Basics & LogarithmsmediumPYQ · JEE Main · 3 Aug 2020 · Shift 1 (Morning)Free

Basics & Logarithms: Let Set Real Roots Equation (JEE Main 2020)

JEE Maths question with a full step-by-step solution.

Question
Let SS be the set of all real roots of the equation 3x(3x−1)+2=∣3x−1∣+∣3x−2∣3^{x}(3^{x}-1)+2=|3^{x}-1|+|3^{x}-2|. Then SS
Acontains at least four elements
Bis an empty set
Ccontains exactly two elements
Dis a singletoncorrect
Solution
Step 1: Let t=3xt=3^{x}, t>0t>0. The equation becomes:
t(t−1)+2=∣t−1∣+∣t−2∣t(t-1)+2=|t-1|+|t-2|
t2−t+2=∣t−1∣+∣t−2∣t^{2}-t+2=|t-1|+|t-2|
 Step 2: Case I: t<1t<1. Then ∣t−1∣=1−t|t-1|=1-t and ∣t−2∣=2−t|t-2|=2-t:
t2−t+2=1−t+2−t=3−2tt^{2}-t+2=1-t+2-t=3-2t
t2−t+2t−1=0  ⇒  t2+t−1=0t^{2}-t+2t-1=0\;\Rightarrow\;t^{2}+t-1=0
t=−1±52t=\dfrac{-1\pm\sqrt{5}}{2}
 Only t=−1+52≈0.618t=\dfrac{-1+\sqrt{5}}{2}\approx 0.618 is positive and <1<1. Accepted. Step 3: Case II: 1≤t≤21\le t\le 2. Then ∣t−1∣=t−1|t-1|=t-1 and ∣t−2∣=2−t|t-2|=2-t:
t2−t+2=t−1+2−t=1t^{2}-t+2=t-1+2-t=1
t2−t+1=0t^{2}-t+1=0
 Discriminant =1−4=−3<0=1-4=-3<0. No real solution. Step 4: Case III: t>2t>2. Then ∣t−1∣=t−1|t-1|=t-1 and ∣t−2∣=t−2|t-2|=t-2:
t2−t+2=t−1+t−2=2t−3t^{2}-t+2=t-1+t-2=2t-3
t2−3t+5=0t^{2}-3t+5=0
 Discriminant =9−20=−11<0=9-20=-11<0. No real solution. Step 5: Only one value of t=5−12t=\dfrac{\sqrt{5}-1}{2} satisfies the equation, giving exactly one value of xx. So SS is a singleton. Answer: (4) is a singleton
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