The maximum area (in sq. units) of a rectangle having its base on the x− axis and its other two vertices on the parabola, y=12−x2 such that the rectangle lies inside the parabola, is :
Held on 12 Jan 2019 · Verified 6 Jul 2026.
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Let $f: [1, \infty) \rightarrow \mathbf{R}$ be a differentiable function defined as $f(x) = \int_1^x f(t)\,dt + (1-x)(\log_e x - 1) + e$. Then the value of $f(f(1))$ is :
Let $f(x)=\int \frac{\mathrm{d} x}{x^{\left(\frac{2}{3}\right)}+2 x^{\left(\frac{1}{2}\right)}}$ be such that $f(0)=-26+24 \log _{\mathrm{e}}(2)$. If $f(1)=\mathrm{a}+\mathrm{b} \log _{\mathrm{e}}(3)$, where $\mathrm{a}, \mathrm{b} \in \mathbf{Z}$, then $\mathrm{a}+\mathrm{b}$ is equal to :
If the solution curve $y=f(x)$ of the differential equation $\left(x^{2}-4\right) y^{\prime}-2 x y+2 x\left(4-x^{2}\right)^{2}=0, x>2$, passes through the point $(3,15)$, then the local maximum value of $f$ is $\_\_\_\_$.
If $\displaystyle\int_{\pi/6}^{\pi/4}\left(\cot\left(x-\dfrac{\pi}{3}\right)\cot\left(x+\dfrac{\pi}{3}\right)+1\right)dx = \alpha\log_e(\sqrt{3}-1)$, then $9\alpha^2$ is equal to ________.
The value of the integral $\displaystyle\int_{0}^{2} \dfrac{\sqrt{x(x^2+x+1)}}{(\sqrt{x+1})(\sqrt{x^4+x^2+1})} \, dx$ is equal to:
Work through every JEE Main Calculus PYQ, year by year.
