Are NCERT Solutions enough to prepare Limits and Derivatives for Class 11 exams?

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9 months ago

Yes, NCERT Solutions for Limits and Derivatives are a good resource and sufficeint to build a strong foundation. Students are aware all of the class 11 maths exams will be taken based on Class 11 CBSE Exams. Our Class 11 Math Limits and Derivatives Solutions are aligned with the latest CBSE 2025 sy

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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 __________

Vishal Baghel

lim (x→1) [f (1)g (x)-f (1)-g (1)f (x)+g (1)] / [f (1)g (x)-f (x)g (1)], form: 0/0
lim (x→1) [f (1)g' (x)-g (1)f' (x)] / [f (1)g' (x)-f' (x)g (1)] = 1

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

lim (n→∞) [n² + 8n] / [n² + 4n] = 1.
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.
lim (n→∞) (1/n) Σ [ 2 (k/n) + 8 ] / [ 2 (k/n) + 4 ]
∫? ¹ (2x+8)/ (2x+4) dx = ∫? ¹ (1 + 4/ (2x+4) dx = [x + 2ln|2x+4|]? ¹
=

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:

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