1. **প্রশ্ন:** $f(y) = \frac{1 + y^2 + y^4}{y^2}$ এবং $g(y) = y^3 - ky^2 + y + 6$। (ক) $f(\frac{1}{2})$ নির্ণয় কর। 2. **সমাধান:** 1. প্রথমে $f(y)$ এর মান নির্ণয় করার জন্য $y=\frac{1}{2}$ বসাও। 2. সূত্র: $$f(y) = \frac{1 + y^2 + y^4}{y^2}$$ 3. $y=\frac{1}{2}$ বসালে, $$f\left(\frac{1}{2}\right) = \frac{1 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^4}{\left(\frac{1}{2}\right)^2}$$ 4. হিসাব করি: $$\left(\frac{1}{2}\right)^2 = \frac{1}{4}, \quad \left(\frac{1}{2}\right)^4 = \left(\frac{1}{4}\right)^2 = \frac{1}{16}$$ 5. তাহলে, $$f\left(\frac{1}{2}\right) = \frac{1 + \frac{1}{4} + \frac{1}{16}}{\frac{1}{4}} = \frac{\frac{16}{16} + \frac{4}{16} + \frac{1}{16}}{\frac{1}{4}} = \frac{\frac{21}{16}}{\frac{1}{4}}$$ 6. ভাগ করলে, $$= \frac{21}{16} \times \frac{4}{1} = \frac{21 \times 4}{16} = \frac{84}{16}$$ 7. সরল করলে, $$= \frac{\cancel{84}^{21} \times 4}{\cancel{16}^{4} \times 4} = \frac{21}{4} = 5.25$$ **উত্তর:** $f\left(\frac{1}{2}\right) = \frac{21}{4}$ বা 5.25 --- **slug:** f_y_evaluation **subject:** algebra **desmos:** {"latex":"f(y)=\frac{1+y^2+y^4}{y^2}","features":{"intercepts":true,"extrema":true}} **q_count:** 12