Application Of Derivatives – JEE Main MathematicsPractice Questions & PYQs

Finding the coordinates of the intersection point $Q$ begins by establishing the tangent line equation for $y=x^{3}$ at $P(t, t^{3})$. The derivative $3x^{2}$ provides the slope $3t^{2}$ at $P$, resulting in the tangent equation $y-t^{3}=3t^{2}(x-t)$. When this tangent intersects the curve, we equate $x^{3}-t^{3}=3t^{2}(x-t)$, which reduces to $(x-t)(x^{2}+xt+t^{2}-3t^{2})=0$. Further factoring the quadratic expression as $(x-t)(x+2t)=0$ identifies the intersection $Q$ at $x=-2t$, which corresponds to the ordinate $y=(-2t)^{3}=-8t^{3}$. With $P$ having an ordinate of $t^{3}$ and $Q$ having an ordinate of $-8t^{3}$, applying the section formula for a 1:2 ratio yields the target ordinate as $\frac{1(-8t^{3}) + 2(t^{3})}{1+2}$. This simplifies to $\frac{-6t^{3}}{3}$, which equals $-2t^{3}$.

Determining the number of real roots for $f(x) = e^{4x}+2e^{3x}-e^{x}-6$ requires analyzing the behavior of the function through its derivative to establish how many times it crosses the x-axis. Calculating the derivative yields $f'(x) = 4e^{4x}+6e^{3x}-e^{x}$, which can be expressed as $e^{x}(4e^{3x}+6e^{2x}-1)$. Since $e^{x}$ is always positive, the sign of $f'(x)$ depends entirely on the term $4e^{3x}+6e^{2x}-1$.

The expression $4e^{3x}+6e^{2x}-1$ is strictly increasing for all real $x$, as its own derivative, $12e^{3x}+12e^{2x}$, is always positive. This term ranges from $-1$ as $x$ approaches $-\infty$ to positive infinity as $x$ increases, meaning $f'(x)$ transitions from negative to positive exactly once. Consequently, $f(x)$ decreases to a single global minimum before increasing indefinitely. Given that $f(x)$ approaches $-6$ as $x$ tends toward $-\infty$ and approaches positive infinity as $x$ tends toward $+\infty$, the function must cross the x-axis at exactly one point to transition between these limits.

This inequality reveals that the rate of change of the function $f$ must be zero everywhere because the difference between function values shrinks faster than the distance between inputs. By dividing both sides of $|f(x) - f(y)| \leq |x - y|^2$ by $|x - y|$ for $x \neq y$, we obtain the following relationship:

$\left| \frac{f(x) - f(y)}{x - y} \right| \leq |x - y|$

Taking the limit as $x$ approaches $y$, the left side becomes the definition of the absolute value of the derivative, $|f'(y)|$, while the right side approaches zero. Because the absolute value cannot be negative, $|f'(y)|$ must equal zero, which confirms that $f'(y) = 0$ for all real numbers $y$. Integrating this result indicates that $f(x)$ is a constant function. Given that $f(0) = 1$, the constant must be $1$ for all real numbers $x$. Since $f(x) = 1$ is always greater than zero, the function consistently takes positive values across its entire domain.

Analyzing the local extrema of a polynomial reveals how many times its graph traverses the x-axis, providing the total count of distinct real roots. For the function $f(x) = 3x^4 + 4x^3 - 12x^2 + 4$, the derivative is $f'(x) = 12x^3 + 12x^2 - 24x$, which factors as $12x(x+2)(x-1) = 0$ to identify critical points at $x = -2, x = 0,$ and $x = 1$. Evaluating the function at these points results in $f(-2) = -28$, $f(0) = 4$, and $f(1) = -1$, showing that the curve oscillates above and below the axis, crossing it four times as it extends toward positive infinity at both extremes.

The function $f(x)=x^{3-}6x^{2+}ax+b$ satisfies $f(2)=8-24+2a+b=0$ and $f(4)=64-96+4a+b=0$, which yields the values $a=8$ and $b=0$. The derivative is $f'(x)=3x^{2-}12x+8$, and setting $f'(x)=0$ gives critical points at $x=2 \pm \frac{2}{\sqrt{3}}$. Since $x=2+\frac{2}{\sqrt{3}} \approx 3.15$ is in $(2, 4)$, we identify this as $x_2$. For $f'(x_1)=-1$, solving $3x^{2-}12x+9=0$ yields $x=1$ and $x=3$, and since $x_1=3$ lies in $(2, 4)$ and $x_1 < x_2$, the first statement is true.

Regarding the second statement, the function $f(x)$ decreases on $(2, x_4)$ and increases on $(x_4, 4)$, meaning $x_4=2+\frac{2}{\sqrt{3}}$ is the local minimum. Evaluating $f(x_4)$ results in $f(2+\frac{2}{\sqrt{3}}) = -\frac{16}{3\sqrt{3}}$. The condition $2f'(x_3)=\sqrt{3}f(x_4)$ simplifies to $2f'(x_3) = -\frac{16}{3}$, which means $f'(x_3)=-\frac{8}{3}$. Solving $3x^{2-}12x+8=-\frac{8}{3}$ leads to $9x^{2-}36x+32=0$, which provides roots $x=\frac{4}{3}$ and $x=\frac{8}{3}$. As $x_3=\frac{8}{3}$ sits within the interval $(2, x_4)$, the second statement is also true.

Minimizing the combined area of two geometric shapes created from a fixed length of wire requires expressing the total area as a function of one variable and determining the value that results in a minimum area using calculus. If we designate the length of the wire used for the square as $x$, then the remaining portion $20-x$ is used for the regular hexagon, resulting in a square side length of $x/4$ and a hexagon side length of $(20-x)/6$. The total area $A$ is the sum of the square's area $(x/4)^2$ and the hexagon's area $\frac{3\sqrt{3}}{2} (\frac{20-x}{6})^2$, which simplifies to $A = \frac{x^2}{16} + \frac{\sqrt{3}(20-x)^2}{24}$.

To find the minimum, we differentiate $A$ with respect to $x$ and set the resulting derivative to zero, which gives the equation $\frac{x}{8} - \frac{\sqrt{3}(20-x)}{12} = 0$. Multiplying this expression by $24$ simplifies the arithmetic to $3x - 2\sqrt{3}(20-x) = 0$ which expands to $3x - 40\sqrt{3} + 2\sqrt{3}x = 0$. Factoring out $x$ yields $x(3 + 2\sqrt{3}) = 40\sqrt{3}$, meaning the length of the wire assigned to the square is $x = \frac{40\sqrt{3}}{3+2\sqrt{3}}$. Because the side length of the hexagon is defined as $s = (20-x)/6$, we substitute $20-x$ by calculating $20 - \frac{40\sqrt{3}}{3+2\sqrt{3}} = \frac{60 + 40\sqrt{3} - 40\sqrt{3}}{3+2\sqrt{3}}$, which simplifies to $\frac{60}{3+2\sqrt{3}}$. Dividing this value by $6$ provides the side length of the hexagon as $\frac{10}{3+2\sqrt{3}}$.

The function $f(x)=x^{3}-3 x^{2}-\frac{3 f''(2)}{2} x+f''(1)$ contains constants determined by its second derivative. Differentiating twice results in the expression
$f''(x)=6x-6$
Substituting the specific points into this expression yields $f''(2)=6(2)-6=6$ and $f''(1)=6(1)-6=0$. When these constants are returned to the original function, it simplifies to
$f(x)=x^{3}-3 x^{2}-9x$
Local minima occur at critical points where the first derivative equals zero:
$f'(x)=3x^{2}-6x-9=3(x-3)(x+1)=0$
The roots at $x=3$ and $x=-1$ represent potential extrema. Applying the second derivative test, $f''(3)=12$, which is positive and indicates a local minimum at $x=3$. The other root, $x=-1$, results in a negative second derivative, corresponding to a local maximum. Evaluating the function at the minimum point yields $f(3)=3^{3}-3(3^{2})-9(3)=-27$. As this is the only local minimum, the sum of all local minimum values is $-27$.

The behavior of the function $f(x) = x^2 + ax + 1$ is governed by its derivative, $f'(x) = 2x + a$, which determines where the function increases or decreases. For the function to be increasing on the interval $[1, 2]$, $f'(x)$ must remain non-negative throughout, and since the smallest value of this derivative occurs at the lower bound $x = 1$, we set $2(1) + a \geq 0$ to find the least value $R = -2$. Conversely, for the function to decrease, $f'(x)$ must stay non-positive, and since the largest value of the derivative occurs at the upper bound $x = 2$, we set $2(2) + a \leq 0$ to determine that the greatest value is $S = -4$. The final value is found by calculating the absolute difference, $|-2 - (-4)|$, which equals 2.

Maximizing the volume of the open box involves expressing the volume as a function of the cutout size $x$ and using calculus to identify the point where the volume curve peaks. After cutting squares of side $x$ from the corners of a rectangular sheet with dimensions $a$ and $b$, the resulting box has length $(a - 2x)$, breadth $(b - 2x)$, and height $x$, leading to the volume formula $V = (a - 2x)(b - 2x)x$, which simplifies to $V = 4x^3 - 2(a + b)x^2 + abx$.

Deriving this function with respect to $x$ yields $V' = 12x^2 - 4(a + b)x + ab$, and setting this to zero provides the critical values through the quadratic formula:
$x = \frac{4(a + b) \pm \sqrt{16(a + b)^2 - 48ab}}{24} = \frac{a + b \pm \sqrt{a^2 + b^2 - ab}}{6}$
The second derivative, $V'' = 24x - 4(a + b)$, is negative for the smaller root, which signifies a local maximum. Thus, the optimal dimension for $x$ that results in the greatest possible volume is:
$x = \frac{a + b - \sqrt{a^2 + b^2 - ab}}{6}$

Minimizing the sum of the areas created by cutting a 36-meter wire into a square and a circle requires expressing the total area as a function of the perimeter assigned to the square. Defining the perimeter of the square as $x$ makes the circumference of the circle $k = 36 - x$, leading to an area function $A = \frac{x^2}{16} + \frac{(36-x)^2}{4\pi}$.

Taking the derivative $\frac{dA}{dx}$ and setting it to zero results in the equation $\frac{x}{8} - \frac{36-x}{2\pi} = 0$. Solving this establishes that the length allocated to the square is $x = \frac{144}{\pi+4}$, which implies the circumference of the circle is $k = \frac{36\pi}{\pi+4}$.

Substituting this value of $k$ into the expression $\left(\frac{4}{\pi}+1\right)k$ simplifies to $\left(\frac{4+\pi}{\pi}\right) \left(\frac{36\pi}{\pi+4}\right)$. Upon cancellation of common factors, this arithmetic yields a final value of 36.

Since the curve $y = ax^2 + bx + c$ has a tangent line defined at the origin, the curve must pass through the point $(0,0)$. Substituting $x=0$ and $y=0$ into the equation gives $0 = a(0)^2 + b(0) + c$, which immediately establishes that $c = 0$. With $c$ determined, we can calculate the slope of the curve at any point using the derivative $\frac{dy}{dx} = 2ax + b$. Because the tangent line at the origin is given as $y=x$, which has a slope of $1$, evaluating the derivative at $x=0$ must equal this value. Therefore, $2a(0) + b = 1$, which confirms that $b = 1$.

Finally, we utilize the fact that the curve passes through the point $(1,2)$ to find the value of $a$. Substituting $x=1$ and $y=2$ into the general equation yields $2 = a(1)^2 + b(1) + c$, or $a + b + c = 2$. By inserting the previously derived values $b=1$ and $c=0$ into this equation, we get $a + 1 + 0 = 2$, which leads to the result $a = 1$.

For a polynomial $f(x)$ of degree 6 with a lead coefficient of $1$, the condition that $\lim_{x \to 0} \frac{f(x)}{x^3} = 1$ implies that all coefficients of terms with powers less than $3$ must be zero, while the coefficient of $x^3$ must be $1$. This establishes the polynomial as $f(x) = x^6 + ax^5 + bx^4 + x^3$. Because the function has extrema at $x = 1$ and $x = -1$, its derivative $f'(x) = 6x^5 + 5ax^4 + 4bx^3 + 3x^2$ must evaluate to zero at these specific values.

Applying $f'(1) = 0$ yields the equation $6 + 5a + 4b + 3 = 0$, which simplifies to $5a + 4b = -9$. Similarly, applying $f'(-1) = 0$ results in $-6 + 5a - 4b + 3 = 0$, simplifying to $5a - 4b = 3$. Solving this system of equations by adding them gives $10a = -6$, so $a = -\frac{3}{5}$, which subsequently leads to $b = -\frac{3}{2}$ when substituted back into either equation. With these constants determined, calculating $5f(2)$ involves substituting $x=2$ into the expression $5(x^6 - \frac{3}{5}x^5 - \frac{3}{2}x^4 + x^3)$, resulting in $5(2^6 - \frac{3}{5} \cdot 2^5 - \frac{3}{2} \cdot 2^4 + 2^3)$. Distributing the factor of $5$ across these terms yields $320 - 96 - 120 + 40$, which equals $144$.

Implicit differentiation allows us to relate the rates of change between $x$ and $y$ directly by treating the equation as a functional dependency. Expanding the product $(a+\sqrt{2}b\cos x)(a-\sqrt{2}b\cos y) = a^{2}-b^{2}$ results in the following equation:
$a^{2} - \sqrt{2}ab\cos y + \sqrt{2}ab\cos x - 2b^{2}\cos x \cos y = a^{2} - b^{2}$
Canceling $a^{2}$ from both sides leaves $-\sqrt{2}ab\cos y + \sqrt{2}ab\cos x - 2b^{2}\cos x \cos y = -b^{2}$. Performing implicit differentiation with respect to $x$ yields the equation $\sqrt{2}ab\sin y\frac{dy}{dx} - \sqrt{2}ab\sin x + 2b^{2}\sin x \cos y + 2b^{2}\cos x \sin y\frac{dy}{dx} = 0$.

Substituting the point $\left(\frac{\pi}{4}, \frac{\pi}{4}\right)$, where the sine and cosine values are all $\frac{1}{\sqrt{2}}$, the equation simplifies to $ab\frac{dy}{dx} - ab + b^{2} + b^{2}\frac{dy}{dx} = 0$. Factoring out $\frac{dy}{dx}$ leads to $(ab + b^{2})\frac{dy}{dx} = ab - b^{2}$, which simplifies to $\frac{dy}{dx} = \frac{a-b}{a+b}$. Taking the reciprocal to find $\frac{dx}{dy}$ results in the final value of $\frac{a+b}{a-b}$.

Rate of change problems involving geometric solids rely on identifying the relationship between the quantities and differentiating them with respect to time using the chain rule. Because the surface area $S$ of a cube relates to its side length $a$ as $S = 6a^{2}$, the rate of change is given by the derivative $\frac{dS}{dt} = 12a \frac{da}{dt}$. Inserting the known rate of 3.6 cm²/sec into this derivative gives $12a \frac{da}{dt} = 3.6$, which simplifies to $a \frac{da}{dt} = 0.3$.

Differentiating the volume formula $V = a^{3}$ with respect to time yields $\frac{dV}{dt} = 3a^{2} \frac{da}{dt}$. This expression can be rewritten as $3a \left( a \frac{da}{dt} \right)$ to utilize the value derived from the surface area calculation. Substituting the side length of 10 cm and the value 0.3 for $a \frac{da}{dt}$ into this equation results in $3 \times 10 \times 0.3$, which equals 9 cm³/sec.

The limit condition $\lim_{x\to 0}\left(2+\frac{f(x)}{x^3}\right)=4$ forces the degree 5 polynomial $f(x)$ to be of the form $f(x) = 2x^3 + ax^4 + bx^5$, as the ratio $\frac{f(x)}{x^3}$ must approach 2 as $x$ approaches 0. Differentiating this gives $f'(x) = 6x^2 + 4ax^3 + 5bx^4$, and because $x = \pm 1$ are critical points, the equations $f'(1) = 6 + 4a + 5b = 0$ and $f'(-1) = 6 - 4a + 5b = 0$ must hold. Solving this system yields $a = 0$ and $b = -\frac{6}{5}$, which defines the function as $f(x) = 2x^3 - \frac{6}{5}x^5$ and its derivative as $f'(x) = 6x^2 - 6x^4 = 6x^2(1-x)(1+x)$. By analyzing the sign of this derivative, it is clear that $f'(x)$ changes from negative to positive at $x = -1$, confirming it as a point of minima, and from positive to negative at $x = 1$, confirming it as a point of maxima. This established behavior directly invalidates the claim that $x = 1$ is a point of minima and $x = -1$ is a point of maxima.

To determine the equation of the normal, we must first identify the coordinates of the point on the curve where $x=0$. By substituting $x=0$ into the equation $y=(1+x)^{2y}+\cos^2(\sin^{-1}x)$, the term $(1+0)^{2y}$ simplifies to $1$, while $\cos^2(\sin^{-1}0)$ becomes $\cos^2(0)$, which also equals $1$. Adding these values confirms that $y=2$, establishing the point of contact as $(0, 2)$.

Next, we find the slope of the tangent line at this point by calculating the derivative $y'$. To simplify the differentiation, we recognize that $\cos^2(\sin^{-1}x)$ is equivalent to $1-x^2$ and express $(1+x)^{2y}$ as $e^{2y\ln(1+x)}$. Differentiating this expression with respect to $x$ yields $y' = e^{2y\ln(1+x)} \cdot \left( \frac{2y}{1+x} + 2\ln(1+x)y' \right) - 2x$. Substituting $x=0$ and $y=2$ into this derivative results in $y' = e^{2(2)\ln(1)} \cdot (\frac{2(2)}{1+0} + 2\ln(1+0)y') - 2(0)$, which simplifies to $y' = 1 \cdot (4 + 0) - 0$, giving a tangent slope of $4$.

Because the slope of the normal is the negative reciprocal of the tangent's slope, the normal slope is $-\frac{1}{4}$. Applying the point-slope form $y - y_1 = m(x - x_1)$ with the point $(0, 2)$ and the slope $-\frac{1}{4}$ gives $y - 2 = -\frac{1}{4}(x - 0)$. Multiplying the entire equation by $4$ leads to $4y - 8 = -x$, which rearranges to the final equation $x+4y=8$.

Determining the perpendicular length begins by identifying the slope of the tangent at (2,2) through implicit differentiation of the curve $x^2 + 2xy - 3y^2 = 0$. Evaluating $2x + 2y + 2x\frac{dy}{dx} - 6y\frac{dy}{dx} = 0$ at the point (2,2) reveals that the tangent has a slope of 1. Because the normal line is perpendicular to the tangent, its slope must be the negative reciprocal, which is $-1$.

With the slope determined, the equation for the normal line passing through (2,2) is $y - 2 = -1(x - 2)$, which rearranges into the standard form $x + y - 4 = 0$. The perpendicular distance from the origin $(0,0)$ to this line is calculated using the formula $\frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$, resulting in $\frac{|0 + 0 - 4|}{\sqrt{1^2 + 1^2}}$, which simplifies to $2\sqrt{2}$.

Parallel lines have identical slopes, so the first step is to determine the slope of the line connecting (0, $\frac{3}{2}$) and (\frac{1}{2}$, 2), which is calculated as$\frac{2 - 3/2}{1/2 - 0} = 1$. Since the tangent at$(a, b)$is parallel to this line, the derivative of the curve$y = x + \sin y$at that point must also equal 1. Differentiating implicitly with respect to$x$yields$\frac{dy}{dx} = 1 + \cos y \frac{dy}{dx}$, and substituting the slope value into this equation gives$1 = 1 + \cos b$, which simplifies to$\cos b = 0$. Given that$\cos b = 0$, the value of$\sin b$must be 1 or$-1$, which, when substituted into the original curve equation$b = a + \sin b$, leads to the conclusion that$|b - a| = 1$.

By applying the Fundamental Theorem of Calculus, we can determine the lower bounds of the function value and its derivative at $x=5$ based on the provided growth rates. The value of a function at any point is simply its initial value plus the accumulated change, which is represented by the definite integral of its derivative over the interval.

Integrating the first derivative inequality $f'(x) \geq 1$ from $x=2$ to $x=5$ provides:
$f(5) - f(2) = \int_{2}^{5} f'(x) \, dx \geq \int_{2}^{5} 1 \, dx = 3$
Substituting $f(2)=8$ leads to $f(5) \geq 11$. Applying the same principle to the second derivative inequality $f''(x) \geq 4$ over the same interval yields:
$f'(5) - f'(2) = \int_{2}^{5} f''(x) \, dx \geq \int_{2}^{5} 4 \, dx = 12$
With $f'(2)=5$, we find that $f'(5) \geq 17$. Combining these results, we conclude $f(5) + f'(5) \geq 11 + 17$, resulting in $f(5) + f'(5) \geq 28$.

Since we are dealing with a third-degree polynomial, its derivative is a quadratic function, and its second derivative is a linear function. The statement that $f'(x)$ has a critical point at $x=1$ implies that the second derivative $f''(x)$ must equal zero at that point, which allows us to define $f''(x) = \lambda(x-1)$ for some constant $\lambda$. By integrating this expression twice, we determine the general form $f'(x) = \frac{\lambda x^2}{2} - \lambda x + C$ and $f(x) = \frac{\lambda x^3}{6} - \frac{\lambda x^2}{2} + Cx + D$. Because $f(x)$ has a critical point at $x=-1$, we know $f'(-1)=0$, which leads us to the value $C = -\frac{3\lambda}{2}$.

Substituting this back into our expression gives $f(x) = \frac{\lambda x^3}{6} - \frac{\lambda x^2}{2} - \frac{3\lambda x}{2} + D$. Applying the known values $f(-1)=10$ and $f(1)=-6$ creates the system of equations $5\lambda + 6D = 60$ and $-11\lambda + 6D = -36$, which solve to $\lambda = 6$ and $D = 5$. With these constants confirmed, the polynomial is fully defined as $f(x) = x^3 - 3x^2 - 9x + 5$. To find the location of the local minima, we set the derivative $f'(x) = 3x^2 - 6x - 9$ equal to zero. Factoring this quadratic expression yields $3(x-3)(x+1) = 0$, identifying stationary points at $x=-1$ and $x=3$. Evaluating the second derivative $f''(x) = 6x - 6$ at these points shows that $f''(3) = 12$, which is positive, confirming the function attains a local minimum at $x=3$.