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Preeti Dabral 2 years, 11 months ago
Since {tex}\alpha \text { and } \beta{/tex} are the zeroes of polynomial 3x2 + 2x + 1.
Hence, {tex}\alpha + \beta = - \frac { 2 } { 3 }{/tex}
and {tex}\alpha \beta = \frac { 1 } { 3 }{/tex}
Now, for the new polynomial,
Sum of zeroes = {tex}\frac { 1 - \alpha } { 1 + \alpha } + \frac { 1 - \beta } { 1 + \beta }{/tex}
{tex}= \frac { ( 1 - \alpha + \beta - \alpha \beta ) + ( 1 + \alpha - \beta - \alpha \beta ) } { ( 1 + \alpha ) ( 1 + \beta ) }{/tex}
{tex}= \frac { 2 - 2 \alpha \beta } { 1 + \alpha + \beta + \alpha \beta } = \frac { 2 - \frac { 2 } { 3 } } { 1 - \frac { 2 } { 3 } + \frac { 1 } { 3 } }{/tex}
{tex}\therefore{/tex} Sum of zeroes = {tex}\frac { 4 / 3 } { 2 / 3 } = 2{/tex}
Product of zeroes = {tex}\left[ \frac { 1 - \alpha } { 1 + \alpha } \right] \left[ \frac { 1 - \beta } { 1 + \beta } \right]{/tex}
{tex}= \frac { ( 1 - \alpha ) ( 1 - \beta ) } { ( 1 + \alpha ) ( 1 + \beta ) }{/tex}
{tex}=\frac{1-(\alpha+\beta)+\alpha \beta}{1+(\alpha+\beta)+\alpha \beta}{/tex}
{tex}{/tex} {tex}={/tex} {tex}\frac { 1 + \frac { 2 } { 3 } + \frac { 1 } { 3 } } { 1 - \frac { 2 } { 3 } + \frac { 1 } { 3 } } = \frac { \frac { 6 } { 3 } } { \frac { 3 } { 3 } } = 3{/tex}
Hence, Required polynomial = x2 - (Sum of zeroes)x + Product of zeroes
= x2 - 2x + 3
Posted by Nayana M S 2 years, 11 months ago
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Preeti Dabral 2 years, 11 months ago
To find the AP if the sum of first n terms of arithmetic progression is 210 and sum of its first (n-1) terms is 171. If the first term is 3 .
As we know that if AP has n terms,then after removing (n-1) terms,we are left with the last term.
ACF, last term is 210-171
l= 39
a= 3

Posted by Shreya M 2 years, 11 months ago
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Preeti Dabral 2 years, 11 months ago
"अटूट ध्वनि में एक स्वर से दूसरे स्वर तक जाने को मींड कहते हैं" मींड को लेते समय स्वरों को इस प्रकार स्पर्श करते हैं की वे अलग - अलग सुनाई नहीं देते। इसको गाने बजाने में लोच और रंजकता आती हैं। मींड को दर्शाने के लिए स्वरों के ऊपर अर्ध - चंद्राकार बनाते है.
3Thank You