Inverse Trigonometric FunctionshardFree

Inverse Trigonometric Functions: Functions Defined 10x Find Number Even Integers Defined

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

Question
Two functions f(x)f(x) and g(x)g(x) are defined as f(x)=log⁡3∣x−2x2−10x+24∣f(x)=\log_{3}\left|\dfrac{x-2}{x^{2}-10x+24}\right| and g(x)=sin⁡−1(2[x]−315)g(x)=\sin^{-1}\left(\dfrac{2[x]-3}{15}\right), then find the number of even integers for which (f(x)+g(x))(f(x)+g(x)) is defined ([⋅][\cdot] is GIF).
Solution
Answer: 5
Step 1: For g(x)g(x): need 2[x]−315∈[−1,1]\dfrac{2[x]-3}{15}\in[-1,1]:
−15≤2[x]−3≤15  ⇒  −12≤2[x]≤18  ⇒  −6≤[x]≤9-15\le 2[x]-3\le 15\;\Rightarrow\;-12\le 2[x]\le 18\;\Rightarrow\;-6\le[x]\le 9
 So x∈[−6,10)x\in[-6,10). Step 2: For f(x)f(x): need x−2x2−10x+24\dfrac{x-2}{x^{2}-10x+24} to be defined and non-zero. Factor: x2−10x+24=(x−4)(x−6)x^{2}-10x+24=(x-4)(x-6). So denominator zero at x=4,6x=4,6. Numerator zero at x=2x=2. Excluded points: x∈{2,4,6}x\in\{2,4,6\}. Step 3: Combined domain: x∈[−6,10)∖{2,4,6}x\in[-6,10)\setminus\{2,4,6\}. Step 4: Even integers in [−6,10)[-6,10): −6,−4,−2,0,2,4,6,8-6,-4,-2,0,2,4,6,8. Excluding {2,4,6}\{2,4,6\}: −6,−4,−2,0,8-6,-4,-2,0,8. Step 5: Count: 55. Answer: 55
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