9. Inequalities

• Theory
  Inequalities Made Easy!
  Inequalities: Tips and hints
  Arithmetic with Inequalities
  Wavy Line Method Application - Complex Algebraic Inequalities
  Solving Quadratic Inequalities: Graphic Approach
  Inequalities - Quadratic Inequalities
  Graphic approach to problems with inequalities
  Inequalities trick
  INEQUATIONS(Inequalities) Part 1
  INEQUATIONS(Inequalities) Part 2
  
  • Questions
    Inequality and absolute value questions from my collection
    DS Questions
    PS Questions

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If y is negative, then it would naturally be less than x, which from (1) is positive.
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. In the area of inequality, the number of equations means the number of inequalities, since the number of equations means the number of conditions.

In the original question, the number of variables is 2 and the number of equation is 0.

Hence, C is most likely the answer.

For the condition (1), we have two cases.
\(x = 1/4\) and \(y = 1/3\) satisfies the condition (1), but \(x > y\) is not true.
\(x = 4\) and \(y = 1\) satisfies the condition (1) and \(x > y\) is true.
Thus, the condition (1) is not sufficient.

For the condition (2), we also have two cases.
\(x = 2\) and \(y = 3\) satisfies the condition (2), but \(x > y\) is not true.
\(x = 3\) and \(y = 2\) satisfies the condition (2) and \(x > y\) is true.
Thus, the condition (2) is not sufficient, either.

For both the conditions (1) and (2) together, we should consider two cases.

Case 1: \(0 \leq x \leq 1\).
Since \(x^3 \leq x\), if \(x^3 > y\) by the condition (2), \(x \geq x^3 > y\), that is, \(x >y\).

Case 2: \(x > 1\)
Since \(x > \sqrt{x}\), if \(\sqrt{x} > y\) by the condition (1), \(x > \sqrt{x} >y\), that is, \(x > y\).

Hence, both the conditions (1) and (2) together are sufficient, as expected earlier.
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Hey VeritasKarishma,
great approach, thanks a lot.
But here I have a doubt (excuse me if silly ) -
When we are combining both graphs - its clear that X>Y when 0>x>1 AND 1>x>+.
But what about when the value of x is between "-1 to 0".....as shown with red color in second graph.

please explain.

regards,
Tamal
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\(x\,\,\,\mathop > \limits^? \,\,\,y\)

\(\left( 1 \right)\,\,\,\sqrt x > y\,\,\,\,\,\left\{ \begin{gathered}
\,Take\,\,\,\left( {x;y} \right) = \left( {1\,;\,0} \right)\,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \hfill \\
\,Take\,\,\,\left( {x;y} \right) = \left( {\frac{1}{4}\,;\,\frac{1}{3}} \right)\,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \hfill \\
\end{gathered} \right.\)

\(\left( 2 \right)\,\,\,{x^3} > y\,\,\,\,\,\left\{ \begin{gathered}
\,(re)\,Take\,\,\,\left( {x;y} \right) = \left( {1;0} \right)\,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \hfill \\
\,Take\,\,\,\left( {x;y} \right) = \left( {2\,;\,7} \right)\,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \hfill \\
\end{gathered} \right.\)


\(\left( {1 + 2} \right)\,\,\,\,\,\)

\(\left. {x = y\,\,\,\,\, \Rightarrow \,\,\,\left\{ \begin{gathered}
\mathop \Rightarrow \limits^{\left( 1 \right)} \,\,\,\,\,\sqrt x > x\,\,\,\,\, \Rightarrow \,\,\,\,\,0 < \sqrt x < 1\,\,\,\,\,\,\,\,\left( {\sqrt x \geqslant 1\,\,\, \Rightarrow \,\,\,\sqrt x > x\,\,{\text{false,}}\,\,\,{\text{impossible}}} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\boxed{0 < x < 1} \hfill \\
\mathop \Rightarrow \limits^{\left( 2 \right)} \,\,\,\,\,{x^3} > x\,\,\,\,\, \Rightarrow \,\,\,\,\,\boxed{0 < x < 1\,\,\,{\text{false}}}\,\,\,\,\,\,\,\,\, \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\,} \right\}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,x \ne y\)

\(x < y\left. {\,\,\,\, \Rightarrow \,\,\,\left\{ \begin{gathered}
\mathop \Rightarrow \limits^{\left( 1 \right)} \,\,\,\,\,x < y < \sqrt x \,\,\,\,\, \Rightarrow \,\,\,\,\,\sqrt x > x\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{above}}} \,\,\,\,\,\,\boxed{0 < x < 1}\,\,\,\, \hfill \\
\mathop \Rightarrow \limits^{\left( 2 \right)} \,\,\,\,\,x < y < {x^3}\,\,\,\, \Rightarrow \,\,\,{x^3} > x\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{again}}} \,\,\,\,\,\,\,\boxed{0 < x < 1\,\,\,{\text{false}}}\,\,\,\,\,\,\,\,\, \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\,} \right\}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,x < y\,\,\,{\text{is}}\,\,{\text{false}}\)

Conclusion: \(\left( {1 + 2} \right)\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\text{YES}}} \right\rangle\)


The above follows the notations and rationale taught in the GMATH method.
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