Basics & LogarithmsmediumPYQ · JEE Main · 10 Apr 2019 · Shift 2 (Afternoon)Free

Basics & Logarithms: Number Real Roots Equation (JEE Main 2019)

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

Question
The number of real roots of the equation 5+∣2x−1∣=2x(2x−2)5+|2^{x}-1|=2^{x}(2^{x}-2) is
A22
B11correct
C33
D44
Solution
Step 1: Let t=2xt=2^{x}, t>0t>0. The equation becomes:
5+∣t−1∣=t(t−2)=t2−2t5+|t-1|=t(t-2)=t^{2}-2t
 Step 2: Case I: t≥1t\ge 1 (i.e., 2x≥12^{x}\ge 1, so x≥0x\ge 0). Then ∣t−1∣=t−1|t-1|=t-1:
5+t−1=t2−2t  ⇒  t2−3t−4=0  ⇒  (t−4)(t+1)=05+t-1=t^{2}-2t\;\Rightarrow\;t^{2}-3t-4=0\;\Rightarrow\;(t-4)(t+1)=0
 t=4t=4 (accepted, since t≥1t\ge 1) or t=−1t=-1 (rejected). t=4  ⇒  2x=4  ⇒  x=2t=4\;\Rightarrow\;2^{x}=4\;\Rightarrow\;x=2. One solution. Step 3: Case II: 0<t<10<t<1 (i.e., x<0x<0). Then ∣t−1∣=1−t|t-1|=1-t:
5+1−t=t2−2t  ⇒  t2−t−6=0  ⇒  (t−3)(t+2)=05+1-t=t^{2}-2t\;\Rightarrow\;t^{2}-t-6=0\;\Rightarrow\;(t-3)(t+2)=0
 t=3t=3 (rejected, not <1<1) or t=−2t=-2 (rejected, not >0>0). No solutions. Step 4: Only one real root: x=2x=2. Answer: (2) 11
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