Application Of Derivatives – JEE Main MathematicsPractice Questions & PYQs

To define the cubic polynomial $f(x) = ax^3 + bx^2 + cx + d$, we determine the coefficients by applying the properties of derivatives at the given critical points. Because $f(x)$ exhibits a local minimum at $x = 1$, the first derivative $f'(x) = 3ax^2 + 2bx + c$ must equal zero at $x = 1$. Similarly, since $f'(x)$ reaches a local minimum at $x = -1$, the second derivative $f''(x) = 6ax + 2b$ must be zero at $x = -1$. Evaluating $f''(-1) = 0$ provides the relationship $-6a + 2b = 0$, so $b = 3a$, and substituting this into $f'(1) = 0$ gives $3a + 2(3a) + c = 0$, which simplifies to $c = -9a$.

We utilize the given points $f(1) = -10$ and $f(-1) = 6$ to create the system $a + b + c + d = -10$ and $-a + b - c + d = 6$. Adding these equations yields $2b + 2d = -4$, or $b + d = -2$, and substituting $b = 3a$ results in $d = -2 - 3a$. Substituting the expressions for $b$, $c$, and $d$ in terms of $a$ into the equation $a + b + c + d = -10$ produces $a + 3a - 9a - 2 - 3a = -10$. This simplifies to $-8a = -8$, confirming that $a = 1$. With $a$ determined, we find $b = 3$, $c = -9$, and $d = -5$, defining the function as $f(x) = x^3 + 3x^2 - 9x - 5$. Calculating the final value, $f(3) = 3^3 + 3(3^2) - 9(3) - 5 = 27 + 27 - 27 - 5$, which results in 22.

The problem requires finding the minimum value of $\alpha$ such that the function $f(x) = \frac{4}{\sin x} + \frac{1}{1-\sin x}$ attains this value at least once in the interval $x \in \left(0, \frac{\pi}{2}\right)$. By letting $t = \sin x$, the range of $t$ becomes $(0, 1)$, and the expression simplifies to $g(t) = \frac{4}{t} + \frac{1}{1-t}$, which can be rewritten as $\frac{4-3t}{t-t^2}$.

To identify the minimum value on this interval, compute the derivative $g'(t)$ and set it to zero:
$g'(t) = \frac{-3t^2 + 8t - 4}{(t-t^2)^2}$
Setting the numerator $-3t^2 + 8t - 4 = 0$ results in the quadratic equation $3t^2 - 8t + 4 = 0$. Factoring this leads to $(3t-2)(t-2) = 0$, yielding roots $t = \frac{2}{3}$ and $t = 2$. Since $t$ is restricted to $(0, 1)$, $t = \frac{2}{3}$ is the only valid critical point. Substituting this value back into the function gives $g\left(\frac{2}{3}\right) = \frac{4}{2/3} + \frac{1}{1-2/3} = 6 + 3 = 9$. Because this represents the lowest point the function reaches within the specified interval, $\alpha$ must be at least 9 for at least one solution to exist.

To determine if a line acts as a tangent to a conic section, we utilize the standard method of replacing variables in the curve equation to find the tangent line at a specific point $(x_1, y_1)$. For the curve defined by $x^2 + 9y^2 = 9$, the equation of the tangent is established by substituting $x^2$ with $x x_1$ and $y^2$ with $y y_1$. Applying this transformation to the point $\left(\frac{3\sqrt{3}}{2}, \frac{1}{2}\right)$ yields the expression $x \left(\frac{3\sqrt{3}}{2}\right) + 9y \left(\frac{1}{2}\right) = 9$. Simplifying this result leads to $\frac{3\sqrt{3}}{2}x + \frac{9}{2}y = 9$. By multiplying both sides of this equation by $\frac{2}{3}$, we arrive at $\sqrt{3}x + 3y = 6$, and dividing the entire equation by $\sqrt{3}$ confirms the line is exactly $x + \sqrt{3}y = 2\sqrt{3}$.

Analyzing the monotonicity of a function relies on the sign of its derivative. Differentiating $f(x) = 3 \sin^{4} x+10 \sin^{3} x+6 \sin^{2} x-3$ results in $f'(x) = 12 \sin^{3} x \cos x+30 \sin^{2} x \cos x+12 \sin x \cos x$. Factoring this expression gives $f'(x) = 6 \sin x \cos x (2 \sin^{2} x+5 \sin x+2)$, which further simplifies to $f'(x) = 6 \sin x \cos x (2 \sin x+1)(\sin x+2)$. Over the interval $(-\frac{\pi}{6}, 0)$, the sine value is negative while the cosine value is positive. Since $\sin x > -\frac{1}{2}$ in this range, the term $(2 \sin x+1)$ is positive, and $(\sin x+2)$ is also positive. Because the derivative consists of a single negative factor multiplied by several positive factors, $f'(x) < 0$ throughout $(-\frac{\pi}{6}, 0)$, confirming the function decreases on this interval.

The polynomial $f(x) = 2 x^{5}+5 x^{4}+10 x^{3}+10 x^{2}+10 x+10$ is of odd degree, which guarantees the existence of at least one real root, and evaluating its derivative confirms that the function is strictly increasing. Calculating the derivative yields $f'(x) = 10 x^{4}+20 x^{3}+30 x^{2}+20 x+10$, which can be expressed as $10 x^{2}(x + \frac{1}{x} + 1)^{2}$. Since this derivative is positive for all real $x$, the function is strictly increasing and thus possesses exactly one real root.

Evaluating the polynomial at specific integers reveals that $f(-2) = -34$ and $f(-1) = 3$. Because the function changes sign from negative to positive as it passes through the interval $(-2, -1)$, the Intermediate Value Theorem ensures that the unique real root lies within these bounds. Comparing this interval to the given $(a, a+1)$, we identify $a = -2$, which means $|a|$ is equal to 2.

A function $f(x)$ increases on an interval where its first derivative $f'(x)$ is strictly positive. For the given function $f(x) = 3 \ln \left| \frac{x-1}{x+1} \right| - \frac{2}{x-1}$, differentiating with respect to $x$ yields $f'(x) = \frac{4(2x-1)}{(x-1)^{2}(x+1)}$. Because the constant $4$ and the squared factor $(x-1)^2$ are always positive for all valid values of $x$, the sign of $f'(x)$ is determined solely by the expression $\frac{2x-1}{x+1}$. Investigating the sign of this quotient across the critical points $x = -1$ and $x = \frac{1}{2}$, the derivative evaluates to a positive value in the regions where $x < -1$ and $x \geq \frac{1}{2}$. Considering the domain of the function where $x$ cannot be $1$, the combined interval for increasing behavior is $(-\infty ,-1) \cup \left( \left[ \frac{1}{2}, \infty \right) - \{1\} \right)$.

A function increases wherever its derivative is strictly positive, so we determine the intervals of increase by finding the derivative for each piece of the function and identifying where each results in a positive value. For the region $x \le 0$, the function is $f(x) = 3xe^x$. Applying the product rule for differentiation yields $f'(x) = 3(e^x + xe^x) = 3e^x(1+x)$. Since $3e^x$ is always positive, the derivative $f'(x)$ is positive whenever $1+x > 0$, or $x > -1$. Within the condition $x \le 0$, this establishes that the function is increasing on the interval $(-1, 0]$.

For the region $x > 0$, the function is $f(x) = -\frac{4}{3}x^3 + 2x^2 + 3x$, which gives the derivative $f'(x) = -4x^2 + 4x + 3$. To find where this quadratic expression is positive, we solve $-4x^2 + 4x + 3 > 0$, which is equivalent to $4x^2 - 4x - 3 < 0$. Factoring the quadratic gives $(2x-3)(2x+1) < 0$, which identifies the roots at $x = -\frac{1}{2}$ and $x = \frac{3}{2}$. Because the leading coefficient of the original derivative is negative, the parabola opens downward and remains positive between these roots. Restricting this to the domain $x > 0$ yields the interval $(0, \frac{3}{2})$. Combining the intervals from both pieces of the function, the union $(-1, 0] \cup (0, \frac{3}{2})$ confirms that the function is increasing on the interval $(-1, \frac{3}{2})$.

A function remains monotonic on an interval only if its derivative does not change sign, as it must be either consistently increasing or decreasing. Separating the absolute value term based on the sign of $x$, the function is defined as $f(x)=(2-\sin(1/x))x$ for $x>0$ and $f(x)=-(2-\sin(1/x))x$ for $x<0$. Differentiating these expressions leads to $f'(x)=2-\sin(1/x)+\frac{1}{x}\cos(1/x)$ when $x>0$, and $f'(x)=-2+\sin(1/x)-\frac{1}{x}\cos(1/x)$ when $x<0$. As $x$ approaches zero, the term $\frac{1}{x}\cos(1/x)$ dominates and oscillates with increasing frequency, causing $f'(x)$ to flip signs infinitely many times. Because the derivative fails to maintain a single sign in any sub-interval of the domain, the function is non-monotonic on both $(-\infty , 0)$ and $(0, \infty )$.

For any quadratic function of the form $f(x)=Ax^{2+}Bx+C$, the turning point where the function transitions between increasing and decreasing behavior is located at the vertex $x=-\frac{B}{2A}$. Applying this property to $f(x)=ax^{2+}6x-15$ with a turning point at $x=\frac{3}{4}$ provides the equation $-\frac{6}{2a}=\frac{3}{4}$, which simplifies to $a=-4$. Substituting this value into $g(x)=ax^{2-}6x+15$ results in the function $g(x)=-4x^{2-}6x+15$. Calculating the vertex for this new function gives $x=-\frac{-6}{2(-4)}$, which simplifies to $-\frac{3}{4}$. Because the leading coefficient is negative, the parabola opens downward, confirming that the function attains a local maximum at this point.

The range of an expression formatted as $A \cos \theta + B \sin \theta$ is $[-\sqrt{A^{2+}B^2}, \sqrt{A^{2+}B^2}]$. Simplifying the determinant involves adding the second column to the first, which uses the identity $\sin^2 x + \cos^2 x = 1$ to produce column entries 2, 2, and 1. The matrix becomes:

$\begin{vmatrix} 2 & 1+\cos^2 x & \cos 2x \\ 2 & \cos^2 x & \cos 2x \\ 1 & \cos^2 x & \sin 2x \end{vmatrix}$

Subtracting the second row from the first row creates a zero-filled row, allowing the determinant to expand into:

$\begin{vmatrix} 0 & 1 & 0 \\ 2 & \cos^2 x & \cos 2x \\ 1 & \cos^2 x & \sin 2x \end{vmatrix}$

This expansion results in $-1(2 \sin 2x - \cos 2x)$, which simplifies to $\cos 2x - 2 \sin 2x$. Applying the range principle with $A=1$ and $B=-2$ gives $\sqrt{1^{2+}(-2)^2}$, resulting in a maximum value of $\sqrt{5}$.

Constructing the $3 \times 3$ matrix $A$ based on the provided conditions for its entries $a_{ij}$ results in
$A = \begin{bmatrix} 1 & -x & 2x+1 \\ -x & 1 & -x \\ 2x+1 & -x & 1 \end{bmatrix}$
The determinant of this matrix provides the function $f(x) = 4x^3 - 4x^2 - 4x$.

Determining the extrema requires finding the derivative $f'(x) = 12x^2 - 8x - 4$ and setting it equal to zero. Solving the equation $4(3x^2 - 2x - 1) = 0$ yields critical points at $x = 1$ and $x = -\frac{1}{3}$. Evaluating the function at these values gives $f(1) = -4$ and $f(-\frac{1}{3}) = \frac{20}{27}$, and their sum equals $-\frac{88}{27}$.

Two curves are defined as cutting at right angles if the product of their slopes at the point of intersection is exactly $-1$. For the curve $x = y^4$, differentiating with respect to $x$ yields $1 = 4y^3 \frac{dy}{dx}$, which means the slope at any point is $\frac{dy}{dx} = \frac{1}{4y^3}$. For the second curve $xy = k$, differentiating implicitly results in $x \frac{dy}{dx} + y = 0$, giving the slope $\frac{dy}{dx} = -\frac{y}{x}$.

At the intersection point, equating the product of these slopes to $-1$ gives $\frac{1}{4y^3} \cdot (-\frac{y}{x}) = -1$. Substituting the relationship $x = y^4$ into this equation produces $\frac{1}{4y^3} \cdot (-\frac{y}{y^4}) = -1$, which simplifies to $\frac{-1}{4y^6} = -1$, revealing that $y^6 = \frac{1}{4}$. Since $k = xy$ and $x = y^4$, we find that $k = y^5$. To determine the value of $(4k)^6$, we expand it to $4^6 \cdot k^6$, which is $4^6 \cdot y^{30}$. Because $y^{30} = (y^6)^5 = (\frac{1}{4})^5$, the expression becomes $4^6 \cdot (\frac{1}{4})^5$, resulting in a final value of $4$.

The slope of the tangent line at any point on the curve is defined by $\frac{dy}{dx} = \frac{x y^2 + y}{x}$, which simplifies to $\frac{dy}{dx} = y^2 + \frac{y}{x}$. Rearranging this equation into $x dy - y dx = x y^2 dx$ and dividing both sides by $x y^2$ produces $\frac{x dy - y dx}{y^2} = x dx$. Integrating this differential form leads to the general solution $-\frac{x}{y} = \frac{x^2}{2} + C$.

The intersection with the line $x + 2y = 4$ at $x = -2$ determines the point $(-2, 3)$, as substituting $x = -2$ into the line equation $-2 + 2y = 4$ simplifies to $y = 3$. Applying the coordinates $(-2, 3)$ to the general solution $-\frac{-2}{3} = \frac{(-2)^2}{2} + C$ gives $\frac{2}{3} = 2 + C$, which results in $C = -\frac{4}{3}$.

The specific curve equation is then $-\frac{x}{y} = \frac{x^2}{2} - \frac{4}{3}$. Evaluating this expression at $x = 3$ produces $-\frac{3}{y} = \frac{3^2}{2} - \frac{4}{3}$. Combining the fractions on the right side as $\frac{9}{2} - \frac{4}{3} = \frac{27 - 8}{6} = \frac{19}{6}$ leads to the equation $-\frac{3}{y} = \frac{19}{6}$, which results in $y = -\frac{18}{19}$.

When two curves intersect at a right angle, their tangent lines at the point of intersection must be perpendicular, meaning the product of their slopes is equal to $-1$. Calculating the slope of the tangent for each curve by differentiating the given equations implicitly with respect to $x$, we find the slopes are $-\frac{bx}{ay}$ and $-\frac{dx}{cy}$ respectively. Requiring the product of these slopes to be $-1$ at the intersection point leads to the condition $bdx^2 = -acy^2$

Because the intersection point satisfies both original equations, subtracting the two equations from one another reveals a relationship between $x^2$ and $y^2$ expressed as $x^2(\frac{1}{a} - \frac{1}{c}) + y^2(\frac{1}{b} - \frac{1}{d}) = 0$. By rearranging this to find the ratio $\frac{x^2}{y^2}$, we can substitute the ratio $\frac{x^2}{y^2} = -\frac{ac}{bd}$ derived from the orthogonality condition. Equating these two expressions allows us to eliminate the coordinate variables, leading to the simplified relation $c - a = d - b$, which rearranges to the final form $a - b = c - d$.

The second derivative of any quadratic function $f(x) = ax^{2}+bx+c$ is a constant value, specifically $f''(x) = 2a$. Equating this to the given maximum of $\frac{1}{2}$ reveals that $a = \frac{1}{4}$. By substituting this value into the expression for the derivative $f'(x) = 2ax + b$ at $x = -1$, we find that $-\frac{1}{2} + b = 1$, which leads to $b = \frac{3}{2}$. Applying these coefficients to the condition $f(-1) = a - b + c = 2$ allows us to solve for the final constant, resulting in $c = \frac{13}{4}$.

With the complete function $f(x) = \frac{1}{4}x^{2} + \frac{3}{2}x + \frac{13}{4}$ defined, we examine its behavior on the interval $[-1, 1]$ to identify the maximum value. Since the parabola opens upward and its vertex is located at $x = -3$, the function is strictly increasing throughout the domain, meaning the maximum must occur at the right endpoint, $x = 1$. Calculating this value yields
$f(1) = \frac{1}{4} + \frac{6}{4} + \frac{13}{4} = 5$
indicating that 5 is the least value for $\alpha$ such that $f(x) \leq \alpha$ for all $x$ in the interval.

To determine the local maximum of a function where both the base and exponent depend on the variable $x$, taking the natural logarithm is the most effective approach for simplifying the differentiation process. By defining the function as $f(x) = (\frac{2}{x})^{x^2}$, we rewrite the expression as $\ln f(x) = x^2(\ln 2 - \ln x)$. Applying the product rule to differentiate both sides with respect to $x$ yields $\frac{f'(x)}{f(x)} = 2x(\ln 2 - \ln x) - x$, which simplifies to $f'(x) = f(x) \cdot x(2\ln 2 - 2\ln x - 1)$.

For the function to reach an extreme value, the derivative must equal zero, requiring the term $2\ln 2 - 2\ln x - 1 = 0$ to be satisfied. This rearranges to $\ln(\frac{4}{x^2}) = 1$, which implies $\frac{4}{x^2} = e$, or $x = \frac{2}{\sqrt{e}}$. Evaluating the original function at this specific value gives $f(\frac{2}{\sqrt{e}}) = (\frac{2}{2/\sqrt{e}})^{(2/\sqrt{e})^2}$, which simplifies to $(\sqrt{e})^{4/e}$. Using exponent properties, this further reduces to $(e^{1/2})^{4/e}$, resulting in the final value of $e^{2/e}$.

Rolle's theorem implies that for a function continuous on $[1, 2]$ and differentiable on $(1, 2)$, the condition $f(1) = f(2)$ must be satisfied if the derivative vanishes within the interval. Evaluating $f(x) = x^3 - ax^2 + bx + 4$ at the endpoints gives $f(1) = 1 - a + b + 4 = 5 - a + b$ and $f(2) = 8 - 4a + 2b + 4 = 12 - 4a + 2b$. Setting these expressions equal yields $5 - a + b = 12 - 4a + 2b$, which simplifies to the linear relation $3a - b = 7$.

The derivative of the function is $f'(x) = 3x^2 - 2ax + b$. Using the given condition $f'\left(\frac{4}{3}\right) = 0$, we substitute $x = \frac{4}{3}$ to obtain $3\left(\frac{4}{3}\right)^2 - 2a\left(\frac{4}{3}\right) + b = 0$. This simplifies to $\frac{16}{3} - \frac{8a}{3} + b = 0$, which rearranges to $16 - 8a + 3b = 0$, or $-8a + 3b = -16$. Substituting $b = 3a - 7$ from the first relation into this second equation results in $-8a + 3(3a - 7) = -16$, leading to $a = 5$ and consequently $b = 8$, providing the ordered pair $(5, 8)$.

The geometric condition that every normal to a curve must pass through a fixed point $(a, b)$ implies that the slope of the normal, given by $-\frac{dx}{dy}$, satisfies the relation $(y - b) = -\frac{dx}{dy} (x - a)$. This relationship can be expressed as the separable differential equation $(y - b) dy = -(x - a) dx$. Integrating both sides of this equation leads to the general form:

$by - \frac{y^2}{2} = -ax + \frac{x^2}{2} + C$

By substituting the known coordinates $(3, -3)$ and $(4, -2\sqrt{2})$ into this equation, we generate a system of linear equations involving $a$, $b$, and the integration constant $C$. The point $(3, -3)$ yields the equation $3a - 3b - C = 9$, while the point $(4, -2\sqrt{2})$ leads to $4a - 2\sqrt{2}b - C = 12$. Subtracting the first equation from the second allows us to eliminate $C$, resulting in $a + (3 - 2\sqrt{2})b = 3$. When compared with the given constraint $a - 2\sqrt{2}b = 3$, it becomes clear that $3b$ must equal $0$, which implies $b = 0$. Substituting $b = 0$ back into the constraint determines $a = 3$. Evaluating the expression $a^2 + b^2 + ab$ with these values gives $3^2 + 0^2 + 0$, which is $9$.

To identify the intervals where the function is increasing, we calculate its derivative $f'(x)$ and seek the regions where $f'(x) > 0$. For the domain $x < -5$, the derivative is $-55$, which is always negative, confirming the function is strictly decreasing in that region. In the segment $-5 \le x < 4$, the derivative is $6(x^2 - x - 20)$, which factors into $6(x - 5)(x + 4)$. Because the term $(x - 5)$ is negative for all $x < 4$, the derivative is positive only when $(x + 4)$ is also negative, which identifies the interval $(-5, -4)$ as a region of increase.

For $x \ge 4$, the derivative becomes $6(x^2 - x - 6)$, which simplifies to $6(x - 3)(x + 2)$. Since any value in this domain is greater than $3$, both $(x - 3)$ and $(x + 2)$ are positive, meaning $f'(x) > 0$ for the entire range $(4, \infty )$. Combining these findings, the function is strictly increasing across the set $(-5, -4) \cup (4, \infty )$.

Determining the point of maximum slope on a curve involves treating the derivative of that curve as a function itself and maximizing it. This requires identifying the critical values where the second derivative of the original function equals zero, then applying the second derivative test to confirm which value yields a local maximum.

The slope of the curve $y = \frac{1}{2} x^{4}-5 x^{3}+18 x^{2}-19 x$ is expressed by $f(x) = 2x^{3}-15x^{2}+36x-19$. Differentiating $f(x)$ to find its extrema results in $6x^{2}-30x+36$, and equating this to zero factors to $6(x-2)(x-3) = 0$, which identifies critical points at $x=2$ and $x=3$. Calculating the derivative of the slope function again gives $12x-30$; evaluating this at $x=2$ produces $-6$, which confirms a local maximum for the slope at that value. Plugging $x=2$ back into the original expression for $y$ yields $\frac{1}{2}(16) - 5(8) + 18(4) - 38$, which simplifies to $8 - 40 + 72 - 38 = 2$, identifying the point as $(2,2)$.