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Question: The function f(x) = \(\frac{\log(1 + ax) - \log ⥄ (1 - bx)}{x}\) is not defined at x = 0. The value ...

The function f(x) = log⁡(1+ax)−log⁡⥄(1−bx)x\frac{\log(1 + ax) - \log ⥄ (1 - bx)}{x} is not defined at x = 0. The value which should be assigned to f at x = 0, so that it is continuous at x = 0 is

A

a – b

B

a + b

C

log a + log b

D

None of these

Answer

a + b

Explanation

Solution

f(x) 6mu= 6mua[log⁡ 6mu(1+ax)ax] 6mu+b[log⁡(1−bx)(−bx)]f(x)\mspace{6mu} = \mspace{6mu} a\left\lbrack \frac{\log\mspace{6mu}(1 + ax)}{ax} \right\rbrack\mspace{6mu} + b\left\lbrack \frac{\log(1 - bx)}{( - bx)} \right\rbrack

so that lim⁡x→0f(x)=a. 6mu1+b.1=a+b 6mu 6mu=f(0)\lim_{x \rightarrow 0}f(x) = a.\mspace{6mu} 1 + b.1 = a + b\mspace{6mu}\mspace{6mu} = f(0).

[lim⁡x→0 6mulog⁡(1+x)x 6mu=1]\left\lbrack \lim_{x \rightarrow 0}\mspace{6mu}\frac{\log(1 + x)}{x}\mspace{6mu} = 1 \right\rbrack

Alternative Solution:

lim⁡x→0a−abx+b+abx(1+ax)(1−bx)\lim_{x \rightarrow 0}\frac{a - abx + b + abx}{(1 + ax)(1 - bx)}= a + b

(by L’Hospital’s Rule)

⇒ f(0) = a + b, if f is continuous