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Question

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


A
121314
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B
32,43,54
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C
12,23,34
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D
12,23,34
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Solution

The correct option is A $$\displaystyle -\frac{1}{2},-\frac{2}{3},-\frac{3}{4}$$
Let the equation whose roots are $$\alpha ,\beta $$ and $$\gamma $$ is $$f(x)=0$$ 
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 $$-1+\dfrac12, -1+\dfrac13, -1+\dfrac14$$

Maths

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