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Mathematics · 2021

JEE Main · 24 February 2021, Shift 1 · Q67

The function f(x)=(4 x^3-3 x^2)/6-2 sin x+(2 x-1) cos x:

The function $\displaystyle f(x)=\frac{4 x^3-3 x^2}{6}-2 \sin x+(2 x-1) \cos x$ :
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JEE Main 2021 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.