Question

If $f(x) = \begin{cases} x+\lambda &x < 3 \\[2ex] 4 & x = 3\\[2ex] 3x - 5 , & x > 3 \end{cases}$
is continuous at $x = 3$, then the value of $\lambda$ is equal to :

• A. 1
• B. -1
• C. 0
• D. does not exist

Answer 1

Answer: (A)

Note: A function $f ( x )$ is said to be continuous at
$x = a$ iff. $R.H . L = L . H . L.= f ( a )$
i.e., $\displaystyle\lim _{x \rightarrow a^{+}} f(x)=\displaystyle\lim _{x \rightarrow a^{-}} f(x)=\displaystyle\lim _{x \rightarrow a} f(x)$
$f(x) = \begin{cases} x+\lambda &x < 3 \\[2ex] 4 & x = 3\\[2ex] 3x - 5 , & x > 3 \end{cases}$
continuous at $x =3$
$\therefore \displaystyle\lim _{x \rightarrow 3^{-}}(x+\lambda)=f(3)=\displaystyle\lim _{x \rightarrow 3^{+}}(3 x-5)$
$\Rightarrow 3+\lambda=4 $
$\Rightarrow \lambda=4-3=1$