Skip to content
dronacharya.ai
← JEE Main 2022 · 25 Jun · Shift 1
MathematicsSingle correct+4 / -1Question 4 of 90

JEE Main 2022 · 25 Jun · Shift 1, Mathematics question 4

Let f:N→Rf:\mathbb N\to\mathbb R be a function such that f(x+y)=2f(x)f(y)f(x+y)=2f(x)f(y) for natural numbers xx and yy. If f(1)=2f(1)=2, then the value of α\alpha for which
∑k=110f(α+k)=5123(220−1)\sum_{k=1}^{10}f(\alpha+k)=\frac{512}3\left(2^{20}-1\right)
holds, is :

Choose one option
Answer and worked solution

Official answer: C

Solve the functional equation. g=2fg=2f satisfies g(x+y)=g(x)g(y)g(x+y)=g(x)g(y) with g(1)=4g(1)=4, so g(n)=4ng(n)=4^n and f(n)=22n−1f(n)=2^{2n-1}.

Sum. ∑k=11022α+2k−1=22α−1⋅4(410−1)3=22α+13(220−1)\displaystyle\sum_{k=1}^{10}2^{2\alpha+2k-1}=2^{2\alpha-1}\cdot\frac{4(4^{10}-1)}3=\frac{2^{2\alpha+1}}3\left(2^{20}-1\right).

Match. 22α+1=512=292^{2\alpha+1}=512=2^9, so α=4\alpha=4.

(C)

Topics:Geometric progressionTypes of functions
← Previous questionNext question →