If the function \( f(x)=\begin{cases}\displaystyle\frac{x^3-1}{x-1} & ,\ x\lt 1\\a & ,\ x=1\\\displaystyle\frac{b\sin(x-1)}{x-1} & ,\ x\gt 1\end{cases} \) is continuous at \( x=1 \) , then which of the following statements are correct ?
A. a = b
B. a + b is a multiple of 4
C. 2a – b is a multiple of 3
D. a + 4b is divisible by 5
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A, C and D only
For continuity at x = 1, the left-hand limit, f(1) = a and the right-hand limit must be equal.
Left-hand limit, factorising \( x^3-1=(x-1)(x^2+x+1) \):
\( \displaystyle \lim_{x\to 1^-}\frac{x^3-1}{x-1}=\lim_{x\to 1^-}(x^2+x+1) \)
\( \displaystyle =3 \)
Right-hand limit, using \( \lim_{h\to 0}\frac{\sin h}{h}=1 \) with h = x − 1:
\( \displaystyle \lim_{x\to 1^+}\frac{b\sin(x-1)}{x-1}=b \)
So a = 3 and b = 3, which give a + b = 6, 2a − b = 3 and a + 4b = 15.
Checking the statements
(A) a = b — Correct. Both are 3.
(B) a + b is a multiple of 4 — Incorrect. a + b = 6.
(C) 2a – b is a multiple of 3 — Correct. 2a − b = 3.
(D) a + 4b is divisible by 5 — Correct. a + 4b = 15.