If a≠0 and −1

Question

If a≠0 and −1a^2 B. aa^3 D. \sqrt {a}

ydmuley · Answer

a≠0 and −1a^2 Take a = -1/2 ==> -1/2 < 1/4 = Hence, Not True B. a 1/2 > 1/4 = Hence, Not True C. a^2>a^3 Take a = -1/2 ==> 1/4 > -1/8 ==> TRUE, Take a = 1/2 ==> 1/4 > 1/8 ==> TRUE =====> Hence, Always True D. \sqrt {a} {1/8}^2 ==> == > 1/2 > 1/64 ==> Hence, Not True E. \sqrt {a} 1/8 ==> 1/2 > 1/8 ==> Hence, Not True Hence, Answer is C

Luckisnoexcuse · Answer

Lets assume two smart values of a (-1/3 and 1/3)

A) a>a^2
Holds for 1/3 but not for -1/3 ..Rejected

B) a<a^2
Holds for -1/3 but not for 1/3 ..Rejected

C) a^2>a^3
Holds for -1/3 and for 1/3..Answer

pushpitkc · Answer

There are two possibilities for the values, one is negative and the other is positive(as a≠0)
Let assume this value to be +1/2 and -1/2

A.  a>a^2 
Though it works for the positive value(+1/2) it doesn't work for the negative value
B.  a<a^2 
Though it works for the negative value(-1/2) it doesn't work for the positive value
C.  a^2>a^3 
This works for both the positive value(+1/2) and the negative value(-1/2)

Hence,  Option C is the correct answer .

swatib28 · Answer

Option 1--> 01 or a<0 does not cover range option 3--> a<1 covers full range , hence true for all values of a Ans:C

Usernamevisible · Answer

For 0 < x < 1:

x^2 < x < cube root of x < 1

For -1 < x < 1, you must split into two cases.

Positive side: 0 < x < 1

x^2 < x < cube root of x < 1

Negative side: -1 < x < 0   
x < cube root of x < x^2

If a≠0 and −1<a<1, then which of the following must be true?

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