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

limx→0−(1+kx)12−(1−kx)12x⇒ltx→0−k21+kx+k21−kx=k2+k2=k⇒ltx→0+2x2+3x¯−2=−2Since f(x)is continuous at x=0⇒k=−2

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

limx→0−(1+kx)12−(1−kx)12x⇒ltx→0−k21+kx+k21−kx=k2+k2=k⇒ltx→0+2x2+3x¯−2=−2Since f(x)is continuous at x=0⇒k=−2

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

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

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$- 4$

$- 3$

$- 2$

$- 1$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} \lim_{x \to 0^{-}} \frac{{(1 + k x)}^{\frac{1}{2}} - {(1 - k x)}^{\frac{1}{2}}}{x} \\ \Rightarrow \underset{x \to 0^{-}}{l t} \frac{k}{2 \sqrt{1 + k x}} + \frac{k}{2 \sqrt{1 - k x}} = \frac{k}{2} + \frac{k}{2} = k \\ \Rightarrow \underset{x \to 0^{+}}{l t} 2 x^{2} + 3 \bar{x} - 2 = - 2 \end{array}$