Question:
The number of positive integral solutions of $ \frac {x^2 (3x-4)^3 (x-2)^4} {(x-5)^5 (2x-7)^6 }\leq 0$ is .
The number of positive integral solutions of $ \frac {x^2 (3x-4)^3 (x-2)^4} {(x-5)^5 (2x-7)^6 }\leq 0$ is .
Updated On: Jul 7, 2022
- $4$
- $3$
- $2$
- $1$
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The Correct Option is B
Solution and Explanation
$ \frac {x^2 (3x-4)^3 (x-2)^4} {(x-5)^5 (2x-7)^6 }\leq 0$
$\Rightarrow x = 0, \frac{4}{3},2,3x -4 < 0, x - 5 >0$
or $3x-4 > 0 , x-5 >0 $
$\left[ \because \, x^{2 }, \left(x-2\right)^{4},\left(2x -7\right)^{6 } >0\right] $
$\Rightarrow x = 0, \frac{4}{3},2 ,x < \frac{4}{3} ,x >5$ or $x > \frac{4}{3} , x <5$
$ \Rightarrow x = 0,2 $ and integral value between
$\frac{4}{3} < x < 5\,$ i.e., $x = 2,3,4$.
Hence positive integral solutions are $2$, $3$, $4$.
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Notes on complex numbers
Concepts Used:
Complex Number
A Complex Number is written in the form
a + ib
where,
- “a” is a real number
- “b” is an imaginary number
The Complex Number consists of a symbol “i” which satisfies the condition i^2 = −1. Complex Numbers are mentioned as the extension of one-dimensional number lines. In a complex plane, a Complex Number indicated as a + bi is usually represented in the form of the point (a, b). We have to pay attention that a Complex Number with absolutely no real part, such as – i, -5i, etc, is called purely imaginary. Also, a Complex Number with perfectly no imaginary part is known as a real number.