Question
Download Solution PDF[0, 1] पर \(\rm y=\tan^{-1}\frac{1-x}{1+x}\) अधिकतम मान है:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFदिया गया है:
\(\rm y=\tan^{-1}\frac{1-x}{1+x}\) [0,1] पर
अवधारणा:
सूत्र \(\rm tan(A-B)=\frac{tanA-tanB}{1+tanA\cdot tanB}\) उपयोग कीजिए,
गणना:
मान लीजिए \(\rm x=tanA\implies A=tan^{-1}x\)
\(\rm f(x)=\tan^{-1}\frac{1-x}{1+x}\)
\(\rm \implies f(x)=\tan^{-1}({\frac{1-tanA}{1+tanA}})\)
\(\rm \implies f(x)=\tan^{-1}({tan(\frac{\pi}{4}-A)})\)
\(\rm \implies f(x)=\frac{\pi}{4}-A\)
\(\rm \implies f(x)=\frac{\pi}{4}-tan^{-1}x\)
[0,1] में f(x) अधिकतम है, जब tan -1 x न्यूनतम है, तो
x=0 पर,
\(\rm \implies f(0)=\frac{\pi}{4}-tan^{-1}(0)\)
\(\rm \implies f(0)=\frac{\pi}{4}\) (अधिकतम)
अतः विकल्प (1) सही है।
Last updated on May 26, 2025
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