TIFR CSE 2023 | Part A | Question: 14

Question

Let $f(x)=a x^{3}+b x^{2}+c x+d$ be a polynomial, where $a, b, c, d$ are unknown real numbers. It is further given that $f(1)=1, f(2)=2, f(3)=9$, and $f^{\prime}(1)=0$. Then, the value of $f^{\prime}(2)$ must be

• A. $1$
• B. $2$
• C. $3$
• D. $4$
• E. $f^{\prime}(2)$ cannot be determined uniquely from the information given in the question.

Answer 1

To solve this problem ,We can take help from Linear algebra.

Given ,

$f(x)=ax^{3}+bx^{2}+cx+d$

$f’(x)=3ax^{2}+2bx+c$

$f(1)=1,f(2)=2,f(3)=9,f’(1)=1$

So, The linear equations looks like,

$a+b+c+d=0$ -----------------------(1)

$3a+2b+c=0$ ------------------------(2)

$8a+4b+2c+d=2$---------------------(3)

$27a+9b+3c+d=9$-------------------(4)

Now we can write this ,

A=$\begin{pmatrix}1 &1 & 1& 1\\ 3&2 & 1&0 \\ 8& 4&2 &1 \\ 27& 9& 3&1 \end{pmatrix}$

$X=\begin{pmatrix} a\\ b\\ c\\ d\end{pmatrix}$

$b=$$\begin{pmatrix} 1\\ 0\\ 2\\ 9\end{pmatrix}$

So,We can write ,

$AX=b$

After solving this System of equation we got unique solution which is ,

$a=1,b=-3,c=3,d=0$ .

Now ,$f’(2)=12a+4b+c$

                  =$12*1+4(-3)+3=3$

Correct answer is $3$.

(I am leaving the calculation part upto you if needed any help in solving please comment )