The number of integers values of 'x', for which (3-(10 -x^(1/2)

Question

The number of integer values of 'x', for which  (3-(10 -x^{\frac{1}{2}})^{\frac{1}{2}})^{\frac{1}{2}}  is a real number, lies between :

(A) 89 and 95
(B) 94 and 100
(C) 99 and 105
(D) 104 and 110
(E) 109 and 115

chetan2u · Accepted Answer

Hi..
attempt this by ensuring the part under square root is NOT <0

so the solution would be

(3-(10 -x^{\frac{1}{2}})^{\frac{1}{2}})^{\frac{1}{2}}

so  \sqrt{10-x^{1/2}}\geq{0}.................\sqrt{x}\leq{10}...............x\leq{100}

Also  \sqrt{10-x^{1/2}}\leq{3}....................10-\sqrt{x}\leq{9}.............\sqrt{x}\geq{1}..........x\geq{1}

so values are from 1 to 100, so 100 values

since we are looking for a choice which includes 100 in the range 
C. between 99 to 104 is the answer

Gagoosh · Answer

the question is testing real numbers.. should I be worried about this one?

zs2 · Answer

\sqrt{3- 10 x}

which means  \sqrt{x} <= 10  , i.e. x<= 100

also  \sqrt{10- x} <=3 , i.e. x>= 1

THIS 1<=X<=100 , X can take 100 value

Hence C

pardhu1212 · Answer

Just using > instead of ≥ cost me this question.

xxdw0007 · Answer

It's a word game, not a math problem.

Adit_ · Answer

Take the innermost root:
x^1/2<=10.
x<=100.

Take the outer root:

(10-x^1/2)^1/2<=3
x>=1.

Combining the two we have 100 values.

Answer:  Option C

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