Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Applications of Derivatives

Question:

Match List-I with List-II

List-I

List-II

(A) Point of minima of $f(x) = |x+1|$

(I) 1

(B) Minimum value of $f(x) = |x|$

(II) -1

(C) Maximum value of $f(x) = 1-x^2$

(III) 2

(D) Minimum value of $f(x) = 2+ \sin^2x$

(IV) 0

Choose the correct answer from the options given below:

Options:

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

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

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

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

Correct Answer:

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

Explanation:

The correct answer is Option (2) → (A)-(II), (B)-(IV), (C)-(I), (D)-(III)

List-I

List-II

(A) Point of minima of $f(x) = |x+1|$

(II) -1

(B) Minimum value of $f(x) = |x|$

(IV) 0

(C) Maximum value of $f(x) = 1-x^2$

(I) 1

(D) Minimum value of $f(x) = 2+ \sin^2x$

(III) 2

(A) Point of minima of $f(x)=|x+1|$

Minimum occurs when $x+1=0$

$x=-1$

$(A)\rightarrow(II)$

(B) Minimum value of $f(x)=|x|$

Minimum value is $0$

$(B)\rightarrow(IV)$

(C) Maximum value of $f(x)=1-x^2$

Maximum occurs at $x=0$

Value $=1$

$(C)\rightarrow(I)$

(D) Minimum value of $f(x)=2+\sin^2 x$

Minimum of $\sin^2 x$ is $0$

Minimum value $=2$

$(D)\rightarrow(III)$

Final Matching: (A)-(II), (B)-(IV), (C)-(I), (D)-(III).