If f x - eft begin matrix frac sqrt 1- kx - sqrt 1- kx x - -for -1 leq x -0 1lendmatrix -right is continuous at x-0- then k -A- -3B- -4C- -2D- -1

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Properties of Inverse Function

If f x = eft ...

Question

If $f (x) = {\begin{matrix} \frac{\sqrt{1 + k x} - \sqrt{1 - k x}}{x}, & f o r - 1 \leq x < 0 2 x^{2} + 3 x - 2, & f o r 0 \leq x \leq 1 \end{matrix}$
is continuous at x = 0 , then k=

-4

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-3

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-2

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-1

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Solution

The correct option is C -2
Since f(x) is continuous at x = 0 the limits about x = 0 should be equal, i.e. L.H.L = R.H.L
$L . H . L = {lim}_{x \to 0^{-}} \frac{\sqrt{1 + k x} - \sqrt{1 - k x}}{x} = k$
$R . H . L = {lim}_{x \to 0^{+}} (2 x^{2} + 3 x - 2) = - 2$
L.H.L = R.H.L
$\Rightarrow$ k = - 2.

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Similar questions

$f (x) = ⎧ ⎨ ⎩ \begin{matrix} \frac{\sqrt{1 + k x} - \sqrt{1 - k x}}{x}, & i f - 1 \leq x < 0 \frac{2 x + 1}{x - 1}, & i f 0 \leq x \leq 1 \end{matrix} a t x = 0$

What is the value of k if the function is continuous at all points?

Q. If

f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} \frac{\sqrt{1 + k x} - \sqrt{1 - k x}}{x} & f o r & - 1 \leq x < 0 2 x^{2} + 3 x - 2 & f o r & 0 \leq x < 1 \end{matrix}

is continuous at

x = 0

, then k

=

Q. If

f (x) = {\begin{matrix} \frac{\sqrt{1 + k x} - \sqrt{1 - k x}}{x} f o r - 1 \leq x < 0 2 x^{2} + 3 x - 2 f o r 0 \leq x < 1 \end{matrix}

is continuous at

x = 0

then

k =

Q. If

f (x) = {\begin{matrix} \frac{x^{2} - 1}{x + 1}, & w h e n x \neq - 1 - 2, & w h e n x = - 1 \end{matrix}

, then

f (x) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{\sqrt{(1 + p x)} - \sqrt{(1 - p x)}}{x}, & - 1 \leq x < 0 \frac{2 x + 1}{x - 2}, & 0 \leq x \leq 1 \end{matrix}

is continuous in the interval

[- 1, 1]

, then

p

is equal to :

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If f(x)={√1+kx−√1−kxx,for −1≤x<02x2+3x−2,for 0≤x≤1 is continuous at x = 0 , then k=

The correct option is C -2Since f(x) is continuous at x = 0 the limits about x = 0 should be equal, i.e. L.H.L = R.H.L L.H.L=limx→0−√1+kx−√1−kxx=k R.H.L=limx→0+(2x2+3x−2)=−2 L.H.L = R.H.L ⇒ k = - 2.

If $f (x) = {\begin{matrix} \frac{\sqrt{1 + k x} - \sqrt{1 - k x}}{x}, & f o r - 1 \leq x < 0 2 x^{2} + 3 x - 2, & f o r 0 \leq x \leq 1 \end{matrix}$
is continuous at x = 0 , then k=

The correct option is C -2
Since f(x) is continuous at x = 0 the limits about x = 0 should be equal, i.e. L.H.L = R.H.L
$L . H . L = {lim}_{x \to 0^{-}} \frac{\sqrt{1 + k x} - \sqrt{1 - k x}}{x} = k$
$R . H . L = {lim}_{x \to 0^{+}} (2 x^{2} + 3 x - 2) = - 2$
L.H.L = R.H.L
$\Rightarrow$ k = - 2.