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

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$ 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}\le

Vedclass · Accepted Answer

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

Vedclass · Answer

$ f(\mathrm{x}) =\mathrm{x}\left\{\sin ^{2} \mathrm{x} \cdot \frac{1}{\mathrm{x}}+\sin 2 \mathrm{x} \cdot \log \mathrm{x}ight\} $

$=\mathrm{x}\left(\sin ^{2} \mathrm{x} \cdot \log \mathrm{x}ight)^{\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 $

$	herefore $ by Rolle's theorem

$g^{\prime}(x)=0$ has at least two root in $(1,2 \pi)$

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

Similar Questions

If Rolle's theorem holds for the function $f(x)=x^{3}-a x^{2}+b x-4, x \in[1,2]$ with $f ^{\prime}\left(\frac{4}{3}\right)=0,$ then ordered pair $( a , b )$ is equal to

Let $\mathrm{f}: \mathbb{R} \rightarrow \mathbb{R}$ be a thrice differentiable function such that $f(0)=0, f(1)=1, f(2)=-1, f(3)=2$ and $f(4)=-2$. Then, the minimum number of zeros of $\left(3 f^{\prime} f^{\prime \prime}+f f^{\prime \prime \prime}\right)(x)$ is....................

Mean value theorem $f(b) -f(a) = (b -a) f '(x_1);$ from $a < x_1 < b,$ if $f(x) = 1/x$ then $x_1 = ?$

Let $f(x) = (x-4)(x-5)(x-6)(x-7)$ then -

From mean value theorem $f(b) - f(a) = $ $(b - a)f'({x_1});$ $a < {x_1} < b$ if $f(x) = {1 \over x}$, then ${x_1} = $