Application Of Derivatives – JEE Main MathematicsPractice Questions & PYQs

The monotonicity of a function is determined by the sign of its derivative. Since the problem states that $f''(x) > 0$ for all $x$ in the interval $(0,3)$, we know that $f'(x)$ is a strictly increasing function, meaning the relationship $f'(a) < f'(b)$ holds if and only if $a < b$.

To find the critical point $\alpha$ where the function $g(x) = 3 f \left(\frac{x}{3}\right) + f(3-x)$ transitions from decreasing to increasing, we differentiate to obtain $g'(x) = f'(\frac{x}{3}) - f'(3-x)$. The transition occurs exactly where $g'(x) = 0$, which requires $f'(\frac{x}{3}) = f'(3-x)$. Because $f'(x)$ is strictly increasing, this equality implies that the arguments must be equal, so we set $\frac{x}{3} = 3-x$. Solving this equation gives $\frac{4x}{3} = 3$, which results in $x = \frac{9}{4}$. Consequently, the boundary point $\alpha$ is $\frac{9}{4}$, and the calculation $8 \alpha$ results in $8 \times \frac{9}{4} = 18$.

The critical points of the function $f(x)=2x^{3-}9ax^{2+}12a^2 x+1$ occur where its derivative $f'(x)=6x^{2-}18ax+12a^2$ equals zero. Dividing this expression by six simplifies the condition to $x^{2-}3ax+2a^2=0$, where the roots of the equation are the given local extrema $\alpha$ and $\alpha^2$. According to Vieta's formulas, the sum of these roots is $\alpha+\alpha^2=3a$ and their product is $\alpha^3=2a^2$. Solving for $a$ by relating these expressions reveals that $a=2$, as the alternative value $a=1/4$ does not satisfy the necessary conditions for the local extrema, and substituting $a=2$ into the quadratic expression yields $x^{2-}6x+8=0$.

The composition $g(f(x)) = x$ implies that $g$ is the inverse function of $f$. Differentiating both sides with respect to $x$ using the chain rule yields $g'(f(x)) \cdot f'(x) = 1$. Solving $x^5 + 2x^3 + 3x + 1 = 7$ reveals $x=1$, which confirms $f(1) = 7$ and $g(7) = 1$. The derivative $f'(x) = 5x^4 + 6x^2 + 3$ evaluates to $f'(1) = 14$, so $g'(7) = \frac{1}{14}$. Calculating the ratio $\frac{g(7)}{g'(7)}$ gives $\frac{1}{1/14}$, which simplifies to 14.

Identifying points of local maxima relies on examining where a function transitions from increasing to decreasing, which is determined by the sign change of the first derivative from positive to negative. Differentiating the given function $f(x) = 4 \cos^3 x + 3 \sqrt{3} \cos^2 x - 10$ yields $f'(x) = -6 \sin x \cos x (2 \cos x + \sqrt{3})$. By setting $f'(x) = 0$, we identify the critical points within the interval $(0, 2\pi)$ by solving for $\sin x = 0$, $\cos x = 0$, and $\cos x = -\frac{\sqrt{3}}{2}$. Analyzing the sign changes of $f'(x)$ around these points reveals that the derivative transitions from positive to negative specifically at $x = \frac{5\pi}{6}$ and $x = \frac{7\pi}{6}$, confirming that the function reaches a local maximum at each of these two values.

Determining the maximum area of a triangle with vertices at $(0,0)$, $(x, y)$, and $(-x, y)$ relies on expressing the area as a function of the horizontal coordinate $x$ and finding its local maximum. Because the points on the curve $y = -2x^2 + 54$ are symmetric with respect to the y-axis, the triangle has a base of length $2x$ and a vertical height equal to the value of $y$. Utilizing the standard area formula $Area = \frac{1}{2} \times \text{base} \times \text{height}$, we define the area function as $A(x) = \frac{1}{2} \times (2x) \times (-2x^2 + 54)$, which simplifies to $A(x) = -2x^3 + 54x$.

To identify the maximum, we calculate the derivative $\frac{dA}{dx} = -6x^2 + 54$ and set it to zero, which reveals a critical point at $x^2 = 9$, or $x = 3$. Substituting this value back into the area function provides $A = 3(-2(3)^2 + 54) = 3(-18 + 54)$, leading to a result of $3 \times 36$, which equals 108.

The monotonicity of a function is determined by the sign of its first derivative, while the behavior of that derivative depends on the sign of the second derivative. Starting with the function $f(x) = \sin x + 3x - \frac{2}{\pi}(x^{2+}x)$, we compute the first derivative to find $f'(x) = \cos x + 3 - \frac{2}{\pi}(2x+1)$. Over the interval $(0, \frac{\pi}{2})$, $\cos x$ is positive and the term $3 - \frac{2}{\pi}(2x+1)$ also remains positive, reaching a minimum at $x = \frac{\pi}{2}$ of $1 - \frac{2}{\pi}$, which is greater than zero; consequently, $f'(x) > 0$, confirming the function is increasing throughout the interval. To evaluate the slope, we compute the second derivative $f''(x) = -\sin x - \frac{4}{\pi}$. Since $\sin x$ is positive for $x \in (0, \frac{\pi}{2})$, $f''(x)$ is strictly negative, which demonstrates that $f'(x)$ is a decreasing function on this interval.

The inequality $f(x) - f(y) \geq \ln x - \ln y + x - y$ provides the key to identifying the function's derivative by comparing difference quotients. By rearranging the terms to isolate the difference quotient, we observe that for $x > y$, dividing by $x - y$ yields $\frac{f(x) - f(y)}{x - y} \geq \frac{\ln x - \ln y}{x - y} + 1$. Taking the limit as $y$ approaches $x$ forces the derivative to satisfy $f'(x) \geq \frac{1}{x} + 1$, and considering $x < y$ reverses the inequality to show $f'(x) \leq \frac{1}{x} + 1$. Since $f$ is differentiable, these bounds force the equality $f'(x) = \frac{1}{x} + 1$.

Substituting $x = \frac{1}{n^2}$ into this result leads to $f' \left( \frac{1}{n^2} \right) = n^2 + 1$. We must compute the sum $\sum_{n=1}^{20} (n^2 + 1)$, which separates into $\sum_{n=1}^{20} n^2 + \sum_{n=1}^{20} 1$. Applying the formula for the sum of squares $\frac{n(n+1)(2n+1)}{6}$ for $n=20$ produces $\frac{20 \cdot 21 \cdot 41}{6} + 20$. Simplifying this calculation results in $2870 + 20$, which equals 2890.

Maximizing the area of an inscribed figure requires expressing the area as a function of one variable, such as a coordinate on the boundary curve, and identifying the stationary point of that function. We place the rectangle such that its vertical height spans from $-b$ to $b$ along the $y$-axis, causing its corners to lie on the parabola $y^2 = 2x$ at the $x$-coordinate $x = \frac{b^2}{2}$. Since the rectangle extends horizontally from this curve to the vertical line $x = 24$, its length is $24 - \frac{b^2}{2}$ and its total vertical height is $2b$, resulting in an area function of $A = 2b(24 - \frac{b^2}{2}) = 48b - b^3$. Taking the derivative with respect to $b$ gives $\frac{dA}{db} = 48 - 3b^2$, and setting this to zero indicates the maximum occurs when $b^2 = 16$, or $b = 4$. Substituting this value back into the area expression provides a maximum area of $A = 2(4)(24 - 8) = 128$.

Finding the intersection points of the curve $y=f(x)$ with the x-axis requires solving $2^x = x^2$. Testing integer values shows that $x=2$ and $x=4$ are roots, and evaluating the function indicates an additional intersection point between -1 and 0, so $m=3$. For the derivative $f'(x) = 2^x \ln 2 - 2x$, identifying the roots involves solving $2^x \ln 2 = 2x$. Observing the function's behavior across specific intervals reveals a sign change between 0 and 1, and another between 2 and 4, which confirms that $n=2$. Summing these values results in $m+n=5$.

Recognizing that the expression describes the squared distance between a point on a circle and a fixed coordinate allows for a straightforward geometric interpretation. Setting $y = \sqrt{8x - x^2 - 12}$ and squaring it yields $y^2 = 8x - x^2 - 12$, which simplifies to the circle equation $(x-4)^2 + y^2 = 4$.

The expression $(y-4)^2 + (x-7)^2$ represents the squared distance from points on this circle to $(7, 4)$, resulting in a minimum value $m = 9$ and a maximum value $M = 41$. The requested difference is $M^2 - m^2 = 41^2 - 9^2$, which equals $1681 - 81 = 1600$.

The differential inequality $(x \ln x) f'(x) + (\ln x) f(x) + f(x) \geq 1$ can be simplified by recognizing it as the derivative of a product. Since the derivative of $f(x) \cdot (x \ln x)$ is $(x \ln x) f'(x) + (\ln x + 1) f(x)$, the given condition is equivalent to $\frac{d}{dx} [x \ln x f(x)] \geq 1$. By defining an auxiliary function $g(x) = x \ln x f(x) - x$, we find that its derivative $g'(x) = \frac{d}{dx} [x \ln x f(x)] - 1$ is non-negative for all $x$ in the interval $[2, 4]$. This confirms that $g(x)$ is a monotonically increasing function on $[2, 4]$.

We determine the range of $g(x)$ by evaluating it at the endpoints of the interval using the values $f(2) = \frac{1}{2}$ and $f(4) = \frac{1}{4}$. Specifically, $g(2) = 2 \ln 2 \cdot \frac{1}{2} - 2 = \ln 2 - 2$ and $g(4) = 4 \ln 4 \cdot \frac{1}{4} - 4 = \ln 4 - 4 = 2 \ln 2 - 4$. Because $g(x)$ is increasing, we have $\ln 2 - 2 \leq x \ln x f(x) - x \leq 2 \ln 2 - 4$. Rearranging this inequality to solve for $f(x)$ results in the bounds $\frac{\ln 2 - 2}{x \ln x} + \frac{1}{\ln x} \leq f(x) \leq \frac{2 \ln 2 - 4}{x \ln x} + \frac{1}{\ln x}$.

Evaluating the upper bound $\frac{2 \ln 2 - 4}{x \ln x} + \frac{1}{\ln x}$, we find it remains strictly less than $1$ for all $x \in [2, 4]$, which confirms $f(x) \leq 1$. Regarding the lower bound, the expression $\frac{\ln 2 - 2}{x \ln x} + \frac{1}{\ln x}$ evaluated at $x=4$ yields $\frac{\ln 2 - 2}{8 \ln 2} + \frac{1}{2 \ln 2} = \frac{\ln 2 + 2}{8 \ln 2} = \frac{1}{8} + \frac{1}{4 \ln 2}$. Since $\frac{1}{4 \ln 2} > 0$, this lower bound is strictly greater than $\frac{1}{8}$, confirming $f(x) \geq \frac{1}{8}$. Consequently, both statements are verified as true.

Functions that are strictly increasing throughout their entire domain are inherently one-one, providing a streamlined way to verify both properties simultaneously. By substituting the definition of $f(x) = \frac{1}{1-e^{-x}}$ into the expression $g(x) = f(-x) - f(x)$, we simplify the function as follows:

$g(x) = \frac{1}{1-e^x} - \frac{1}{1-e^{-x}} = \frac{1+e^x}{1-e^x}$

To determine the behavior of this function on the interval $(0,1)$, we evaluate its derivative. Applying the derivative rules, we obtain:

$g'(x) = \frac{2e^x}{(1-e^x)^2}$

Because the squared term $(1-e^x)^2$ is always positive for any real $x$ and the numerator $2e^x$ is also strictly positive, the derivative $g'(x)$ is consistently greater than zero across the entire domain. Since the derivative is strictly positive, the function $g$ is strictly increasing. Consequently, the function must be one-one, as any strictly monotonic function maintains a unique output for every input, confirming that both statements are true.

The differential equation exhibits homogeneity because both $x$ and $y$ are raised to a total power of two in every term of the ratio, making $y = vx$ an appropriate substitution. Applying this substitution results in $v + x \frac{dv}{dx} = \frac{x^2 + (vx)^2}{2x(vx)}$, which simplifies by canceling $x^2$ to $\frac{1+v^2}{2v}$. Isolating the derivative term gives $x \frac{dv}{dx} = \frac{1+v^2}{2v} - v = \frac{1-v^2}{2v}$. Separating the variables leads to the integral expression $\int \frac{2v}{1-v^2} dv = \int \frac{1}{x} dx$. Performing the integration results in $-\ln|1-v^2| = \ln x + C$, which simplifies through exponentiation to $x(1-v^2) = K$. Resubstituting $v = \frac{y}{x}$ into this equation yields $x(1 - \frac{y^2}{x^2}) = K$, or more simply $x^2 - y^2 = Kx$. Using the condition $y(2)=0$, we substitute $x=2$ and $y=0$ to find $2^2 - 0 = K(2)$, which gives $K=2$. The final curve equation is $x^2 - y^2 = 2x$. Setting $x=8$ in this equation results in $64 - y^2 = 16$, meaning $y^2 = 48$, which gives $y = 4\sqrt{3}$.

A quadratic function that crosses the x-axis at $x = -1$ can be expressed in the form $f(x) = (x+1)(ax+b)$, which expands to $f(x) = ax^2 + (a+b)x + b$. The condition that the curve passes through $(1,1)$ implies $1 = (1+1)(a+b)$, establishing that $a+b = 0.5$. Because the curve is tangent to the line $y=x$ at $x=1$, its derivative $f'(x) = 2ax + (a+b)$ must equal the slope of the line, which is 1, at that point. Solving $2a(1) + (a+b) = 1$ with the substitution $a+b=0.5$ gives $2a = 0.5$, resulting in $a=0.25$ and $b=0.25$, which defines the curve as $f(x) = \frac{(x+1)^2}{4}$.

To find the location $(\alpha, \alpha+1)$ on this curve, we set the function equal to the y-coordinate: $\alpha+1 = \frac{(\alpha+1)^2}{4}$. Excluding the root at $x=-1$, this simplifies to $1 = \frac{\alpha+1}{4}$, yielding $\alpha = 3$. At this point $(3, 4)$, the derivative $f'(3) = \frac{3+1}{2} = 2$ provides the slope of the tangent line. The normal line at this point therefore has a slope of $-\frac{1}{2}$, expressed by the equation $y - 4 = -\frac{1}{2}(x - 3)$. Setting $y=0$ to determine the x-intercept results in $-4 = -\frac{1}{2}(x - 3)$, which simplifies to $8 = x - 3$, giving a final intercept of $11$.

To determine how many times the curve $y = x^5 - 20x^3 + 50x + 2$ intersects the $x$-axis, we analyze the function's critical points by setting its derivative to zero. Calculating the derivative gives $\frac{dy}{dx} = 5(x^4 - 12x^2 + 10)$, which implies the critical points occur where $x^2 = 6 \pm \sqrt{26}$. These calculations identify four turning points at approximately $x \approx -3.3, -0.95, 0.95,$ and $3.3$. Because the function is a fifth-degree polynomial, its end behavior dictates that it extends toward $-\infty$ as $x \to -\infty$ and toward $+\infty$ as $x \to +\infty$. By evaluating the function's values at these critical points, we observe that the curve alternates signs, forcing it to cross the $x$-axis exactly 5 times.

To determine the greatest term of the sequence defined by $a_n = \frac{n^3}{n^{4+}147}$, we analyze the behavior of the corresponding continuous function $f(x) = \frac{x^3}{x^{4+}147}$ by finding its stationary points. Calculating the derivative using the quotient rule gives $f'(x) = \frac{3x^2(x^{4+}147) - x^3(4x^3)}{(x^{4+}147)^2}$, which simplifies to $f'(x) = \frac{441x^2 - x^6}{(x^{4+}147)^2}$. Setting the derivative to zero yields the critical point $x^4 = 441$, or $x = \sqrt{21} \approx 4.58$.

Because the function increases for $x < 4.58$ and decreases for $x > 4.58$, the maximum term of the sequence must occur at the integer nearest to this critical value, either $n=4$ or $n=5$. Evaluating the function at these two points provides $f(4) = \frac{64}{403} \approx 0.158$ and $f(5) = \frac{125}{772} \approx 0.161$. Since $f(5)$ results in a larger value than $f(4)$, the greatest term in the sequence occurs at $n=5$, thus $a=5$.

A function's extreme points are located where its derivative is equal to zero, so if two functions share an extreme point, their derivatives must both vanish at the same value of $x$. Differentiating the given equations leads to $f'(x) = x^2 + ax + 2b$ and $g'(x) = x^2 + 2bx + a$. Equating these derivatives allows us to isolate the shared root:
$x^2 + ax + 2b = x^2 + 2bx + a$
Subtracting $x^2$ from both sides and regrouping terms gives $x(a - 2b) = a - 2b$, which simplifies to $(x - 1)(a - 2b) = 0$. Given that $a \neq 2b$, the only possible solution is $x = 1$. Substituting $x = 1$ into the derivative expression $f'(x) = 0$ provides the relationship $1 + a + 2b = 0$, which implies $a + 2b = -1$. Applying this to the final expression, $a + 2b + 7$ becomes $-1 + 7$, resulting in 6.

The core of this problem lies in translating the geometric property of the tangent line into a differential equation. Any tangent to the curve $y=f(x)$ at a point $(b, f(b))$ follows the linear equation $y - f(b) = f'(b)(x - b)$. Setting $x=0$ allows us to determine the $y$-intercept $c$ as $c = f(b) - b f'(b)$. By incorporating the given condition $bc=3$, which implies $c = 3/b$, we equate these expressions to obtain $3/b = f(b) - b f'(b)$, which rearranges into the form $b f'(b) - f(b) = -3/b$. Dividing the entire equation by $b^2$ reveals the derivative of a quotient, specifically $\frac{b f'(b) - f(b)}{b^2} = -\frac{3}{b^3}$, which simplifies to $\frac{d}{db} \left(\frac{f(b)}{b}\right) = -3b^{-3}$. Integrating both sides with respect to $b$ yields $\frac{f(b)}{b} = \frac{3}{2b^2} + \lambda$. Because the curve passes through $P(1, 3/2)$, substituting these coordinates results in $\frac{3/2}{1} = \frac{3}{2(1)^2} + \lambda$, confirming that $\lambda = 0$. Consequently, the function simplifies to $f(b) = \frac{3}{2b}$. Given that the curve also passes through $Q(a, 1/2)$, we solve the equation $\frac{1}{2} = \frac{3}{2a}$ to find $a = 3$, placing point $Q$ at $(3, 1/2)$. With $P(1, 3/2)$ and $Q(3, 1/2)$, the squared distance $(\text{PQ})^2$ is calculated using the distance formula as $(3 - 1)^2 + (1/2 - 3/2)^2$, which results in $2^2 + (-1)^2 = 4 + 1 = 5$.

Finding the global maximum of a function over a closed interval typically involves identifying stationary points where the derivative vanishes and testing those candidates alongside the boundaries of the interval. The expression $f(x) = x - 2 \sin x \cos x + \frac{1}{3} \sin 3x$ simplifies using the double-angle identity $2 \sin x \cos x = \sin 2x$, which yields $f(x) = x - \sin 2x + \frac{1}{3} \sin 3x$. Computing the derivative results in $f'(x) = 1 - 2 \cos 2x + \cos 3x$, and setting this to zero allows us to determine the stationary points $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$ within the interval $0 \le x \le \pi$.

To classify these points, the second derivative $f''(x) = 4 \sin 2x - 3 \sin 3x$ reveals that $f''\left(\frac{5\pi}{6}\right) < 0$, confirming a local maximum at $x = \frac{5\pi}{6}$. Evaluating the original function at this coordinate provides the maximum value $\frac{5\pi}{6} + \frac{\sqrt{3}}{2} + \frac{1}{3}$, which simplifies to $\frac{5\pi + 2 + 3\sqrt{3}}{6}$.

Optimization problems of this type rely on expressing the total value to be minimized as a function of one variable and finding the point where the derivative equals zero. Given the total length of 20 m, we represent the square side as $\frac{\ell_1}{4}$ to get area $A_1 = \frac{\ell_1^2}{16}$ and the circle radius as $\frac{\ell_2}{2\pi}$ to get area $A_2 = \frac{\ell_2^2}{4\pi}$. The objective function $S = 2A_1 + 3A_2$ then simplifies to $S = \frac{\ell_1^2}{8} + \frac{3\ell_2^2}{4\pi}$.

Since the sum of the two segments is constant, the derivative of $\ell_2$ with respect to $\ell_1$ is $-1$. Differentiating the expression for $S$ with respect to $\ell_1$ and setting the derivative to zero yields $\frac{\ell_1}{4} - \frac{6\ell_2}{4\pi} = 0$. Solving this equation by isolating the target term results in $\frac{\pi\ell_1}{\ell_2} = 6$.