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f(x) and g(x) are two differentiable functions on [0, 2] such that f''(x) - g''(x) = 0, f'(1) = 2, g'(1) = 4, f(2) = 3, g(2) = 9 then f(x) - g(x) at x = $${3 \over 2}$$ is

JEE · Math · previous-year question

  1. A.0
  2. B.2
  3. C.10
  4. D.-5correct

Answer

D. -5

Explanation

To find the value of $$f(x) - g(x)$$ at $$x = \frac{3}{2}$$, we need to use the given conditions and properties of differentiable functions. First, we are told that: $$f''(x) - g''(x) = 0$$ This implies that: $$f''(x) = g''(x)$$ Since the second derivatives of both functions are equal, their difference, $$f'(x) - g'(x)$$, must be a linear function. Let’s denote it as: $$f'(x) - g'(x) = k$$ We'll find the constant $$k$$ using the initial conditions of the derivatives: $$f'(1) = 2$$ $$g'(1) = 4$$ Thus, $$f'(1) - g'(1) = 2 - 4 = -2$$ Therefore, $$f'(x) - g'(x) = -2$$ Integrating the above result, we get: $$f(x) - g(x) = -2x + C$$ To determine the constant $$C$$, we use the values of the functions at $$x = 2$$: $$f(2) = 3$$ $$g(2) = 9$$ Thus, $$f(2) - g(2) = 3 - 9 = -6$$ Therefore, $$-2 \cdot 2 + C = -6$$ $$-4 + C = -6$$ $$C = -2$$ So the expression for $$f(x) - g(x)$$ is: $$f(x) - g(x) = -2x - 2$$ We need to find $$f\left( \frac{3}{2} \right) - g\left( \frac{3}{2} \right)$$: $$f\left( \frac{3}{2} \right) - g\left( \frac{3}{2} \right) = -2 \cdot \frac{3}{2} - 2$$ $$= -3 - 2$$ $$= -5$$ Thus, the value of $$f(x) - g(x)$$ at $$x = \frac{3}{2}$$ is -5.

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