A) No real roots
B) 1real root
C) 2real roots
D) more than 2real roots
Correct Answer: C
Solution :
\[\begin{align} & f'(x)=3{{x}^{2}}+6x+6+2\operatorname{cosx} \\ & \\ \end{align}\], |
\[=3{{\left( x+1 \right)}^{2}}+3+\cos x>0\]for all x, |
So\[f\left( x \right)\]is an increasing function. |
Thus \[f(1)<f(2)<f(3)\] |
Let\[f(1)=a,f(2)=b\,and\,f(3)=c,\] |
Then \[a<b<c\] |
Give equation is \[\frac{1}{x-a}+\frac{2}{x-b}+\frac{3}{x-c}=0\] |
\[\begin{align} & \Rightarrow \left( x-b \right)\left( x-c \right)+2\left( x-a \right)\left( x-c \right) \\ & +3\left( x-a \right)\left( x-b \right)=0 \\ \end{align}\] |
\[\text{g }\left( a \right)=\left( a-b \right)\left( a-c \right)+2.0+3.0>0;\] |
\[g(b)=2\left( b-a \right)\left( b-c \right)<0and\] And |
\[g(c)=3\left( c-a \right)\left( c-b \right)>0\] |
Hence, the equation g(x) =0 has a root in (a, b) and another in (b, c). |
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