QJEEM17Maf195: JEE Main 2017 Mathematics - Class 12 JEEM17Maf195 Solution: Let f be a polynomial function such that f (3x) = f ' (x) . f…

Let $f(x) = {a_0}{x^n} + {a_1}{x^{n - 1}} + {a_2}{x^{n - 1}} + \,\,....\,\, + {a_{n - 1}}x + {a_n}$

$f'(x) = {a_0}n{x^{n - 1}} + {a_1}(n - 1){x^{n - 2}} + \,\,.....\,\, + {a_{n - 1}}$

$f''(x) = {a_0}n(n - 1){x^{n - 2}} + {a_1}(n - 1)(n - 2){x^{n - 3}} + \,\,....\,\, + {a_{n - 2}}$

Now,

$f(3x) = {3^n}{a_0}{x^n} + {3^{n - 1}}{a_1}{x^{n - 1}} + {3^{n - 2}}{a_2}{x^{n - 2}} + \,\,....\,\, + 3{a_{n - 1}} + {a_n}$

$f'(x)\,.\,f''(x) = [{a_0}n{x^{n - 1}} + {a_1}(n - 1){x^{n - 2}} + \,\,....\,\, + {a_{n - 1}}]$

$[{a_0}n(n - 1){x^{n - 2}} + {a_1}(n - 1)(n - 2){x^{n - 3}} + \,\,....\,\, + {a_{n - 2}}]$

Comparing highest powers of x, we get

${3^n}{a_0}{x_n} = a_0^2(n - 1){x^{n - 1 + n - 2}} = a_0^2{n^2}(n - 1){x^{2n - 3}}$

Therefore, $2n - 3 = n$

$\Rightarrow$ n = 3 and ${3^n}{a_0} = a_0^2{n^2}(n - 1)$

$\Rightarrow {a_0} = 27 = {3 \over 2}$

Therefore, $f(x) = {3 \over 2}{x^3} + {a_1}{x^2} + {a_2}x + {a_3}$

$f'(x) = {9 \over 2}{x^2} + 2{a_1}x + {a_2}$

$f''(x) = 9x + 2{a_1}$

$f(3x) = {{81} \over 2}{x^3} + 9{a_1}{x^2} + 3{a_2}x + {a_3}$

Now, $f(3x) = f'(x)\,.\,f''(x)$

$\Rightarrow {{81} \over 2}{x^3} + 9{a_1}{x^2} + 3{a_2}x + {a_3}$

$\Rightarrow \left( {{9 \over 2}{x^2} + 2{a_1}x + {a_2}} \right)(9x + 2{a_1})$

$\Rightarrow {{81} \over 2}{x^3} + 9{a_1}{x^2} + 3{a_2}x + {a_3} = {{81} \over 2}{x^3} + [9{a_1} + 18{a_1}]{x^2} + [4a_1^2 + 9{a_2}]x + 2{a_1}{a_2}$

Comparing the coefficients, we get

$9{a_1} = 27{a_1}$

$\Rightarrow {a_1} = 0,\,3{a_2} = 4a_1^2 + 9{a_2} = 9{a_2}$

$\Rightarrow {a_2} = 0$

Therefore, $f(x) = {3 \over 2}{x^3}$

$f'(x) = {9 \over 2}{x^2}$

$f''(x) = 9x$

Hence, $f''(2) - f'(x) = 18 - 18 = 0$