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JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations

જો \(a \in R\) હોય અને સમીકરણ \( - 3{\left( {x - \left[ x \right]} \right)^2} + 2\left( {x - \left[ x \right]} \right) + {a^2} = 0\) ને પૂર્ણાંક ઉકેલ ન હોય તો \(a\) શકય કિંમતો . . . અંતરાલમાં હોય . .

  1. A \(\left( { - 1,0} \right) \cup \left( {0,1} \right)\)
  2. B \(\left( {1,2} \right)\)
  3. C \(\left( { - 2, - 1} \right)\)
  4. D \(\left( { - \infty , - 2} \right) \cup \left( {2,\infty } \right)\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\left( { - 1,0} \right) \cup \left( {0,1} \right)\)

Step-by-step Solution

Detailed explanation

Here, \(a \in R\) and equation is \(-3\{x-[x]\}^{2}+2\{x-[x]\}+a^{2}=0\) \(\text { Let } \quad t=x-[x]\) \(\therefore-3 t^{2}+2 t+a^{2}=0\) \(\Rightarrow t=\frac{1 \pm \sqrt{1+3 a^{2}}}{3}\) \(\because \quad t=x-[x]=\{x\} \quad[\text { fractional part }]\)…
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