Match List-I with List-II
|
List-I Function |
List-II Points of discontinuity |
|
(A) $f(x) =\frac{x^2+1}{x}$ |
(I) $x=4$ |
|
(B) $f(x)=\frac{|x-1|}{x-1}$ |
(II) $x = 2$ |
|
(C) $f(x)=\left\{\begin{matrix}x-1,&x<2\\x+1,&x≥2\end{matrix}\right.$ |
(III) $x = 0$ |
|
(D) $f(x)=\frac{1-x}{(x-4)}$ |
(IV) $x = 1$ |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
|
List-I Function |
List-II Points of discontinuity |
|
(A) $f(x) =\frac{x^2+1}{x}$ |
(III) $x = 0$ |
|
(B) $f(x)=\frac{|x-1|}{x-1}$ |
(IV) $x = 1$ |
|
(C) $f(x)=\left\{\begin{matrix}x-1,&x<2\\x+1,&x≥2\end{matrix}\right.$ |
(II) $x = 2$ |
|
(D) $f(x)=\frac{1-x}{(x-4)}$ |
(I) $x=4$ |
(A) $f(x)=\frac{x^{2}+1}{x}$
Undefined at $x=0$ → point of discontinuity $x=0$ → (III)
(B) $f(x)=\frac{|x-1|}{x-1}$
Undefined at $x=1$ → discontinuous at $x=1$ → (IV)
(C) $f(x)=\begin{cases}x-1,&x<2\\x+1,&x\ge2\end{cases}$
Left limit at $x=2$: $2-1=1$; right limit: $2+1=3$ → not equal → discontinuous at $x=2$ → (II)
(D) $f(x)=\frac{1-x}{x-4}$
Denominator zero at $x=4$ → discontinuous at $x=4$ → (I)
Matching:
(A) → (III), (B) → (IV), (C) → (II), (D) → (I)