Let f(x) be a polynomial of degree 4 having extreme values at x=1 and x=2.
If \[\mathop {lim}\limits_{x \to 0} \left( {{{f\left( x \right)} \over {{x^2}}} + 1} \right) = 3\] then f(-1) is equal to :
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Official solution
\[ \because \] f(x) has extremum values at x = 1, and x = 2
\[ \because \] f'(1) = 0 and f'(2) = 0
As, f(x) is a polynomial of degree 4.
Suppose f(x) = Ax4 + Bx3 + cx2 + Dx + E
\[ \because \] \[\mathop {\lim }\limits_{x \to 0} \] \[\left( {{{f\left( x \right)} \over {{x^2}}} + 1} \right)\] = 3
\[ \Rightarrow \] \[\mathop {\lim }\limits\_{x \to 0} \left( {{{A{x\^4} + B{x\^3} + C{x\^2} + Dx + E} \over {{x\^2}}} + 1} \right)\] = 3
\[ \Rightarrow \] \[\mathop {\lim }\limits_{x \to 0} \] \[\left( {A{x^2} + Bx + C + {D \over x} + {E \over {{x^2}}} + 1} \right)\] = 3
As limit has finite value, so D = 0 and E = 0
Now A(0)2 + B(0) + C + 0 + 0 + 1 = 3
\[ \Rightarrow \] c + 1 = 3 \[ \Rightarrow \] c = 2
f'(x) = 4Ax3 + 3Bx2 + 2Cx + D
f'(1) = 0 \[ \Rightarrow \] 4A(1) + 3B(1) + 2C(1) + D = 0
\[ \Rightarrow \] 4A + 3B =−4 . . . .(1)
f'(2) = 0⇒4A(8) + 3B(4) + 2C(2) + D = 0⇒8A + 3B =-\[ 2 . . . . .(2)
From equations (1) and (2), we get
A = \]{1 \over 2}and B =−2
So, f(x) =2x4−2x3+2x2Therefore, f(−1) =2(−1)4−2(−1)3+2(−1)2=21+2+2=29Hence f(−1) =29