Mathematics · 2019
JEE Main · 9 April 2019, Shift 1 · Q68
Let Σ_k =1^10 f( a + k )=16(2^10-1), where the function f satisfies f(x+y)=f(x) f(y) for all natural numbers x, y and f(1)=2. Then the natural number…
Let $\displaystyle \sum_{\mathrm{k}=1}^{10} f(\mathrm{a}+\mathrm{k})=16\left(2^{10}-1\right)$, where the function $\displaystyle f$ satisfies $\displaystyle f(x+y)=f(x) f(y)$ for all natural numbers $\displaystyle x, y$ and $\displaystyle f(1)=2$. Then the natural number ' $\displaystyle a$ ' is :
Official answer
(2)
$\displaystyle 3$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.