No products in the cart.

Ask questions which are clear, concise and easy to understand.

Ask Question
  • 0 answers
  • 0 answers
  • 0 answers
  • 1 answers

Makhan Meena 1 year, 3 months ago

Chapter 2 solutions
  • 0 answers
  • 0 answers
  • 1 answers

Saksham Sharma 1 year, 4 months ago

Hh
  • 0 answers
  • 1 answers

Preeti Dabral 1 year, 8 months ago

The Venn diagram for {tex}(A \cap B)'{/tex}The shaded portion represents {tex}(A \cap B)'{/tex}

  • 0 answers
  • 0 answers
  • 1 answers

Manmohan Barmaiya 1 year, 11 months ago

G
  • 1 answers

Preeti Dabral 1 year, 11 months ago

{tex}\begin{aligned} & \text { Let } y=f(x)=\operatorname{cosec} x \\ & \therefore f(x+h)=\operatorname{cosec}(x+h) \\ & \therefore \frac{d y}{d t}=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ & =\lim _{h \rightarrow 0} \frac{\operatorname{cosec}(x+h)-\operatorname{cosec} x}{h} \\ & =\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{1}{\sin (x+h)}-\frac{1}{\sin x}\right] \\ & =\lim _{h \rightarrow 0} \frac{\sin x-\sin (x+h)}{h \cdot \sin x \cdot \sin (x+h)} \\ & =\lim _{h \rightarrow 0} \frac{2 \cos \left(\frac{2 x+h}{2}\right) \sin \left(-\frac{h}{2}\right)}{h \cdot \sin x \sin (x+h)} \\ & =-\lim _{h \rightarrow 0} \frac{\cos \left(x+\frac{h}{h}\right)}{\sin x \cdot \sin (u+h)} \cdot \lim _{h \rightarrow 0} \frac{\sin \frac{h}{2}}{h / 2} \\ & =-\frac{1 \cos x}{\sin x \cdot \sin x} \cdot \lim _{z \rightarrow 0} \frac{\sin z}{z} \\ & {\left[z=\frac{h}{2} \text {; Then, } z \rightarrow 0 \text { when } \Rightarrow h \rightarrow 0\right]} \\ & =\frac{-\cos x}{\sin x \cdot \sin x} \cdot 1 \\ & =-\frac{\cos x}{\sin x} \cdot \frac{1}{\sin x} \\ & =-\operatorname{cosec} x . \cot x \\ & \therefore \frac{d y}{d x}=-1 \operatorname{csec} x \cdot \cot x \\ & \end{aligned}{/tex}

  • 1 answers

Abhishek Namdev 1 year, 11 months ago

(x = 2x-1),(y+1 = -5) x = 1 , y = -6
  • 1 answers

Preeti Dabral 2 years ago

Here total number of digits = 10
Number of digits used (no digit is repeated) = 4
Since 0 cannot be filled in the fourth place, so number of permutations for fourth place = 9
Now the remaining three places can be filled with 9 digits.
{tex}\therefore {/tex} Number of permutations = {tex}^9{P_3}{/tex}
{tex}= \frac{{9!}}{{6!}} = \frac{{9 \times 8 \times 7 \times 6!}}{{6!}} = 504{/tex}
Hence total number of permutations {tex}= 9 \times 504 = 4536{/tex}

1+2
  • 1 answers

Nilesh Yadav 1 year, 9 months ago

3

myCBSEguide App

myCBSEguide

Trusted by 1 Crore+ Students

Test Generator

Test Generator

Create papers online. It's FREE.

CUET Mock Tests

CUET Mock Tests

75,000+ questions to practice only on myCBSEguide app

Download myCBSEguide App