Question

Let \(\rm f(x) = \left\{\begin{matrix} 1+\frac{x}{2k}, & 0 < x < 2\\\ kx, & 2 \le x < 4 \end {matrix}\right.\)

If \(\displaystyle\lim_{x\rightarrow 2}\) f(x) exists, then what is the value of k?

1. -2
2. -1
3. 0
4. 1

Answer 1

Correct Answer - Option 4 : 1

Concept:

Step: 1

Check f(a) is defined. If it is not defined then no need to go further.  The function is not continuous at a. If f(a) is defined then

Step: 2

Check Left-hand limit (LHL) and Right-hand limit (RHL) 

\(LHL = \rm \lim_{x\rightarrow a^{-}}f(x) \)

\(RHL= \lim_{x\rightarrow a^{+}} f(x) \)

If LHL = RHL then limit exist.

Calculation:​

\(\rm \lim_{x\rightarrow 2^{-}} f(x) = \lim_{x\rightarrow 2}(1 + \frac{x}{2k} )\)

⇒ \(\rm \lim_{x\rightarrow 2^{-}}f(x)\) = \(\rm 1 + \frac{2}{2k} = 1 + \frac{1}{k}\)

⇒​ \(\rm \lim_{x\rightarrow 2^{-}}f(x)\) = \( 1 + \frac{1}{k}\)         ----(1)

Also,

\(\rm \lim_{x\rightarrow 2^{+}}f(x)\) = \(\rm \lim_{x\rightarrow 2^{+}}(kx)\)

⇒ \(\rm \lim_{x\rightarrow 2^{+}}f(x)\) =   2k    ----(2)

If \(\displaystyle\lim_{x\rightarrow 2}\) f(x) exists, then 

\(\rm 1 + \frac{1}{k} = 2k\)

⇒ 2k² - k - 1 = 0

⇒ 2k2 - 2k  + k - 1 = 0

⇒ 2k(k - 1) +1(k - 1) = 0

⇒ (k - 1)(2k + 1) = 0

⇒ k = 1 & -1/2

∴ The value of k is 1.

 Compare f(a), LHL & RHL . 

If \(\rm \lim_{x\rightarrow a^{-}}f(x) = \lim_{x\rightarrow a^{+}} f(x) = f(a)\) then,

The function is continuous at f(a).