Application Of Derivatives – JEE Main MathematicsPractice Questions & PYQs

The volume of a sphere is defined as $V = \frac{4}{3}\pi r^3$, where the total radius $r$ encompasses both the iron core and the outer ice layer. Given that the iron ball has a 10 cm radius and the ice layer is 5 cm thick, the total radius $r$ at this instant is 15 cm. Differentiating the volume formula with respect to time yields the relationship $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$, which links the rate at which the volume melts to the rate at which the thickness decreases. Substituting the known rate of volume change, 50 cm$^3$/min, and the radius $r = 15$ cm into the equation gives $50 = 4\pi(15)^2 \frac{dr}{dt}$. Simplifying this calculation results in $50 = 900\pi \frac{dr}{dt}$, confirming that the rate at which the ice thickness decreases is $\frac{1}{18\pi}$ cm/min.

Lagrange's Mean Value Theorem allows us to relate the average rate of change of a differentiable function over an interval to its derivative at an interior point. Applying this theorem to the interval $[1, 6]$ gives the expression
$\frac{f(6) - f(1)}{6 - 1} = f'(c)$
where $c$ is a value in the interval. Since $f'(c) \geq 2$ and $f(1) = -2$, substituting these values leads to $\frac{f(6) + 2}{5} \geq 2$. Simplifying this inequality results in $f(6) + 2 \geq 10$, which confirms $f(6) \geq 8$.

Rolle's Theorem guarantees that for any continuous and differentiable function $f(x)$ where $f(0) = f(1)$, the derivative $f'(x)$ must have at least one root in the interval $(0, 1)$. By constructing the auxiliary function $f(x) = \frac{ax^3}{3} + \frac{bx^2}{2} + cx$, we observe that $f'(x) = ax^2 + bx + c$, which is the quadratic expression in the problem. Evaluating the endpoints yields $f(0) = 0$ and $f(1) = \frac{a}{3} + \frac{b}{2} + c = \frac{2a + 3b + 6c}{6}$. Given the condition $2a + 3b + 6c = 0$, it follows that $f(1) = 0$, creating the condition $f(0) = f(1) = 0$ required by Rolle's Theorem. Consequently, the equation $ax^2 + bx + c = 0$ must possess at least one root within the interval $(0, 1)$.

The geometry of a normal to a curve is defined by a slope perpendicular to the tangent, and a fixed point on such a normal is one that satisfies the line's equation regardless of the parameter $\theta$. Determining the slope of the curve at any given $\theta$ involves finding the derivatives with respect to the parameter, where $\frac{dx}{d\theta} = -a\sin\theta$ and $\frac{dy}{d\theta} = a\cos\theta$, which yields a tangent slope of $\frac{dy}{dx} = -\cot\theta$. Since the normal is perpendicular to the tangent, its slope is the negative reciprocal, $\tan\theta$.

The equation of the normal passing through the point $(a(1+\cos\theta), a\sin\theta)$ is $y - a\sin\theta = \tan\theta(x - a - a\cos\theta)$. By expressing the tangent as $\frac{\sin\theta}{\cos\theta}$ and multiplying both sides by $\cos\theta$, the equation simplifies to $y\cos\theta - a\sin\theta\cos\theta = x\sin\theta - a\sin\theta - a\sin\theta\cos\theta$. Canceling the common term $-a\sin\theta\cos\theta$ from both sides results in $y\cos\theta = x\sin\theta - a\sin\theta$, which can be rearranged as $y\cos\theta = \sin\theta(x-a)$. Substituting $x=a$ into this equation forces the right side to become zero, which implies $y$ must also be zero, confirming that the point $(a,0)$ lies on the normal for any value of $\theta$.

The rate at which the ordinate $y$ changes with respect to the abscissa $x$ is represented by the derivative $\frac{dy}{dx}$, which is specified as 2. Differentiating the parabola $y^2 = 18x$ with respect to $x$ yields $2y \frac{dy}{dx} = 18$, which simplifies to $\frac{dy}{dx} = \frac{9}{y}$. By equating the derivative to 2, we determine that $y = \frac{9}{2}$, and substituting this value into the original parabola equation provides $x = \frac{81/4}{18} = \frac{9}{8}$, pinpointing the location at $\left( \frac{9}{8}, \frac{9}{2} \right)$.

To recover a function from its second derivative, we perform successive integrations, determining the integration constants by applying the known properties of the tangent line and a specific point on the curve. Since the second derivative is $f''(x)=6(x-1)$, we integrate once to find the first derivative $f'(x)=3x^{2-}6x+c$. Because the tangent line at the point (2, 1) is $y=3x-5$, the derivative $f'(2)$ must equal the slope of this line, which is 3. Solving the equation $3(2)^{2-}6(2)+c=3$ yields $c=3$, meaning the derivative simplifies to $f'(x)=3(x-1)^2$.

We then integrate $f'(x)=3(x-1)^2$ with respect to $x$ to obtain the general form $f(x)=(x-1)^{3+}D$. To find the constant $D$, we substitute the coordinates of the point (2, 1) into the equation, leading to $1=(2-1)^{3+}D$. This calculation results in $D=0$, confirming the final function is $f(x)=(x-1)^3$.

Extrema for a polynomial function occur where its first derivative is zero, and the classification of these points depends on the sign of the second derivative. Differentiating $f(x)=2 x^{3-}9 a x^{2+}12 a^2 x+1$ yields $f'(x)=6 x^{2-}18 a x+12 a^2$. Setting this expression equal to zero produces $6(x-a)(x-2 a)=0$, which places the critical points at $x=a$ and $x=2 a$.

Testing these points with the second derivative $f''(x)=12 x-18 a$ determines the nature of each stationary point. Evaluating at $x=a$ gives $f''(a)=-6 a$, which is negative because $a>0$, identifying a maximum at $p=a$. Conversely, evaluating at $x=2 a$ yields $f''(2 a)=6 a$, which is positive, identifying a minimum at $q=2 a$. Applying the condition $p^2=q$ leads to the equation $a^2=2 a$, which simplifies to $a(a-2)=0$. Rejecting the root $a=0$ because $a$ must be positive leaves $a=2$ as the required value.

To find the value of $x$ that minimizes the expression $f(x) = x + \frac{1}{x}$, one must first determine the critical points by setting the derivative of the function, $f'(x) = 1 - \frac{1}{x^2}$, equal to zero. Solving this equation results in $x^2 = 1$, which gives the two potential candidates $x = 1$ and $x = -1$. Applying the second derivative test, $f''(x) = \frac{2}{x^3}$, allows us to confirm the nature of these points; substituting $x = 1$ yields a positive value of $2$, verifying that the function attains a minimum at $x = 1$.

Applying Rolle's theorem provides a straightforward approach to identifying the interval containing the roots of a quadratic equation. By defining a primitive function $f(x) = \frac{ax^3}{3} + \frac{bx^2}{2} + cx$, we establish a polynomial that is continuous and differentiable across any interval. Evaluating this function at the endpoints $x=0$ and $x=1$ yields $f(0)=0$ and $f(1) = \frac{a}{3} + \frac{b}{2} + c = \frac{2a+3b+6c}{6}$. Substituting the given condition $2a+3b+6c=0$ confirms that $f(1)=0$, resulting in $f(0)=f(1)$. Consequently, Rolle's theorem guarantees at least one value in the interval $(0,1)$ where the derivative $f'(x) = ax^2 + bx + c$ equals zero, meaning the quadratic equation must have at least one root within the interval $[0,1]$.

The distance $D$ of any point $(x, y)$ on the curve from the origin is given by the expression $\sqrt{x^2 + y^2}$. By substituting the parametric equations $x = a \sin t - b \sin\left(\frac{at}{b}\right)$ and $y = a \cos t - b \cos\left(\frac{at}{b}\right)$ and applying the trigonometric identity $\cos(A - B) = \cos A \cos B + \sin A \sin B$, the squared distance simplifies to $D^2 = a^2 + b^2 - 2ab \cos\left(t - \frac{at}{b}\right)$. Since the cosine function has a minimum value of $-1$, the distance reaches its maximum when the term $\cos\left(t - \frac{at}{b}\right)$ equals $-1$. This yields $D^2 = a^2 + b^2 + 2ab$, which is the perfect square $(a + b)^2$, resulting in a maximum distance of $a + b$.

The slope of a tangent line to a function at any given point is determined by its first derivative, while the normal line at that same point is perpendicular to the tangent, meaning its slope is the negative reciprocal of the tangent's slope. Differentiating the function $y(x) = 1 + \sqrt{4x-3}$ gives the slope of the tangent as $\frac{dy}{dx} = \frac{2}{\sqrt{4x-3}}$. Setting this derivative equal to the provided tangent slope of $\frac{2}{3}$ allows us to find the x-coordinate of point P, where $\frac{2}{\sqrt{4x-3}} = \frac{2}{3}$, which simplifies to $\sqrt{4x-3} = 3$. Squaring both sides yields $4x-3 = 9$, resulting in $x = 3$. Substituting this x-value back into the original function gives $y(3) = 1 + \sqrt{4(3)-3} = 4$, identifying P as $(3, 4)$. Because the tangent slope is $\frac{2}{3}$, the slope of the normal line must be $-\frac{3}{2}$. Applying the point-slope form $y - y_1 = m(x - x_1)$, the equation of the normal becomes $y - 4 = -\frac{3}{2}(x - 3)$, which simplifies to $3x + 2y = 17$. Testing the provided options against this equation, the point $(1, 7)$ is the only one that satisfies the condition:
$3(1) + 2(7) = 17$

The function $f(x) = \left|(x-1)(x^{2-}2x-3)\right| + x - 3$ simplifies significantly by analyzing the sign of the expression inside the absolute value, which is $(x-1)(x+1)(x-3)$. Within the intervals $(0, 1)$ and $(3, 4)$, the product is positive, so the function behaves as $f(x) = x^3 - 3x^2$. In the interval $(1, 3)$, the product is negative, causing the absolute value term to flip its sign, resulting in $f(x) = -x^3 + 3x^2 + 2x - 6$.

To determine the local extrema, we examine the derivative $f'(x)$ across these intervals:
$

$Within the intervals$(0, 1)$and$(3, 4)$, the derivative$3x^2 - 6x$does not equal zero, indicating that no stationary points occur there. Within$(1, 3)$, the derivative$-3x^2 + 6x + 2$is equal to zero at a single point, where the function reaches a local maximum. At the transition points$x=1$and$x=3$, the slope changes from negative to positive, confirming that both values are points of local minimum. Given two points of local minimum and one point of local maximum, the sum$m+M$ is 3.

Calculating the behavior of this exponential function requires applying the chain rule to determine its first and second derivatives. The first derivative $f'(x)$ involves the product of the original function, the logarithmic constant, and the polynomial factor $6x(x^{2-}2)^2$. Since the exponential component and the squared term $(x^{2-}2)^2$ are always non-negative, the sign of $f'(x)$ is determined solely by $x$. Consequently, $f'(x)$ is negative for $x < 0$ and positive for $x > 0$, confirming that $x=0$ is a point of local minima.

The second derivative $f''(x)$ allows for the identification of inflection points and intervals of monotonicity. Evaluating this derivative reveals that $f''(x)$ vanishes at $x=\sqrt{2}$. Because the second derivative changes sign from negative to positive at this specific value, $x=\sqrt{2}$ qualifies as a point of inflection. Furthermore, since $f''(x) > 0$ for all values where $x > \sqrt{2}$, the derivative $f'(x)$ is strictly increasing on that interval, confirming the validity of the final statement.

Because the derivative $f'(x)=12 \sqrt{2} x^{2}-3 \sqrt{2}$ is non-negative for all $x \in \left[\frac{1}{2}, 1\right]$, the function $f(x)=4 \sqrt{2} x^{3}-3 \sqrt{2} x-1$ is strictly increasing, allowing the use of the Intermediate Value Theorem to verify the number of roots. Evaluating the function at the boundary points yields $f\left(\frac{1}{2}\right) = -\sqrt{2} - 1 < 0$ and $f(1) = \sqrt{2} - 1 > 0$. Since the function is continuous and changes sign exactly once within this interval, the curve must intersect the x-axis at exactly one point.

Setting $f(x)=0$ leads to the identity $\sqrt{2}(4x^3 - 3x) = 1$, which simplifies to $4x^3 - 3x = \frac{1}{\sqrt{2}}$. Applying the trigonometric substitution $x = \cos \theta$ transforms the equation into the triple-angle identity $\cos(3\theta) = \frac{1}{\sqrt{2}}$. Since $\frac{1}{\sqrt{2}}$ is equivalent to $\cos \frac{\pi}{4}$, equating the angles gives $3\theta = \frac{\pi}{4}$, or $\theta = \frac{\pi}{12}$. This confirms that $x = \cos \frac{\pi}{12}$ is indeed the root of the equation.