Original JEE Main and CUET-style questions on application of derivatives, with a worked solution and the usual trap for each. Tap an option, or type a numerical answer, to check it.
Learn the chapters first: Class 12, Application of Derivatives.
The greatest value of \(f(x) = x^3 - 6x^2 + 9x + 2\) on \([0, 4]\) is
\(f'(x) = 3(x - 1)(x - 3)\). Check critical points and end points: \(f(0) = 2\), \(f(1) = 6\), \(f(3) = 2\), \(f(4) = 6\).
Greatest value 6.
Trap: Looking only at critical points misses that an end point can tie with, or beat, a local maximum.
\(f(x) = x^2 - 4x + 3\) is strictly increasing on
\(f'(x) = 2x - 4 > 0\) exactly when \(x > 2\).
Two positive numbers \(x\) and \(y\) add up to 12. Find the greatest possible value of \(xy^2\).
\(P = (12 - y)y^2 = 12y^2 - y^3\), \(P' = 24y - 3y^2 = 3y(8 - y)\), zero at \(y = 8\); \(P'' = 24 - 6y < 0\) there.
\(x = 4\), \(y = 8\): \(P = 4 \times 64 = 256\).
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