NEET UG Calculus — practice questions
15 free MCQs with worked solutions. Tap any question for the answer + explanation, or practice them all in the app.
Practice NEET UG Calculus in the app →The minimum value of $4e^{2x} + 9e^{-2x}$ isFor real $x$, let $f(x) = x^3 + 5x + 1$, thenThe function $f(x) = \log(1 + x) - \frac{2x}{2 + x}$ is increasing on$\lim_{x \to \infty} \left( \frac{x^3}{3x^2 - 4} - \frac{x^2}{3x + 2} \right)$ is equal to$\int \frac{ qrt{x} \, dx}{1 + x}$ equalsThe domain of definition of the function
$$
y = \frac{1}{\log_{10}(1 - x)} + qrt{x + 2}
$$If $x = \log p$ and $y = \frac{1}{p}$, thenThe equation $y^{2} \mathrm{e}^{xy} = 9 \mathrm{e}^{-3} \cdot x^{2}$ defines $y$ as a differentiable function If $\frac{dv}{dx} = y + 3 > 0$ and $y(0) = 2$, then $y(\ln 2)$ is equal toLet $p(x)$ be a function defined on $R$ such that $p'(x) = p'(1 - x)$, for all $x \in [0, 1]$, $p(0) = 1$ and $\lim_{n\to \infty} um_{r = 1}^{n}1 - \frac{1}{n} e^{r}$ is$\int_{-2}^{2}|1 - x^2|dx =$The value of $f(0)$ so that the function $f(x) = \frac{ qrt{1 + x} - qrt[3]{1 + x}}{x}$ becomes continuous isA value of $C$ for which the conclusion of mean value theorem holds for the function $f(x) = \log_{e}x$ on theThe angle made by the tangent of the curve $x = a$ ($t + in t$ cost); $y = a(1 + in t)^2$ with the $x$-axis