Given that the function f(x) is continuo

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Target Exam

CUET

Subject

-- Mathematics - Section A

Chapter

Continuity and Differentiability

Question:

Given that the function f(x) is continuous on R. Match List-I with List-II.

List-I		List-II
(A)	$f(x)=\left\{\begin{matrix}kx^2, & if & x < 3\\3, &if & x ≥ 3\end{matrix}\right.$	(I)	$k=\frac{5}{4}$
(B)	$f(x)=\left\{\begin{matrix}kx+1, & if & x ≥ 4\\x+2 &if & x < 4\end{matrix}\right.$	(II)	$k=-1$
(C)	$f(x)=\left\{\begin{matrix}\frac{5+x}{2}, & if & x = 0\\3k, &if & x ≠0\end{matrix}\right.$	(III)	$k=\frac{1}{3}$
(D)	$f(x)=\left\{\begin{matrix}x+k, & if & x >-2\\-3, &if & x ≤-2 \end{matrix}\right.$	(IV)	$k=\frac{5}{6}$

Choose the correct answer from the options given below :

Options:

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

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

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

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

Correct Answer:

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

Explanation:

The correct answer is Option (4) → (A)-(III), (B)-(I), (C)-(IV), (D)-(II)

(A) $f(3)=3$

$\underset{x→3}{\lim}kx^2=9k$

$9k=3⇒k=\frac{1}{3}$ (III)

(B) $f(4)=4k+1$

$\underset{x→4}{\lim}(x+2)=6$

$6=4k+1⇒k=\frac{5}{4}$ (I)

$\underset{x→0}{\lim}3k=3k$

$⇒k=\frac{5}{6}$ (IV)

(D) $f(-2)=-3$

$\underset{x→0}{\lim}(x+k)=k-2$

so $k-2=-3⇒k=-1$ (II)