Register

NIELIT 2019 Feb Scientist D - Section D: 18

• recategorized by
1,817 views

1 Answer

1 1 vote
Given : $x^{a}=y^{b}=z^{c},y^2=zx$

Let  $x^{a} = y^{b} = z^{c}= k$

$\Rightarrow x=k^{\frac{1}{a}}$

$\Rightarrow y=k^{\frac{1}{b}}$

$\Rightarrow z=k^{\frac{1}{c}}$

$\because y^{2}=zx$

$\therefore (k^{\frac{1}{b}})^2$ $=$ $(k^{\frac{1}{c}})*$ $(k^{\frac{1}{b}})$

$\therefore \frac{1}{a}+\frac{1}{b}=\frac{2}{b}$

Option $(C)$ is correct.
• edited by
Voting & Accepting an answer helps improve the community
← Previous · NIELIT 2019 Feb Scie...Next → · NIELIT 2019 Feb Scie...
← Previous in category · NIELIT 2019 Feb Scie...Next in category → · NIELIT 2019 Feb Scie...
Position:
Show:

Related questions

0 0 votes
1 1 answer
1.9k
1.9k views
Lakshman Bhaiya asked Apr 3, 2020
1,926 views
NIELIT 2019 Feb Scientist D - Section D: 7
Find all the polynomials with real coefficients $P\left(x \right)$ such that $P\left(x^{2}+x+1 \right)$ divides $P\left(x^{3}-1 \right)$.$ax^{n}$$ax^{n+2}$$ax$$2ax$
0 0 votes
1 1 answer
2.3k
2.3k views
Lakshman Bhaiya asked Apr 3, 2020
2,322 views
NIELIT 2019 Feb Scientist D - Section D: 15
The roots of the equation $x^{2/3}+x^{1/3}-2=0$ are :$1, -8$$-1, -2$$\frac{2}{3}, \frac{1}{3}$$-2, -7$
0 0 votes
1 1 answer
1.8k
1.8k views
Lakshman Bhaiya asked Apr 3, 2020
1,834 views
NIELIT 2019 Feb Scientist D - Section D: 28
$\left [\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^{2}}+\frac{4}{1+x^{4}}+\frac{8}{1+x^{8}} \right ]$ equal to :$1$$0$$\frac{8}{1-x^{8}}$$\frac{16}{1-x^{16}}$
1 1 vote
1 1 answer
1.9k
1.9k views
Lakshman Bhaiya asked Apr 3, 2020
1,878 views
NIELIT 2019 Feb Scientist D - Section C: 3
If $t^{2}-4t+1=0$, then the value of $\left[t^{3}+1/t^{3} \right]$ is :$44$$48$$52$$64$
1 1 vote
1 1 answer
2.1k
2.1k views
Lakshman Bhaiya asked Apr 3, 2020
2,051 views
NIELIT 2019 Feb Scientist D - Section C: 26
If $a^{x}=b$, $b^{y}=c$ and $c^{z}=a$, then the value of $xyz$ is :$0$$1$$\frac{1}{3}$$\frac{1}{2}$