\int_{0}^{x}{f(t)d(t)} ----> 5 as |x| ---> 1, then value of 'a' so that the equation
2x + \int_{0}^{x}{f(t)d(t)} = a has at least two roots of opposite signs in (-1,1) is
a) 0<a<1
b) 0<a<3
c) -1<a<∞
d) 3<a<∞
Answer: b
1708\hspace{-16}$Observation , Not Solution.....\\\\\\Let If We take $\bf{\int_{0}^{x}f(t)dt=x^2+4}.$(Which satisfy Given Condition.)\\\\\\ Then $\bf{2x+\int_{0}^{x}f(t)dt=a\Leftrightarrow 2x+x^2+4=a}$\\\\\\ $\bf{(x+1)^2=a-3\Leftrightarrow a-3\geq 0\Leftrightarrow a\geq 3}$
1Thanks sir..
It wud be gr8 if I cud have a solution too..
Anyways I liked this method too...
1708\hspace{-16}$Let $\bf{g(x)=2x+\int_{0}^{x}f(x)dx-a}$\\\\\\ Then $\bf{g(-1)=\lim_{x\rightarrow -1}2x+\int_{0}^{x}f(x)dx-a=-2+5-a=3-a}$\\\\\\ Bcz $\bf{\int_{0}^{x}f(x)dx\rightarrow 5}$ as $\bf{\mid x \mid \rightarrow 1}$\\\\\\ So $\bf{g(1)=\lim_{x\rightarrow 1}2x+\int_{0}^{x}f(x)dx-a=2+5-a=7-a}$\\\\\\ and $\bf{g(0)=0-a=-a}$\\\\\\ Now If the equation $\bf{g(x)=0}$ has at least $\bf{2}$ Real Roots, Then
262continuing from previous post ,
then g(0)*g(1) <0
and g(0)*g(-1)<0 (for solutions with opposite sign)
taking intersection of the above two,
we get 0<a<3
1708Thanks Adiyata For completing that Question. (My Latex Image is not Completely Inserted)