KVPY2013MathematicsApplication of Derivatives
Let f(x)=1+ x 1 ! + x² 2 ! + x³ 3 ! + x⁴ 4 ! . The number of real roots of f(x)=0 is
Options
- A0
- B1
- C2
- D4
Correct answer
A. 0
Step-by-step solution
f(x)=1+x+ x² 2 + x³ 6 + x⁴ 24 f(x)=1+x+ x² 2 + x³ 6 f^ (x)=1+x+ x² 2 >0 f(x) is an increasing f^ f(x)=0 at x=x₀ f (x₀ )=0 1+x₀+ x₀² 2 + x₀³ 6 =0 ...(1) f(-2) f(-1) 0