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### $3^a=5^b=225$, $\frac {ab}{a+b}=?$

While trying to find $\frac {ab}{a+b}$ given $3^a=5^b=225$, a student follows the method below:

$3^a=5^b=225=9\times 25=3^2\times 5^2$

So, $3^a\times 5^b=(3^2\times 5^2)\times (3^2\times 5^2)$

Or, $3^a\times 5^b=3^4\times 5^4$

Since 3 & 5 are prime numbers, a=4 & b=4

Now, $\frac {ab}{a+b}=\frac {4\times 4}{4+4}=2$

Considering above solution, select correct options. More that 1 option may be correct.

(A) a = 4 is correct
(B) b = 4 is correct
(C) $\frac {ab}{a+b}=2$ is correct
(D) $\frac {1}{a}+\frac {1}{b}=\frac {1}{2}$ is correct

Key: (C) & (D)

### $f(x)=x^6+2x^4+x^3+2x+3 $$\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$$n=?$

Let $f(x)=x^6+2x^4+x^3+2x+3,x \in R$. Then the natural number n for which $\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$ is _ _ _ _ . Solution Since the limit has $\left[ {\frac{0}{0}} \right]$ form, L.H. Rule is applicable. Thus, $\mathop {\lim }\limits_{x \to 1} n{x^{n - 1}}f(1) - f'(x) = 44$ $\therefore nf(1) - f'(1) = 44$ $\therefore n.9 - ({6.1^5} + {8.1^3} + {3.1^2} + 2.1) = 44$ $\Rightarrow 9n - 19 = 44$ $\Rightarrow n=7$