Application Of Derivatives – JEE Main MathematicsPractice Questions & PYQs

The functional equation $f\left(\frac{x+y}{3}\right) = \frac{f(x)+f(y)}{3}$ represents a linear relationship, which becomes apparent when differentiating the expression partially with respect to $x$ to obtain $f'\left(\frac{x+y}{3}\right) = f'(x)$. This identity indicates that the derivative is constant for all values. Evaluating this derivative at $x=0$ using the given condition $f'(0) = 3$ confirms that $f'(x) = 3$ everywhere, and solving for the constant of integration using $f(0) = 0$ leads directly to the function $f(x) = 3x$.

Substituting this result into the definition of $g(x) = 3 + 3x e^x$ allows us to find the minimum by setting the derivative $g'(x) = 3e^x(1+x)$ equal to zero. This derivative vanishes at $x = -1$, where the function reaches its minimum value. Calculating $g(-1)$ results in $3 + 3(-1)e^{-1}$, which simplifies to $3\left(\frac{e-1}{e}\right)$.

The minimum value of a continuous function on a closed interval occurs at either the critical points where the first derivative is zero or at the boundaries of the interval. The derivative of the function is $f'(x) = 2025x^{2024} - 2000x^{1999} = 25x^{1999}(81x^{25} - 80)$. Setting $f'(x) = 0$ provides the critical point $x^{25} = \frac{80}{81}$. Substituting this value into the expression $f(x) = x^{2000}(x^{25} - 1)$ yields $f(x) = (\frac{80}{81})^{80}(\frac{80}{81} - 1) = (\frac{80}{81})^{80}(-\frac{1}{81})$. This expression simplifies to $-(80)^{80} (81)^{-81}$, which is equivalent to $(80)^{80}(-81)^{-81}$ because the odd exponent preserves the negative sign. Comparing this result to the given form $(80)^{80}(n)^{-81}$, it follows that $n = -81$.

The trigonometric expressions inside the function $f(\theta)$ are simplified using reduction formulas to obtain $\sin^4(\frac{7\pi}{2}-\theta) = \cos^4\theta$, $\sin^4(11\pi+\theta) = \sin^4\theta$, $\sin^6(\frac{3\pi}{2}-\theta) = \cos^6\theta$, and $\sin^6(9\pi-\theta) = \sin^6\theta$. Substituting these values into the function gives $f(\theta) = 4(\cos^4\theta + \sin^4\theta) - 2(\cos^6\theta + \sin^6\theta)$. Applying the identities $\sin^4\theta + \cos^4\theta = 1 - 2\sin^2\theta\cos^2\theta$ and $\sin^6\theta + \cos^6\theta = 1 - 3\sin^2\theta\cos^2\theta$, the expression becomes $f(\theta) = 4(1 - 2\sin^2\theta\cos^2\theta) - 2(1 - 3\sin^2\theta\cos^2\theta)$, which simplifies to $f(\theta) = 2 - 2\sin^2\theta\cos^2\theta$.

By expressing $\sin^2\theta\cos^2\theta$ as $\frac{1}{4}\sin^2(2\theta)$, the function is written as $f(\theta) = 2 - \frac{1}{2}\sin^2(2\theta)$. Since the range of $\sin^2(2\theta)$ is $[0, 1]$, the maximum value $\alpha$ occurs when $\sin^2(2\theta) = 0$, yielding $\alpha = 2$, and the minimum value $\beta$ occurs when $\sin^2(2\theta) = 1$, yielding $\beta = 2 - \frac{1}{2} = 1.5$. Thus, $\alpha + 2\beta = 2 + 2(1.5) = 5$.

The function $f(x) = 16\sin\left(\frac{x}{2}\right)\cos^{3}\left(\frac{x}{2}\right)$ can be simplified for $0 \le x \le \pi$ by using trigonometric identities to express it in terms of $\sin(x)$ and $\cos(x)$. Recognizing that $2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right) = \sin(x)$ and $\cos^{2}\left(\frac{x}{2}\right) = \frac{1+\cos(x)}{2}$ allows the expression to be rewritten as $4\sin(x)(1+\cos(x))$. Setting the derivative $4(\cos(x) + \cos(2x))$ to zero results in the equation $8\cos\left(\frac{3x}{2}\right)\cos\left(\frac{x}{2}\right) = 0$, yielding critical points at $x = \frac{\pi}{3}$ and $x = \pi$. Evaluating the function at the critical points and boundaries, specifically $f\left(\frac{\pi}{3}\right) = 16\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)^{3}$, reveals that the maximum value is $3\sqrt{3}$.

A quadratic equation $A m^2 + B m + C = 0$ has two equal roots if its discriminant $B^2 - 4AC$ equals zero. For the equation $f(x) m^2 - 2 f'(x) m + f''(x) = 0$, this condition requires $(-2 f'(x))^2 - 4 f(x) f''(x) = 0$, which simplifies to $(f'(x))^2 = f(x) f''(x)$. Dividing by $f(x) f'(x)$ gives $\frac{f'(x)}{f(x)} = \frac{f''(x)}{f'(x)}$, and integrating both sides results in $\ln |f'(x)| = \ln |f(x)| + C$, leading to $f'(x) = k f(x)$. Given $f(0)=1$ and $f'(0)=2$, it follows that $f(x) = e^{2x}$.

Substituting $f(x) = e^{2x}$ into the expression $f(\ln x - x)$ yields the function $g(x)$:
$g(x) = e^{2(\ln x - x)} = x^2 e^{-2x}$
The derivative $g'(x) = 2x e^{-2x} - 2x^2 e^{-2x} = 2x e^{-2x} (1-x)$ determines the interval of increase. Since $x > 0$ is required for the domain of the logarithm, the expression $2x e^{-2x}$ is always positive, making $g'(x) > 0$ only when $1-x > 0$, or $x < 1$. Thus, the largest interval of increase is $(0, 1)$, where $\alpha = 0$ and $\beta = 1$, and $\alpha + \beta = 1$.

The monotonicity of a function is determined by the sign of its first derivative, where a negative derivative corresponds to a strictly decreasing interval. For $f(t) = \frac{|t+1|}{t^2}$ with $t < 0$, we evaluate two cases based on the modulus. When $-1 < t < 0$, $f(t) = \frac{t+1}{t^2} = t^{-1} + t^{-2}$, so $f'(t) = -\frac{1}{t^2} - \frac{2}{t^3} = -\frac{t+2}{t^3}$, which is positive since $t+2 > 0$ and $t^3 < 0$. When $t < -1$, $f(t) = -\frac{t+1}{t^2} = -t^{-1} - t^{-2}$, leading to $f'(t) = \frac{1}{t^2} + \frac{2}{t^3} = \frac{t+2}{t^3}$, which is negative when $t+2$ and $t^3$ have opposite signs, occurring exactly in the interval $(-2, -1)$. Equating this to $(2\alpha, \alpha)$ gives $2\alpha = -2$, hence $\alpha = -1$.

Substituting $\alpha = -1$ into the function $g(x) = 2\ln(x-2) - x^2 + 4x + 1$ yields the derivative $g'(x) = \frac{2}{x-2} - 2x + 4 = \frac{2}{x-2} - 2(x-2)$. Setting $g'(x) = 0$ implies $(x-2)^2 = 1$, resulting in $x = 3$ for the domain $x > 2$. The second derivative $g''(x) = -\frac{2}{(x-2)^2} - 2$ is negative at $x = 3$, confirming a local maximum. Evaluating the function at this point gives $g(3) = 2\ln(3-2) - (3)^2 + 4(3) + 1 = 0 - 9 + 12 + 1 = 4$.

The function $f(x) = |\ln x| - |x-1|$ is defined piecewise as $f(x) = -\ln x + x - 1$ for $0 < x < 1$ and $f(x) = \ln x - x + 1$ for $x \geq 1$. Determining the differentiability at $x=1$ requires evaluating the one-sided derivatives: $f'(1^-) = \lim_{x \to 1^-} (-\frac{1}{x} + 1) = 0$ and $f'(1^+) = \lim_{x \to 1^+} (\frac{1}{x} - 1) = 0$. Since the left-hand and right-hand derivatives are equal, the function is differentiable at $x=1$, and as it is differentiable for all other $x > 0$, statement (I) is true.

For monotonicity, the derivative of $f(x)$ on the interval $(0, 1)$ is $f'(x) = \frac{x-1}{x}$. Because $x-1 < 0$ and $x > 0$ for $x \in (0, 1)$, the derivative $f'(x)$ is negative, which means the function is decreasing rather than increasing in this interval, making statement (II) false. On the interval $(1, \infty )$, the derivative is $f'(x) = \frac{1-x}{x}$, which is negative because $1-x < 0$ and $x > 0$, confirming that the function is decreasing in $(1, \infty )$ and making statement (III) true. Therefore, only statements (I) and (III) are true.

The derivative of $g(x) = f((\tan x - 1)^2 + a - 1)$ is calculated as $g'(x) = f'((\tan x - 1)^2 + a - 1) \cdot 2(\tan x - 1)\sec^2 x$. Given $f''(x) > 0$, the derivative $f'(x)$ is a strictly increasing function. Since $f'(a-1) = 0$, it follows that $f'(y) > 0$ for $y > a-1$ and $f'(y) < 0$ for $y < a-1$. Because $(\tan x - 1)^2 \ge 0$, the argument $(\tan x - 1)^2 + a - 1$ is always greater than or equal to $a-1$, ensuring $f'((\tan x - 1)^2 + a - 1) \ge 0$ for all $x \in (0, \pi/2)$.

Consequently, the sign of $g'(x)$ depends solely on the factor $(\tan x - 1)$ and the term $2\sec^2 x$, which is always positive. For $x \in (0, \pi/4)$, $\tan x < 1$, which implies $g'(x) < 0$, so $g$ is decreasing, meaning statement (I) is false. For $x \in (\pi/4, \pi/2)$, $\tan x > 1$, which implies $g'(x) > 0$, so $g$ is increasing, meaning statement (II) is false.

Critical points occur where the derivative of a function is either undefined or zero. In the interval $(-2\pi, 2\pi)$, the function $|\frac{\sin x}{x}|$ possesses sharp cusps where the internal expression equals zero, specifically at $x = \pi$ and $x = -\pi$, which are two locations where the derivative does not exist. Stationary points occur where the derivative of the underlying function $\frac{\sin x}{x}$ vanishes, which simplifies to the condition $\tan x = x$. The intersection of the graphs of $y = \tan x$ and $y = x$ yields three distinct solutions within the interval $(-2\pi, 2\pi)$, including $x=0$. Combining the two points of non-differentiability and the three stationary points results in a total of 5 critical points.

A polynomial of degree 5 having extrema at $x=1$ and $x=-1$ implies that the derivative $f'(x)$ must be zero at these points, meaning $(x-1)$ and $(x+1)$ are factors of $f'(x)$. Since the derivative of a fifth-degree polynomial is a fourth-degree polynomial, we can represent $f'(x)$ as $(x^{2-}1)(ax^{2+}bx+c)$. Expanding this gives:
$f'(x) = ax^4 + bx^3 + (c-a)x^2 - bx - c$

Integrating this derivative provides the general expression for the polynomial:
$f(x) = \frac{ax^5}{5} + \frac{bx^4}{4} + \frac{(c-a)x^3}{3} - \frac{bx^2}{2} - cx + d$
The condition $\lim_{x \to 0} \frac{f(x)}{x^3} = -5$ requires that the function behaves like $-5x^3$ near the origin, necessitating that the constant term $d$ and the coefficients of $x$ and $x^2$ are zero, specifically $d=0$, $c=0$, and $b=0$. Applying these constraints to the $x^3$ coefficient yields the relation $\frac{c-a}{3} = -5$, and substituting $c=0$ results in $a=15$.

Substituting these coefficients back into the polynomial expression, we obtain $f(x) = 3x^5 - 5x^3$. Evaluating this at $x=2$ gives $f(2) = 3(2)^5 - 5(2)^3 = 3(32) - 5(8) = 56$, and since $f(x)$ is an odd function, $f(-2) = -56$. The final required value is $f(2) - f(-2) = 56 - (-56) = 112$.

Trigonometric reduction relies on the double-angle identities $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$, $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$, and $2 \sin \theta \cos \theta = \sin 2\theta$ to transform the given expression into:
$E = \frac{1 + \cos 2\theta}{2} - 3 \sin 2\theta + 3 \left( \frac{1 - \cos 2\theta}{2} \right) + 2 = 4 - \cos 2\theta - 3 \sin 2\theta$
The expression $a \cos x + b \sin x$ varies within the interval $[-\sqrt{a^{2+}b^2}, \sqrt{a^{2+}b^2}]$. For the term $(\cos 2\theta + 3 \sin 2\theta)$, the range is $[-\sqrt{1^{2+}3^2}, \sqrt{1^{2+}3^2}]$, which is $[-\sqrt{10}, \sqrt{10}]$. Subtracting the maximum value of this trigonometric part from 4 yields the minimum value of $4 - \sqrt{10}$.

The shortest distance between a parabola and a circle is measured along their common normal, which connects the center of the circle to the nearest point on the parabola. By rearranging the circle equation $x^2 + y^2 + 12y + 35 = 0$ into standard form $x^2 + (y+6)^2 = 1$, the center is identified as $(0, -6)$ with a radius of 1. For the parabola $y^2 = 8x$, the parameter $a$ is 2, and the equation for a normal with slope $m$ is given by $y = mx - 4m - 2m^3$.

Since the normal must pass through the circle's center, substituting $(0, -6)$ into the normal equation yields $-6 = -4m - 2m^3$, which simplifies to $m^3 + 2m - 3 = 0$. Solving this equation reveals that $m=1$ is the real root. Using this slope, the coordinates of the contact point $P$ on the parabola are found using $(am^2, -2am)$, which evaluates to $(2, -4)$. The distance from the circle center $(0, -6)$ to the point $P(2, -4)$ is calculated as $\sqrt{(2-0)^2 + (-4+6)^2} = \sqrt{2^2 + 2^2} = 2\sqrt{2}$. Subtracting the circle's radius of 1 from this segment length provides the shortest distance of $2\sqrt{2}-1$.

A function is strictly increasing where its first derivative is positive and strictly decreasing where its derivative is negative. For the function $f(x)=2\ln(x-2)-x^{2+}ax+1$, the derivative is $f'(x)=\frac{2}{x-2}-2x+a$. Since the second derivative $f''(x)=-\frac{2}{(x-2)^2}-2$ is always negative, $f'(x)$ is a strictly decreasing function. Because $(2,3)$ is the largest interval where the function increases, the derivative must be non-negative in this range and equal to zero at the boundary $x=3$. Setting $f'(3)=0$ yields $\frac{2}{3-2}-6+a=0$, which determines that $a=4$.

Substituting $a=4$ into $g(x)=(x-1)^3(x+2-a)^2$ simplifies the expression to $g(x)=(x-1)^3(x-2)^2$. Differentiating this using the product rule produces $g'(x)=3(x-1)^2(x-2)^{2+}2(x-1)^3(x-2)$. Factoring out the common terms results in $g'(x)=(x-1)^2(x-2)(3(x-2)+2(x-1))$, which simplifies further to $g'(x)=(x-1)^2(x-2)(5x-8)$. For the function to be strictly decreasing, $g'(x)<0$ must hold. Since $(x-1)^2$ is always positive, we solve $(x-2)(5x-8)<0$ to find the interval $(\frac{8}{5},2)$, identifying $b=\frac{8}{5}$ and $c=2$. Finally, evaluating the expression $100(a+b-c)$ gives $100(4+\frac{8}{5}-2)$, which simplifies to $100(2+1.6)=360$.

The maximum area of a parallelogram $AFDE$ with vertices on the sides of a triangle $ABC$ is always equal to half the area of the triangle itself. The area of the triangle with vertices A(4, -2), B(1, 1), and C(9, -3) is calculated using the coordinate formula:
$\text{Area} = \frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$
Substituting the specific vertex coordinates leads to the following step:
$\text{Area} = \frac{1}{2} |4(1 - (-3)) + 1((-3) - (-2)) + 9((-2) - 1)|$
This expression simplifies to $\frac{1}{2} |16 - 1 - 27|$, which is 6. With the area of the triangle determined, the maximum area of the parallelogram is $\frac{1}{2} \times 6 = 3$.

The optimization of an inscribed area relies on establishing a function that defines the shape's dimensions relative to a coordinate boundary and applying differential calculus to locate the maximum value. For the region $R = \{(x, y) : x \leq y \leq 9 - \frac{11}{3} x^2, x \geq 0\}$, an inscribed rectangle with sides parallel to the axes has a height equal to the vertical distance between the bounding parabola and the line $y=x$. At any horizontal position $x_1$, the dimensions of this rectangle are a height of $9 - \frac{11}{3} x_1^2 - x_1$ and a width of $x_1$.

The resulting area function is given by:
$A = 9x_1 - x_1^2 - \frac{11}{3}x_1^3$
To find the maximum, we differentiate with respect to $x_1$ to obtain the slope function:
$\frac{dA}{dx_1} = 9 - 2x_1 - 11x_1^2$
Setting this derivative to zero, we solve $9 - 2x_1 - 11x_1^2 = 0$ to identify the optimal horizontal coordinate $x_1 = \frac{9}{11}$. Substituting this coordinate back into the area expression produces the values $\frac{81}{11} - \frac{81}{121} - \frac{243}{121}$, which yields a final area of $\frac{567}{121}$ square units.

The derivative of the function $f'(x) = 18x^2 - 90ax + 108a^2$ equals zero at the local maximum and minimum points $x_1$ and $x_2$. Dividing the equation $f'(x) = 0$ by 18 shows that $x_1$ and $x_2$ are the roots of the quadratic equation $x^2 - 5ax + 6a^2 = 0$. By Vieta's formulas, the product of the roots is $x_1 x_2 = 6a^2$, which implies $6a^2 = 54$, resulting in $a^2 = 9$ and $a = 3$ since $a > 0$. The sum of the roots is $x_1 + x_2 = 5a$, which evaluates to $x_1 + x_2 = 5(3) = 15$. Adding these values, $a + x_1 + x_2 = 3 + 15 = 18$.

Critical points occur where the derivative of a function vanishes, a principle that allows us to determine the unknown parameters in $f(x)=x^{3+}a x^{2+}b \ln |x|+1$. By differentiating the function, we obtain $f'(x)=3 x^{2+}2 a x+\frac{b}{x}$. Since the function has critical points at $x=-1$ and $x=2$, the derivative must equal zero at these values, resulting in the equations $3-2a-b=0$ and $12+4a+\frac{b}{2}=0$. Solving this system simultaneously leads to $a=-\frac{9}{2}$ and $b=12$.

With the function fully determined as $f(x)=x^{3-}\frac{9}{2} x^{2+}12 \ln |x|+1$, we identify the absolute extrema on the interval $[-2, -0.5]$ by evaluating the function at the boundary points and any critical points contained within that range. The critical point $x=-1$ falls inside the interval, yielding $f(-1)=-4.5$. Evaluating at the boundary points provides $f(-2)=-16.6$ and $f(-0.5)=-8.5$. Comparing these values, the maximum $M$ is $-4.5$ and the minimum $m$ is $-16.6$. The final calculation is $|M+m|=|-16.6-4.5|=21.1$.

A cubic function possesses three distinct real roots if its local maximum and minimum values lie on opposite sides of the x-axis, forcing the product of the function evaluated at its critical points to be negative. For the function $f(x) = 5 x^3 - 15 x - a$, the critical points occur where the derivative $f'(x) = 15 x^2 - 15$ equals zero, yielding $x = 1$ and $x = -1$. The condition $f(1) \cdot f(-1) < 0$ simplifies to $(-10 - a)(10 - a) < 0$, and rearranging this inequality results in $(a + 10)(a - 10) < 0$. This indicates that $a$ must lie in the interval $(-10, 10)$, identifying $\alpha = -10$ and $\beta = 10$. Consequently, the value of $\beta - 2\alpha$ is $10 - 2(-10) = 30$.

Local extrema of a cubic function are found by examining the roots of its derivative, where the slope of the curve becomes zero. For the given function $f(x)=2x^{3-}9ax^{2+}12a^2x+1$, we determine its stationary points by calculating the first derivative $f'(x)=6x^{2-}18ax+12a^2$. Setting this expression to zero allows us to factorize it as $6(x-a)(x-2a)=0$, which identifies $x=a$ and $x=2a$ as the critical points. Analyzing the behavior of this cubic function with a positive leading coefficient, we recognize that the local maximum occurs at the smaller value, $p=a$, and the local minimum occurs at the larger value, $q=2a$.

Applying the condition $p^2=q$ directly leads to the equation $a^2=2a$. Given that $a>0$, we divide by $a$ to find that $a=2$. Substituting this value back into the original function definition updates it to $f(x)=2x^{3-}18x^{2+}48x+1$. Evaluating this function at $x=3$ yields $f(3)=2(3)^{3-}18(3)^{2+}48(3)+1$, which simplifies to $54-162+144+1$, resulting in 37.

Identifying local minima requires analyzing the trend of the function across its defined intervals. For the first portion $x < -1$, the function $1-2x$ continuously decreases, providing no local minimum. Within the range $-1 \le x \le 2$, the absolute value term $\frac{1}{3}(7+2|x|)$ acts as a V-shaped graph: it decreases as $x$ approaches $0$ from the left and increases as $x$ moves to the right. This behavior confirms a local minimum at $x = 0$ with a value of $f(0) = \frac{7}{3}$.