Ask & Answer Question

Ncert Solutions Maths class 11th Maths Class 11th Mathematical Reasoning

Show that the statement p: “If x is a real number such that x³ + 4x = 0, then x is 0” is true by

(i) Direct method, (ii) Method of contradiction, (iii) Method of contrapositive

2 Views | Posted 3 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

3 months ago

This is a Mathematical Reasoning Solutions Type Questions as classified in NCERT Exemplar

(i) Direct method:

Let $x^{3} + 4 x = 0, x \in R$

$\Rightarrow x (x^{2} + 4) = 0$

$\Rightarrow$ Either x=0 or x²+4=0

But x² + 4 > 4 because $x \in R$ and hence $\neq 0 .$

Therefore x=0

$∴ x^{3} + 4 x = 0, x \in R \Rightarrow x \neq 0$

This, p is a true statement.

(ii) Method of contradiction

Let $x^{3} + 4 x = 0, x \in R$

Suppose $x \neq 0$

$\Rightarrow x^{2} > 0$

$\Rightarrow x^{2} + 4 > 0$

$\Rightarrow x^{2} + 4 \neq 0$

Now $x \neq 0$ and $x^{2} + 4 \neq 0$

$\Rightarrow x (x^{2} + 4) \neq 0$

$\Rightarrow x^{3} + 4 x \neq 0$ which is a contradiction to given.

$∴$ Our supposition is wrong and hence x = 0

This, p is a true statement.

(iii) Method of contrapositive: The components of the give if… then statement p are:

Let $q : x \in R$ andx³ + 4x = 0 and r:x = 0

$∴$ The given statement p is q $\Rightarrow$ r

Its contra positive is $\sim r \Rightarrow \sim q .$

Let $\sim r$ be true i.e

...more

This is a Mathematical Reasoning Solutions Type Questions as classified in NCERT Exemplar

(i) Direct method:

Let $x^{3} + 4 x = 0, x \in R$

$\Rightarrow x (x^{2} + 4) = 0$

$\Rightarrow$ Either x=0 or x²+4=0

But x² + 4 > 4 because $x \in R$ and hence $\neq 0 .$

Therefore x=0

$∴ x^{3} + 4 x = 0, x \in R \Rightarrow x \neq 0$

This, p is a true statement.

(ii) Method of contradiction

Let $x^{3} + 4 x = 0, x \in R$

Suppose $x \neq 0$

$\Rightarrow x^{2} > 0$

$\Rightarrow x^{2} + 4 > 0$

$\Rightarrow x^{2} + 4 \neq 0$

Now $x \neq 0$ and $x^{2} + 4 \neq 0$

$\Rightarrow x (x^{2} + 4) \neq 0$

$\Rightarrow x^{3} + 4 x \neq 0$ which is a contradiction to given.

$∴$ Our supposition is wrong and hence x = 0

This, p is a true statement.

(iii) Method of contrapositive: The components of the give if… then statement p are:

Let $q : x \in R$ andx³ + 4x = 0 and r:x = 0

$∴$ The given statement p is q $\Rightarrow$ r

Its contra positive is $\sim r \Rightarrow \sim q .$

Let $\sim r$ be true i.e., x is a non-zero real number

Now $x \neq 0, x \in R$

$\Rightarrow x^{2} > 0$

$\Rightarrow x^{2} + 4 > 4$

$\Rightarrow x^{2} + 4 \neq 0$

$∴ x (x^{2} + 4) \neq 0 \Rightarrow x^{3} + 4 x \neq 0$

$∴ \sim q$ is true

i.e., $\sim r \Rightarrow \sim q$

$∴ q \Rightarrow r$ is true

This, p is a true statement.

less

0 Upvotes 0 Downvotes

Show that the statement p: “If x is a real number such that x³ + 4x = 0, then x is 0” is true by

(i) Direct method, (ii) Method of contradiction, (iii) Method of contrapositive

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

Show that the statement p: “If x is a real number such that x3 + 4x = 0, then x is 0” is true by (i) Direct method, (ii) Method of contradiction, (iii) Method of contrapositive

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

Show that the statement p: “If x is a real number such that x³ + 4x = 0, then x is 0” is true by

(i) Direct method, (ii) Method of contradiction, (iii) Method of contrapositive