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 :
Answer & explanation
Correct answer: option 4
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)
(C) $f(0)=\frac{5}{2}$
$\underset{x→0}{\lim}3k=3k$
$⇒k=\frac{5}{6}$ (IV)
(D) $f(-2)=-3$
$\underset{x→0}{\lim}(x+k)=k-2$
so $k-2=-3⇒k=-1$ (II)