through $A$ and $B_1$ has a slope $m = \frac{8-1}{0-2} = -\frac{7}{2}$, and its equation is $y - 8 = -\frac{7}{2}x$, or $7x + 2y = 16$. Setting $y=4$ to find the intersection, we get $7x + 8 = 16$, which yields $x = \frac{8}{7}$, so the minimum occurs at $\beta = \frac{8}{7}$.

For the maximum perimeter, we evaluate the function $f(t) = BC + AC = \sqrt{t^2 + 4^2} + \sqrt{(t-2)^2 + 3^2}$ on the closed interval $t \in [0,4]$. Since this function represents the sum of distances from a point on a line to two fixed points, it is convex, meaning its maximum must occur at one of the interval's endpoints. Evaluating at $t=0$, we get $4 + \sqrt{2^2 + 3^2} = 4 + \sqrt{13}$, and evaluating at $t=4$, we get $\sqrt{4^2 + 4^2} + \sqrt{(4-2)^2 + 3^2} = 4\sqrt{2} + \sqrt{13}$. Because $4\sqrt{2} \approx 5.66$ is greater than $4$, the maximum perimeter is attained at the endpoint $\alpha = 4$. Consequently, the final value is $6\alpha + 21\beta = 6(4) + 21\left(\frac{8}{7}\right) = 24 + 3(8) = 48$.

The function $g(x)=f(x)+f(1-x)$ possesses a derivative $g'(x)=f'(x)-f'(1-x)$. Because $f''(x)>0$, the derivative $f'$ is strictly increasing, ensuring $g'(x)=0$ occurs only when $x=1-x$, or $x=1/2$. As $g$ decreases on the interval $(0,\alpha)$ and increases on $(\alpha,1)$, it is clear that $\alpha=1/2$. Evaluating the expression $\tan^{-1}(2\alpha)+\tan^{-1}\left(\frac{\alpha+1}{\alpha}\right)$ with $\alpha=1/2$ gives $\tan^{-1}(1)+\tan^{-1}(3)$. Using the established trigonometric identity $\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\pi$, the final result is $\pi$.

To find the local maximum of the function $f(x)= \left( \frac{\sqrt{3e}}{2 \sin x} \right)^{\sin^2 x}$ for $x \in (0, \frac{\pi}{2})$, it is effective to examine the natural logarithm of the function, $y = \ln f(x) = \sin^2 x \cdot \ln \left( \frac{\sqrt{3e}}{2 \sin x} \right)$. By substituting $u = \sin^2 x$, the expression simplifies to $y = u \left( \ln \left( \frac{\sqrt{3e}}{2} \right) - \frac{1}{2} \ln u \right)$. Differentiating $y$ with respect to $u$ and setting the result equal to zero identifies the critical point: $\frac{d}{du} (u \ln \left( \frac{\sqrt{3e}}{2} \right) - \frac{1}{2} u \ln u) = \ln \left( \frac{\sqrt{3e}}{2} \right) - \frac{1}{2} \ln u - \frac{1}{2} = 0$.

This simplifies to $\ln \left( \frac{\sqrt{3e}}{2 \sqrt{u}} \right) = \frac{1}{2}$. Exponentiating both sides yields the equation $\frac{\sqrt{3e}}{2 \sqrt{u}} = e^{1/2} = \sqrt{e}$. Dividing both sides by $\sqrt{e}$ leaves $\frac{\sqrt{3}}{2 \sqrt{u}} = 1$, which implies $\sqrt{u} = \frac{\sqrt{3}}{2}$, or $\sin^2 x = \frac{3}{4}$. Substituting $\sin^2 x = \frac{3}{4}$ back into the original function gives the local maximum value as follows:

$\left( \frac{\sqrt{3e}}{2 \cdot \frac{\sqrt{3}}{2}} \right)^{3/4} = \left( \frac{\sqrt{3e}}{\sqrt{3}} \right)^{3/4} = (\sqrt{e})^{3/4} = e^{3/8}$

Given that this value corresponds to $\frac{k}{e}$, we have $\frac{k}{e} = e^{3/8}$, which implies $k = e \cdot e^{3/8} = e^{11/8}$. Consequently, $k^8 = (e^{11/8})^8 = e^{11}$. The required expression $\left( \frac{k}{e} \right)^8 + \frac{k^8}{e^5} + k^8$ can then be computed by substituting these values:

$(e^{3/8})^8 + \frac{e^{11}}{e^5} + e^{11} = e^3 + e^6 + e^{11}$

The equation of a curve passing through a point must hold true for those coordinates, while the slope of the tangent at that point is given by the negative reciprocal of the slope of the normal line. Substituting the coordinates 1 and -3 into $y = \frac{x-a}{(x+b)(x-2)}$ results in $-3 = \frac{1-a}{(1+b)(1-2)}$, which simplifies to the relationship $1-a = 3(1+b)$.

Rearranging the normal equation $x - 4y = 13$ shows the normal slope is $\frac{1}{4}$, making the tangent slope -4. Differentiation of the function yields:
$\frac{dy}{dx} = \frac{(x+b)(x-2) - (x-a)(2x+b-2)}{(x+b)^2(x-2)^2}$
Setting this derivative to -4 at $x=1$ and utilizing the relation $1-a = 3(1+b)$ leads to $b = -3$. Substituting this back into the first relation provides $a = 7$, thus satisfying $a + b = 4$.

To determine the local maximum of the function $f(x)=2 x^{4-}18 x^{2+}8 x+12$, we first identify the stationary points where the derivative $f'(x)=8 x^{3-}36 x+8$ equals zero. Factoring the equation $2 x^{3-}9 x+2=0$ using the known root $x=2$ yields $(x-2)(2 x^{2+}4 x-1)=0$, which identifies the other critical points at $x = -1 \pm \frac{\sqrt{6}}{2}$. We then simplify the evaluation of $f(x)$ at $x = \frac{\sqrt{6}-2}{2}$ by expressing the function through polynomial division as $f(x) = \left(x^{2-}2 x-\frac{9}{2}\right)(2 x^{2+}4 x-1) + 24 x + \frac{15}{2}$. Substituting this critical point into the remainder term, we calculate $24\left(\frac{\sqrt{6}-2}{2}\right) + \frac{15}{2} = 12 \sqrt{6} - 24 + \frac{15}{2}$, which results in $12 \sqrt{6} - \frac{33}{2}$.

A cubic function with a local maximum at a negative coordinate and a local minimum at a positive coordinate must have a derivative $f'(x)=6 x^{2}+2(2 p-7) x+3(2 p-9)$ that possesses two real roots of opposite signs. Because the quadratic coefficient of the derivative is positive, this condition is satisfied if and only if the y-intercept of the derivative is negative, which is expressed as $f'(0) < 0$. Substituting zero into the derivative yields $3(2 p-9) < 0$, a condition that simplifies to $2 p < 9$ or $p < \frac{9}{2}$. This results in the set of all valid values for $p$ being $\left(-\infty , \frac{9}{2}\right)$.

The given line $x + 90y + 2 = 0$ has a slope $m = -1/90$, so any parallel normal line must also have a slope $m = -1/90$. Because the tangent line at any point is perpendicular to the normal, the tangent's slope must be the negative reciprocal, which is 90. Equating the derivative $y' = 270x^4 - 540x^3 - 210x^2 + 360x + 210$ to 90 produces the equation $270x^4 - 540x^3 - 210x^2 + 360x + 120 = 0$. This polynomial yields four distinct real roots for $x$ where $x = 1, 2, -2/3, -1/3$, each corresponding to a point on the curve where the normal line is parallel to the given line.

To find the points where tangents are parallel to the coordinate axes, we perform implicit differentiation on the curve $y^{5-}9xy+2x=0$ with respect to $x$. By applying the product rule to the $xy$ term and the chain rule to the $y^5$ term, we obtain $5y^4 \frac{dy}{dx}-9y-9x \frac{dy}{dx}+2=0$, which rearranges to isolate the derivative as $\frac{dy}{dx}=\frac{9y-2}{5y^{4-}9x}$.

A tangent line is parallel to the $x$-axis when the derivative $\frac{dy}{dx}$ is zero, which occurs when the numerator $9y-2=0$. This condition identifies $M=1$ such point on the curve. Conversely, a tangent is parallel to the $y$-axis when the slope becomes undefined, which requires the denominator $5y^{4-}9x=0$. By solving this condition in conjunction with the original curve equation, we determine $N=1$ valid point. Consequently, the sum of these values is $M+N=2$.

To determine the extreme values of the function $f(x) = |2x^{2+}3x-2| + \sin x \cos x$ on the interval $[0,1]$, we start by simplifying the trigonometric component using the double-angle identity $\sin x \cos x = \frac{1}{2}\sin(2x)$. The quadratic expression inside the absolute value, $2x^{2+}3x-2$, factors into $(2x-1)(x+2)$, which changes sign at $x=\frac{1}{2}$. This naturally divides our analysis into two intervals. When $x \in [0, \frac{1}{2})$, the expression is negative, so $f(x) = -(2x^{2+}3x-2) + \frac{1}{2}\sin(2x)$. The derivative here, $f'(x) = -4x-3+\cos(2x)$, is strictly negative because $-4x-3$ is at most $-3$ and $\cos(2x)$ is at most $1$, meaning the function decreases throughout this segment.

For $x \in [\frac{1}{2}, 1]$, the expression is positive, making $f(x) = 2x^{2+}3x-2 + \frac{1}{2}\sin(2x)$. The derivative, $f'(x) = 4x+3+\cos(2x)$, is strictly positive because $4x+3$ is at least $5$ and $\cos(2x)$ is at least $-1$, so the function increases throughout this segment. Given these behaviors, the absolute minimum must occur at the critical point $x=\frac{1}{2}$, where $f(\frac{1}{2}) = |0| + \frac{1}{2}\sin(1) = \frac{1}{2}\sin(1)$. The absolute maximum must occur at one of the interval endpoints, $x=0$ or $x=1$. Calculating these gives $f(0) = |-2| + 0 = 2$ and $f(1) = |3| + \frac{1}{2}\sin(2) = 3 + \frac{1}{2}\sin(2)$. Since $3 + \frac{1}{2}\sin(2)$ is clearly greater than $2$, the maximum value is $3 + \frac{1}{2}\sin(2)$. Combining these results, the sum is $3 + \frac{1}{2}\sin(1) + \frac{1}{2}\sin(2)$, which simplifies to $3 + \frac{1}{2}(\sin(1) + \sin(2))$. Using the identity $\sin(2) = 2\sin(1)\cos(1)$, this expression matches $3+\frac{1}{2}(1+2\cos(1))\sin(1)$.

Determining the intervals where a function increases requires calculating its derivative and identifying where that expression remains positive. For the function $f(x) = x e^{x(1-x)}$, the product rule is used to find the derivative: $f'(x) = e^{x(1-x)} + x e^{x(1-x)}(1-2x)$. Factoring out the exponential term simplifies this to $f'(x) = e^{x(1-x)}(1 + x - 2x^2)$.

Since the exponential term $e^{x(1-x)}$ is always positive for any real number, the sign of the derivative depends entirely on the quadratic factor $-2x^2 + x + 1$. This quadratic can be factored as $(1-x)(2x+1)$. The expression $(1-x)(2x+1)$ is positive when $x$ is between the roots $-\frac{1}{2}$ and $1$. Because the derivative is positive within this interval, the function maintains an increasing slope throughout $(-\frac{1}{2}, 1)$.

The cone's geometry is defined by the semivertical angle $\theta$ where $\tan \theta = \frac{3}{4}$. This establishes a constant ratio between the radius $r$ and the height $h$, specifically $h = \frac{4}{3} r$. The volume of the water in the cone is given by $V = \frac{1}{3} \pi r^2 h$, which can be written entirely in terms of the radius as $V = \frac{4}{9} \pi r^3$. Differentiating this expression with respect to time $t$ yields the rate equation $\frac{dV}{dt} = \frac{4}{3} \pi r^2 \frac{dr}{dt}$.

When the water depth is $4$ meters, the radius is $r = \frac{3}{4} (4) = 3$ meters. Using the given constant rate of $6$ cubic meters per hour for $\frac{dV}{dt}$, we solve $6 = \frac{4}{3} \pi (3)^2 \frac{dr}{dt}$ to find that the radius is increasing at a rate of $\frac{dr}{dt} = \frac{1}{2\pi}$ meters per hour.

The wet curved surface area of the cone is $A = \pi r \sqrt{r^2 + h^2}$. Substituting $h = \frac{4}{3} r$ into this formula gives $A = \pi r \sqrt{r^2 + \frac{16}{9} r^2} = \frac{5}{3} \pi r^2$. Differentiating this with respect to time leads to $\frac{dA}{dt} = \frac{10}{3} \pi r \frac{dr}{dt}$. Plugging in the values $r = 3$ and $\frac{dr}{dt} = \frac{1}{2\pi}$, we find the rate of increase of the curved surface area is $\frac{10}{3} \pi (3) \left( \frac{1}{2\pi} \right) = 5$ square meters per hour.

The minimum value of a function occurs at the critical point where its first derivative equals zero. For the function $f(x) = \frac{5x^2}{2} + \frac{\alpha}{x^5}$, the derivative is $f'(x) = 5x - \frac{5\alpha}{x^6}$. Setting this derivative equal to zero yields the equation $5x = \frac{5\alpha}{x^6}$, which simplifies to $x^7 = \alpha$, or $x = \alpha^{1/7}$. Substituting this value back into the original function results in $f(x) = \frac{5(\alpha^{1/7})^2}{2} + \frac{\alpha}{(\alpha^{1/7})^5} = \frac{5}{2}\alpha^{2/7} + \alpha^{2/7}$. Given the minimum value is 14, we solve the equation $\frac{7}{2}\alpha^{2/7} = 14$. This simplifies to $\alpha^{2/7} = 4$, and calculating $\alpha$ yields $4^{7/2}$, which equals 128.

The point $P(x_1, y_1)$ on the curve $y=2x^{2+}x+2$ is determined by the condition that the normal line at this point passes through $Q(6,4)$. Differentiating the curve yields the tangent slope $4x+1$, so the normal slope at $P$ is $-\frac{1}{4x_1+1}$. Equating this to the slope of line segment $PQ$ gives the relationship $\frac{y_1-4}{x_1-6} = -\frac{1}{4x_1+1}$. Substituting $y_1=2x_1^{2+}x_1+2$ transforms this into $\frac{2x_1^{2+}x_1-2}{x_1-6} = -\frac{1}{4x_1+1}$, which simplifies to the cubic equation $8x_1^{3+}6x_1^{2-}6x_1-8=0$.

Inspecting this equation reveals $x_1=1$ is a valid solution, yielding $y_1=2(1)^{2+}1+2=5$. The vertices of the triangle are $O(0,0)$, $Q(6,4)$, and $P(1,5)$. Applying the coordinate area formula $\frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$ with these coordinates results in $\frac{1}{2} |0(4-5) + 6(5-0) + 1(0-4)|$. This simplifies to $\frac{1}{2} |30 - 4|$, which gives an area of 13.

Determining how the wet surface area of a conical vessel changes requires relating the geometry of the cone to the rate at which its internal volume increases. Because the cone's dimensions are fixed, the ratio between the radius $r$ and the height $h$ remains constant at $r/h = 1/5$, allowing both the volume $V = \frac{1}{3} \pi r^2 h$ and the lateral surface area $S = \pi r \sqrt{r^2 + h^2}$ to be expressed entirely in terms of $h$.

By substituting $r = \frac{h}{5}$ into the volume formula, we obtain $V = \frac{\pi h^3}{75}$. Differentiating this with respect to time $t$ gives $\frac{dV}{dt} = \frac{\pi h^2}{25} \frac{dh}{dt}$. Given that the water fills at 1 cm$^3$/sec, we can rearrange this to solve for the rate of height increase as $\frac{dh}{dt} = \frac{25}{\pi h^2}$.

For the surface area, substituting $r = \frac{h}{5}$ into the expression for $S$ yields $S = \frac{\sqrt{26} \pi h^2}{25}$. Differentiating this expression with respect to $t$ results in $\frac{dS}{dt} = \frac{2\sqrt{26} \pi h}{25} \frac{dh}{dt}$. Plugging in the previously found expression for $\frac{dh}{dt}$ simplifies the rate of surface area growth to $\frac{dS}{dt} = \frac{2\sqrt{26}}{h}$. Evaluating this at a height of 10 cm yields $\frac{\sqrt{26}}{5}$ cm$^2$/sec.

The slope of a normal at any point is the negative reciprocal of the slope of the tangent, which implies that the derivative $\frac{dy}{dx}$ is equal to the negative reciprocal of the given slope $\frac{x^2}{xy - x^2y^2 - 1}$. This simplifies to $\frac{dy}{dx} = \frac{x^2y^2 - xy + 1}{x^2}$, which can be rewritten as $\frac{dy}{dx} = y^2 - \frac{y}{x} + \frac{1}{x^2}$. By substituting $v = xy$, we find that $\frac{dy}{dx} = \frac{1}{x}\frac{dv}{dx} - \frac{v}{x^2}$. Substituting this expression back into the differential equation leads to $\frac{1}{x}\frac{dv}{dx} - \frac{v}{x^2} = \frac{v^2}{x^2} - \frac{v}{x^2} + \frac{1}{x^2}$, which simplifies to the separable form $\frac{dv}{dx} = \frac{v^2 + 1}{x}$.

Integrating both sides yields $\tan^{-1}(v) = \ln(x) + C$. Using the condition that the curve passes through the point $(1, 1)$, we substitute $x = 1$ and $v = 1(1) = 1$ to find $C = \tan^{-1}(1)$. At the point where $x = e$, the equation becomes $\tan^{-1}(e \cdot y(e)) = \ln(e) + \tan^{-1}(1)$, which simplifies to $\tan^{-1}(e \cdot y(e)) = 1 + \tan^{-1}(1)$. Taking the tangent of both sides and applying the identity $\tan(A+B) = \frac{\tan(A) + \tan(B)}{1 - \tan(A)\tan(B)}$ results in $e \cdot y(e) = \frac{\tan(1) + 1}{1 - \tan(1)}$.

Two curves share a common tangent if their derivatives at the respective points of contact are equal, meaning the slope of the tangent line must be consistent for both functions at those specific locations. For the curve $y=5x^{2+}2x-25$, the derivative is $10x+2$, which yields a slope of $22$ at $x=2$. Given this slope and the point $(2, -1)$, the equation of the tangent line is $y+1=22(x-2)$, which simplifies to $y=22x-45$.

For the curve $y=x^{3-}x^{2+}x$ to share this tangent at some point $(a, b)$, its derivative $3a^{2-}2a+1$ must also equal $22$. This requirement leads to the quadratic equation $3a^{2-}2a-21=0$, which factors as $(3a+7)(a-3)=0$, indicating potential $x$-coordinates of $a=3$ and $a=-7/3$. Evaluating the curve at $a=3$ produces $b=3^{3-}3^{2+}3=21$. Checking the point $(3, 21)$ against the tangent equation $y=22x-45$ confirms that $21 = 22(3)-45$ is satisfied. Because the other potential value for $a$ leads to a point that does not lie on the tangent line, the only valid point is $(3, 21)$. Calculating the final expression $|2a+9b|$ with these coordinates gives $|2(3)+9(21)|$, which equals $195$.

To maximize a product expression constrained by a linear relationship, one can express one variable in terms of the other using the line's equation and then apply calculus to identify the point that yields the maximum value. The line passing through the points $(50+\alpha, 0)$ and $(0, 50+\alpha)$ follows the intercept form equation
$\frac{x}{50+\alpha} + \frac{y}{50+\alpha} = 1$
which simplifies to $x+y = 50+\alpha$. Substituting $x = 50+\alpha - y$ into the expression $xy^4$ results in the function $f(y) = (50+\alpha)y^4 - y^5$, and calculating the derivative
$\frac{df}{dy} = 4(50+\alpha)y^3 - 5y^4 = 0$
reveals the relevant critical point $y = \frac{4}{5}(50+\alpha)$, since the trivial solution $y=0$ would not provide a maximum.

Determining the corresponding value of $x$ using the constraint $x+y = 50+\alpha$ leads to $x = 50+\alpha - \frac{4}{5}(50+\alpha) = \frac{1}{5}(50+\alpha)$. Dividing $y$ by $x$ shows that $y$ is exactly four times $x$, confirming that the point $(x, y)$ resides on the line $y=4x$.

The rate of viral transmission depends on both the infected population and the non-infected population, modeled by the differential equation $\frac{dx}{dt} = kx(100-x)$, where $x$ is the number of infected students. Separating the variables and integrating both sides yields $\int \frac{dx}{x(100-x)} = \int k dt$, which simplifies using partial fractions to $\frac{1}{100} \ln \left( \frac{x}{100-x} \right) = kt + c$.

At the starting point of day zero, with two students infected, the constant of integration is determined by $\frac{1}{100} \ln \left( \frac{2}{98} \right) = c$. Substituting this back gives the governing relationship $\ln \left( \frac{x}{100-x} \right) = 100kt + \ln \left( \frac{1}{49} \right)$, which rearranges to $\ln \left( \frac{49x}{100-x} \right) = 100kt$.

When day four arrives, thirty students are infected, allowing for the calculation of the growth parameter: $\ln \left( \frac{49 \times 30}{70} \right) = \ln(21) = 100k(4)$, meaning $100k = \frac{\ln(21)}{4}$. To determine the number of infections on the eighth day, we substitute $t=8$ into the equation $\ln \left( \frac{49x}{100-x} \right) = \left( \frac{\ln(21)}{4} \right) \times 8$. This simplifies to $\ln \left( \frac{49x}{100-x} \right) = 2 \ln(21) = \ln(21^2)$. By removing the logarithms, we get $\frac{49x}{100-x} = 441$, which reduces to $\frac{x}{100-x} = \frac{441}{49} = 9$. Solving for $x$ leads to $x = 900 - 9x$, resulting in $10x = 900$, which confirms that the number of infected students on the eighth day is 90.