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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

\frac{(x-2)^{100}(2+x)^{101}(x-1)^{10}}{x^{41}(x+1)^{39}(4-x)^{11}} \leq 0

Solution

Answer: 3

Step 1: Poles (undefined):

x = 0, -1, 4

Zeros:

x = 2

(even power, expression

= 0

x = -2

(odd power, sign change),

x = 1

(even power, expression

= 0

). Step 2: Since

(x-2)^{100}

(x-1)^{10}

are even powers (

\geq 0

), the sign of the expression is determined by:

\text{sgn}\left(\frac{(x+2)}{x(x+1)(4-x)}\right)

Sign analysis with critical points

-2, -1, 0, 4

x < -2

: expression

< 0

(satisfies

\leq 0

)

-2 < x < -1

: expression

> 0

-1 < x < 0

: expression

< 0

0 < x < 4

: expression

> 0

x > 4

: expression

< 0

Including zeros

x = 1

and

x = 2

(even powers, expression

= 0

). Step 3: Solution set:

(-\infty,-2] \cup (-1,0) \cup \{1\} \cup \{2\} \cup (4,+\infty)

. Single digit primes:

\{2, 3, 5, 7\}

x = 2

: in

\{2\}

. Satisfies.

x = 3

: in

(0,4)

, expression

> 0

. Does not satisfy.

x = 5

: in

(4,+\infty)

. Satisfies.

x = 7

: in

(4,+\infty)

. Satisfies. Count

= 3

Answer: 3

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