Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Continuity and Differentiability

Question:

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:

Options:

(A)-(IV), (B)-(III), (C)-(II), (D)-(I)

(A)-(III), (B)-(IV), (C)-(I), (D)-(II)

(A)-(III), (B)-(IV), (C)-(II), (D)-(I)

(A)-(I), (B)-(III), (C)-(IV), (D)-(II)

Correct Answer:

(A)-(III), (B)-(IV), (C)-(II), (D)-(I)

Explanation:

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)