If f(x) = sin^2x + xsin2x.logx, then f(x) = 0 has - Vidyadip

MCQ

If $f(x)$ = $sin^2x + xsin2x.logx$, then $f(x)$ = $0$ has

A
exactly one root in $\left( {0,2\pi } \right]$
✓
atleast two roots in $\left( {0,2\pi } \right]$
C
atmost one root in $\left( {0,2\pi } \right]$
D
no root in $\left( {0,2\pi } \right]$

✓

Answer

Correct option: B.

atleast two roots in $\left( {0,2\pi } \right]$

b
$ f(\mathrm{x}) =\mathrm{x}\left\{\sin ^{2} \mathrm{x} \cdot \frac{1}{\mathrm{x}}+\sin 2 \mathrm{x} \cdot \log \mathrm{x}\right\} $

$=\mathrm{x}\left(\sin ^{2} \mathrm{x} \cdot \log \mathrm{x}\right)^{\prime} $

$\mathrm{g}(\mathrm{x}) =\sin ^{2} \mathrm{x} \cdot \log \mathrm{x} $

$ \mathrm{g}(1) =0, \mathrm{g}(\pi)=0, \mathrm{g}(2 \pi)=0 $

$\therefore $ by Rolle's theorem

$g^{\prime}(x)=0$ has at least two root in $(1,2 \pi)$

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