JEE Main 2017 — Differentiation Question with Solution

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

$$f^{\prime }(x)=a_{0}n{x}^{n-1}+a_1(n-1)x^{n-2}+.....+a_{n-1}$$

$$f^{\prime \prime }(x)=a_{0}n(n-1)x^{n-2}+a_1(n-1)(n-2)x^{n-3}+....+a_{n-2}$$

Now,

$$f(3x)=3^n{a}_0{x}^n+3^{n-1}{a}_1{x}^{n-1}+3^{n-2}{a}_2{x}^{n-2}+....+3a_{n-1}+a_n$$

$$f^{\prime }(x).f^{\prime \prime }(x)=[a_{0}n{x}^{n-1}+a_1(n-1)x^{n-2}+....+a_{n-1}]$$

$$[a_{0}n(n-1)x^{n-2}+a_1(n-1)(n-2)x^{n-3}+....+a_{n-2}]$$

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 $$3^n{a}_0=a_0^2{n}^2(n-1)$$

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

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

$$f^{\prime }(x)=\frac{9}{2}{x}^2+2a_{1}x+a_2$$

$$f^{\prime \prime }(x)=9x+2a_1$$

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

Now, $$f(3x)=f^{\prime }(x).f^{\prime \prime }(x)$$

$$\Rightarrow \frac{81}{2}{x}^3+9a_1{x}^2+3a_{2}x+a_3$$

$$\Rightarrow \left(\frac{9}{2}{x}^2+2a_{1}x+a_2\right)(9x+2a_1)$$

$$\Rightarrow \frac{81}{2}{x}^3+9a_1{x}^2+3a_{2}x+a_3=\frac{81}{2}{x}^3+[9a_1+18a_1]x^2+[4a_1^2+9a_2]x+2a_1{a}_2$$

Comparing the coefficients, we get

$$9a_1=27a_1$$

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

$$\Rightarrow a_2=0$$

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

$$f^{\prime }(x)=\frac{9}{2}{x}^2$$

$$f^{\prime \prime }(x)=9x$$

Hence, $$f^{\prime \prime }(2)-f^{\prime }(x)=18-18=0$$