CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Definite Integral as Limit of Sum

Solve for x...

Question

Solve for $x$ .
$| x^{2} - 5 x | > | x^{2} | - | 5 x |$

Open in App

Solution

At $x = 0, x = 5, | x^{2} - 5 x | = | x^{2} | - | 5 x | = 0$
For $x \in (0, 5), | x^{2} | - | 5 x | < 0$ but, $| x^{2} - 5 x | > 0$ .
For $x > 5, | x^{2} - 5 x | = | x^{2} | - | 5 x |$ as all terms are positive.
For $x < 0, | x^{2} | - | 5 x | < 0$ but, $| x^{2} - 5 x | > 0$ .
Hence, $| x^{2} - 5 x | > | x^{2} | - | 5 x | \forall x \in (- \infty, 0) \cup (0, 5)$

Suggest Corrections

thumbs-up

0

Similar questions

x : ∣ ∣ ∣ \frac{x^{2} + 6}{5 x} ∣ ∣ ∣ \geq 1

\frac{| x - 3 |}{x^{2} - 5 x + 6} \geq 2

Q. Solve for x:

x^{2}

5 x

8

0

x^{2} - 5 x + 6 = x^{2} + 5 x - 8

x^{2} - (1 + \sqrt{5}) x + \sqrt{5} = 0

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Definite Integral as Limit of Sum

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Definite Integral as Limit of Sum

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon