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If a, b, c be three distinct real numbers in G.P. and a + b + c = xb , then x cannot be

JEE · Math · previous-year question

  1. A.2correct
  2. B.-3
  3. C.4
  4. D.-2

Answer

A. 2

Explanation

a, b, c are in G.P. So, b = ar and c = ar2 given a + b + c = xb $$ \Rightarrow $$ a + br + ar2 = x(ar) $$ \Rightarrow $$ 1 + r + r2 = xr $$ \Rightarrow $$ x = 1 + r + $${1 \over r}$$ let sum of r + $${1 \over r}$$ = M $$ \therefore $$ r2 + 1 = Mr $$ \Rightarrow $$ r2 $$-$$ Mr + 1 = 0 this quadratic equation will have real solution when discriminant is $$ \ge $$ 0 $$ \therefore $$ b2 $$-$$ 4ac $$ \ge $$ 0 M2 $$-$$ 4.1.1 $$ \ge $$ 0 $$ \Rightarrow $$ M2 $$ \ge $$ 4 M $$ \ge $$ 2 or M $$ \le $$ $$-$$ 2 $$ \therefore $$ M $$ \in $$ ($$-$$ $$ \propto $$, $$-$$ 2] $$ \cup $$ [2, $$ \propto $$) As x = 1 + r + $${1 \over r}$$ = 1 + M $$ \therefore $$ x $$ \in $$ ($$-$$ $$ \propto $$, $$-$$ 1] $$ \cup $$ [3, $$ \propto $$) $$ \therefore $$ x can't be 0, 1, 2.

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