What are the concepts covered in Class 11 Limits and Derivatives NCERT Solutions?

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9 months ago

The NCERT Solutions for Class 11 Limits and Derivatives provided by shiksha, cover all important concepts, including the definition of a limit, techniques to evaluate limits, left-hand and right-hand limits, and derivatives. Our NCERT Solutions provide step-by-step explanations to questions . Our S

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f(1) = 2, g(1) = 2, f'(x), g'(x) exist, then the value of lim(x→1) [f(1)g(x)-f(1)-g(1)f(x)+g(1)] / [f(1)g(x)-f(x)g(1)] is __________

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The value of lim(n→∞) [ Σ(k=1 to n) (2k-1) + 8n ] / [ Σ(k=1 to n) (2k-1) + 4n ] is equal to:


Vishal Baghel

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The question is likely a Riemann sum.
lim (n→∞) (1/n) Σ [ (2k/n - 1/n) / (2k/n - 1/n + 4) ]
This is too complex. Let's follow the image solution.
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Let the lines y + 2x = $\sqrt[]{11} + 7 \sqrt[]{7}$ and $2 y + x = 2 \sqrt[]{11} + 6 \sqrt[]{7}$ be normal to a circle C : ${(x - h)}^{2} + {(y - k)}^{2} = r^{2} .$ If the line $\sqrt[]{11} y - 3 x = \frac{5 \sqrt[]{77}}{3} + 11$ is tangent to the circle C, then the value of ${(5 h - 8 k)}^{2} + 5 r^{2}$ is equal to

Payal Gupta

$y + 2 x = \sqrt[]{11} + 7 \sqrt[]{7}$ ……… (i)

$2 y + x = 2 \sqrt[]{11} + 6 \sqrt[]{7}$ ……… (ii)

$\Rightarrow x + y = \sqrt[]{11} + \frac{13}{3} \sqrt[]{7}$ ……… (iii)

Centre of the circle given by solving (i) & (ii)

$a s (\frac{8 \sqrt[]{7}}{3}, \sqrt[]{11} + \frac{5 \sqrt[]{7}}{3})$

Again $\sqrt[]{11} y - 3 x = \frac{5 \sqrt[]{77}}{3}$ is tangent to the circle.

$∴ r = | \frac{\sqrt[]{11} (\sqrt[]{11} + \frac{5 \sqrt[]{7}}{3}) - 8 \sqrt[]{7} - \frac{5 \sqrt[]{77}}{3}}{\sqrt[]{11 + 9}} | = | \frac{11 - 8 \sqrt[]{7}}{\sqrt[]{20}} |$

$∴ {(5 h - 8 k)}^{2} + 5 r^{2} = 816$

Kindly consider the following

$\underset{x \to 0}{l i m} \frac{{(1 + x)}^{n} - 1}{x}$ is

(a) $n$

(b)   1

(c)   – $n$

(d)   0

Alok Kumar Singh

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 0}{l i m} \frac{{(1 + x)}^{n} - 1}{x} \\ = \underset{x \to 0}{l i m} \frac{{(1 + x)}^{n} - {(1)}^{n}}{(1 + x) - (1)} = \underset{1 + x \to 1}{l i m} \frac{{(1 + x)}^{n} - {(1)}^{n}}{(1 + x) - (1)} = n {(1)}^{n - 1} = n [? \underset{x \to a}{l i m} \frac{x^{n} - a^{n}}{x - a} = n . a^{n - 1}] \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

Kindly consider the following
$l i m_{x \to 3} \frac{x}{[x]} =_________$ .

Alok Kumar Singh

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n \underset{x \to 3^{+}}{l i m} \frac{x}{[x]} \\ = \underset{h \to 0}{l i m} \frac{(3 + h)}{[3 + h]} = 1 \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s 1 . \end{array}$