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1.Relation and Function
hard
माना $\mathrm{f}(\mathrm{x})=2 \mathrm{x}^{\mathrm{n}}+\lambda, \lambda \in \mathbb{R}, \mathrm{n} \in \mathbb{N}$ और $\mathrm{f}(4)=133, \mathrm{f}(5)=255$ है। तो $(\mathrm{f}(3)-\mathrm{f}(2))$ के सभी धनात्मक पूर्णांक भाजकों का योग है -
A
$61$
B
$60$
C
$58$
D
$59$
(JEE MAIN-2023)
Solution
$f(x)=2 x^{ n }+\lambda$
$f(4)=133$
$f(5)=255$
$133=2 \times 4^{ n }+\lambda……(1)$
$255=2 \times 5^{ n }+\lambda……(2)$
$(2) -(1)$
$122=2\left(5^{ n }-4^{ n }\right)$
$\Rightarrow 5^{ n }-4^{ n }=61$
$\therefore n =3\, and\, \lambda=5$
Now, $f(3)-f(2)=2\left(3^3-2^3\right)=38$
Number of Divisors is $1,2,19,38$; and their sum is $60$.
Standard 12
Mathematics