JEE WhatsApp Group
Sign in
Please select an account to continue using cracku.in
↓ →
JEE WhatsApp Group
Join Our JEE Preparation Group
Prep with like-minded aspirants; Get access to free daily tests and study material.
Let $$f(x) = 3\sqrt{x - 2} + \sqrt{4 - x}$$ be a real valued function. If $$\alpha$$ and $$\beta$$ are respectively the minimum and the maximum values of $$f$$, then $$\alpha^2 + 2\beta^2$$ is equal to
f(x)=3√(x-2)+√(4-x). Domain [2,4].
f'(x)=3/(2√(x-2))-1/(2√(4-x))=0. 3√(4-x)=√(x-2). 9(4-x)=x-2. 36-9x=x-2. 10x=38. x=3.8.
f(3.8)=3√1.8+√0.2=3·1.342+0.447=4.472=2√5. β=2√5.
f(2)=0+√2=√2. f(4)=3√2+0=3√2. Check: f(2)=√2≈1.41, f(4)=3√2≈4.24, f(3.8)≈4.47.
α=√2 (minimum), β=2√5 (maximum).
α²+2β²=2+40=42.
The answer is Option (1): 42.
Create a FREE account and get:
Educational materials for JEE preparation
Ask our AI anything
AI can make mistakes. Please verify important information.
AI can make mistakes. Please verify important information.
