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Ask QuestionPosted by Shashank Shekhar 7 years, 5 months ago
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Posted by Uttam Shetty 7 years, 5 months ago
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Posted by Wahith Meeran 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
We have,
f(x) = 4x2 - 4x + 1
= 4x2 - 2x -2x + 1
= 2x (2x - 1) - 1 (2x - 1)
= (2x - 1)(2x - 1)
Now f(x)=0 if 2x-1 =0
hence zeros of f(x) are {tex}\frac12{/tex} and {tex}\frac12{/tex}
sum of zeros = {tex} \frac { 1 } { 2 } + \frac { 1 } { 2 } = 1 {/tex}
product of zeros = {tex}\frac12{/tex} {tex}\times \frac12 =\frac14 {/tex}
In f(x) = 4x2 - 4x + 1
a=4, b=-4,c=1
{tex}\style{font-family:Arial}{\style{font-size:14px}{\begin{array}{l}\text{sum of zeros=-}\frac ba=-\frac{-4}4=1\\\text{Product of zeros=}\frac ca=\frac14\end{array}}}{/tex}
Hence the relation of zeros and coefficients is verified.
Posted by Sargun Kaur 7 years, 5 months ago
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Wahith Meeran 7 years, 5 months ago
Posted by Raghul M 7 years, 5 months ago
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Posted by Akash Goswami 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
f(x) = 6x2 + x - 2
a = 6, b = 1, c = -2
Let zeroes be {tex}\alpha{/tex} and β.Then
Sum of zeroes= {tex}\alpha{/tex} + β {tex}=\;-\frac ba\;=-\frac16{/tex}
Product of zeroes {tex}\alpha{/tex}× β {tex}=\;\;\frac ca\;=\;\frac{-2}6\;=\;-\frac13{/tex}
{tex}\frac { \alpha } { \beta } + \frac { \beta } { \alpha } = \frac { \alpha ^ { 2 } + \beta ^ { 2 } } { \alpha \beta }{/tex}
{tex}= \frac { ( \alpha + \beta ) ^ { 2 } - 2 \alpha \beta } { \alpha \beta } \left[ \because ( \alpha + \beta ) ^ { 2 } = \alpha ^ { 2 } + \beta ^ { 2 } + 2 \alpha \beta \right]{/tex}
{tex}= \frac { \left[- \frac { 1 } { 6 } \right] ^ { 2 } - 2 \left[ - \frac { 1 } { 3 } \right] } { \left[ - \frac { 1 } { 3 } \right] }{/tex}
{tex}= \frac { \frac { 1 } { 36 } + \frac { 2 } { 3 } } { - \frac { 1 } { 3 } }{/tex}
{tex}= \frac { \frac { 1 + 24 } { 36 } } { - \frac { 1 } { 3 } }{/tex}
{tex}= \frac { 25 } { 36 } \times \frac { - 3 } { 1 }{/tex}
{tex}= \frac { - 25 } { 12 }{/tex}
Posted by Shikayna Panday 7 years, 5 months ago
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Khushi Chahal 7 years, 5 months ago
Posted by Aayush Jaiswal 7 years, 5 months ago
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Posted by Vikash Kumar 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Let a be the first term and d be the common difference of the given AP. Then,
{tex}T _ { 7 } = \frac { 1 } { 9 } \Rightarrow a + 6 d = \frac { 1 } { 9 }{/tex} ...(i)
{tex}T _ { 9 } = \frac { 1 } { 7 } \Rightarrow a + 8 d = \frac { 1 } { 7 }{/tex} ...(i)
On subtracting (i) from (ii), we get
{tex}2 d = \left( \frac { 1 } { 7 } - \frac { 1 } { 9 } \right) = \frac { 2 } { 63 } \Rightarrow d = \left( \frac { 1 } { 2 } \times \frac { 2 } { 63 } \right) = \frac { 1 } { 63 }.{/tex}
Putting d = {tex}\frac { 1 } { 63 }{/tex}in (i), we get
{tex}a + \left( 6 \times \frac { 1 } { 63 } \right) = \frac { 1 } { 9 } \Rightarrow a + \frac { 2 } { 21 } = \frac { 1 } { 9 } \Rightarrow a = \left( \frac { 1 } { 9 } - \frac { 2 } { 21 } \right) = \left( \frac { 7 - 6 } { 63 } \right) = \frac { 1 } { 63 }.{/tex}
Thus, a = {tex}\frac { 1 } { 63 }{/tex} and d = {tex}\frac { 1 } { 63 }.{/tex}
{tex}\therefore{/tex} T63 = a + (63-1)d = (a + 62d)
{tex}= \left( \frac { 1 } { 63 } + 62 \times \frac { 1 } { 63 } \right) = \left( \frac { 1 } { 63 } + \frac { 62 } { 63 } \right) = 1.{/tex}
Hence, 63rd term of the given AP is 1.
Posted by Kaif Khan 7 years, 5 months ago
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Posted by Kartik Chawla 7 years, 5 months ago
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Posted by Unique Queen 7 years, 5 months ago
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Posted by Aman Bedi 7 years, 5 months ago
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Aman Singh 5 years ago
Posted by Rafath Parveen 7 years, 5 months ago
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Posted by Aashish Relwani 7 years, 5 months ago
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Anusree Unnikrishnan 6 years, 9 months ago
an=1500. d=175
Amount paid in 21st installment= a21
a21=1500+20(175). =1500+3500. = ₹ 5000
Total amount paid till 21st installment = a21
S21= 21/2 [2(1500)+20(175)]
= 21/2 [3000+5000]
= 21/2 (8000). = 21×4000. =₹84000
Amount of loan to be paid after the 21st installment = 1,75,000-84,000
= ₹ 91,000
Posted by Beenu Solanki 7 years, 5 months ago
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Bhargav Singh 7 years, 5 months ago
3Thank You