Question

If \[ \Delta= \begin{vmatrix} f(x) & f\!\left(\dfrac{1}{x}\right)+f(x)\\ 1 & f\!\left(\dfrac{1}{x}\right) \end{vmatrix} =0 \] where \(f(x)\) is a polynomial and \(f(2)=17\), then \(f(5)=\) ?

• A. 624
• B. 626
• C. 82
• D. 79

Hint:

When a polynomial satisfies a functional relation involving \(x\) and \(\frac{1}{x}\), try assuming the lowest-degree polynomial (linear) to eliminate variable dependence.

Answer 1

The Correct Option is C

Step 1: Evaluate the determinant condition. \[ \Delta = f(x)\,f\!\left(\frac{1}{x}\right) -1\big[f\!\left(\frac{1}{x}\right)+f(x)\big] =0 \] \[ \Rightarrow f(x)f\!\left(\frac{1}{x}\right) - f\!\left(\frac{1}{x}\right) - f(x)=0 \]
Step 2: Rearrange the expression. \[ f(x)f\!\left(\frac{1}{x}\right) - f\!\left(\frac{1}{x}\right) - f(x)=0 \] Add 1 to both sides: \[ \big(f(x)-1\big)\big(f\!\left(\tfrac{1}{x}\right)-1\big)=1 \]
Step 3: Use the fact that \(f(x)\) is a polynomial. Since \(f(x)\) is a polynomial, \(f\!\left(\frac{1}{x}\right)\) is also finite only if \[ f(x)=ax+b \] (a constant or linear polynomial).
Step 4: Assume \(f(x)=ax+b\). Then: \[ f\!\left(\frac{1}{x}\right)=\frac{a}{x}+b \] Substitute into: \[ \big(f(x)-1\big)\big(f\!\left(\tfrac{1}{x}\right)-1\big)=1 \] \[ (ax+b-1)\left(\frac{a}{x}+b-1\right)=1 \] For this to be independent of \(x\), we must have: \[ a=b-1 \]
Step 5: Use the given value \(f(2)=17\). \[ f(2)=2a+b=17 \] Substitute \(a=b-1\): \[ 2(b-1)+b=17 \Rightarrow 3b=19 \Rightarrow b=\frac{19}{3} \] \[ a=\frac{16}{3} \]
Step 6: Find \(f(5)\). \[ f(5)=5a+b =5\left(\frac{16}{3}\right)+\frac{19}{3} =\frac{80+19}{3} =\frac{99}{3} =82 \]
Hence, the required value is \[ \boxed{82} \]