Question

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– Non-copied Answer
– Well-explained Answer

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Answer 1

Answer:

To find:-

The value of k

Solution:-

$Now , f(0) = \frac{2(0) + 1}{0 - 1} = \frac{1}{ - 1} = - 1$

$and \: LHL = lim \: f(0 - h) \\ \: \: \: \: \: \: \: \: h \rightarrow0$

$= \frac{lim}{ h \rightarrow0} \frac{ \sqrt{1 - kh} - \sqrt{1 + kh} }{ - h}$

$\sf= \frac{lim}{h \rightarrow0} \frac{ \sqrt{1 - kh } + \sqrt{1 + kh} }{ - h} \times \frac{ \sqrt{1 - kh} + \sqrt{1 + kh} }{ \sqrt{1 - kh} + \sqrt{1 + kh} }$

$\sf \: \frac{lim}{ h \rightarrow0} \: \frac{(1 - kh) - (1 + kh)}{ - h( \sqrt{1 - kh} + \sqrt{1 + kh} ) } \\ \\ \sf[ (a + b)(a - b) = {a}^{2} - {b}^{2} ]$

$\sf \: = \frac{lim}{ h \rightarrow0} \frac{ - 2kh}{ - h( \sqrt{1 - kh} + \sqrt{1 + kh}) }$

$\sf \: = \frac{lim}{ h \rightarrow0} \frac{2k}{ \sqrt{ 1 - kh} + \sqrt{1 + kh} } = \frac{2k}{1 + 1} = \frac{2k}{2} = 2$

Since , f(x) is constant at x= 0.

$\therefore \: f(0) = LHL ⇒ - 1 = k$

$⇒k = - 1$

Answer 2

GIVEN :–

$\\ \implies \bf F(x) = \bigg\{ \dfrac{ \sqrt{1 + kx} - \sqrt{1 - kx} }{x} \:,\: if - 1 \leqslant x < 0$

$\\ \implies \bf F(x) = \bigg\{ \dfrac{2x + 1}{x - 1} \:,\: if \: \: 0 \leqslant x <1$

• Function F(x) continuous at x = 0.

TO FIND :–

• Value of 'k' = ?

SOLUTION :–

• Let's find the value of F(x) at x=0 –

$\\ \implies \bf F(0) =\dfrac{2(0) + 1}{(0)- 1}$

$\\ \implies \bf F(0) =\dfrac{1}{- 1}$

$\\ \longrightarrow \large \red{ \boxed{ \bf F(0) = - 1}}$

• Now let's find R.H.L. –

$\\ \implies \bf R.H.L. = \lim_{x \to {0}^{ + }}\dfrac{2x + 1}{x - 1}$

$\\ \implies \bf R.H.L. = \lim_{h \to 0}\dfrac{2( 0+ h) + 1}{(0 + h)- 1}$

$\\ \implies \bf R.H.L. = \lim_{h \to 0}\dfrac{2h+ 1}{h- 1}$

$\\ \implies \bf R.H.L. =\dfrac{2(0)+ 1}{(0)- 1}$

$\\ \implies \bf R.H.L. =\dfrac{ 1}{- 1}$

$\\ \longrightarrow \large \red{ \boxed{ \bf R.H.L. = - 1}}$

• Now let's find L.H.L. –

$\\ \implies \bf L.H.L. = \lim_{x \to {0}^{ - }} \dfrac{ \sqrt{1 + kx} - \sqrt{1 - kx} }{x}$

$\\ \implies \bf L.H.L. = \lim_{h\to 0} \dfrac{ \sqrt{1 + k(0 - h)} - \sqrt{1 - k(0 - h)} }{(0 - h)}$

$\\ \implies \bf L.H.L. = \lim_{h\to 0} \dfrac{ \sqrt{1 - kh} - \sqrt{1 + kh} }{- h}$

$\\ \implies \bf L.H.L. = \lim_{h\to 0} \dfrac{ \sqrt{1 - kh} - \sqrt{1 + kh} }{- h} \times \dfrac{\sqrt{1 - kh} + \sqrt{1 + kh}}{\sqrt{1 - kh} + \sqrt{1 + kh}}$

$\\ \implies \bf L.H.L. = \lim_{h\to 0} \dfrac{ (\sqrt{1 - kh})^{2} - ( \sqrt{1 + kh}) ^{2} }{- h(\sqrt{1 - kh} + \sqrt{1 + kh})}$

$\\ \implies \bf L.H.L. = \lim_{h\to 0} \dfrac{(1 - kh) - (1 + kh)}{- h(\sqrt{1 - kh} + \sqrt{1 + kh})}$

$\\ \implies \bf L.H.L. = \lim_{h\to 0} \dfrac{1 - kh - 1 - kh}{- h(\sqrt{1 - kh} + \sqrt{1 + kh})}$

$\\ \implies \bf L.H.L. = \lim_{h\to 0} \dfrac{- 2kh}{- h(\sqrt{1 - kh} + \sqrt{1 + kh})}$

$\\ \implies \bf L.H.L. = \lim_{h\to 0} \dfrac{2k}{(\sqrt{1 - kh} + \sqrt{1 + kh})}$

$\\ \implies \bf L.H.L. =\dfrac{2k}{(\sqrt{1 - k(0)} + \sqrt{1 + k(0)})}$

$\\ \implies \bf L.H.L. =\dfrac{2k}{(\sqrt{1} + \sqrt{1})}$

$\\ \implies \bf L.H.L. =\dfrac{2k}{2}$

$\\ \implies \bf L.H.L. =k$

• By the condition of continuous function –

$\\ \implies \red{\bf L.H.L. =R.H.L. = F(0)}$

$\\ \implies\bf k= - 1 = - 1$

• By first & second condition –

$\\ \implies\bf k= - 1$

$\\ \implies \large\red{ \boxed{\bf k=-1}}$