Question

If $\frac{1}{2},\frac{1}{3},\frac{1}{4}$ are the roots of $a{x}^3+b{x}^2+cx+d=0$, then the roots of $a(x+1{)}^3+b(x+1{)}^2+c(x+1)+d=0$ are

• A. $-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}$
• B. $\frac{3}{2},\frac{4}{3},\frac{5}{4}$
• C. $-\frac{1}{2},-\frac{2}{3},-\frac{3}{4}$
• D. $\frac{1}{2},\frac{2}{3},\frac{3}{4}$

Answer 1

Answer: (C)

$-\frac{1}{2},-\frac{2}{3},-\frac{3}{4}$

Let the equation whose roots are $\alpha ,\beta$ and $\gamma$ is $f\left(x\right)=0$
Then,
The equation whose roots are $\alpha -1,\beta -1$ and $\gamma -1$ is $f(x+1)=0$
Thus the roots of above equation [which is $f(x+1)=0$] will be $-1+\frac{1}{2},-1+\frac{1}{3},-1+\frac{1}{4}$