SATHEE: Limit Continuity and Differentiability 2 Question 1

Limit Continuity and Differentiability 2 Question 1

1. Let $f : R \to R$ be a differentiable function satisfying $f^{'} (3) + f^{'} (2) = 0$ . Then $lim_{x \to 0} {\frac{1 + f (3 + x) - f (3)}{1 + f (2 - x) - f (2)}}^{\frac{1}{x}}$ is equal

(a) $e$

(b) $e^{- 1}$

(d) 1

(2019 Main, 8 April II)

Show Answer

Answer:

Correct Answer: 1. (d)

Solution:

Given $α$ and $β$ are roots of quadratic equation $375 x^{2} - 25 x - 2 = 0$

$∴ α + β = \frac{25}{375} = \frac{1}{15}$

and $α β = - \frac{2}{375}$

Now, $lim_{n \to \infty} \sum_{r = 1}^{n} α^{r} + lim_{n \to \infty} \sum_{r = 1}^{n} β^{r}$

$= (α + α^{2} + α^{3} + \dots +$ upto infinite terms $) +$

$(β + β^{2} + β^{3} + \dots$ +upto infinite terms)

$= \frac{α}{1 - α} + \frac{β}{1 - β} ∵ S_{\infty} = \frac{a}{1 - r}$ for GP

$= \frac{α (1 - β) + β (1 - α)}{(1 - α) (1 - β)} = \frac{α - α β + β - α β}{1 - α - β + α β}$

$= \frac{(α + β) - 2 α β}{1 - (α + β) + α β}$

On substituting the value $α + β = \frac{1}{15}$ and $α β = \frac{- 2}{375}$ from

Eqs. (i) and (ii) respectively,

we get

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

$\sec^{2} x$

Limit Continuity and Differentiability 2 Question 1

1. Let $f : R \to R$ be a differentiable function satisfying $f^{'} (3) + f^{'} (2) = 0$ . Then $lim_{x \to 0} {\frac{1 + f (3 + x) - f (3)}{1 + f (2 - x) - f (2)}}^{\frac{1}{x}}$ is equal

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Limit Continuity and Differentiability 2 Question 1

1. Let f:R→R be a differentiable function satisfying f′(3)+f′(2)=0. Then limx→01+f(3+x)−f(3)1+f(2−x)−f(2)1x is equal

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu

1. Let $f : R \to R$ be a differentiable function satisfying $f^{'} (3) + f^{'} (2) = 0$ . Then $lim_{x \to 0} {\frac{1 + f (3 + x) - f (3)}{1 + f (2 - x) - f (2)}}^{\frac{1}{x}}$ is equal