| (i) The value of \[2+\frac{1}{2+\frac{1}{2+......\infty }}\] is \[\sqrt{2}\]. |
| (ii) A line segment AB of length 2 m is divided at C into two parts such that\[A{{C}^{2}}=AB-CB\] The length of the part CB is\[3+\sqrt{5}\]. |
| (iii) Every quadratic equation can have at most two real roots. |
| (iv) A real number a is said to be root of the quadratic equation \[a{{x}^{2}}+bx+c=0,\] if\[a{{\alpha }^{2}}+b\alpha +c=0\]. |
A) i-F ii-T iii-T iv-T
B) i-F ii-T iii-T iv-F
C) i-T ii-F iii-F iv-T
D) i-F ii-F iii-T iv-T
Correct Answer: D
Solution :
(i) Let \[x=2+\frac{1}{2+\frac{1}{2+....\infty }}\] \[\Rightarrow \] \[x=2+\frac{1}{x}\] \[\Rightarrow \] \[{{x}^{2}}-2x-1=0\] \[\Rightarrow \] \[x=1+\sqrt{2}\] or \[1-\sqrt{2}\] (ii) Let \[AC=x,\] then \[CB=2-x\] Since, \[A{{C}^{2}}=AB\times CB\] \[\Rightarrow \] \[{{x}^{2}}=2(2-x)\] \[\Rightarrow \] \[{{x}^{2}}+2x-4=0\] \[\Rightarrow \] \[x=-1+\sqrt{5}\] or \[-1-\sqrt{5}\] (Not possible) \[\therefore \] \[CB=2-x=2-(\sqrt{5}-1)=(3-\sqrt{5})\,m\] (iii) True, every quadratic equation has maximum of two real roots. (iv) True.
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