+91-8076753736 support@wiredfaculty.com
logo
logo

A PHP Error was encountered

Severity: Notice

Message: Undefined variable: temp_qds

Filename: Questions_Page/Ncert_Question.php

Line Number: 320

Backtrace:

File: /home/wiredfa1/public_html/application/views/final/Questions_Page/Ncert_Question.php
Line: 320
Function: _error_handler

File: /home/wiredfa1/public_html/application/controllers/Home.php
Line: 235
Function: view

File: /home/wiredfa1/public_html/index.php
Line: 315
Function: require_once

Sponsor Area

Continuity And Differentiability

Question
CBSEENMA12035449
Wired Faculty App

If space space space straight y equals left parenthesis sin to the power of negative 1 end exponent straight x right parenthesis squared comma space prove space that space left parenthesis 1 minus straight x 2 right parenthesis straight y subscript 2 minus straight x space straight y subscript 1 minus 2 equals 0. space

Solution
space straight y equals left parenthesis sin to the power of negative 1 end exponent straight x right parenthesis squared space space space space rightwards double arrow space space straight y subscript 1 equals 2 space sin to the power of negative 1 end exponent straight x. space fraction numerator 1 over denominator square root of 1 minus straight x squared end root end fraction therefore space square root of 1 minus straight x squared end root space straight y subscript 1 equals 2 space sin to the power of negative 1 end exponent straight x space space space rightwards double arrow space left parenthesis 1 minus straight x squared right parenthesis straight y subscript 1 squared equals 4 left parenthesis sin to the power of negative 1 end exponent straight x right parenthesis squared therefore space space space left parenthesis 1 minus straight x squared right parenthesis straight y subscript 1 squared equals 4 space straight y space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space left square bracket because space straight y equals left parenthesis sin to the power of negative 1 end exponent straight x right parenthesis squared right square bracket Differentiating space both space sides space straight w. straight r. straight t. straight x comma space we space get comma left parenthesis 1 minus straight x squared right parenthesis. space 2 space straight y subscript 1 space straight y subscript 2 plus straight y 1 squared left parenthesis negative 2 space straight x right parenthesis equals 4 space straight y subscript 1 space space Dividing space both space sides space by space 2 space straight y 1 comma space we space get comma space left parenthesis 1 minus straight x squared right parenthesis straight y subscript 2 minus straight x space straight y subscript 1 equals 2 space or space left parenthesis 1 minus straight x squared right parenthesis straight y subscript 2 space minus straight x space straight y subscript 1 minus 2 equals 0

Some More Questions From Continuity and Differentiability Chapter

Mock Test Series

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation