Limit Continuity and Differentiability 2 Question 1
1. Let $f: R \rightarrow R$ be a differentiable function satisfying $f^{\prime}(3)+f^{\prime}(2)=0$. Then $\lim _{x \rightarrow 0} \frac{1+f(3+x)-f(3)}{1+f(2-x)-f(2)}^{\frac{1}{x}}$ is equal
to
(a) $e$
(b) $e^{-1}$
(c) $e^{2}$
(d) 1
(2019 Main, 8 April II)
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Answer:
Correct Answer: 1. (d)
Solution:
- Given $\alpha$ and $\beta$ are roots of quadratic equation $375 x^{2}-25 x-2=0$
$\therefore \quad \alpha+\beta=\frac{25}{375}=\frac{1}{15}$
and $\quad \alpha \beta=-\frac{2}{375}$
Now, $\lim _{n \rightarrow \infty} \sum _{r=1}^{n} \alpha^{r}+\lim _{n \rightarrow \infty} \sum _{r=1}^{n} \beta^{r}$
$=\left(\alpha+\alpha^{2}+\alpha^{3}+\ldots+\right.$ upto infinite terms $)+$
$\left(\beta+\beta^{2}+\beta^{3}+\ldots\right.$ +upto infinite terms)
$=\frac{\alpha}{1-\alpha}+\frac{\beta}{1-\beta} \quad \because S _{\infty}=\frac{a}{1-r}$ for GP
$=\frac{\alpha(1-\beta)+\beta(1-\alpha)}{(1-\alpha)(1-\beta)}=\frac{\alpha-\alpha \beta+\beta-\alpha \beta}{1-\alpha-\beta+\alpha \beta}$
$=\frac{(\alpha+\beta)-2 \alpha \beta}{1-(\alpha+\beta)+\alpha \beta}$
On substituting the value $\alpha+\beta=\frac{1}{15}$ and $\alpha \beta=\frac{-2}{375}$ from
Eqs. (i) and (ii) respectively,
we get
$$ =\frac{\frac{1}{15}+\frac{4}{375}}{1-\frac{1}{15}-\frac{2}{375}}=\frac{29}{375-25-2}=\frac{29}{348}=\frac{1}{12} $$
$\sec ^{2} x$
