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Ask QuestionPosted by Anushka Choudhary 7 years, 6 months ago
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Vivek Kohli 7 years, 6 months ago
Hemant Hemant Kumar 7 years, 6 months ago
Posted by Roshini Rajesh 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Given polynomial is f(x) = x3 - 3x2 + x + 1
Let {tex} \alpha{/tex} = (a - b), {tex} \beta{/tex} = a and {tex} \gamma{/tex} = (a + b)
Now, {tex} \alpha + \beta + \gamma{/tex} = {tex} - \frac { ( - 3 ) } { 1 }{/tex}
⇒ (a - b) + a + ( a + b ) = 3
⇒ a - b + a + a+ b = 3
⇒ a + a + a = 3
⇒ 3a = 3
⇒ a = 3/3
⇒ a = 1
Also, {tex} \alpha \beta + \beta y + \gamma \alpha = \frac { 1 } { 1 }{/tex}
⇒ (a - b)a + a (a + b) + (a + b)(a - b) = 1
⇒ a2 - ab + a2 +ab + a2 - b2 = 1
⇒ 3a2 - b2 = 1 ( ∵ a = 1)
⇒ 3(1)2 - b2 = 1( ∵ a = 1)
⇒ 3 - b2 = 1
⇒ b2 = 2
⇒ b = {tex} \pm \sqrt{2}{/tex}
Hence, a = 1 and b = {tex} \pm \sqrt{2}{/tex}
Posted by Dev Chaudhary 7 years, 6 months ago
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Posted by Anandhu Anandhu.S.D 7 years, 6 months ago
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Yash Gupta 7 years, 6 months ago
Posted by Varun Kumar 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
By Euclid's division algorithm,
a = bq + r = 4q + r
Take b = 4.
Since, 0 {tex}\leqslant{/tex} r < 4, r = 0,1, 2, 3
{tex} a=4q,4q+1,4q+2 ,4q+3{/tex}
Clearly, a =4q=2(2q) and
4q+2=2×(2q+1)
So 4q and 4q+2 are even
Therefore 4q + 1, 4q + 3 are odd, as they are proceeding numbers of even numbers 4q and 4q+2.
{tex}\therefore{/tex} Any positive odd integer is of form 4q+1 or 4q+3 .Where q is a positive integer.
Posted by Sonu Gupta 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
Points are (4, 7) and (0, 7)
Distance {tex}= \sqrt { ( 0 - 4 ) ^ { 2 } + ( 7 - 7 ) ^ { 2 } }{/tex}
= {tex}\sqrt { 4 ^ { 2 } + 0 } = \sqrt { 16 }{/tex} = 4 units
Posted by Chelsy Chenya Maravi 7 years, 6 months ago
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Sahil Prreet 7 years, 6 months ago
Posted by Dheeraj Verma 7 years, 6 months ago
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Posted by Kaviya Ravi 7 years, 6 months ago
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Posted by Priyanshu Patalvanshi Priyanshu 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Let {tex} \alpha{/tex} and {tex} \frac { 1 } { \alpha }{/tex} be the zeros of (a2 + 9)x2 + 13x + 6a.
Then, we have
{tex} \alpha \times \frac { 1 } { \alpha } = \frac { 6 a } { a ^ { 2 } + 9 }{/tex}
⇒ 1 = {tex} \frac { 6 a } { a ^ { 2 } + 9 }{/tex}
⇒ a2 + 9 = 6a
⇒ a2 - 6a + 9 = 0
⇒ a2 - 3a - 3a + 9 = 0
⇒ a(a - 3) - 3(a - 3) = 0
⇒ (a - 3) (a - 3) = 0
⇒ (a - 3)2 = 0
⇒ a - 3 = 0
⇒ a = 3
So, the value of a in given polynomial is 3.
Posted by Shailendra Singh Rajput 7 years, 6 months ago
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Posted by Deepak Kumar 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Let n-1 and n be consecutive positive integers,Let P be their product
Then {tex}\style{font-family:Arial}{\style{font-size:12px}{\mathrm n(\mathrm n-1)=\mathrm n^2-1\;.................(1)}}{/tex}
We know that any positive integers is of the form 2q or 2q + 1, where q is a positive integer
Case I: When n = 2q, then
P=n2 - n = (2q)2 - 2q = 4q2 - 2q = 2q(2q - 1)-----(2)
Case II: When n = 2q + 1, then
P=n2 - n = (2q + 1)2 - (2q + 1)
= 4q2 + 4q + 1 - 2q - 1
= 4q2 + 2q
P = 2q(2q + 1)..........(3)
From (2) and (3) we conclude that the product of n-1 and n is divisible by 2
Posted by Lakshita Kataria 7 years, 6 months ago
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Posted by Khushu Jangra 7 years, 6 months ago
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Ananya Tyagi 7 years, 6 months ago
Posted by Rahul Verma 7 years, 6 months ago
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Posted by Priya Kumari 7 years, 6 months ago
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Maneesh Chaudhary 7 years, 6 months ago
Posted by Sukhpreet Kaur 7 years, 6 months ago
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Posted by Krishna Gupta 7 years, 6 months ago
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Posted by Sahil Yadav 7 years, 6 months ago
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Posted by Prashant Deep 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
LHS {tex} \frac{{1 + \cos \theta + \sin \theta }}{{1 + \cos \theta - \sin \theta }}{/tex}
Dividing numerator and denominator by cos{tex} \theta {/tex}
{tex}= \frac{{\frac{1}{{\cos \theta }} + \frac{{\cos \theta }}{{\cos \theta }} + \frac{{\sin \theta }}{{\cos \theta }}}}{{\frac{1}{{\cos \theta }} + \frac{{\cos \theta }}{{\cos \theta }} - \frac{{\sin \theta }}{{\cos \theta }}}}{/tex}
{tex}= \frac{{\sec \theta + 1 + \tan \theta }}{{\sec \theta + 1 - \tan \theta }}{/tex}
Multiplying and dividing by {tex} \sec \theta + 1 + \tan \theta {/tex}
{tex}= \frac{{\sec \theta + 1 + \tan \theta }}{{\sec \theta + 1 - \tan \theta }} \times \frac{{\sec \theta + 1 + \tan \theta }}{{\sec \theta + 1 + \tan \theta }}{/tex}
{tex}= \frac{{{{(\sec \theta + 1 + \tan \theta )}^2}}}{{{{(\sec \theta + 1)}^2} - {{\tan }^2}\theta }}{/tex}
{tex}= \frac{{{{\sec }^2}\theta + 1 + {{\tan }^2}\theta + 2\sec \theta + 2\tan \theta + 2\sec \theta \tan \theta }}{{1 + {{\sec }^2}\theta + 2\sec \theta - {{\tan }^2}\theta }}{/tex}
Now, {tex} 1 + {\tan ^2}\theta = {\sec ^2}\theta {/tex}
{tex}= \frac{{2{{\sec }^2}\theta + 2\sec \theta + 2\tan \theta + 2\sec \theta \tan \theta }}{{2 + 2\sec \theta }}{/tex}
{tex} = \frac{{2\left[ {\sec \theta (\sec \theta + 1) + \tan \theta (1 + \sec \theta } \right]}}{{2(1 + \sec \theta )}}{/tex}
{tex} = \frac{{(\sec + \tan \theta )(\sec \theta + 1)}}{{(1 + \sec \theta )}}{/tex}
{tex} = \sec \theta + \tan \theta = \frac{1}{{\cos \theta }} + \frac{{\sin \theta }}{{\cos \theta }}{/tex}
{tex}= \frac{{1 + \sin \theta }}{{\cos \theta }} = RHS{/tex}
Posted by Thejasprasad Bk 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
We have,f(x) = 3x2 - x - 4
= 3x2 - 4x + 3x - 4
= x(3x - 4) + 1(3x - 4)
= (3x - 4) )(x + 1)
{tex}\therefore{/tex} f(x) = 0
{tex}\Rightarrow{/tex} (3x - 4)(x +1) = 0
{tex}\Rightarrow{/tex} 3x - 4 = 0 or x + 1 = 0
{tex}\Rightarrow x = \frac { 4 } { 3 }{/tex} or x = -1
So, the zeros of f(x) are {tex}\frac { 4 } { 3 } \text { and } - 1{/tex}
Now sum of zeros {tex}= \frac { 4 } { 3 } + ( - 1 ) = \frac { 1 } { 3 } = \frac { - ( \text { coefficient of } x ) } { \left( \text { coefficient of } x ^ { 2 } \right) }{/tex}.
And product of zeros {tex}= \frac { 4 } { 3 } \times ( - 1 ) = \frac { - 4 } { 3 } = \frac { \text { constant term } } { \left( \text { coefficient of } x ^ { 2 } \right) }{/tex}
Posted by Chandrakanth Nirna 7 years, 6 months ago
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Posted by Suryansh Tripathi 7 years, 6 months ago
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Posted by Amit Kumar 7 years, 6 months ago
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Amit Kumar 7 years, 6 months ago
Posted by Pradipti Pandey 7 years, 6 months ago
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Vivek Kohli 7 years, 6 months ago
1Thank You