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If m and n are the roots of the equation \( \large ax^{2}\)+\(bx \)+\( c\) =0 then the equation whose roots are \( \Large \frac{\left(m ^{2}+ 1\right)}{m} \) and \( \Large \frac{\left(n^{2}+1\right)}{n} \) is
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A) \( acx^{2}\)+\( \left( ab+bc\right)\)\( x\)+\( b^{2}\)+\( \left( a-c\right)^{2}\)=\( 0\) |
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B) \( acx^{2}\)+\( \left( ab-bc\right)\)\( x\)+\( b^{2}\)+\( \left( a-c\right)^{2}\)=\( 0\) |
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C) \( acx^{2}\)+\( \left( ab-bc\right)\)\( x\)+\( b^{2}\)-\( \left( a-c\right)^{2}\)=\( 0\) |
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D) \( acx^{2}\)+\( \left( ab+bc\right)\)\( x\)+\( b^{2}\)-\( \left( a-c\right)^{2}\)=\( 0\) |
Correct Answer:
| A) \( acx^{2}\)+\( \left( ab+bc\right)\)\( x\)+\( b^{2}\)+\( \left( a-c\right)^{2}\)=\( 0\) |
Description for Correct answer:
Let m = n = 1
Equation with root 1 and 1 is \( \Large (x - 1)(x - 1)= x^{2} - 2x + -1 \) ...(i) Comparing equation (i) with \( \Large ax^{2} + bx + c = 0 \)
We get a = 1, b = -2 and c = 1.
Now \( \Large \frac{m^{2} + 1}{m} = 2 \) and \( \Large \frac{n^{2} + 1}{n} = 2 \)
Now from option (a)
\( \Large acx^{2} +(ab + bc)x + b^{2} +(a - c)^{2} \)
\( \Large = x^{2} + (-2-2)x + 4 + (0) \)
\( \Large = x^{2} - 4x + 4 \)
\( \Large = (x - 2)(x - 2) \)
x = 2,3
Hence equation is satisfied
Option a is correct
Let m = n = 1
Equation with root 1 and 1 is \( \Large (x - 1)(x - 1)= x^{2} - 2x + -1 \) ...(i) Comparing equation (i) with \( \Large ax^{2} + bx + c = 0 \)
We get a = 1, b = -2 and c = 1.
Now \( \Large \frac{m^{2} + 1}{m} = 2 \) and \( \Large \frac{n^{2} + 1}{n} = 2 \)
Now from option (a)
\( \Large acx^{2} +(ab + bc)x + b^{2} +(a - c)^{2} \)
\( \Large = x^{2} + (-2-2)x + 4 + (0) \)
\( \Large = x^{2} - 4x + 4 \)
\( \Large = (x - 2)(x - 2) \)
x = 2,3
Hence equation is satisfied
Option a is correct
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