If {f}(x)={10^{x}+7^{x}-14^{x}-5^{x}}{1- x}, x ≠ 0 is continuous at x=0, then the value of {f}(0) is

Correct Answer is : log⁡4[log⁡(57)] 4≤ft[ ≤ft(5/7)]log4[log(75)]

Correct Answer is : log⁡4[log⁡(57)] 4≤ft[ ≤ft(5/7)]log4[log(75)]

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MHTCET 23 Apr 2025 Shift 1 Paper

Question : 109 of 150

Marks: +1, -0

\mathrm{f}(x)=\frac{10^{x}+7^{x}-14^{x}-5^{x}}{1-\cos x}, x \neq 0

x=0

\mathrm{f}(0)

Solution:

Concept:

Since

f(x)

is continuous at

x=0

, we must have

f(0)=\lim_{x\to 0} f(x)

Explanation:

Rewrite

10^x

and

14^x

using prime factors:

10^x = 2^x 5^x,\quad 14^x = 2^x 7^x

So the numerator becomes:

2^x5^x + 7^x - 2^x7^x - 5^x = (2^x-1)(5^x-7^x)

Now divide numerator and denominator by

x^2

to use standard limits:

\lim_{x\to0}\frac{2^x-1}{x}=\log 2,\quad \lim_{x\to0}\frac{5^x-1}{x}=\log 5,\quad \lim_{x\to0}\frac{7^x-1}{x}=\log 7

Also,

\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac{1}{2}

Therefore,

f(0)=\lim_{x\to0}\frac{(2^x-1)(5^x-7^x)}{1-\cos x}

=\frac{(\log 2)(\log 5-\log 7)}{\frac{1}{2}}

=2\log 2\log\left(\frac{5}{7}\right)

Since

2\log 2=\log 4

, we get:

f(0)=\log 4\log\left(\frac{5}{7}\right)

Answer:

f(0)=\log 4\left[\log\left(\frac{5}{7}\right)\right]

Therefore, the correct option is B.

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