If f(x) = begincases (√1+ kx -√1- kx / X ) textfor -1 ≤ x

Question

If f(x) = begincases (√1+ kx -√1- kx / X ) textfor -1 ≤ x<0 [2ex] 2 x2+3 x-2 textfor 0 ≤ x ≤ 1 endcases continuous at x =0, then k is equal to (A) -4 (B) -3 (C) -2 (D) -1

Tardigrade Admin · Accepted Answer

LHL = displaystyle lim x arrow 0- (√1+ kx -√1- kx / x ) = displaystyle lim x arrow 0- (2 kx /x(√1+ kx +√1- kx ))= k RHL = displaystyle lim x arrow 0+(2 x 2+3 x -2)=-2 f (0)=-2 ∵ It is given that f ( x ) is continuous ∴ LHL = RHL = f (0) ⇒ k =-2

Q. If $f (x) = ⎩ ⎨ ⎧ \frac{1 + k x - 1 - k x}{X} 2 x^{2} + 3 x - 2 for - 1 \leq x < 0 for 0 \leq x \leq 1$
continuous at $x = 0$ , then $k$ is equal to

-4

-3

-2

-1

$L H L = x \to 0^{-} lim \frac{1 + k x - 1 - k x}{x}$
$= x \to 0^{-} lim \frac{2 k x}{x ( 1 + k x + 1 - k x )} = k$
$R H L = x \to 0^{+} lim (2 x^{2} + 3 x - 2) = - 2& f (0) = - 2$
$∵$ It is given that $f (x)$ is continuous
$∴ L H L = R H L = f (0)$
$\Rightarrow k = - 2$

Q. If f(x)=⎩⎨⎧​X1+kx​−1−kx​​2x2+3x−2​for −1≤x<0for 0≤x≤1​ continuous at x=0, then k is equal to

-4

-3

-2

-1

LHL=x→0−lim​x1+kx​−1−kx​​ =x→0−lim​x(1+kx​+1−kx​)2kx​=k RHL=x→0+lim​(2x2+3x−2)=−2&f(0)=−2 ∵ It is given that f(x) is continuous ∴LHL=RHL=f(0) ⇒k=−2

Q. If $f (x) = ⎩ ⎨ ⎧ \frac{1 + k x - 1 - k x}{X} 2 x^{2} + 3 x - 2 for - 1 \leq x < 0 for 0 \leq x \leq 1$
continuous at $x = 0$ , then $k$ is equal to

$L H L = x \to 0^{-} lim \frac{1 + k x - 1 - k x}{x}$
$= x \to 0^{-} lim \frac{2 k x}{x ( 1 + k x + 1 - k x )} = k$
$R H L = x \to 0^{+} lim (2 x^{2} + 3 x - 2) = - 2& f (0) = - 2$
$∵$ It is given that $f (x)$ is continuous
$∴ L H L = R H L = f (0)$
$\Rightarrow k = - 2$