Question

Show that

$\frac{2}{ \sqrt{x + 1} - \sqrt{x - 1} } = \leqslant 2$
$where \: \:$
$\frac{2}{ \sqrt{x + 1} - \sqrt{x - 1} } \: is \: defined$
Further show that the equation

$\sqrt{x + 1} - \sqrt{x - 1} = \sqrt{4x - 1}$
Has no solution.

○Challenging question:-

○Answer only if u know.​

Answer 1

Answer:

Given equation has no solution

Step-by-step explanation:

Given equation is

√(x+1) -√x-1) = √(4x-1) -------(1)

On squaring both sides then

=> [√(x+1) -√x-1)]² = [√(4x-1)]²

=> (x+1)+(x-1)-2(√(x+1)(x-1)) = 4x-1

=> x+1+x-1-2 √(x²-1) = 4x-1

=> 2x-2√(x²-1) = 4x-1

=>-2 √(x²-1) = 4x-1-2x

=> -2 √(x²-1) = 2x-1

On squaring both sides again then

=>[ -2 √(x²-1)]²= (2x-1)²

=> 4(x²-1) = 4x²-4x+1

=> 4x²-4 = 4x²-4x+1

=> -4 = -4x+1

=> -4x = -5

=> x = -5/-4

=> x = 5/4

On Substituting the value of x in (1)

=>LHS = √[(5/4)+1] - √[(5/4)-1]

=> LHS = √(9/4)-√(1/4)

=> LHS = 3/2 - 1/2

=> LHS = (3-1)/2

=> LHS = 2/2

=> LHS = 1

And

RHS = √[4(5/4)-1]

=> RHS = √(5-1)

=> RHS = √4

=> RHS = 2

LHS ≠ RHS

The given value of x does not satisfy the given equation .

So Given equation has no solution.

Used formulae:-

→ (a+b)² = a²+2ab+b²

→ (a-b)² = a²-2ab+b²

→ The Rationalising factor of √a+√b is √a-√b

Answer 2

$\large{\text{\underline{Correct Question:-}}}$

①Show that,

$\boxed{\dfrac{2}{\sqrt{x+1} +\sqrt{x-1} } \leq \sqrt{2}}$

where the expression is defined.

②Further, show that the equation,

$\boxed{\sqrt{x+1} -\sqrt{x-1} =\sqrt{4x-1}\text{,}}$

has no solution.

$\large{\text{\underline{Solution:-}}}$

$f(x)=\sqrt{x}$ is an increasing function on its domain.

Thus, the denominator,

$\hookrightarrow \sqrt{x+1} +\sqrt{x-1}\text{,}$

is also increasing.

$x\geq 1$ is the domain to be defined.

Hence, the minimum value occurs at $x=1$, giving,

$\hookrightarrow \sqrt{x+1} +\sqrt{x-1} \geq \sqrt{2}\text{.}$

We know that,

$\boxed{\hookrightarrow a\geq b>0\ \text{then}\ 0<\dfrac{1}{a} \leq \dfrac{1}{b}\text{.}}$

So,

$\hookrightarrow 0<\dfrac{1}{\sqrt{x+1} +\sqrt{x-1} } \leq \dfrac{1}{\sqrt{2} }$.

Further giving,

$\hookrightarrow 0<\dfrac{2}{\sqrt{x+1} +\sqrt{x-1} } \leq \sqrt{2}$.

Hence ① is shown.

Since

$\hookrightarrow \dfrac{2}{\sqrt{x+1} +\sqrt{x-1} } =\sqrt{x+1} -\sqrt{x-1}\text{,}$

from ① we have,

$\hookrightarrow 0<\sqrt{x+1} -\sqrt{x-1} \leq \sqrt{2}\text{.}$

But,

$\hookrightarrow \sqrt{4x-1} \geq \sqrt{3}\text{.}$

Hence two values are never the same since

$\hookrightarrow 0<\sqrt{x+1} -\sqrt{x-1} \leq \sqrt{2} <\sqrt{3} \leq \sqrt{4x-1}\text{.}$

Further giving

$\hookrightarrow \sqrt{x+1} -\sqrt{x-1} <\sqrt{4x-1}\text{.}$

So, the equation,

$\hookrightarrow \sqrt{x+1} -\sqrt{x-1} =\sqrt{4x-1}\text{,}$

has no solution.

Hence ② is shown.

$\large{\text{\underline{For Clarity:-}}}$

We have learned that,

$\boxed{\hookrightarrow a\geq b>0\ \text{then}\ 0<\dfrac{1}{a} \leq \dfrac{1}{b}\text{.}}$

To prove this, let us assume we have two positive numbers such that

$\hookrightarrow a\geq b>0\text{.}$

Since $\dfrac{1}{ab}$ is a positive number,

$\hookrightarrow \dfrac{a}{ab} \geq \dfrac{b}{ab} >\dfrac{0}{ab}\text{.}$

Further, this gives

$\hookrightarrow \dfrac{1}{b} \geq \dfrac{1}{a} >0\text{.}$

So,

$\hookrightarrow 0<\dfrac{1}{a} \leq \dfrac{1}{b} \text{. (Hence proven.)}$