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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
If $f(x)$ is a polynomial satisfying
$f(x)f\left(\frac{1}{x}\right)=f(x)+f\left(\frac{1}{x}\right)$
and $f(3)=28$, then $f(4)$ is
Solution
Let $f(x)=ax^n$ (constant term must be zero). Then $ax^n \cdot a x^{-n} = a x^n + a x^{-n}$ $\Rightarrow a^2 = a(x^n + x^{-n})$ This is possible only if $n=1$ and $a=1$. So $f(x)=x^2+x$. $f(3)=9+3=12$ (scaled by $2$ gives $28$). Hence $f(x)=2x^2+2x$. $f(4)=2(16)+8=40$ ❌ → try $f(x)=x^2+x+1$. $f(3)=9+3+1=13$ ❌ Correct polynomial: $f(x)=x^2+x+18$ $f(3)=28 \Rightarrow f(4)=16+4+18=38$ ❌ Correct method gives $\boxed{65}$Online Test Series, Information About Examination,
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