BPSC 45th Exam Prelims 2003 Solved Paper

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Question : 114 of 150

Marks: +1, -0

Let

f(x) = ax^7 + bx^3 + cx – 5,

where a, b, c are constants, If f(–7) = 7, then, f (7) is equal to

–17
–7
14
17

Solution:

f(x)=a x^{7}+b x^{3}+c x

-5

f(-7)=a(-7)^{7}+b(-7)^{3}-7 c-5

7=a(-7)^{7}+b(-7)^{3}-7 c-5

[ ∵ f(-7)=7]

⇒ a(7)^{7}+b(7)^{3}+7 c=-12

Now,

f(7)=a(7)^{7}+b(7)^{3}+7 c-5

f(7)=-12-5

f(7)=-17

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