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Posted by Khushi Patel 7 years, 6 months ago
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Posted by Subhashree Behera 7 years, 6 months ago
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Rupesh Kumar Singh 7 years, 6 months ago
Let x be an odd integer.
On dividing x by 4, let q be the quotient and r be the remainder.
So, by Euclid's division lemma, we have:
x = 4q + r, where 0 <_ r < 4.
Therefore, x2 = ( 4q + r )2 .
= 16q2 + r2 + 8qr .....(i).where 0 <_ r < 4.
Case l. When r = 0
Putting r = 0 in (I), we get:
x2 = 16q2 = 4 ( 4q2 ) = 4 Q, where Q = 4q2 is an integer.
Case ll. When r = 1
Putting r = 1 in (I), we get:
x2 = ( 16q2 + 1 + 8q ) = 4 ( 4q2 + 2q ) + 1 = ( 4 Q + 1),
where Q = ( 4q2 + 2q ) is an integer.
Clearly, 4Q is even and since x is odd. So x is not equal to 4Q.
Hence, the square of an odd integer is of form ( 4Q + 1) for some integer Q.
Posted by Munish Kumar 7 years, 6 months ago
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Posted by Jasmeen Kaur 7 years, 6 months ago
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Posted by Mamon Nath 7 years, 6 months ago
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Posted by Diana Charles 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
We have, 4 tan {tex}\theta{/tex} = 3
{tex}\Rightarrow{/tex} tan {tex}\theta{/tex} = {tex}\frac{3}{4}{/tex}
{tex}\because sec\theta=\sqrt{1+tan^2\theta}{/tex} = {tex}\sqrt{1+\frac{9}{16}}{/tex} ={tex}\sqrt{\frac{16+9}{16}} {/tex} {tex} = \sqrt{\frac{25}{16}}{/tex} {tex} = \frac{5}{4}{/tex}
Now, {tex}\frac{4 \sin \theta-\cos \theta+1}{4 \sin \theta+\cos \theta-1}{/tex}
= {tex}\frac{cos\theta(4 \frac{\sin \theta}{cos\theta}-\frac{\cos \theta}{cos\theta}+\frac{1}{cos\theta})}{cos\theta(4 \frac{\sin \theta}{{cos\theta}{}}+\frac{\cos \theta}{cos\theta}-\frac{1}{cos\theta})}{/tex}
= {tex}\frac{4tan\theta-1+sec\theta}{4tan\theta+1-sec\theta}{/tex}
Substituting the values, we get,
={tex}\frac{4(\frac{3}{4})-1+(\frac{5}{4})}{4(\frac{3}{4})+1-(\frac{5}{4})}{/tex} = {tex}\frac{3-1+(\frac{5}{4})}{3+1-(\frac{5}{4})}{/tex} = {tex}\frac{2+(\frac{5}{4})}{4-(\frac{5}{4})}{/tex} = {tex}\frac{\frac{8+5}{4}}{\frac{16-5}{4}}{/tex} = {tex}\frac{\frac{13}{4}}{\frac{11}{4}}{/tex} = {tex}{\frac{13}{4}}\times{\frac{4}{11}}{/tex} = {tex}\frac{13}{11}{/tex}
Posted by Sakshi Singh 7 years, 6 months ago
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Posted by Sakshi Singh 7 years, 6 months ago
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Posted by Shramana Nandy 7 years, 6 months ago
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Posted by Jasmeen Kaur 7 years, 6 months ago
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Ragul Sv 7 years, 6 months ago
Shrajjal Prakash 7 years, 6 months ago
Posted by Himani Patidar 7 years, 6 months ago
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Posted by Himani Patidar 7 years, 6 months ago
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Posted by Ayush Toppo 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Here the given polynomial f(x)=2x2 + 2ax + 5x +10
If x+a is a factor of f(x) then x+a=0 or x = -a
{tex}\Rightarrow{/tex}f(-a) = 0
{tex}\Rightarrow{/tex}2(-a)2 + 2a.(-a) + 5(-a) + 10 =0
{tex}\Rightarrow{/tex}2a2 - 2a2 -5a +10 =0
{tex}\Rightarrow{/tex}-5a = -10
{tex}\Rightarrow{/tex}a=2.
Posted by Kusum Lohar 7 years, 6 months ago
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Priyanshu Kumar 7 years, 6 months ago
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Amit Kumar 7 years, 6 months ago
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