If x be real and positive- then the value of y - x - 1-x satisfies

The correct option is C y ≥ 2 #160;Given y = x+ 1 / x =(√( x ) )^ 2 +( 1 /√( x ) #160; #160;) ^ 2 =( √( x ) 1 /√( x ) ...

484

You visited us 484 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard IX

Mathematics

Absolute Value

If x be rea...

Question

If $x$ be real and positive, then the value of
$y = x + \frac{1}{x}$ satisfies

0 < y \leq 0.5

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

0.5 < y \leq 1

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

1 < y < 2

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

y \geq 2

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

Open in App

Solution

The correct option is C $y \geq 2$
$G i v e n y = x + \frac{1}{x} = (\sqrt{x})^{2} + {(\frac{1}{\sqrt{x}})}^{2} = {(\sqrt{x} - \frac{1}{\sqrt{x}})}^{2} + 2 \sqrt{x} \frac{1}{\sqrt{x}} = {(\sqrt{x} - \frac{1}{\sqrt{x}})}^{2} + 2 F i r s t t e r m i s a s q u a r e d t e r m s o i t i s p o s i t i v e . ∴ {(\sqrt{x} - \frac{1}{\sqrt{x}})}^{2} + 2 > 2 w h e n {(\sqrt{x} - \frac{1}{\sqrt{x}})}^{2} h a s a f i n i t e v a l u e a n d {(\sqrt{x} - \frac{1}{\sqrt{x}})}^{2} + 2 = 2 w h e n {(\sqrt{x} - \frac{1}{\sqrt{x}})}^{2} i s z e r o . ∴ y \geq 2 (A n s)$

Suggest Corrections

Similar questions

Q. If

y = f (x)

satisfies the property

(x - y) f (x + y) - (x + y) f (x - y) = 4 x y (x^{2} - y^{2}), f (1) = 1

, then the number of real roots of

f (x) = 4

will be

Q. Let

a, b, x

and

y

be real numbers such that

a - b = 1

and

y \neq 0

. If the complex numbers

z = x + i y

satisfies Im

(\frac{a z + b}{z + 1}) = y

, then which of the following is possible value of

x

Q. If

y = y (x)

is the solution of

x \int 0 t y d t = x^{2} + y^{2}

and

y (0) = 1,

where point

(a, 3)

satisfies the curve

y = y (x),

then the value of

a^{2} =

Q. If

\frac{d y}{d x} = \frac{x y}{x^{2} + y^{2}}; y (1) = 1;

then a value of

x

satisfying

y (x) = e

is:

Q. Let

a, b, x

and

y

be real numbers such that

a - b = 1

and

y \neq 0

. If the complex number

z = x + i y

satisfies

I m (\frac{a z + b}{z + 1}) = y

, then which of the following is(are) possible value(s) of

x

Join BYJU'S Learning Program

If x be real and positive, then the value of y=x+1x satisfies

The correct option is C y≥2Giveny=x+1x=(√x)2+(1√x)2=(√x−1√x)2+2√x1√x=(√x−1√x)2+2Firsttermisasquaredtermsoitispositive.∴(√x−1√x)2+2>2when(√x−1√x)2hasafinitevalueand(√x−1√x)2+2=2when(√x−1√x)2iszero.∴y≥2(Ans)

If $x$ be real and positive, then the value of
$y = x + \frac{1}{x}$ satisfies