f′(x)=cosx−x+1f′(x)=−sinx−1 f is decreasing ∀x∈R f(x)=0f(0)=2,f(π)=−π f is strictly decreasing in [0,π] and f(0).f(π)<0 ⇒ only one solution of f(x)=0 S1 is correct and S2 is incorrect.
JEE Main 2024 — Mathematics Calculus
For the function f(x)=(cosx)−x+1,x∈R, between the following two statements (S1) f(x)=0 for only one value of x in [0,π]. (S2) f(x) is decreasing in [0,2π] and increasing in [2π,π].
Held on 8 Apr 2024 · Verified 6 Jul 2026.
Both (S1) and (S2) are correct.
Both (S1) and (S2) are incorrect.
Only (S2) is correct.
Only (S1) is correct.
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
The value of $\int_{-\pi / 6}^{\pi / 6}\left(\frac{\pi+4 x^{11}}{1-\sin (|x|+\pi / 6)}\right) d x$ is equal to:
The product of all possible values of $\alpha$, for which $\displaystyle\lim_{x \to 0}\left(\dfrac{1 - \cos(\alpha x)\cos((\alpha+1)x)\cos((\alpha+2)x)}{\sin^2((\alpha+1)x)}\right) = 2$, is:
The value of the integral $\displaystyle\int_0^\infty \dfrac{\log_e(x)}{x^2 + 4}\,dx$ is:
Let $f$ be a differentiable function satisfying $f(x)=1-2 x+\int_{0}^{x} \mathrm{e}^{(x-t)} f(t) \mathrm{dt}, x \in \mathbf{R}$ and let $\mathrm{g}(x)=\int_{0}^{x}(f(\mathrm{t})+2)^{15}(\mathrm{t}-4)^{6}(\mathrm{t}+12)^{17} \mathrm{dt}, x \in \mathbf{R}$. If p and q are respectively the points of local minima and local maxima of g, then the value of $|\mathrm{p}+\mathrm{q}|$ is equal to $\_\_\_\_$.
Let the area of the region bounded by the curve $y=\max \{\sin x, \cos x\}$, lines $x=0, x=\frac{3 \pi}{2}$, and the $x$-axis be A. Then, $\mathrm{A}+\mathrm{A}^{2}$ is equal to $\_\_\_\_$.
Work through every JEE Main Calculus PYQ, year by year.