IPMAT Indore 2023 · Quantitative Ability · Question 31

Step 1 — strip out the factors that cannot change sign. {x}^{2}\ge 0 always, so it never affects the sign — it only excludes x=0, where the expression is zero rather than positive.

Similarly {\left( 2x+1 \right)}^{3} has an odd exponent, so its sign matches that of \left( 2x+1 \right); it excludes x=-1/2, where the denominator vanishes.

Step 2 — reduce to the governing expression. For x\ne 0 and x\ne -1/2,

\text{sign}\left[ \frac{{x}^{2}\left( x+1 \right)}{\left( x-1 \right){\left( 2x+1 \right)}^{3}}\right]=\text{sign}\left[ \frac{x+1}{\left( x-1 \right)\left( 2x+1 \right)}\right]

Step 3 — mark the critical points at x=-1,-1/2,1 and test each interval.

Step 4 — assemble, remembering the exclusions. The expression is positive on

\left( -1,-1/2 \right)\cup \left( 1,\infty \right)

and x=0 does not lie in either piece, so nothing further is removed.

Answer: Option 2 — \left( -1,-1/2 \right)\cup \left( 1,\infty \right)

Note on the source file. In the supplied .docx the exponent on \left( 2x+1 \right) is rendered as a square. With a square, that factor is always positive and the solution set becomes \left( -\infty ,-1 \right)\cup \left( 1,\infty \right) — which matches none of the four options. The official paper has a cube, as used above, and that reproduces the printed answer. The square is a transcription error in the equation image, not an error in the key.

Shortcut. Even exponents punch holes; odd exponents carry sign. That single rule does the whole question:

{x}^{2} — even, so it only deletes the point x=0.

{\left( 2x+1 \right)}^{3} — odd, so it behaves exactly like \left( 2x+1 \right).

What remains is \frac{x+1}{\left( x-1 \right)\left( 2x+1 \right)}. Mark -1,-1/2,1 on a line, note the rightmost interval is positive (all factors positive), then alternate signs leftwards across each simple root:

+ | - | + | -

Pick out the plus intervals: \left( -1,-1/2 \right)\cup \left( 1,\infty \right).

Alternation only works for roots of odd multiplicity — which is why identifying the even powers first is step one, not an afterthought.