Let defined by then function f(x) is

JEE MainMathematicsEasyFunctions

Let $f:R \rightarrow R$ defined by
$f(x) = x^{3} + x^{2} + 100x + 5sinx$ then function f(x) is

Show answer and solution

f^{'}(x) = 3x^{2} + 2x + 100 + 5cosx

= 3x^{2} + 2x + 94 + (6 + 5cosx) > 0

\therefore

f is strictly increasing hence one one also

Source: Grass-G / M-17 Mathematics Paper 26 Nov Second.docx

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