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:

  1. 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

$$ =\frac{\frac{1}{15}+\frac{4}{375}}{1-\frac{1}{15}-\frac{2}{375}}=\frac{29}{375-25-2}=\frac{29}{348}=\frac{1}{12} $$

$\sec ^{2} x$



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