Power Trouble

Question

As always, timing is of paramount importance.

If m denotes a number to the left of 0 on the number line such that m^{4} is less than \frac{1}{81} , then the the reciprocal of m^{2} must be less than -3 between -\frac{1}{3} and 0 between 0 and \frac{1}{9} between \frac{1}{9} and 1 greater than 9

sudhir18n · Answer

Left of 0 means a negative number 
m^4< 1/81 
(-1/3)^ 4 = 1/81

so recirprocal will be (-3) 
m^2 = 9 
so its E

vinodmallapu · Answer

Answer is E Given m^4 < 1/81 and m < 0 => values of m = ( -1/3 .... 0 ) So , m^2 = ( 1/9 .... 0 ) Reciprocal of m^2 = ( 9 ..... undefined ) => 1/(m^2) > 9

fluke · Answer

Sol:

m^{4} < \frac{1}{81}

(m^2)^2 < \frac{1}{9^2}

Taking square root on both sides:
 \sqrt{(m^2)^2} < \sqrt{\frac{1}{9^2}}

|m^2| < \frac{1}{9}

m^2 < \frac{1}{9}

As we know m is not 0 and m^2 is a +ve value, we can cross multiply:

9 < \frac{1}{m^2}

Flip the sides and reverse the symbol:
 \frac{1}{m^2}>9

Ans: "E"

samramandy · Answer

yes m^2 is 1/9 so 1/m^2 is 9.

So the answer is E.

BR
Mandeep

jamifahad · Answer

OA is indeed E. 
I followed an approach similar to fluke's. But got confused a bit in absolute value.

Spidy001 · Answer

m^4 < 1/81

(m^2)^2 < (1/9)^2

take square root on both sides

mod(m^2) < 1/9 ( note sqrt(x^2) = mod(x) )

=> -1/9<m^2<1/9

=> 0<m^2<1/9 ( as m <0 , m^2 is always >0)

m^2<1/9

multiplying both sides by 9/(m^2) we have

9 < 1/(m^2)

=> 1/(m^2) > 9

Answer is E.

Capricorn369 · Answer

Time - 1:49m. 
Answer - E.

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