f(x) is given as
f(x)=2xn+λ,λ∈R,n∈N
∵f(4)=133
⇒133=2×4n+λ...(1)
∵f(5)=255
⇒255=2×5n+λ...(2)
Now, (2)−(1)
122=2(5n−4n)
⇒5n−4n=61
\therefore n=3&\lambda =5
Now f(3)−f(2)=2(33−23)=38
Number of divisors =1,2,19,38 and their sum is 60
Let f(x)=2xn+λ,λ∈R,n∈N, and f(4)=133, f(5)=255. Then the sum of all the positive integer divisors of (f(3)−f(2)) is
Held on 25 Jan 2023 · Verified 6 Jul 2026.
61
60
58
59
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
If the roots of x² - 5x + k = 0 are in the ratio 2:3, then k equals:
Let $\alpha = 3+4+8+9+13+14+\ldots$ upto 40 terms. If $(\tan\beta)^{\frac{\alpha}{1020}}$ is a root of the equation $x^2+x-2=0$, $\beta \in \left(0, \dfrac{\pi}{2}\right)$, then $\sin^2\beta + 3\cos^2\beta$ is equal to:
If the set of all solutions of $|x^2 + x - 9| = |x| + |x^2 - 9|$ is $[\alpha, \beta] \cup [\gamma, \infty)$, then $(\alpha^2 + \beta^2 + \gamma^2)$ is equal to:
The sum of squares of all the real solutions of the equation $\log_{(x+1)}(2x^2+5x+3) = 4 - \log_{(2x+3)}(x^2+2x+1)$ is equal to ________.
Let $f:(1,\infty)\to\mathbb{R}$ be a function defined as $f(x) = \dfrac{x-1}{x+1}$. Let $f^{i+1}(x) = f(f^i(x))$, $i=1, 2, \ldots, 25$, where $f^1(x)=f(x)$. If $g(x) + f^{26}(x) = 0$, $x \in (1, \infty)$, then the area of the region bounded by the curves $y=g(x)$, $2y=2x-3$, $y=0$ and $x=4$ is:
Work through every JEE Main Algebra PYQ, year by year.