Application Of Derivatives – JEE Main MathematicsPractice Questions & PYQs

Local extrema occur at points where the derivative of a function equals zero. The derivative of the function is $f'(x) = \frac{1}{x} + 2bx + a$. Setting this expression to zero at $x = -1$ leads to $-1 - 2b + a = 0$, or $a - 2b = 1$, while evaluating at $x = 2$ results in $\frac{1}{2} + 4b + a = 0$, or $a + 4b = -\frac{1}{2}$. Solving this system of linear equations produces the values $a = \frac{1}{2}$ and $b = -\frac{1}{4}$. Using these parameters, the derivative simplifies to $f'(x) = \frac{-(x+1)(x-2)}{2x}$, which indicates that the derivative shifts from positive to negative sign values when passing through both $x = -1$ and $x = 2$, identifying both as local maxima. Because these parameter values directly enable the verification of the function's behavior at these critical points, both statements are true, and the second statement correctly explains the first.

A line passing through $(1, 2)$ with slope $m$ is represented by the equation $(y - 2) = m(x - 1)$. Intersecting this line with the coordinate axes defines a triangle $O P Q$ where the $x$-intercept is $1 - \frac{2}{m}$ and the $y$-intercept is $2 - m$. The area of this triangle is calculated as $\frac{1}{2} (1 - \frac{2}{m})(2 - m)$, which simplifies to the expression $2 - \frac{m}{2} - \frac{2}{m}$. Finding the derivative of this area function with respect to $m$ and setting it to zero, $-\frac{1}{2} + \frac{2}{m^2} = 0$, yields $m^2 = 4$. The second derivative test confirms the area is minimized when the slope is $-2$.

By the Fundamental Theorem of Calculus, the derivative of the integral function $f(x) = \int_{0}^{x} \sqrt{t} \sin t \, dt$ is given by $f'(x) = \sqrt{x} \sin x$. The critical points occur where $f'(x) = 0$, which, within the interval $\left(0, \frac{5\pi}{2}\right)$, corresponds to $x = \pi$ and $x = 2\pi$ since $\sin x = 0$ at these values. To classify these extrema, we apply the second derivative test by differentiating $f'(x)$ to obtain $f''(x) = \sqrt{x} \cos x + \frac{\sin x}{2\sqrt{x}}$. Evaluating this at $x = \pi$ gives $f''(\pi) = \sqrt{\pi} \cos \pi + 0 = -\sqrt{\pi}$, which is negative and confirms a local maximum. Evaluating at $x = 2\pi$ gives $f''(2\pi) = \sqrt{2\pi} \cos 2\pi + 0 = \sqrt{2\pi}$, which is positive and confirms a local minimum.

The shortest distance between a line and a curve occurs at the point on the curve where the tangent is parallel to the given line. We can rewrite the line $y-x=1$ as $x-y+1=0$, which clearly has a slope of $1$. For the curve $x=y^2$, we determine the slope of the tangent by differentiating with respect to $y$, giving $\frac{dx}{dy} = 2y$, which implies the slope $\frac{dy}{dx} = \frac{1}{2y}$. Setting this tangent slope equal to the line's slope of $1$ allows us to find the point of interest, $\frac{1}{2y} = 1$, which simplifies to $y = \frac{1}{2}$.

With the $y$-coordinate identified as $\frac{1}{2}$, we find the corresponding $x$-coordinate by substituting back into the curve equation $x = (\frac{1}{2})^2 = \frac{1}{4}$, identifying the point $(\frac{1}{4}, \frac{1}{2})$ as the location of the shortest distance. The perpendicular distance $d$ from this point to the line $x-y+1=0$ is calculated using the distance formula
$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$
Substituting the values gives $d = \frac{|\frac{1}{4} - \frac{1}{2} + 1|}{\sqrt{1^2 + (-1)^2}}$, which simplifies to $d = \frac{|3/4|}{\sqrt{2}}$, resulting in the final value $d = \frac{3 \sqrt{2}}{8}$.

The function $f(x) = \frac{1}{e^x + 2e^{-x}}$ reaches its maximum value when its denominator is at its minimum. Applying the Arithmetic Mean-Geometric Mean inequality, $e^x + 2e^{-x} \geq 2\sqrt{e^x \cdot 2e^{-x}} = 2\sqrt{2}$, which establishes that the function is bounded by $0 < f(x) \leq \frac{1}{2\sqrt{2}}$. Given this range and the continuity of the function, the Intermediate Value Theorem implies that $f(x)$ attains every value between its limit of $0$ as $x$ approaches $\pm\infty$ and its supremum of $\frac{1}{2\sqrt{2}}$. Since $1/3$ is less than $\frac{1}{2\sqrt{2}}$ and greater than $0$, there must exist some $c \in \mathbb{R}$ such that $f(c) = \frac{1}{3}$, confirming that Statement 2 provides the necessary bound to validate Statement 1.

A line parallel to the $x$-axis has a slope of zero, which means the derivative of the curve must be zero at the point of tangency. By differentiating $y = x + \frac{4}{x^2}$ with respect to $x$, we obtain the slope function $\frac{dy}{dx} = 1 - \frac{8}{x^3}$. Setting this derivative equal to zero leads to $1 - \frac{8}{x^3} = 0$, which simplifies to $x^3 = 8$, giving $x = 2$. Substituting this back into the original curve equation yields the height $y = 2 + \frac{4}{4} = 3$, so the resulting equation for the horizontal tangent line is $y = 3$.

A function possesses a local minimum at a point if the value at that point is less than or equal to the values in its immediate neighborhood. Evaluating the function at the boundary $x = -1$ yields $f(-1) = k - 2(-1) = k + 2$. As $x$ approaches $-1$ from the right, the function $f(x) = 2x + 3$ approaches a limit of $2(-1) + 3 = 1$. Requiring the function value $f(-1)$ to be less than or equal to this right-hand limit results in the inequality $k + 2 \le 1$, which simplifies to $k \le -1$. Among the choices provided, $k = -1$ is the value that satisfies this condition.

Given the polynomial $P(x)=x^{4+}a x^{3+}b x^{2+}c x+d$, the derivative is calculated as $P'(x)=4 x^{3+}3 a x^{2+}2 b x+c$. Since $x=0$ is the only real root of $P'(x)=0$, we must have $c=0$. Factoring the derivative as $P'(x)=x(4 x^{2+}3 a x+2 b)$, the existence of a single real root implies that the quadratic expression $4 x^{2+}3 a x+2 b$ cannot have any real roots. This necessitates that its discriminant is negative, which ensures that $b$ is positive. Substituting the condition $P(-1)<P(1)$ into the original polynomial results in the inequality $1-a+b+d < 1+a+b+d$, which simplifies to $2a > 0$, or $a > 0$.

With $a > 0$ and $b > 0$, the quadratic term $4 x^{2+}3 a x+2 b$ remains strictly positive for all real $x$ because it is an upward-opening parabola with no real roots. Consequently, the sign of $P'(x)$ is determined entirely by $x$. For all $x < 0$, $P'(x) < 0$, indicating the function is strictly decreasing on the interval $[-1, 0]$. For all $x > 0$, $P'(x) > 0$, indicating the function is strictly increasing on the interval $[0, 1]$. Because the function decreases until $x=0$ and then increases, the absolute minimum on the interval $[-1, 1]$ occurs at $x=0$, meaning $P(-1)$ cannot be the minimum. Comparing the values at the endpoints, the given condition $P(-1) < P(1)$ ensures that $P(1)$ is the maximum value of the function.

The behavior of the equation $x^{7+}14x^{5+}16x^{3+}30x-560=0$ is best understood by examining the derivative of the function, which is given by:
$f'(x) = 7x^{6+}70x^{4+}48x^{2+}30$
Because all powers of $x$ have positive coefficients and the constant term is also positive, $f'(x) > 0$ holds true for every real number. This confirms that the function is strictly increasing, climbing continuously from negative infinity to positive infinity across the entire real line. Consequently, the curve must cross the horizontal axis at exactly one point, indicating that the equation possesses only one real solution.

To identify the turning points of a function, we analyze its slope by calculating the first derivative and then classify those points using the second derivative test. For the cubic function $f(x) = x^3 - px + q$, the first derivative is $f'(x) = 3x^2 - p$. Setting this derivative to zero to find where the slope is horizontal gives us $3x^2 = p$, resulting in the critical values $x = \pm \sqrt{\frac{p}{3}}$. The second derivative, $f''(x) = 6x$, helps distinguish between the relative maximum and minimum. Evaluating this at the positive critical point $x = \sqrt{\frac{p}{3}}$ yields a positive result, which indicates a local minimum, while evaluating it at the negative critical point $x = -\sqrt{\frac{p}{3}}$ produces a negative result, confirming a local maximum.

The constraint $p^{2+}q^2=1$ allows us to substitute $p = \cos \theta$ and $q = \sin \theta$ for some angle $\theta$, transforming the target expression $p+q$ into $\cos \theta + \sin \theta$. Based on the property that any linear combination $a \cos \theta + b \sin \theta$ has a maximum value of $\sqrt{a^{2+}b^2}$, we identify the coefficients as $a=1$ and $b=1$. Applying this property yields $\sqrt{1^{2+}1^2}$, which results in a maximum value of $\sqrt{2}$.

The Mean Value Theorem states that for a differentiable function on an interval, there exists a point $c$ such that $f'(c)$ equals the average slope over that interval. For the function $f(x) = \log_e x$ on the interval $[1, 3]$, the derivative $f'(x) = 1/x$ must equal the average slope $\frac{f(3) - f(1)}{3 - 1}$. Substituting the values yields the equation:

$\frac{1}{c} = \frac{\log_e 3 - \log_e 1}{2}$

Since the natural logarithm of 1 is 0, this simplifies to $1/c = \frac{\log_e 3}{2}$, which rearranges to $c = \frac{2}{\log_e 3}$. By applying the logarithmic property $\frac{1}{\log_e 3} = \log_3 e$, we find that $c = 2\log_3 e$.

The function $f(x)=\tan^{-1}(\sin x+\cos x)$ is increasing when its derivative $f'(x)$ is positive. Differentiating the function gives
$f'(x)=\frac{\cos x-\sin x}{1+(\sin x+\cos x)^2}$
which simplifies to
$f'(x)=\frac{\sqrt{2}\cos\left(x+\frac{\pi}{4}\right)}{1+(\sin x+\cos x)^2}$
by applying trigonometric identities. Since the denominator is always positive, the condition $f'(x)>0$ is satisfied when $\cos(x+\frac{\pi}{4})>0$, occurring for $-\frac{\pi}{2} < x+\frac{\pi}{4} < \frac{\pi}{2}$. Solving this inequality leads to $-\frac{3\pi}{4} < x < \frac{\pi}{4}$, indicating that the function increases on the interval $(-\frac{\pi}{2}, \frac{\pi}{4})$.

The area of a triangle defined by two sides of length $x$ and an included angle $\theta$ is calculated using the formula $Area = \frac{1}{2} x^2 \sin \theta$ Since the two fence lengths are constant, the area depends entirely on the value of $\sin \theta$. This trigonometric function reaches its maximum value of 1 when the angle is $\frac{\pi}{2}$, leading to the largest possible area of $\frac{1}{2} x^2$

To find the local minimum of a function, we must first locate its critical points where the derivative vanishes and then assess the curvature at those points using the second derivative test. For the given function $f(x) = \frac{x}{2} + \frac{2}{x}$, calculating the derivative yields $f'(x) = \frac{1}{2} - \frac{2}{x^2}$. Setting this derivative equal to zero leads to $x^2 = 4$, which identifies potential extrema at $x=2$ and $x=-2$. By determining the second derivative, $f''(x) = \frac{4}{x^3}$, and evaluating it at $x=2$, we obtain a positive value; this confirms that the function is concave up at this location and therefore possesses a local minimum at $x=2$.

The angle between two lines is determined by their slopes, where a product of $-1$ confirms that the lines are perpendicular. By differentiating $y=x^{2}-5x+6$, we obtain the slope function $\frac{dy}{dx} = 2x-5$. Substituting the $x$-coordinates of the points $(2,0)$ and $(3,0)$ into this derivative gives $m_1 = 2(2)-5 = -1$ and $m_2 = 2(3)-5 = 1$. Because $m_1 \cdot m_2 = -1$, the tangents meet at a right angle, resulting in an angle of $\frac{\pi}{2}$.

The equation of any line can be expressed in the normal form $x\cos\alpha + y\sin\alpha = p$, where $p$ represents the perpendicular distance of the line from the origin. To find the equation of the normal to the given curve, we first determine the slope of the tangent by differentiating $x$ and $y$ with respect to $\theta$, yielding $\frac{dx}{d\theta} = a\theta\cos\theta$ and $\frac{dy}{d\theta} = a\theta\sin\theta$. Dividing these results gives the slope of the tangent as $\frac{dy}{dx} = \tan\theta$, which implies that the slope of the normal line is the negative reciprocal, $-\cot\theta$.

Using the point-slope form with the point $(a(\cos\theta+\theta \sin\theta), a(\sin\theta-\theta \cos\theta))$, the equation of the normal becomes $y - a(\sin\theta - \theta\cos\theta) = -\cot\theta(x - a(\cos\theta + \theta\sin\theta))$. Expanding and simplifying this equation leads to $y\sin\theta - a\sin^2\theta + a\theta\cos\theta\sin\theta = -x\cos\theta + a\cos^2\theta + a\theta\sin\theta\cos\theta$, which further reduces to $x\cos\theta + y\sin\theta = a(\cos^2\theta + \sin^2\theta)$. Since $\cos^2\theta + \sin^2\theta = 1$, the equation simplifies to $x\cos\theta + y\sin\theta = a$. Because this takes the form $x\cos\alpha + y\sin\alpha = p$, the constant $a$ represents the perpendicular distance from the origin, confirming the line remains at a fixed distance regardless of the chosen point $\theta$.

Any point on an ellipse can be represented using parametric coordinates as $(a \cos \theta, b \sin \theta)$, which allows the vertices of a symmetric inscribed rectangle to be defined at $(\pm a \cos \theta, \pm b \sin \theta)$. The area of this rectangle is determined by the product of its dimensions, $2a \cos \theta$ and $2b \sin \theta$, which simplifies to the expression $Area = 2ab \sin 2\theta$. Given that the maximum value of the sine function is $1$, the greatest area that such a rectangle can enclose is $2ab$.

Rolle's Theorem guarantees that if a differentiable function takes equal values at two points, its derivative must vanish at some point between them. For the polynomial defined as
$f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x = 0$
the condition $a_1 \neq 0$ implies that $f(0) = 0$, and we are given that $f(\alpha) = 0$. Applying Rolle's Theorem to the interval $[0, \alpha]$ ensures there is at least one root of the derivative
$f'(x) = n a_n x^{n-1} + (n-1) a_{n-1} x^{n-2} + \cdots + a_1 = 0$
lying strictly between $0$ and $\alpha$. Therefore, the derivative equation must possess a positive root that is smaller than $\alpha$.

The core requirement for a function to be increasing is that its first derivative must be non-negative, meaning $f'(x) \ge 0$. Evaluating the derivative of $f(x) = 3x^2 - 2x + 1$ yields $f'(x) = 6x - 2$. Setting $6x - 2 \ge 0$ shows that the function increases specifically where $x \ge \frac{1}{3}$, which corresponds to the interval $[\frac{1}{3}, \infty )$. Consequently, labeling the interval of increase as $(-\infty , \frac{1}{3}]$ is incorrect.