f(x) is defined if −1≤2x−1≤1 and cosx>0 or 0≤x≤4 and −2π<x<2π ∴x∈[0,2π).
The largest interval lying in (- π 2, π 2 ) for which the function [f(x)=4^-x^2+cos ^-1 ( x 2-1 )+log (cos x) ] is defined, is
JEE Main 2007 — Mathematics Algebra
The largest interval lying in (−2π,2π) for which the function [f(x)=4−x2+cos−1(2x−1)+log(cosx)] is defined, is
Held on 30 Apr 2007 · Verified 6 Jul 2026.
Options
- A
[0,π]
- B
(−2π,2π)
- C
[−4π,2π)
- D
[0,2π)
Did you get this right?
Sign in to track your attempts and accuracy.
Your note
Sign in to keep a private note on this question. Nothing you write is ever public.
JEE Main Algebra in other years
More JEE Main Algebra questions
If the quadratic equation $(\lambda+2)x^2-3\lambda x+4\lambda=0$, $\lambda \neq -2$, has two positive roots, then the number of possible integral values of $\lambda$ is:
The value of $\frac{{ }^{100} \mathrm{C}_{50}}{51}+\frac{{ }^{100} \mathrm{C}_{51}}{52}+\ldots.+\frac{{ }^{100} \mathrm{C}_{100}}{101}$ is :
Let $e_1$ and $e_2$ be two distinct roots of the equation $x^2 - ax + 2 = 0$. Let the sets $\{a \in \mathbb{R} : e_1 \text{ and } e_2 \text{ are the eccentricities of hyperbolas}\} = (\alpha, \beta)$, and $\{a \in \mathbb{R} : e_1 \text{ and } e_2 \text{ are the eccentricities of an ellipse and a hyperbola, respectively}\} = (\gamma, \infty)$. Then $\alpha^2 + \beta^2 + \gamma^2$ is equal to:
Among the statements : I: If $\left|\begin{array}{ccc}1 & \cos \alpha & \cos \beta \\ \cos \alpha & 1 & \cos \gamma \\ \cos \beta & \cos \gamma & 1\end{array}\right|=\left|\begin{array}{ccc}0 & \cos \alpha & \cos \beta \\ \cos \alpha & 0 & \cos \gamma \\ \cos \beta & \cos \gamma & 0\end{array}\right|$, then $\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=\frac{3}{2}$, and II : If $\left|\begin{array}{ccc}x^{2}+x & x+1 & x-2 \\ 2 x^{2}+3 x-1 & 3 x & 3 x-3 \\ x^{2}+2 x+3 & 2 x-1 & 2 x-1\end{array}\right|=\mathrm{p} x+\mathrm{q}$, then $\mathrm{p}^{2}=196 \mathrm{q}^{2}$,
If $\left(\frac{1}{{ }^{15} \mathrm{C}_{0}}+\frac{1}{{ }^{15} \mathrm{C}_{1}}\right)\left(\frac{1}{{ }^{15} \mathrm{C}_{1}}+\frac{1}{{ }^{15} \mathrm{C}_{2}}\right) \cdots\left(\frac{1}{{ }^{15} \mathrm{C}_{12}}+\frac{1}{{ }^{15} \mathrm{C}_{13}}\right)=\frac{\alpha^{13}}{{ }^{14} \mathrm{C}_{0}{ }^{14} \mathrm{C}_{1} \cdots{ }^{14} \mathrm{C}_{12}}$, then $30 \alpha$ is equal to $\_\_\_\_$.
Other Mathematics topics for JEE Main
Keep practicing Algebra
Work through every JEE Main Algebra PYQ, year by year.