Basics & LogarithmsmediumFree

Solving sqrt(log_2 x - 1) + (1/2) log_(1/2) x^3 + 2 > 0 | JEE Advanced

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

Question
The solution set of the inequality
log⁡2x−1+12log⁡1/2x3+2>0\sqrt{\log_2x-1}+\frac12\log_{1/2}x^3+2>0
 may be
A[2,3)\left[2,3\right)correct
B(2,3]\left(2,3\right]correct
C[2,4)\left[2,4\right)correct
D(2,4]\left(2,4\right]
Solution
Step 1:
log⁡1/2x3=log⁡2x3log⁡2(1/2)=3log⁡2x−1=−3log⁡2x\log_{1/2}x^3 = \frac{\log_2x^3}{\log_2\left(1/2\right)} = \frac{3\log_2x}{-1} = -3\log_2x
 With u=log⁡2xu = \log_2x the inequality becomes
u−1−32u+2>0\sqrt{u-1}-\frac32u+2>0
 Step 2: The square root needs u≥1u \ge 1, i.e. x≥2x \ge 2, which is the domain. Putting t=u−1≥0t = \sqrt{u-1} \ge 0, so u=t2+1u = t^2+1,
t−32(t2+1)+2>0⟹−32t2+t+12>0t-\frac32\left(t^2+1\right)+2>0 \quad\Longrightarrow\quad -\frac32t^2+t+\frac12>0
 Step 3: Multiplying by −2-2 and reversing the sign,
3t2−2t−1<0⟹(3t+1)(t−1)<0⟹−13<t<13t^2-2t-1<0 \quad\Longrightarrow\quad \left(3t+1\right)\left(t-1\right)<0 \quad\Longrightarrow\quad -\frac13<t<1
 Combined with t≥0t \ge 0,
0≤t<10 \le t<1
 Step 4: Squaring is increasing for t≥0t \ge 0, so each step here is reversible:
0≤u−1<1⟹0≤u−1<1⟹1≤u<20 \le \sqrt{u-1}<1 \quad\Longrightarrow\quad 0 \le u-1<1 \quad\Longrightarrow\quad 1 \le u<2
1≤log⁡2x<2⟹2≤x<41 \le \log_2x<2 \quad\Longrightarrow\quad 2 \le x<4
 Step 5: The solution set is [2,4)\left[2,4\right). The words "may be" mean the question accepts any option contained in it: (1) [2,3)⊂[2,4)\left[2,3\right) \subset \left[2,4\right) (2) (2,3]⊂[2,4)\left(2,3\right] \subset \left[2,4\right) (3) [2,4)\left[2,4\right), the set itself (4) (2,4]⊄[2,4)\left(2,4\right] \not\subset \left[2,4\right), since 44 is not a solution (at x=4x = 4: 2−1−32⋅2+2=1−3+2=0\sqrt{2-1}-\tfrac32\cdot2+2 = 1-3+2 = 0, not >0>0). Answer: (1), (2), (3)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Basics & Logarithms · easy
If x=∏n=12026nx = \prod_{n=1}^{2026} n, then the value of the expression 11log⁡2x+1log⁡3x+⋯+1log⁡2026x\frac{1}{\dfrac{1}{\log_2 x} + \dfrac{1}{\log_3 x} + \cdots + \dfrac{1}{\log_{2026} x}} is
Basics & Logarithms · easy
Number of integral values of xx satisfying the inequality (34)6x+10−x2<2764\left(\dfrac{3}{4}\right)^{6x+10-x^2} < \dfrac{27}{64} is:
Basics & Logarithms · easy
Number of natural numbers for which the number log⁡4−x(x2−14x+45)\log_{4-x}(x^2-14x+45) is defined is:
Basics & Logarithms · easy
The product of all solutions of the equation e5(log⁡ex)2+3=x8e^{5(\log_{e}x)^{2}+3}=x^{8}, x>0x>0, is

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath