Number of solution of equation 3tanx + x^3 = 2 in left( {0,frac{π }{4}} right) is

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5. Continuity and Differentiation

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Number of solution of the equation $ 3tanx + x^3 = 2 $ in $ \left( {0,\frac{\pi }{4}} \right)$ is

A

$0$

B

$1$

C

$2$

D

$3$

Solution

$f(x) = 3tanx + x^3$

then $f '(x) = 3sec^2x + 3x^2 > 0$

hence $f(x) \uparrow $.

Thus $ f(x)$ assume the value $2$ exactly once.

Also $f(0) = 0$ and $f\left( {\frac{\pi }{4}} \right)> 2$

so by intermediate value theorem $f(c) = 2$ for some $c$ in $(0,\frac{\pi}{4})$

Standard 12

Mathematics

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