If abc 0, which of the following is definitely positive?

Question

If  abc  < 0 and  acd  > 0, which of the following is definitely positive?

A.  a^4b^4c^3d^5 
B.  a^4b^4c^4d^5 
C.  a^5b^5c^4d^3 
D.  a^5b^3c^5d^5 
E.  a^5b^2c^5d^5

gemministorm · Answer

a*c*b -> -ve
case1: + + -
case2: - - -

a*c*d -> +ve
case1: + + +
case2: - - +

Since abc<0 and acd>0, b and d have opposite signs. Make b even-powered to remove its sign, and ensure the combined parity of a,c,d is odd so the expression includes (acd) an odd number of times, which is positive

E: a5 b2 c5 d5 -> b even -> a *c*d >0. Definitely positive

MaitriD · Answer

Why can't it be  B.   a  4  b  4  c  4  d  5

Bunuel · Answer

B.  a^4b^4c^4d^5

In B,  a^4b^4c^4  will be positive, so the sign of the whole expression will be determined by  d , which can be both positive and negative.

For example, a, b, and c are positive, and d is negative, or a and c are positive, and b and d are negative.

arushi118 · Answer

acb<0
++-
--+
-++
+-+

acd>0
+++
--+ 
-+-
+--

acbd
++-+ -> power of b should be even for this combination to be positive
--++ -> power of a and c should be same -> that is both powers should either be odd or both be even
-++- -> power of a and d should be same -> that is both powers should either be odd or both be even
+-+- -> power of c and d should be same -> that is both powers should either be odd or both be even

So from above we know that powers of a,c,d -> all should be either odd or even and power of b should be even.
Hence, correct answer is E.

Dereno · Answer

Given that abc <0 and acd >0 abc <0 , there are following ways this can be true. - + + + - + + + - - - - For the condition acd >0 , the following cases can be true - + - + + + + - - - - + Combining both we get a b c d - + + - + - + + + + - - - - - + Let’s look at the options : A. a^4b^4c^3d^5 when we substitute the first case : (-)^4 * (+)^4 * (+)^4 * (-)^5 = + * + * + * - = -ve option. Hence, False. B. a^4b^4c^4d^5 Substitute case 1, we get (-)^4 * (+)^4 *(+)^4*(-)^5 = + * + * + * - = -ve. Hence, False . C. a^5b^5c^4d^3 substitute case 1, we get (-)^5 * (+)^5 *(+)^5 *(-)^3 = -*+*+*- = +ve substitute case 2, we get (+)^5 * (-)^5 * (+)^4* (+)^3 = +*-*+*+ = -ve Hence, False . D. a^5b^3c^5d^5 Substitute case 1, we get (-^5) * (+^3) * (+ ^5) * ( -^5) = -*+*+*- = +ve Substitute case 2, we get (+^5) * (-^3) * (+^5) * (+^5) = + * - * + * + = -ve Hence, False. E. a^5b^2c^5d^5 We are left with this option, and substituting all the cases, we get a definite positive. Hence, DEFINITELY POSITIVE. OPTION E.

Bunuel · Answer

Experts' Global Video Explanation:

https://www.youtube.com/watch?v=xKZItIS-Rag

If abc < 0 and acd > 0, which of the following is definitely positive?

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