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NDA 2017
Exam
NDA
Year
2017
Subject
Mathematics
Chapter
Quadratic Equation

NDA 2017 · Mathematics · Quadratic Equation

If the difference between the roots of the equation x^2+k x+1=0 is...

If the difference between the roots of the equation x2+kx+1=0x^2+k x+1=0 is strictly less than 5\sqrt{5}, where ∣k∣≥2|\mathrm{k}| \geq 2, then k can be any element of the interval

Options

  1. A[−3,−2]∪[2,3][-3,-2] \cup[2,3]
  2. B(−3,3)(-3,3)
  3. C[−3,−2]∪[2,3][-3,-2] \cup[2,3]
  4. DNone of the above

Correct answer

A. [−3,−2]∪[2,3][-3,-2] \cup[2,3]

Step-by-step solution

Let the roots of the equation x2+Kx+1=0\mathrm{x}^2+\mathrm{Kx}+1=0 be α,β\alpha, \beta. Then, sum of roots, α+β=−Kα⋅β=+1\alpha+\beta=-\mathrm{K}\alpha \cdot \beta=+1 (Since, sum of roots of ax2+bx+c=0a x^2+b x+c=0 is −ba\frac{-b}{a}; Product of roots is ca\frac{c}{a} ) Given that, α−β<5\alpha-\beta < \sqrt{5}⇒(α−β)2<5⇒(α+β)2−4αβ<5⇒(−K)2−4(1)<5\begin{array}{ll}\Rightarrow & (\alpha-\beta)^2 < 5 \\ \Rightarrow & (\alpha+\beta)^2-4 \alpha \beta < 5 \\ \Rightarrow & (-K)^2-4(1) < 5\end{array}⇒K2−4−5<0⇒ and K2−9<0⇒K2−32<0⇒(K+3)(K−3)<0⇒ and −3<K<3 and  Also, it is given that ∣K∣≥2⇒ and K≤−2 (i)  from K≥2 (i), (ii), K∈(−3,−2]∪[2,3)\begin{array}{ll} \Rightarrow& \mathrm{K}^2-4-5 < 0 \\ \Rightarrow \text{ and } \mathrm{ K}^2-9 < 0 \Rightarrow \mathrm{ K}^2-3^2 < 0 \Rightarrow(\mathrm{ K}+3)(\mathrm{K}-3) < 0 \\ \Rightarrow \text{ and } -3 < \mathrm{K} <3 \\ \text{ and } \text{ Also, it is given that }|\mathrm{K}| \geq2 \\ \Rightarrow \text{ and } \mathrm{ K} \leq-2 \text{ (i) }\\ & \text{ from } \mathrm{K} \geq2 \\ \text{ (i), (ii), } \mathrm{K} \in(-3,-2] \cup[2,3) \end{array}$$
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