Basics & LogarithmsmediumFree

Basics & Logarithms — JEE Maths practice question

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

Question
If SS is the set of all real xx such that x2(5−x)(1−2x)(5x+1)(x+2)\dfrac{x^2(5-x)(1-2x)}{(5x+1)(x+2)} is negative and 3x+16x3+x2−x\dfrac{3x+1}{6x^3+x^2-x} is positive, then SS contains:
A(1,4)(1, 4)correct
B(5,11)(5, 11)
C(−32,12)\left(-\dfrac{3}{2}, \dfrac{1}{2}\right)
D(−10,−4)(-10, -4)
Solution
Step 1: Solve the second condition
3x+1x(3x−1)(2x+1)>0  ⟹  x∈(−∞,−12)∪(−13,0)∪(13,∞)\frac{3x+1}{x(3x-1)(2x+1)} > 0 \implies x \in \left(-\infty, -\frac{1}{2}\right) \cup \left(-\frac{1}{3}, 0\right) \cup \left(\frac{1}{3}, \infty\right)
 Step 2: Solve the first condition
x2(5−x)(1−2x)(5x+1)(x+2)<0  ⟹  x∈(−2,−15)∪(12,5)\frac{x^2(5-x)(1-2x)}{(5x+1)(x+2)} < 0 \implies x \in (-2, -\tfrac{1}{5}) \cup (\tfrac{1}{2}, 5)
 Step 3: Intersect
S=(−2,−12)∪(−13,−15)∪(12,5)S = (-2, -\tfrac{1}{2}) \cup (-\tfrac{1}{3}, -\tfrac{1}{5}) \cup (\tfrac{1}{2}, 5)
 The interval (1,4)(1, 4) lies inside (12,5)(\tfrac{1}{2}, 5), so SS contains (1,4)(1, 4). Answer: (1)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Basics & Logarithms · easy
The product of all solutions of the equation e 5( e x) 2 +3 =x 8 , x>0 , is
Basics & Logarithms · easy
The number of real solutions of the equation x (x 2 +3|x|+5|x-1|+6|x-2| )=0 is
Basics & Logarithms · easy
If a+b+c=1 , ab+bc+ca=2 and abc=3 , then the value of a 4 +b 4 +c 4 is equal to
Basics & Logarithms · easy
Sum of all the values of x satisfying the equation 17 11 ( x+11 + x ) = 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