Application Of Derivatives – JEE Main MathematicsPractice Questions & PYQs

The behavior of the function across its defined intervals reveals the points of local extrema. By examining the piecewise definition $f(x) = 2-x$ for $x \le -2$, $f(x) = -3x-2$ for $-2 < x \le -\frac{2}{3}$, $f(x) = 3x+2$ for $-\frac{2}{3} < x \le 0$, $f(x) = 2-x$ for $0 < x < 2$, and $f(x) = x-2$ for $x \ge 2$, we can track the slope changes. At $x = -\frac{2}{3}$, the function shifts from a decreasing slope to an increasing one, signifying a local minimum. At $x = 0$, the function transitions from increasing to decreasing, which marks a local maximum. At $x = 2$, the function switches back from decreasing to increasing, creating a second local minimum. These findings indicate two points of local minima and one point of local maxima, meaning $m = 2$ and $n = 1$, which results in $m+n = 3$.

Finding the local extrema of an integral function begins by calculating its derivative via the Leibniz Integral Rule. Differentiating $f(x)=\int_{0}^{x^2} \frac{t^{2-}8t+15}{e^t} dt$ yields:

$f'(x) = \frac{(x^2)^2 - 8(x^2) + 15}{e^{x^2}} \cdot 2x = \frac{2x(x^{2-}5)(x^{2-}3)}{e^{x^2}}$

Setting this derivative to zero identifies the stationary points at $x = 0$, $x = \pm\sqrt{3}$, and $x = \pm\sqrt{5}$.

The sign of $f'(x)$ fluctuates across these critical values, defining the local behavior of the function. Moving from left to right, the derivative changes from negative to positive at $x = -\sqrt{5}$, indicating a local minimum. At $x = -\sqrt{3}$, the sign shifts from positive to negative, signifying a local maximum, while at $x = 0$, it changes from negative to positive, confirming a local minimum. The pattern persists as the derivative flips from positive to negative at $x = \sqrt{3}$, marking the second local maximum, and from negative to positive at $x = \sqrt{5}$, defining the third local minimum. Consequently, the function exhibits two local maximum points and three local minimum points.

A function increases when its derivative is positive and decreases when it is negative, making the derivative the primary tool for identifying intervals of monotonicity. Differentiating $f(x)=\frac{x}{3}+\frac{3}{x}+3$ yields $f'(x)=\frac{x^{2-}9}{3x^2}$, where the numerator determines the sign since $3x^2$ is always positive for $x \neq 0$. The derivative is positive for $x \in (-\infty , -3) \cup (3, \infty )$, identifying $\alpha_1=-3$ and $\alpha_2=3$, and it is negative for $x \in (-3, 0) \cup (0, 3)$, which provides $\alpha_3=-3$, $\alpha_4=0$, and $\alpha_5=3$. Squaring and summing these values results in $(-3)^{2+}(3)^{2+}(-3)^{2+}0^{2+}(3)^2$, which equals 36.

The volume of a spherical shell, representing the layer of ice-cream, is the difference between the volume of the outer sphere, with radius $r+x$, and the inner chocolate ball of radius $r$. As the ice-cream melts, the rate of change of its volume is directly linked to the outer surface area and the rate at which the thickness $x$ decreases.

Differentiating the volume $V = \frac{4}{3}\pi((r+x)^3 - r^3)$ with respect to time yields the relationship:
$\frac{dV}{dt} = 4\pi(r+x)^2 \frac{dx}{dt}$
Substituting $x = 1$ cm, $\frac{dV}{dt} = 81$ $\text{cm}^3/\text{min}$, and $\frac{dx}{dt} = \frac{1}{4\pi}$ $\text{cm}/\text{min}$ into the formula gives $81 = 4\pi(r+1)^2 \cdot \frac{1}{4\pi}$. Simplifying this leads to $(r+1)^2 = 81$, which results in $r+1 = 9$, or $r = 8$ cm. The surface area of the chocolate ball, calculated as $4\pi r^2$, is $4\pi(8^2)$, which equals $256\pi$ $\text{cm}^2$.

The condition that the limit of $\frac{f(x)}{x^2}$ as $x \to 0$ equals $5$ implies that the degree 4 polynomial $f(x)$ lacks a constant and linear term, with the coefficient of the $x^2$ term equal to $5$. We can express the function as $f(x) = ax^4 + bx^3 + 5x^2$. Since the function possesses extreme values at $x=4$ and $x=5$, its derivative $f'(x) = 4ax^3 + 3bx^2 + 10x$ must be zero at these points.

Applying the conditions $f'(4) = 0$ and $f'(5) = 0$ results in the system $256a + 48b + 40 = 0$ and $500a + 75b + 50 = 0$. Solving this system provides $a = \frac{1}{8}$ and $b = -\frac{3}{2}$. Substituting these coefficients back into the expression yields $f(x) = \frac{1}{8}x^4 - \frac{3}{2}x^3 + 5x^2$. Evaluating this at $x=2$ results in $f(2) = 2 - 12 + 20 = 10$.

The sides $a$ and $b$ of the outer rectangle $PQRS$ depend on the tilt angle $\theta$ of the inscribed rectangle $ABCD$, leading to the expressions $a = 4 \cos\theta + 2 \sin\theta$ and $b = 2 \cos\theta + 4 \sin\theta$. The area of the outer rectangle is $a \times b$, which simplifies to $8 + 10 \sin 2\theta$ and reaches its maximum when $\sin 2\theta = 1$, meaning $\theta = 45^{\circ}$. At this angle, the sum of the sides $a + b$ becomes $6 \cos 45^{\circ} + 6 \sin 45^{\circ}$, which simplifies to $6\sqrt{2}$. Squaring this result provides the final value:

$(a+b)^2 = (6\sqrt{2})^2 = 72$

The local extrema of $f(x)=2x+3(x)^{\frac{2}{3}}$ occur where its derivative $f'(x)=2+2x^{-\frac{1}{3}}$ equals zero or remains undefined. Factoring this expression into $f'(x)=2\left(\frac{x^{\frac{1}{3}}+1}{x^{\frac{1}{3}}}\right)$ identifies a stationary point at $x=-1$ and a point of non-differentiability at $x=0$.

Investigating the sign of the derivative around these points shows that $f'(x)$ transitions from positive to negative at $x=-1$, confirming a local maximum. Similarly, moving across $x=0$, the derivative shifts from negative to positive, which characterizes a local minimum. Thus, the function exhibits exactly one point of local maxima and exactly one point of local minima.

The behavior of the function $f(x)=\cos x-x+1$ is determined by its derivative $f'(x)=-\sin x-1$. Because $\sin x$ is bounded by -1 and 1, the derivative $f'(x)$ remains less than or equal to 0 across the entire domain, confirming the function never increases and thus contradicts statement (S2). The values $f(0)=2$ and $f(\pi)=-\pi$ demonstrate that statement (S1) is correct, as the continuous and strictly decreasing function must transition from a positive value to a negative value, crossing the horizontal axis exactly once within the interval $[0,\pi]$.

To determine the intervals where a function increases or decreases, we examine the sign of its derivative $f'(x)$. Calculating the derivative of $f(x) = \frac{x}{x^{2-}6x-16}$ using the quotient rule results in $f'(x) = -\frac{x^{2+}16}{(x^{2-}6x-16)^2}$. Because the numerator $x^{2+}16$ is always positive for any real number $x$ and the squared denominator is always positive within the function's domain, the value of $f'(x)$ is strictly negative for all $x \in \mathbb{R} \setminus \{-2, 8\}$. Consequently, since the derivative is always negative, the function is decreasing throughout its entire defined domain of $(-\infty ,-2) \cup (-2,8) \cup (8,\infty )$.

Locating the local minimum of a function requires identifying where its derivative is zero and its second derivative is positive, while the specified inequality defines the allowed interval for that point. The numerator of the expression $\frac{x^{2+}x+2}{x^{2+}5x+6} < 0$ is always positive because its discriminant is negative and its leading coefficient is positive; therefore, the inequality simplifies to $(x+2)(x+3) < 0$, which is satisfied for $x \in (-3, -2)$.

For the function $f(x) = 1 + x(\lambda^{2-}x^2) = 1 + \lambda^2 x - x^3$, the derivative is $f'(x) = \lambda^2 - 3x^2$. Setting $f'(x) = 0$ yields critical points at $x = \pm \frac{\lambda}{\sqrt{3}}$. Since the second derivative $f''(x) = -6x$ must be positive for a local minimum, $x$ must be negative, identifying the local minimum at $x = -\frac{\lambda}{\sqrt{3}}$.

Equating this location to the valid interval results in $-3 < -\frac{\lambda}{\sqrt{3}} < -2$. Multiplying throughout by $-\sqrt{3}$ reverses the inequality signs, leading to $2\sqrt{3} < \lambda < 3\sqrt{3}$. This defines the interval $(\alpha, \beta)$ as $(2\sqrt{3}, 3\sqrt{3})$, where $\alpha = 2\sqrt{3}$ and $\beta = 3\sqrt{3}$. Consequently, the sum of their squares is:
$\alpha^2 + \beta^2 = (2\sqrt{3})^2 + (3\sqrt{3})^2 = 12 + 27 = 39$

The interval of strict increase for $f(x) = x^x$ is identified by finding where its derivative is positive. Applying logarithmic differentiation by taking the natural log of both sides, $\ln y = x \ln x$, and differentiating with respect to $x$ results in $\frac{dy}{dx} = x^x(1 + \ln x)$. Since $x^x$ is always positive for $x > 0$, the derivative is non-negative when the term $(1 + \ln x) \ge 0$. Solving this inequality for $x$ gives $\ln x \ge -1$, which implies $x \ge e^{-1}$, or $x \ge \frac{1}{e}$. The function is therefore strictly increasing on the interval $[\frac{1}{e}, \infty )$.

The range of the function $f(x)=\frac{2 x^{2}-3 x+8}{2 x^{2}+3 x+8}$ is determined by treating $y=f(x)$ as a quadratic equation in $x$ after cross-multiplying, which leads to $x^2(2y-2) + x(3y+3) + 8y-8 = 0$. For $x$ to be a real value, the discriminant of this quadratic must be non-negative, satisfying $D = (3y+3)^2 - 4(2y-2)(8y-8) \geq 0$. Expanding this expression results in $9(y+1)^2 - 64(y-1)^2 \geq 0$, which simplifies to the inequality $(11y-5)(5y-11) \leq 0$, confirming the range is the closed interval $y \in \left[ \frac{5}{11}, \frac{11}{5} \right]$. The minimum value is $\frac{5}{11}$ and the maximum value is $\frac{11}{5}$, so their sum is $\frac{5}{11} + \frac{11}{5} = \frac{25+121}{55} = \frac{146}{55}$. Since $m=146$ and $n=55$ are coprime, the sum $m+n$ equals $146 + 55 = 201$.

Functional equations involving reciprocal arguments are effectively resolved by generating a secondary equation through the substitution of $1/x$ for $x$, which creates a system that allows for the isolation of $f(x)$. Substituting $x$ with $1/x$ in the equation
$5 f(x)+4 f\left(\frac{1}{x}\right)=x^{2-}2$
produces the secondary relation
$4 f(x) + 5 f\left(\frac{1}{x}\right) = \frac{1}{x^2} - 2$
Solving this system for $f(x)$ yields the expression $f(x) = \frac{5 x^4 - 2 x^2 - 4}{9 x^2}$, which simplifies the definition of $y$ to the polynomial
$y = 5 x^4 - 2 x^2 - 4$

Determining the intervals where $y$ is strictly increasing requires identifying where its derivative is greater than zero. Calculating the derivative gives
$\frac{dy}{dx} = 20 x^3 - 4 x = 4x(5x^2 - 1)$
Investigating the sign of this derivative reveals that it remains positive when $x$ falls within the union of the intervals $\left(-\frac{1}{\sqrt{5}}, 0\right)$ and $\left(\frac{1}{\sqrt{5}}, \infty \right)$.

The behavior of the function $f(x)=e^{x^{3-}3x+1}$ on the interval $(-\infty , -1]$ is governed by its derivative, $f'(x)=3e^{x^{3-}3x+1}(x^{2-}1)$. For all $x \le -1$, the factor $(x^{2-}1)$ remains non-negative, confirming the function is monotonically increasing. The minimum value $a$ is found as $x$ approaches negative infinity, resulting in $a=0$, while the maximum value $b$ occurs at the right endpoint $x=-1$, yielding $b=e^{(-1)^{3-}3(-1)+1}=e^3$.

With these parameters, the coordinates of the point $P(2b+4, a+2)$ are $(2e^{3+}4, 2)$. The perpendicular distance from this point to the line $x+e^{-3}y-4=0$ is calculated using the standard distance formula, giving $\frac{|1(2e^{3+}4) + e^{-3}(2) - 4|}{\sqrt{1^2 + (e^{-3})^2}}$. Simplifying the numerator leaves $|2e^3 + 2e^{-3}|$, which is equivalent to $\frac{2(e^{6+}1)}{e^3}$. Since the denominator simplifies to $\sqrt{\frac{e^{6+}1}{e^6}} = \frac{\sqrt{e^{6+}1}}{e^3}$, the ratio becomes $\frac{2(e^{6+}1)}{e^3} \cdot \frac{e^3}{\sqrt{e^{6+}1}}$, ultimately simplifying to $2\sqrt{1+e^6}$.

Since the function $f$ is strictly increasing, for any positive $x$ we observe that $f(x) < f(5x) < f(7x)$. Dividing each part of this inequality by the positive value $f(x)$ results in $1 < \frac{f(5x)}{f(x)} < \frac{f(7x)}{f(x)}$. Because the limit of the rightmost term $\frac{f(7x)}{f(x)}$ approaches $1$ as $x$ tends to infinity, the squeeze theorem forces the limit of the middle term $\frac{f(5x)}{f(x)}$ to also be $1$. Applying this result to the expression $\lim_{x \to \infty } \left[ \frac{f(5x)}{f(x)} - 1 \right]$ gives a final value of $1 - 1$, which simplifies to $0$.

The existence of critical points for a differentiable function is determined by whether its derivative equals zero within the specified domain. To evaluate this for $f(x) = (p^{2-}6p+8)(\sin^2 2x - \cos^2 2x) + 2(2-p)x + 7$, use the identity $\cos 4x = \cos^2 2x - \sin^2 2x$ to rewrite the function as $f(x) = -(p^{2-}6p+8)\cos 4x + 2(2-p)x + 7$. Differentiation with respect to $x$ yields $f'(x) = 4(p^{2-}6p+8)\sin 4x + 4 - 2p$. For $f(x)$ to have no critical points, the derivative $f'(x)$ must never be zero, implying $4(p^{2-}6p+8)\sin 4x \neq 2p - 4$.

Factoring $p^{2-}6p+8$ as $(p-2)(p-4)$ allows the condition to be written as $4(p-2)(p-4)\sin 4x \neq 2(p-2)$. Excluding $p=2$, because $f'(x)$ would be zero everywhere in that case, the inequality simplifies to $\sin 4x \neq \frac{1}{2(p-4)}$. Since the range of $\sin 4x$ is $[-1, 1]$, the function possesses no critical points if $\left| \frac{1}{2(p-4)} \right| > 1$, which is equivalent to $|2(p-4)| < 1$. This inequality simplifies to $|p-4| < 0.5$, resulting in $3.5 < p < 4.5$. With $a = 7/2$ and $b = 9/2$, the value of $16ab$ becomes $16 \times \frac{7}{2} \times \frac{9}{2} = 252$.

The critical points of a function occur where its derivative $f'(x)$ is either zero or fails to exist within the function's domain. Differentiating $f(x)=(x-2)^{\frac{2}{3}}(2x+1)$ yields the derivative $\frac{3x-1}{(x-2)^{\frac{1}{3}}}$. Setting the numerator equal to zero identifies a critical point at $x=\frac{1}{3}$ where the derivative vanishes, while the derivative is undefined at $x=2$ where the denominator becomes zero. Since both these values are valid critical points, the function has 2 such points in total.

To find the absolute maximum and minimum of a continuous function on a closed interval, one must evaluate the function at the critical points where its derivative is zero as well as at the interval endpoints. For the function $f(x)=(x+3)^2(x-2)^3$, we determine the derivative using the product rule, which simplifies to $f'(x)=5(x+3)(x-2)^2(x+1)$. Solving $f'(x)=0$ identifies the critical values $x=-3$, $x=-1$, and $x=2$, all of which lie within the interval $[-4, 4]$.

Calculating the function's value at these critical points and the endpoints provides the set of candidates for the global extremes. At the boundaries, $f(-4)=(-4+3)^2(-4-2)^3=-216$ and $f(4)=(4+3)^2(4-2)^3=392$. At the critical points, $f(-3)=0$, $f(-1)=(-1+3)^2(-1-2)^3=-108$, and $f(2)=0$. Comparing these values reveals that the maximum $M$ is $392$ and the minimum $m$ is $-216$. Subtracting the minimum from the maximum, $M-m$ results in $392 - (-216)$, giving a final value of $608$.

The derivative of a determinant $f(x)$ is calculated by taking the sum of three separate determinants, where in each term, only one row is differentiated while the other two remain unchanged. Substituting $x=0$ into the original function gives $f(0) = 12$. Differentiating row by row at $x=0$ produces three distinct determinant components. The first component is

, which evaluates to $24$. The second component is

, which equals $0$ because the first column consists entirely of zeros. The third component is

, which simplifies to $-6$. Summing these three results gives $f'(0) = 24 + 0 - 6 = 18$. The final expression $2f(0) + f'(0)$ is calculated as $2(12) + 18$, which equals 42.

To determine the range of the function $f(x)=3\sqrt{x-2}+\sqrt{4-x}$, we first establish the domain constraints, as the expressions under the square roots must be non-negative. This requires $x-2 \ge 0$ and $4-x \ge 0$, which restricts $x$ to the interval $[2, 4]$. We can find the maximum value of this function by applying the Cauchy-Schwarz inequality, which states that for any real numbers $a, b, u, v$, the inequality $(a\sqrt{u}+b\sqrt{v})^2 \leq (a^{2+}b^2)(u+v)$ holds. Setting $a=3$, $b=1$, $u=x-2$, and $v=4-x$, the sum of the radicands becomes $(x-2)+(4-x)=2$. Substituting these values into the inequality gives $(f(x))^2 \leq (3^{2+}1^2)(2) = 20$, implying that the maximum value $\beta$ is $\sqrt{20}$.

The minimum value $\alpha$ occurs at the boundaries of the domain since the function is continuous. Evaluating $f(x)$ at the endpoints, we find $f(2) = 3\sqrt{0} + \sqrt{4-2} = \sqrt{2}$ and $f(4) = 3\sqrt{4-2} + \sqrt{0} = 3\sqrt{2}$. Comparing these results, the minimum value $\alpha$ is $\sqrt{2}$. With $\alpha=\sqrt{2}$ and $\beta=\sqrt{20}$, we calculate $\alpha^{2+}2\beta^2$ by substituting these values into the expression to get $2 + 2(20)$, which results in 42.