Mathematics · 2022
JEE Main · 24 June 2022, Shift 1 · Q27
Let 0 ≤ x ≤ 2 Max {(9-x^2)/(5-x)}=α and 0 ≤ x ≤ 2 Min {(9-x^2)/(5-x)}=β. If ∫_β-8/3^2 α-1 Max {(9-x^2)/(5-x), x} d x=α_1+α_2 log_e (8/15) then…
Let $\displaystyle \underset{0 \leqslant x \leqslant 2}{\operatorname{Max}}\left\{\frac{9-x^2}{5-x}\right\}=\alpha$ and $\displaystyle \underset{0 \leqslant x \leqslant 2}{\operatorname{Min}}\left\{\frac{9-x^2}{5-x}\right\}=\beta$.If $\displaystyle \int_{\beta-\frac{8}{3}}^{2 \alpha-1} \operatorname{Max}\left\{\frac{9-x^2}{5-x}, x\right\} \mathrm{d} x=\alpha_1+\alpha_2 \log _{\mathrm{e}}\left(\frac{8}{15}\right)$ then $\displaystyle \alpha_1+\alpha_2$ is equal to \_\_\_\_.
Official answer
34
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.