KVPY SX SB Exam 2013 Question Paper

Question : 86 of 120

Marks: +1, -0

Let

f(x)=1+x/1!+x^2/2!+x^3/3!+x^4/4!

. The number of real roots of

f ( x ) = 0

Solution:

f(x)=1+x +x^2/2 +x^3/6 +x^4/24

f'(x)=1+x +x^2/2 +x^3/6

f"(x)=1+x+x^2/2 > 0

⇒ f'(x) is an increasing

\f^\n

⇒ f'(x) = 0 at

\x = \x_0

⇒\f'(x_0)=0

⇒1+x_0+x^2_0/2+x_0^3/6=0

.... (1)

f'(-2) f'(-1) < 0

⇒ x_0 ∈ (-2, -1)

f(x_0)=1+x_0+x_0^2/2+x_0^3/6+x_0^4/24=x_0^4/24 > 0

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