can you please solve.......no spam answer needed....
this is my exam questions.
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Answer........................ 5th question
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Solution:
Question 5 -
a + b√3 = ( √3 + 1)/( √3 - 1)
We need to first rationalise the RHS .
This can be done by multiplying both sides ( the numerator and denominator ) with (√3+1) .
a + b√3
=> [ √3 + 1]/[ √3 - 1] × [ √3 +1 ]/[ √3 + 1]
=> ( √3 + 1)²/2
=> ( 3 + 1 + 2√3 )/2
=> 4 + 2√3/2
=> 2 + √3
Comparing this , we get :
a = 2
b = 1.
Question 6 :
x = (√3+√2 )/( √3 - √2 )
y = ( √3 - √2 )/( √3 + √2 )
x = (√3+√2 )/( √3 - √2 )
=> [ √3 + √2 ]^2
y = inverse of x
=> [ √3 - √2 ]^2
x + y
=> ( √3 + √2)² + (√3 - √2 )²
=> 5 + 5
=> 10 .
This is the required answer .
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