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Let $$f$$ and $$g$$ be the twice differentiable functions on $$\mathbb{R}$$ such that $$f''(x)=g''(x)+6x$$ $$f'(1)=4g'(1)-3=9$$ $$f(2)=3g(2)=12$$. Then which of the following is NOT true?

JEE · Math · previous-year question

  1. A.$$g(-2)-f(-2)=20$$
  2. B.There exists $$x_0\in(1,3/2)$$ such that $$f(x_0)=g(x_0)$$
  3. C.$$|f'(x)-g'(x)| < 6\Rightarrow -1 < x < 1$$
  4. D.If $$-1 < x < 2$$, then $$|f(x)-g(x)| < 8$$correct

Answer

D. If $$-1 < x < 2$$, then $$|f(x)-g(x)| < 8$$

Explanation

$$f''(x) = g''(x) + 6x$$ $$ \Rightarrow f'(x) = g'(x) + 3{x^2} + C$$ $$f'(1) = g'(1) + 3 + C$$ $$ \Rightarrow g = 3 + 3 + C \Rightarrow C = 3$$ $$ \Rightarrow f'(x) = g'(x) + 3{x^2} + 3$$ $$ \Rightarrow f(x) = g(x) + {x^2} + 3x + C'$$ $$x = 2$$ $$f(2) = g(2) + 14 + C'$$ $$12 = 4 + 14 + C'$$ $$ \Rightarrow C' = - 6$$ $$ \Rightarrow f(x) = g(2) + {x^3} + 3x - 6$$ $$f( - 2) = g( - 2) - 8 - 6 - 6$$ $$g( - 2) - f( - 2) = 20$$ $$f'(x) - g'(x) = 3{x^2} + 3$$ $$x \in ( - 1,1)$$ $$3{x^2} + 3 \in (0,6)$$ $$ \Rightarrow f'(x) - g'(x) \in (0,6)$$ $$f(x) - g(x) = {x^3} + 3x - 6$$ At $$x = - 1$$ $$|f( - 1) - g( - 1)| = 10$$ $$\therefore$$ Option (4) is false.

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