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MATHEMATICAL REASONING — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSMATHEMATICAL REASONING3 Marks
Question
Show that the statement p: “If x is a real number such that x3 + 4x = 0, then x is 0” is true by
  1. direct method,
  2. method of contradiction,
  3. method of contrapositive
✓

Answer

p: “If x is a real number such that $\text{x}^3+4\text{x}=0,$ then x is 0”. Let q: x is a real number such that $\text{x}^3+4\text{x}=0$ r: x is 0.
  1. To show that statement p is true, we assume that q is true and then show that r is true.
Therefore, let statement q be true.
$\therefore\text{x}^3+4\text{x}=0$
$\text{x}(\text{x}^2+4)=0$
$\Rightarrow\text{x}=0$ or $\text{x}^2+4=0$
However, since x is real, it is 0.
Thus, statement r is true.
Therefore, the given statement is true.
  1. To show statement p to be true by contradiction, we assume that p is not true.
Let x be a real number such that $\text{x}^3+4\text{x}=0$ and let x is not 0.
Therefore, $\text{x}^3+4\text{x}=0$
$\text{x}(\text{x}^2+4)=0$
$\text{x}=0$ or $\text{x}^2+4=0$
$\text{x}=0$ or $\text{x}^2=-4$
However, x is real. Therefore, $\text{x}=0,$ which is a contradiction since we have assumed that x is not 0.
Thus, the given statement p is true.
  1. To prove statement p to be true by contra positive method, we assume that r is false and prove that q must be false.
Here, r is false implies that it is required to consider the negation of statement r. This obtains the following statement.
$\sim\text{r . x}$ is not 0.
It can be seen that $(\text{x}^2+4)$ will always be positive.
$\text{x}\neq0$ implies that the product of any positive real number with x is not zero.
Let us consider the product of x with $(\text{x}^2+4)$.
$\therefore\text{x}(\text{x}^2+4)\neq0$
$\Rightarrow\text{x}^3+4\text{x}\neq0$
This shows that statement q is not true.
Thus, it has been proved that
$\sim\text{r}\Rightarrow\sim\text{q}$
Therefore, the given statement p is true.

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