Question

Let \[P(x)\] be a polynomial of degree \[4\], with \[P(2) = 1\], \[P'(2) = P''(2) = 0\], \[P'''(2) =
12\] and \[P''''(2) = 24\]. Then the value of \[P''(1)\] is
A. \[0\]
B. \[26\]
C. \[2\]
D. \[12\]

Answer 1

Hint:We are asked to find the value of \[P''(x)\] putting \[x = 1\]. It is given that the polynomial is of degree \[4\], so try to form a general equation for the polynomial of degree \[4\]. Then differentiate the polynomial to find the value of the constants using the given information. Then find the value of \[P''(1)\].

Complete step by step solution:
Given a polynomial \[P(x)\] of degree \[4\]
Such that \[P(2) = 1\], \[P'(2) = P''(2) = 0\], \[P'''(2) = 12\] and \[P''''(2) = 24\].
A polynomial of degree \[n\] can written in the form
\[a{x^n} + b{x^{n - 1}} + c{x^{n - 2}} + ........ + q = 0\]
where \[a\], \[b\], \[c\] and \[q\] are constants.

Let us assume the polynomial \[P(x)\] of degree \[4\] to be in the form
\[P(x) = a{\left( {x - 2} \right)^4} + b{(x - 2)^3} + c{(x - 2)^2} + d(x - 2) + e\] (i)
Differentiating equation (i) with respect to \[x\], we get
\[P'(x) = 4a{\left( {x - 2} \right)^3} + 3b{(x - 2)^2} + 2c(x - 2) + d\] (ii)
Differentiating equation (ii) with respect to \[x\], we get
\[P''(x) = 12a{(x - 2)^2} + 6b(x - 2) + 2c\] (iii)
Differentiating equation (iii) with respect to \[x\], we get
\[P'''(x) = 24a(x - 2) + 6b\] (iv)
Differentiating equation (iv) with respect to \[x\], we get
\[P''''(x) = 24a\] (v)
Now, putting \[x = 2\] in equations (i), (ii), (iii),(iv) and (v) we get
Equation (i) \[ \Rightarrow P(2) = e\] (vi)
Equation (ii) \[ \Rightarrow P'(2) = d\] (vii)
Equation (iii) \[ \Rightarrow P''(2) = 2c\] (ix)
Equation (iv) \[ \Rightarrow P'''(2) = 2b\] (x)
Equation (v) \[ \Rightarrow P''''(2) = 24a\] (xi)
We are given that \[P(2) = 1\] comparing it with equation (vi) we get,
\[e = 1\]
We are given that \[P'(2) = 0\] comparing it with equation (vii) we get,

\[d = 0\]
We are given that \[P''(2) = 0\] comparing it with equation (viii) we get,
\[2c = 0\]
\[ \Rightarrow c = 0\]
We are given that \[P'''(2) = 12\] comparing it with equation (ix) we get,
\[2b = 12\]
\[ \Rightarrow b = 6\]

We are given that \[P''''(2) = 24\] comparing it with equation (x) we get,
\[24a = 24\]
\[ \Rightarrow a = 1\]

Now putting the values of \[a\], \[b\], \[c\], \[d\] and \[e\] in equation (i) we get,
\[P(x) = 1{\left( {x - 2} \right)^4} + 2{(x - 2)^3} + 0{(x - 2)^2} + 0(x - 2) + 1\]
\[ \Rightarrow P(x) = {\left( {x - 2} \right)^4} + 2{(x - 2)^3} + 1\]

Now putting the values of \[a\], \[b\] and \[c\] in equation (iii) we get
\[P''(x) = 12 \times 1 \times {(x - 2)^2} + 6 \times 6 \times (x - 2) + 2 \times 0\]
\[ \Rightarrow P''(x) = 12{(x - 2)^2} + 12(x - 2)\] (xii)

Putting \[x = 1\] in equation (xii) we get,
\[P''(1) = 12{(1 - 2)^2} + 12(1 - 2)\]
\[ \Rightarrow P''(1) = 12 - 12 = 0\]

Therefore the value of \[P''(1)\] is zero.

Hence, the correct answer is option (A) \[\]

Note:Remember the general formula for a polynomial of degree \[n\] using this you can form a polynomial of any degree. Whenever such a question is given where you need to find the polynomial of given degree then check for the information using which you can find the constants of the polynomial. And after finding the constants you can easily form the polynomial.