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

S

be the set of all real roots of the equation

3^{x}(3^{x}-1)+2=|3^{x}-1|+|3^{x}-2|

. Then

S

Acontains at least four elements

Bis an empty set

Ccontains exactly two elements

Dis a singletoncorrect

Solution

Step 1: Let

t=3^{x}

t>0

. The equation becomes:

t(t-1)+2=|t-1|+|t-2|

t^{2}-t+2=|t-1|+|t-2|

Step 2: Case I:

t<1

. Then

|t-1|=1-t

and

|t-2|=2-t

t^{2}-t+2=1-t+2-t=3-2t

t^{2}-t+2t-1=0\;\Rightarrow\;t^{2}+t-1=0

t=\dfrac{-1\pm\sqrt{5}}{2}

Only

t=\dfrac{-1+\sqrt{5}}{2}\approx 0.618

is positive and

<1

. Accepted. Step 3: Case II:

1\le t\le 2

. Then

|t-1|=t-1

and

|t-2|=2-t

t^{2}-t+2=t-1+2-t=1

t^{2}-t+1=0

Discriminant

=1-4=-3<0

. No real solution. Step 4: Case III:

t>2

. Then

|t-1|=t-1

and

|t-2|=t-2

t^{2}-t+2=t-1+t-2=2t-3

t^{2}-3t+5=0

Discriminant

=9-20=-11<0

. No real solution. Step 5: Only one value of

t=\dfrac{\sqrt{5}-1}{2}

satisfies the equation, giving exactly one value of

x

. So

S

is a singleton. Answer: (4) is a singleton

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