Which one of the following is correct about the given graph of f x -A- -x y-1B- - x - y -1C- -x-y-1D- x-y-1

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

Byju's Answer

Other

Quantitative Aptitude

Functions

Which one of ...

Question

Which one of the following is correct about the given graph of f(x) :

|xy| = 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

|x| y = 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x|y| = 1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

|x||y| = 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C x|y| = 1
Since x is always positive and y can assume both positive and negative values therefore (c) is the correct choice.
$x | y | = 1 \Rightarrow x = \frac{1}{| y |}$ for x > 0

Suggest Corrections

Similar questions

Which of the following is a linear expression ?

a) $x^{2} \times y + x y + 1$

b) $x + y + 1$

c) $x^{3} \times y + x y + 1$

d) $4 x^{2} + 1$

$x = 1 + a + a^{2} + . . . \infty (a < 1)$
$y = 1 + b + b^{2} + . . . \infty (b < 1)$
Then the value of $1 + a b + a^{2} b^{2} + . . . . . \infty$ is
[MNR 1980; MP PET 1985]

Using the given graph which of the following could be the equation of line $L_{2}$ .

Q. If

x = 1 + a + a^{2} + . . . . \infty

and

y = 1 + b + b^{2} + . . . . \infty,

then the value of

1 + a b + + a^{2} b^{2} + . . . . \infty

(0 < a < 1, 0 < b < 1)

Q. The function

y = \frac{1}{x}

shifted 1 unit down and 1 unit right is given by?

Join BYJU'S Learning Program

Which one of the following is correct about the given graph of f(x) :

The correct option is C x|y| = 1Since x is always positive and y can assume both positive and negative values therefore (c) is the correct choice. x|y|=1⇒x=1|y| for x > 0

The correct option is C x|y| = 1
Since x is always positive and y can assume both positive and negative values therefore (c) is the correct choice.
$x | y | = 1 \Rightarrow x = \frac{1}{| y |}$ for x > 0