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Let \( f(x)=\displaystyle\int_{0}^{x}t(t^2-9t+20)\,dt,\; 1\le x\le 5 \). If the range of \( f \) is \( [\alpha,\beta] \), then \( 4(\alpha+\beta) \) equals:

Let \( f:\mathbb{R} \to \mathbb{R} \) be a function defined by \( f(x) = \big||x+2| - 2|x|\big| \). If \( m \) is the number of points of local minima and \( n \) is the number of points of local maxima of \( f \), then \( m + n \) is:

The sum of all local minimum values of the function

If the function \( f(x)=2x^3-9ax^2+12a^2x+1 \), where \( a>0 \), attains its local maximum and local minimum values at \( p \) and \( q \), respectively, such that \( p^2=q \), then \( f(3) \) is equal to:

A spherical chocolate ball has a layer of ice-cream of uniform thickness around it. When the thickness of the ice-cream layer is \( 1 \, \text{cm} \), the ice-cream melts at the rate of \( 81 \, \text{cm}^{3}/\text{min} \) and the thickness of the ice-cream layer decreases at the rate of \( \frac{1}{4\pi} \, \text{cm} \ / \text{min} \). The surface area (in \( \text{cm}^{2} \)) of the chocolate ball (without the ice-cream layer) is:

Let \( f,g:(1,\infty)\to \mathbb{R} \) be defined as \( f(x)=\frac{2x+3}{5x+2} \) and \( g(x)=\frac{2-3x}{1-x} \). If the range of the function \( f\circ g : [2,4]\to \mathbb{R} \) is \( [\alpha,\beta] \), then \( \frac{1}{\beta-\alpha} \) is equal to:

Let \( f : \mathbb{R} \to \mathbb{R} \) be a twice differentiable function such that \( (\sin x \cos y)(f(2x+2y)-f(2x-2y))=(\cos x \sin y)(f(2x+2y)+f(2x-2y)) \), for all \( x, y \in \mathbb{R} \). If \( f'(0)=\frac{1}{2} \), then the value of \( 24\, f'\!\left(\frac{5\pi}{3}\right) \) is:

Let \(\displaystyle f(x) = \int_{0}^{x^{2}} \frac{t^{2}-8t+15}{e^{t}} \, dt , \; x \in \mathbb{R} \). Then the numbers of local maximum and local minimum points of \( f \), respectively, are:

Let \( x = -1 \) and \( x = 2 \) be the critical points of the function \( f(x) = x^3 + ax^2 + b \log_e |x|+1, \, x \neq 0 \). Let \( m \) and \( M \) respectively be the absolute minimum and the absolute maximum values of \( f \) in the interval \( \left[-2, -\frac{1}{2}\right] \). Then \( |M + m| \) is equal to: \( \text{(Take } \log_e 2 = 0.7 \text{)} \)

Let \( (2, 3) \) be the largest open interval in which the function \( f(x) = 2 \, \log_e (x - 2) - x^2 + a x + 1 \) is strictly increasing and \( (b, c) \) be the largest open interval, in which the function \( g(x) = (x - 1)^3 (x + 2 - a)^2 \) is strictly decreasing. Then \( 100(a + b - c) \) is equal to :

Consider the region \( R = \left\{ (x, y) : x \le y \le 9 - \frac{11}{3} x^{2}, \ x \ge 0 \right\} \). The area, of the largest rectangle of sides parallel to the coordinate axes and inscribed in \( R \), is:

\( x \in \) \(\mathbb{R}\) , then \( \frac{d^{2}y}{dx^{2}} + y \) is equal to:

Let the function \( f(x) = \frac{x}{3} + \frac{3}{x} + 3, \ x \neq 0 \) be strictly increasing in \( (-\infty, \alpha_{1}) \cup (\alpha_{2}, \infty) \) and strictly decreasing in \( (\alpha_{3}, \alpha_{4}) \cup (\alpha_{4}, \alpha_{5}) \). Then \( \displaystyle \sum_{i=1}^{5}\alpha_{i}^{2} \) is equal to:

Let \( a>0 \). If the function \( f(x)=6x^3-45ax^2+108a^2x+1 \) attains its local maximum and minimum values at the points \( x_1 \) and \( x_2 \) respectively such that \( x_1x_2=54 \), then \( a+x_1+x_2 \) is equal to:

Let \( f:\mathbb{R}\to\mathbb{R} \) be a polynomial function of degree from having extreme values at \( x = 4 \) and \( x = 5 \). If \( \displaystyle \lim_{x\to 0} \dfrac{f(x)}{x^{2}} = 5 \), then \( f(2) \) is equal to:

Question 1 of 15Next →

JEE Main 2025 29th Jan Evening Shift

Let \( f(x)=\displaystyle\int_{0}^{x}t(t^2-9t+20)\,dt,\; 1\le x\le 5 \). If the range of \( f \) is \( [\alpha,\beta] \), then \( 4(\alpha+\beta) \) equals: