TG EAMCET 2 May 2025 Shift 2 Paper

Let$\;f(x)=x^4-10 x^3+37 x^2-60 x+36=0$$\;\;\text{ At }\; x=1, f(1)=1-10+37-60+36 \neq 0$$\;\;\text{ At }\; x=2, f(2)= (2^4-10 (2^3+37 (2^2...$$\;-60(2+36=0$$\; \;\; =16-80+148-120+36$$\;=200-200=0$So, $x=2$ is a factor of $f(x)$Similarly $x=3$ is a factor of $f(x)$$\therefore f(x)=(x-2)^2(x-3)^2$$\therefore$ Original roots are $2,2,3,3$The distinct roots are 2 and 3There are three possible pairs of distinct roots(i) Increase the two $2 s$ by 1(ii) Increase the two $3 s$ by 1(iii) Increase one 2 and one 3 by 1Let us assume we choose one root 2 and one root 3 to increase by 1Since, these distinct roots. Increase one ' 2 ' by $1: 2+1=3$Increase one ' 3 ' by $1: 3+1=4$Other two roots ( 2 and 3 ) are kept fixed$\therefore$ New sets of roots of he transformed equation is $3,4,2,3$Let the transformed equation be $g(x)$. its roots are $2,3,3,4$$\therefore g(x)\;=(x-2)(x-3)(x-3)(x-4)$$\;=(x-2)(x-3)^2(x-4)$The common roots are value of $x$ for which $f(x)=0$ and $g(x)=0$$\therefore$ Common roots are $2,3,3$$\therefore$ Number of common roots $=3$