The equation $2x^2-2(2a+1)x+a^2+a=0$ has one root less than $a$ and other root greater than $a$,if
$(A)0<a<1\hspace{1 cm}(B)-1<a<0\hspace{1 cm}(C)a>0\hspace{1 cm}(D)a<-1$
As one root is less than $a$ and other root greater than $a$,so $f(a)<0$
$$2a^2-2(2a+1)a+a^2+a<0\implies a>0 or a<-1$$
But the book says answer is $(A),(C),(D)$,i dont know how $(A)$ is possible as answer.
$(A)$ is a sub-case of $(C)$: $0 < a < 1 \Rightarrow a > 0$.
You've proved that $a>0 \Rightarrow f(a)<0$ and hence $a$ lies between the roots.
Combine these two together and you get $0 < a < 1 \Rightarrow f(a)<0$. So $(A)$ is a valid answer too.
If $0<a<1$ we have that
$$f(0)=a^2+a>0$$ and $$f(a)=-a^2-a<0.$$ Thus there exists a root $r\in (0,a).$ Now, from $f(a)<0$ and $$\lim_{x\to\infty} f(x)=+\infty$$ we get that there exists a root $s\in (a,\infty).$
Your solution is right and from your solution follows that $B$ is impossible because for $A$ $$(0,1)\subset (-\infty,-1)\cup(0,\infty),$$ for $C$ $$(0,+\infty)\subset (-\infty,-1)\cup(0,\infty),$$ for $D$ $$(-\infty,-1)\subset (-\infty,-1)\cup(0,\infty),$$ but for $B$ $$(-1,0)\cap\left((-\infty,-1)\cup(0,\infty)\right)=\oslash$$
If you factor $a$ out, you have the necessary and sufficient condition: $$f(a)=a\bigl(2a-2(2a+1)+a+1\bigr)=-a(a+1)<0$$ hence $a(a+1)>0$, i.e. $\;a<-1\;$ or $\; a>0$.
A), C) and D) are sufficient conditions for this to happen.
Hint: Use the Quadratic Formula to express the roots in terms of $a$.
Can you take it from here?
