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Question: The number of real values of the parameter k for which \(b = \frac{9}{4}\)with real coefficients wil...

The number of real values of the parameter k for which b=94b = \frac{9}{4}with real coefficients will have exactly one solution is.

A

2

B

1

C

4

D

None of these

Answer

1

Explanation

Solution

Let ⇒\Rightarrow

This quadratic equation will have exactly one solution if its discriminant vanishes.

x3=93+112=63+33+92+22x^{3} = 9\sqrt{3} + 11\sqrt{2} = 6\sqrt{3} + 3\sqrt{3} + 9\sqrt{2} + 2\sqrt{2}

=33+22+63+92=33+22+3(23+32)= 3\sqrt{3} + 2\sqrt{2} + 6\sqrt{3} + 9\sqrt{2} = 3\sqrt{3} + 2\sqrt{2} + 3(2\sqrt{3} + 3\sqrt{2}) =33+22+32.3(2+3)= 3\sqrt{3} + 2\sqrt{2} + 3\sqrt{2}.\sqrt{3}(\sqrt{2} + \sqrt{3}) =(3)3+(2)3+3.2.2(3+2)=(3+2)3= (\sqrt{3})^{3} + (\sqrt{2})^{3} + 3.\sqrt{2}.\sqrt{2}(\sqrt{3} + \sqrt{2}) = (\sqrt{3} + \sqrt{2})^{3} x3=(3+2)3x^{3} = (\sqrt{3} + \sqrt{2})^{3} x=3+2x = \sqrt{3} + \sqrt{2}.

But x+x2+1=a⇒x2+1=a−xx + \sqrt{x^{2} + 1} = a \Rightarrow \sqrt{x^{2} + 1} = a - x is not defined ⇒\Rightarrow, x2+1=(a−x)2=x2−2ax+a2⇒x^{2} + 1 = (a - x)^{2} = x^{2} - 2ax + a^{2} \Rightarrow.

x=1−a2−2a=a2−12a=12(a−1a)x = \frac{1 - a^{2}}{- 2a} = \frac{a^{2} - 1}{2a} = \frac{1}{2}\left( a - \frac{1}{a} \right)Number of real values of x=7+3,xy=4x = \sqrt{7} + \sqrt{3},xy = 4.