If $R$ is the set of all real numbers and $f: R \rightarrow R$ is defined by $f(x) = 3x^2 + 1$, then the set $f^{-1}([1, 6])$ is
- A$\{ -\sqrt{\frac{5}{3}}, 0, \sqrt{\frac{5}{3}} \}$
- B$[ -\sqrt{\frac{5}{3}}, \sqrt{\frac{5}{3}} ]$
- C$[ -\sqrt{\frac{1}{3}}, \sqrt{\frac{1}{3}} ]$
- D$( -\sqrt{\frac{5}{3}}, \sqrt{\frac{5}{3}} )$
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Let $f(x) = \frac{x^2-6x+5}{x^2-5x+6}$. Match the conditions / expressions in Column $I$ with statements in Column $II$.
Column $I$ Column $II$ $(A)$ If $-1 < x < 1$,then $f(x)$ satisfies $(p)$ $0 < f(x) < 1$ $(B)$ If $1 < x < 2$,then $f(x)$ satisfies $(q)$ $f(x) < 0$ $(C)$ If $3 < x < 5$,then $f(x)$ satisfies $(r)$ $f(x) > 0$ $(D)$ If $x > 5$,then $f(x)$ satisfies $(s)$ $f(x) < 1$
| Column $I$ | Column $II$ |
| $(A)$ If $-1 < x < 1$,then $f(x)$ satisfies | $(p)$ $0 < f(x) < 1$ |
| $(B)$ If $1 < x < 2$,then $f(x)$ satisfies | $(q)$ $f(x) < 0$ |
| $(C)$ If $3 < x < 5$,then $f(x)$ satisfies | $(r)$ $f(x) > 0$ |
| $(D)$ If $x > 5$,then $f(x)$ satisfies | $(s)$ $f(x) < 1$ |
Let the domains of the functions $f(x) = \log_4 \log_3 \log_7(8 - \log_2(x^2 + 4x + 5))$ and $g(x) = \sin^{-1}(\frac{7x + 10}{x - 2})$ be $(\alpha, \beta)$ and $[\gamma, \delta]$,respectively. Then $\alpha^2 + \beta^2 + \gamma^2 + \delta^2$ is equal to:
DifficultJEE MAIN 2025
View SolutionThe domain of the real valued function $f(x) = \sqrt{9 - \sqrt{x^2 - 144}}$ is
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