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Basics & Logarithms: Find Number Single Digit Prime Numbers Satisfying

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

Question
Find the number of single digit prime numbers satisfying
(x−2)100(2+x)101(x−1)10x41(x+1)39(4−x)11≤0\frac{(x-2)^{100}(2+x)^{101}(x-1)^{10}}{x^{41}(x+1)^{39}(4-x)^{11}} \leq 0
 is
Solution
Answer: 3
Step 1: Poles (undefined): x=0,−1,4x = 0, -1, 4 Zeros: x=2x = 2 (even power, expression =0= 0), x=−2x = -2 (odd power, sign change), x=1x = 1 (even power, expression =0= 0). Step 2: Since (x−2)100(x-2)^{100}, (x−1)10(x-1)^{10} are even powers (≥0\geq 0), the sign of the expression is determined by:
sgn((x+2)x(x+1)(4−x))\text{sgn}\left(\frac{(x+2)}{x(x+1)(4-x)}\right)
 Sign analysis with critical points −2,−1,0,4-2, -1, 0, 4: x<−2x < -2: expression <0< 0 (satisfies ≤0\leq 0) −2<x<−1-2 < x < -1: expression >0> 0 −1<x<0-1 < x < 0: expression <0< 0 0<x<40 < x < 4: expression >0> 0 x>4x > 4: expression <0< 0 Including zeros x=1x = 1 and x=2x = 2 (even powers, expression =0= 0). Step 3: Solution set: (−∞,−2]∪(−1,0)∪{1}∪{2}∪(4,+∞)(-\infty,-2] \cup (-1,0) \cup \{1\} \cup \{2\} \cup (4,+\infty). Single digit primes: {2,3,5,7}\{2, 3, 5, 7\}. x=2x = 2: in {2}\{2\}. Satisfies. x=3x = 3: in (0,4)(0,4), expression >0> 0. Does not satisfy. x=5x = 5: in (4,+∞)(4,+\infty). Satisfies. x=7x = 7: in (4,+∞)(4,+\infty). Satisfies. Count =3= 3 Answer: 3
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