If f x = (√(4+ax) − √(4−ax))/x , −1 ≤x<0 (3x+2)/(x−8) , 0 ≤x ≤1 is continuous in [ −1,1 ], then the value of a is

Correct answer: (D) -1/2. Given that fx is continuous in the interval [-1, 1], so, it will be continuous at x=0. Now f0^(+)= (-1)/4 f0^(-)= limx →0^(-)⁡2ax/(x (√(4+ax) + √(4-ax) )) (rationalizing) = limx→0^(-)⁡2a/4=a/2 Now f0^(+)=f(0^(-)) (∵ f(x) is continuous at point 0) ⇒ a/2=(-1)/4 ⇒ a=(-1)/2

Q23 / 93 · NTA Abhyas (JEE Main) 2020 · Mathematics · Continuity and Differentiability

If $f (x) = \{\begin{matrix} \frac{\sqrt{4 + a x} - \sqrt{4 - a x}}{x}, & - 1 \leq x < 0 \\ \frac{3 x + 2}{x - 8}, & 0 \leq x \leq 1 \end{matrix}$ is continuous in $[- 1, 1],$ then the value of $a$ is

A
$1$
B
$- 1$
C
$\frac{1}{2}$
Correct answer
D
$- \frac{1}{2}$

Solution

Given that

f (x)

is continuous in the interval

[- 1, 1]

, so, it will be continuous at

x = 0

.

Now

f (0^{+}) = \frac{- 1}{4}

f (0^{-}) = \lim_{x \to 0^{-}} \frac{2 a x}{x (\sqrt{4 + a x} + \sqrt{4 - a x})}

(rationalizing)

= \lim_{x \to 0^{-}} \frac{2 a}{4} = \frac{a}{2}

Now

f (0^{+}) = f (0^{-})

(

∵

f (x)

is continuous at point 0)

\Rightarrow \frac{a}{2} = \frac{- 1}{4}

\Rightarrow a = \frac{- 1}{2}

Hence, the correct answer is option: 'D'.

If f x = 4+ax − 4−ax x , −1 ≤x<0 3x+2 x−8 , 0 ≤x ≤1 is continuous in [ −1,1 ], then the value of a is

Solution

More Continuity and Differentiability Questions

If $f (x) = \{\begin{matrix} \frac{\sqrt{4 + a x} - \sqrt{4 - a x}}{x}, & - 1 \leq x < 0 \\ \frac{3 x + 2}{x - 8}, & 0 \leq x \leq 1 \end{matrix}$ is continuous in $[- 1, 1],$ then the value of $a$ is