Mathematics · 2021
JEE Main · 4 August 2021, Shift 1 · Q90
If tan A is equal to the integral solution of the inequality 4 x^2-24 x+35<0 and sin B is equal to the slope of the angle bisector of the second…
If $\displaystyle \tan \mathrm{A}$ is equal to the integral solution of the inequality $\displaystyle 4 x^2-24 x+35<0$ and $\displaystyle \sin \mathrm{B}$ is equal to the slope of the angle bisector of the second quadrant, then $\displaystyle 40 \sin (A+B) \sin (B-A)$ is \_\_\_\_.
Official answer
4
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JEE Main 2021 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.