Application Of Derivatives – JEE Main MathematicsPractice Questions & PYQs

For the function $f_a(x) = \tan^{-1} 2x - 3ax + 7$ to be non-decreasing on the interval $\left(-\frac{\pi}{6}, \frac{\pi}{6}\right)$, its derivative $f_a'(x) = \frac{2}{1+4x^2} - 3a$ must be non-negative for all $x$ in that range. This requires $3a \le \frac{2}{1+4x^2}$, meaning $3a$ must be less than or equal to the minimum value of the expression on the right side over the given interval. Since the expression $\frac{2}{1+4x^2}$ is minimized at the endpoints $x = \pm \frac{\pi}{6}$, we calculate $3\overline{a} = \frac{2}{1+4(\frac{\pi^2}{36})} = \frac{2}{1+\frac{\pi^2}{9}} = \frac{18}{9+\pi^2}$, resulting in $\overline{a} = \frac{6}{9+\pi^2}$. Evaluating the function at $x = \frac{\pi}{8}$ with this value of $\overline{a}$ leads to $f_{\overline{a}}\left(\frac{\pi}{8}\right) = 1 - 3\left(\frac{\pi}{8}\right)\left(\frac{6}{9+\pi^2}\right) + 7$, which simplifies to $8 - \frac{9\pi}{4(9+\pi^2)}$.

Determining the absolute extrema of a function containing an absolute value term begins by partitioning the domain $[-1, 2]$ into intervals where the expression inside the modulus maintains a constant sign. This simplification allows us to analyze the function as distinct polynomial segments, where critical points are found by setting the derivative to zero and considering the boundaries of each interval. Evaluating the function at these critical points and the overall endpoints reveals the candidate values for the global maximum and minimum.

For this specific function, comparing these values identifies an absolute maximum of $3$ and an absolute minimum of $\frac{\sqrt{17}-3}{2}$. Adding these two values together results in a sum of

$\frac{\sqrt{17}+3}{2}$

The behavior of the function $f(x)=4 x^{3}-11 x^{2}+8 x-5$ depends on its derivative $f'(x)=12 x^{2}-22 x+8$. Factoring this expression as $2(2 x-1)(3 x-4)$ identifies the critical points at $x = \frac{1}{2}$ and $x = \frac{4}{3}$, where the slope of the curve is zero.

Testing any value within the open interval $\left(\frac{1}{2}, \frac{4}{3}\right)$ yields a negative result for $f'(x)$, confirming that the function is decreasing throughout this range. Because the derivative is positive for values less than $\frac{1}{2}$ and greater than $\frac{4}{3}$, the function is strictly increasing in those regions. Therefore, the function decreases specifically in the interval $\left(\frac{1}{2}, \frac{4}{3}\right)$.

To determine the parameter $a$, we analyze the derivative of $f(x) = 2x^2 - \ln x$. Calculating the derivative yields $f'(x) = 4x - \frac{1}{x} = \frac{4x^2 - 1}{x}$, which shows the function decreases when $4x^2 < 1$ and increases when $4x^2 > 1$. Given the interval $(0, a)$ for decrease and $(a, 4)$ for increase with $x > 0$, we identify $a = \frac{1}{2}$. This makes the parabola equation $y^2 = 4ax = 2x$.

A line $y = mx + c$ tangent to this parabola satisfies $c = \frac{a}{m} = \frac{1}{2m}$. We know the tangent passes through $(8a, 8a-1)$, which simplifies to $(4, 3)$. Substituting this point into the tangent equation $y = mx + \frac{1}{2m}$ gives $3 = 4m + \frac{1}{2m}$, which rearranges to $8m^2 - 6m + 1 = 0$. Solving this quadratic leads to $m = \frac{1}{2}$ or $m = \frac{1}{4}$. The condition that the tangent must not pass through $(-\frac{1}{a}, 0)$, which is $(-2, 0)$, allows us to exclude the case $m = \frac{1}{2}$ because the tangent $y = \frac{1}{2}x + 1$ contains the point $(-2, 0)$. Consequently, we select the slope $m = \frac{1}{4}$ for the tangent line.

With the slope $m = \frac{1}{4}$, the point of tangency corresponds to the parameter $t = \frac{1}{m} = 4$ for the parabola $y^2 = 2x$. The equation of the normal to the parabola at a point corresponding to parameter $t$ is $y = -tx + 2at + at^3$. Substituting $a = \frac{1}{2}$ and $t = 4$ provides $y = -4x + 2\left(\frac{1}{2}\right)(4) + \left(\frac{1}{2}\right)(4^3)$, which simplifies to $y = -4x + 4 + 32$, or $4x + y = 36$. Dividing by 36 yields $\frac{x}{9} + \frac{y}{36} = 1$. Comparing this to $\frac{x}{\alpha} + \frac{y}{\beta} = 1$, we find $\alpha = 9$ and $\beta = 36$, resulting in a sum $\alpha + \beta = 45$.

Determining the range of a continuous function on a closed interval relies on analyzing its monotonicity through its derivative. Differentiating $f(x)=2\cos^{-1}x+4\cot^{-1}x-3x^{2-}2x+10$ yields $f'(x)=-\frac{2}{\sqrt{1-x^2}}-\frac{4}{1+x^2}-6x-2$. Because each component of this derivative is negative across the domain $x\in[-1,1]$, the function is strictly decreasing throughout the entire interval.

Given this downward trend, the minimum value occurs at the right endpoint $x=1$, while the maximum value occurs at the left endpoint $x=-1$. Evaluating the function at these boundaries provides $f(1)=2\cos^{-1}(1)+4\cot^{-1}(1)-3(1)^{2-}2(1)+10$, which simplifies to $0+\pi-3-2+10=\pi+5$. Computing $f(-1)=2\cos^{-1}(-1)+4\cot^{-1}(-1)-3(-1)^{2-}2(-1)+10$ results in $2\pi+3\pi-3+2+10=5\pi+9$.

With the range defined as $[a,b] = [\pi+5, 5\pi+9]$, we identify $a=\pi+5$ and $b=5\pi+9$. Substituting these values into the expression $4a-b$ leads to $4(\pi+5)-(5\pi+9)$, which simplifies to $4\pi+20-5\pi-9$, resulting in $11-\pi$.

To determine the absolute maximum of the product $f(x)=(x^{2}-2x+7)e^{4x^{3}-12x^{2}-180x+31}$ on the closed interval $[-3,0]$, consider the two factors separately. The first factor, $f_{1}(x)=x^{2}-2x+7$, has a derivative $f_{1}'(x)=2x-2$, which is negative for all $x \in [-3,0]$, confirming that $f_{1}(x)$ is a positive, decreasing function throughout the interval.

Next, examine the second factor, $f_{2}(x)=e^{4x^{3}-12x^{2}-180x+31}$. Since the base $e$ is greater than $1$, the behavior of $f_{2}(x)$ is determined by its exponent, $g(x)=4x^{3}-12x^{2}-180x+31$. The derivative $g'(x)=12x^{2}-24x-180$ simplifies to $12(x-5)(x+3)$. Throughout the interval $[-3,0]$, the term $(x-5)$ is negative and the term $(x+3)$ is non-negative, which results in $g'(x) \le 0$. Consequently, the exponent $g(x)$ is a decreasing function, making $f_{2}(x)$ a decreasing and positive function on the given interval. Because $f(x)$ is the product of two positive, decreasing functions, $f(x)$ is also decreasing on $[-3,0]$, ensuring the absolute maximum occurs at the left endpoint, $x=-3$.

The slope of the tangent line at any point on a curve is defined as the tangent of the angle it forms with the positive $x$-axis, which in this case is $\tan(\frac{\pi}{3}) = \sqrt{3}$. We determine this slope using the parametric derivative $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$. Differentiating $y = 12(1+\sin t)^2$ with respect to $t$ gives $24(1+\sin t)\cos t$, and differentiating $x = 12(t+\sin t \cos t)$ with respect to $t$ gives $12(1+\cos 2t)$. Simplifying this quotient results in $\frac{dy}{dx} = \frac{1+\sin t}{\cos t}$, which is equivalent to the trigonometric identity $\tan(\frac{\pi}{4}+\frac{t}{2})$. Setting this equal to $\sqrt{3}$ leads to the equation $\frac{\pi}{4}+\frac{t}{2} = \frac{\pi}{3}$, which solves for $t = \frac{\pi}{6}$. Evaluating the original equation for $y$ at this value gives $y_0 = 12(1+\sin \frac{\pi}{6})^2 = 12(1+\frac{1}{2})^2 = 12(2.25) = 27$.

Determining the nature of local extrema for a function of the form $f(x) = (x-a)^{n_1}(x-b)^{n_2}$ requires evaluating the first derivative, which is given by $f'(x) = (x-a)^{n_1-1}(x-b)^{n_2-1} [(n_1+n_2)x - (n_1 b + n_2 a)]$. The critical point within the interval $(a, b)$ is located at $x_c = \frac{n_1 b + n_2 a}{n_1 + n_2}$. A local maximum occurs when the derivative changes sign from positive to negative as $x$ increases through $x_c$, whereas a local minimum occurs when the sign shifts from negative to positive.

In the case of $n_1=3$ and $n_2=4$, the critical point is $x_c = \frac{3(5) + 4(3)}{3+4} = \frac{27}{7} \approx 3.86$, where the derivative $f'(x) = (x-3)^2(x-5)^3(7x-27)$ shifts from positive to negative, indicating a local maximum. For $n_1=4$ and $n_2=3$, the critical point is $x_c = \frac{4(5) + 3(3)}{4+3} = \frac{29}{7} \approx 4.14$, and the derivative $f'(x) = (x-3)^3(x-5)^2(7x-29)$ changes from negative to positive, confirming a local minimum. Examining $n_1=3$ and $n_2=5$ yields $x_c = \frac{3(5) + 5(3)}{8} = 3.75$; the derivative $f'(x) = (x-3)^2(x-5)^4(8x-30)$ transitions from negative to positive at this point, which signifies a local minimum rather than the local maximum claimed. Finally, for $n_1=4$ and $n_2=6$, the critical point is $x_c = \frac{4(5) + 6(3)}{10} = 3.8$, where the derivative $f'(x) = (x-3)^3(x-5)^5(10x-38)$ changes from positive to negative, denoting a local maximum.

To determine the extreme values of the function $f(x) = |5x-7| + [x^{2+}2x]$ within the interval $[5/4, 2]$, we analyze the behavior of the absolute value and the floor component. We first check the boundary $x = 5/4$, yielding $f(5/4) = |5(1.25) - 7| + [1.5625 + 2.5] = 0.75 + [4.0625] = 19/4$. Next, we evaluate the critical point where the absolute value vanishes, $x = 7/5$, which results in f(7/5) = |5(7/5) - 7| + \[(7/5)^2 + 2(7/5)]$ = 0 + [4.76] = 4$. Because both components are non-decreasing for$x > 7/5$, the maximum must occur at the upper boundary$x = 2$, where$f(2) = |5(2) - 7| + $[2^2 + 2(2)]$ = 3 + 8 = 11$. Summing these findings, the minimum value of 4 and the maximum value of 11 produce a total sum of 15.

Determining the number of real solutions for a polynomial equation relies on assessing its monotonicity. By defining $f(x) = x^7 + 5x^3 + 3x + 1$, we find its derivative $f'(x) = 7x^6 + 15x^2 + 3$ is strictly positive for all real $x$, which indicates the function is strictly increasing. Because the function is continuous and maps the output range from negative infinity to positive infinity, it must cross the horizontal axis exactly once.

The requirement for a line to be tangent to a curve at a specific point is that the derivative of the curve at that point must equal the slope of the line. Starting with the curve equation $\left(\frac{x}{a}\right)^n + \left(\frac{y}{b}\right)^n = 2$, we apply implicit differentiation with respect to $x$ to obtain $\frac{n}{a}\left(\frac{x}{a}\right)^{n-1} + \frac{n}{b}\left(\frac{y}{b}\right)^{n-1} \frac{dy}{dx} = 0$. Rearranging this equation reveals that the slope of the curve at any point $(x, y)$ is given by $\frac{dy}{dx} = -\frac{b}{a} \left( \frac{x/a}{y/b} \right)^{n-1}$.

Substituting the point of tangency $(a, b)$ into this derivative expression simplifies the ratio $\frac{x/a}{y/b}$ to $\frac{a/a}{b/b} = 1$, which results in a slope of $-\frac{b}{a}$ regardless of the value of $n$. The equation of the line, $\frac{x}{a} + \frac{y}{b} = 2$, can be rearranged into the slope-intercept form $\frac{y}{b} = -\frac{x}{a} + 2$, confirming that the line also possesses a slope of $-\frac{b}{a}$. Since the point $(a, b)$ lies on both the curve and the line, and the slopes match perfectly for every natural number $n$, the line remains a tangent for the entire set of natural numbers, meaning $S = \mathbb{N}$.

To minimize the combined area of a square and an equilateral triangle formed from a wire of fixed length, we must express the total area as a function of a single variable and identify its critical point using calculus. Let the side length of the square be $a$ and the side length of the equilateral triangle be $b$. Since the total length of the wire is 22 m, the perimeter constraint is $4a + 3b = 22$, which allows us to express the square's side length as $a = \frac{22 - 3b}{4}$. The combined area $A$ is the sum of the square's area, $a^2$, and the equilateral triangle's area, $\frac{\sqrt{3}}{4}b^2$. Substituting the expression for $a$ into this formula gives $A(b) = \left(\frac{22 - 3b}{4}\right)^2 + \frac{\sqrt{3}}{4}b^2$.

Finding the minimum area requires differentiating $A(b)$ with respect to $b$ and setting the result to zero. Differentiating the area function yields $\frac{dA}{db} = 2 \left(\frac{22 - 3b}{4}\right) \left(-\frac{3}{4}\right) + \frac{2\sqrt{3}}{4}b$, which simplifies to $-\frac{3(22 - 3b)}{8} + \frac{\sqrt{3}}{2}b = 0$. Multiplying this equation by 8 to eliminate the denominator results in $-3(22 - 3b) + 4\sqrt{3}b = 0$, or $-66 + 9b + 4\sqrt{3}b = 0$. Solving for $b$ by grouping the terms involving $b$ leads to $b(9 + 4\sqrt{3}) = 66$, which gives the optimal side length for the equilateral triangle as $b = \frac{66}{9 + 4\sqrt{3}}$.

For a tangent to the curve $y=x^{3+}3 x^{2+}5$ to pass through the origin, the slope of the tangent at the point of contact $(x_1, y_1)$ must be equal to the slope of the line connecting that point to the origin $(0,0)$. The derivative of the curve, which provides the tangent's slope, is $\frac{dy}{dx}=3 x^{2+}6 x$, meaning at $(x_1, y_1)$ the slope is $3 x_1^{2+}6 x_1$. Since the slope between $(x_1, y_1)$ and the origin is simply $\frac{y_1}{x_1}$, equating these two expressions gives:
$y_1=3 x_1^{3+}6 x_1^2$

Because the point $(x_1, y_1)$ must also reside on the original curve, we set $3 x_1^{3+}6 x_1^2$ equal to $x_1^{3+}3 x_1^{2+}5$. Simplifying this leads to the polynomial:
$2 x_1^{3+}3 x_1^{2-}5=0$
Testing for roots reveals that $x_1=1$ is a valid solution, and substituting this value into the original curve equation gives $y_1=1^{3+}3(1)^{2+}5=9$. Substituting the point $(1,9)$ into the provided equations confirms that it satisfies every option except $\frac{x}{3}-y^2=2$, as the calculation $\frac{1}{3}-81$ results in a value that is not equal to 2.

For a hyperbola defined by the equation $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$, the condition for a line $y = mx + c$ to be a tangent is given by $c^2 = A^2 m^2 - B^2$. The given hyperbola $a^2 x^2 - y^2 = b^2$ can be rewritten in the standard form $\frac{x^2}{b^2/a^2} - \frac{y^2}{b^2} = 1$, revealing that $A^2 = \frac{b^2}{a^2}$ and $B^2 = b^2$. Rearranging the line equation $\lambda x - 2y = \mu$ into slope-intercept form gives $y = \frac{\lambda}{2}x - \frac{\mu}{2}$, which identifies the slope $m = \frac{\lambda}{2}$ and the constant term $c = -\frac{\mu}{2}$. Substituting these parameters into the tangency condition yields $(-\frac{\mu}{2})^2 = (\frac{b^2}{a^2})(\frac{\lambda}{2})^2 - b^2$, which simplifies to $\frac{\mu^2}{4} = \frac{b^2 \lambda^2}{4a^2} - b^2$. Multiplying the entire equation by $\frac{4}{b^2}$ isolates the expression to produce $\frac{\mu^2}{b^2} = \frac{\lambda^2}{a^2} - 4$, which leads directly to the required value of $\frac{\lambda^2}{a^2} - \frac{\mu^2}{b^2} = 4$.

Maximizing the total volume of a composite system under a fixed total surface area requires relating the two variables via the constraint and finding the point where the derivative of the volume function equals zero. The total surface area is the sum of the cuboid's surface area, calculated as $76x^2$, and the closed hemisphere's surface area, which is $3\pi r^2$, yielding the constraint $76x^2 + 3\pi r^2 = k$. Differentiating this with respect to $r$ gives $152x \frac{dx}{dr} + 6\pi r = 0$, which defines the derivative $\frac{dx}{dr} = -\frac{6\pi r}{152x}$.

The total volume $V = 40x^3 + \frac{2}{3}\pi r^3$ is optimized when its derivative with respect to $r$ vanishes, expressed as $\frac{dV}{dr} = 120x^2 \frac{dx}{dr} + 2\pi r^2 = 0$. Substituting the previously derived expression for $\frac{dx}{dr}$ results in $120x^2 \left(-\frac{6\pi r}{152x}\right) + 2\pi r^2 = 0$. By rearranging this algebraic equation to solve for the ratio $\frac{x}{r}$, we isolate $\frac{x}{r} = \frac{2\pi \cdot 152}{120 \cdot 6\pi}$, which leads to the final ratio $19:45$.

When a curve is tangent to the $x$-axis at a specific point, both the value of the function and its first derivative must vanish at that coordinate, which provides the necessary conditions to determine the unknown coefficients. For the cubic function $y(x) = ax^3 + bx^2 + cx + 5$, the $y$-intercept at $x=0$ immediately shows that the constant term is $5$, and taking the derivative $y'(x) = 3ax^2 + 2bx + c$ evaluated at $x=0$ gives $y'(0) = c = 3$. Applying the conditions at the tangency point $P(-2,0)$, we have $y(-2) = -8a + 4b - 6 + 5 = 0$, leading to $-8a + 4b = 1$, and $y'(-2) = 12a - 4b + 3 = 0$, leading to $12a - 4b = -3$. Adding these two equations yields $4a = -2$, resulting in $a = -\frac{1}{2}$, and substituting this back into either equation provides $b = -\frac{3}{4}$.

With the complete function $y(x) = -\frac{1}{2}x^3 - \frac{3}{4}x^2 + 3x + 5$ established, we identify the stationary points by setting the derivative $y'(x) = -\frac{3}{2}x^2 - \frac{3}{2}x + 3$ equal to zero. Factoring this expression gives $-\frac{3}{2}(x+2)(x-1) = 0$, indicating critical points at $x = -2$ and $x = 1$. Because $x = -2$ is the point of tangency, the local maximum must occur at $x = 1$. Evaluating the function at this point gives $y(1) = -\frac{1}{2} - \frac{3}{4} + 3 + 5$, which simplifies to $\frac{27}{4}$.

The function $f(x)=4 \log_e(x-1)-2x^{2+}4x+5$ has a first derivative $f'(x)=\frac{4}{x-1}-4(x-1)$, which confirms it increases for $x$ in the interval $(1, 2)$ and decreases for $x > 2$. Since the local maximum at $x=2$ is 5 and the function approaches negative infinity at the boundaries, the equation $f(x)=-1$ necessarily possesses two solutions. Further validation via the intermediate value theorem shows that since $f(e) > 0$ and $f(e+1) < 0$, a root must exist between $e$ and $e+1$.

Calculating the values for the final option, the second derivative at $x=2$ is $f''(2)=-8$, leading to $f'(e)-f''(2)=\frac{4}{e-1}-4(e-1)+8$, which is clearly positive. Because this expression yields a positive result, the provided inequality $f'(e)-f''(2) < 0$ is incorrect.

The surface area $S$ of a sphere depends on its radius according to the formula $S = 4\pi r^2$. Because the surface area increases at a constant rate, we can model the surface area at any time $t$ using the linear equation $S(t) = kt + C$, where $k$ represents the rate of increase and $C$ is the initial surface area. At the start time $t = 0$ with a radius $r = 3$, the initial surface area is $S = 4\pi(3)^2 = 36\pi$, which establishes that $C = 36\pi$.

After 5 seconds, the radius grows to $r = 7$, meaning the surface area becomes $S = 4\pi(7)^2 = 196\pi$. We determine the rate $k$ by substituting these values into the linear equation $196\pi = k(5) + 36\pi$, which simplifies to $160\pi = 5k$, yielding $k = 32\pi$. To find the radius at 9 seconds, we calculate the surface area at that time: $S = 32\pi(9) + 36\pi = 288\pi + 36\pi = 324\pi$. By setting this result equal to the geometric formula $4\pi r^2 = 324\pi$, we find that $r^2 = 81$, which means the radius at 9 seconds is 9.

For a polynomial to be monotonically increasing over its entire domain, its first derivative must be non-negative for all real $x$. Differentiating $f_\lambda(x) = 4\lambda x^3 - 36\lambda x^2 + 36x + 48$ gives:
$f'(x) = 12(\lambda x^2 - 6\lambda x + 3)$
This derivative remains non-negative for all $x$ if the quadratic factor possesses a non-positive discriminant, which ensures the function does not decrease. The condition for the discriminant is:
$D = (-6\lambda)^2 - 4(\lambda)(3) = 36\lambda^2 - 12\lambda \leq 0$
This inequality simplifies to $12\lambda(3\lambda - 1) \leq 0$, restricting $\lambda$ to the interval $[0, 1/3]$, which identifies the largest parameter value as $\lambda^* = 1/3$. Substituting $\lambda^* = 1/3$ into the original function yields $f_{\lambda^*}(x) = \frac{4}{3}x^3 - 12x^2 + 36x + 48$. Evaluating this expression at $x=1$ gives $f_{\lambda^*}(1) = \frac{220}{3}$, and at $x=-1$ gives $f_{\lambda^*}(-1) = -\frac{4}{3}$. Summing these values, we find the final result is 72.

A function increases on an interval whenever its first derivative remains non-negative throughout that region. Differentiating the given function term-by-term yields $f'(x) = 2x^2 - x - 2\cos x + 2\cos x - (2x-1)\sin x$, which simplifies upon grouping terms to $f'(x) = (2x-1)(x-\sin x)$. Within the interval $[\frac{1}{2}, \infty )$, the term $(2x-1)$ is non-negative, and because $x \ge \sin x$ for all $x \ge 0$, the factor $(x-\sin x)$ is also non-negative. Since the derivative is the product of two non-negative terms, it follows that $f'(x) \ge 0$ for $x \ge \frac{1}{2}$, confirming the function is increasing in this domain.