If functions f(x) and g(x) are continuous in [a, b] and differentiable in (a, b) then which of the following is correct if there exists at least one point c, a < c < b, such that \(\begin{vmatrix}f(a) & f(b) \\g(a) & g(b) \end {vmatrix}\)?(a) (b + a)\(\begin{vmatrix}f(a) & f”(c) \\g(a) & g”(c) \end {vmatrix}\)(b) (b – a)\(\begin{vmatrix}f(a) & f”(c) \\g(a) & g”(c) \end {vmatrix}\)(c) (b + a)\(\begin{vmatrix}f(a) & f'(c) \\g(a) & g'(c) \end {vmatrix}\)(d) (b – a)\(\begin{vmatrix}f(a) & f'(c) \\g(a) & g'(c) \end {vmatrix}\)The question was posed to me in a national level competition.Question is taken from First Order Derivative in portion Limits and Derivatives of Mathematics – Class 11

If functions f(x) and g(x) are continuous in [a, b] and differentiable

1.	If functions f(x) and g(x) are continuous in [a, b] and differentiable in (a, b) then which of the following is correct if there exists at least one point c, a < c < b, such that \(\begin{vmatrix}f(a) & f(b) \\g(a) & g(b) \end {vmatrix}\)?(a) (b + a)\(\begin{vmatrix}f(a) & f”(c) \\g(a) & g”(c) \end {vmatrix}\)(b) (b – a)\(\begin{vmatrix}f(a) & f”(c) \\g(a) & g”(c) \end {vmatrix}\)(c) (b + a)\(\begin{vmatrix}f(a) & f'(c) \\g(a) & g'(c) \end {vmatrix}\)(d) (b – a)\(\begin{vmatrix}f(a) & f'(c) \\g(a) & g'(c) \end {vmatrix}\)The question was posed to me in a national level competition.Question is taken from First Order Derivative in portion Limits and Derivatives of Mathematics – Class 11
Answer» The correct answer is (d) (b – a)\(\begin{vmatrix}f(a) & f'(c) \\g(a) & g'(c) \end {vmatrix}\) The BEST explanation: Let, F(X) = \(\begin{vmatrix}f(a) & f(b) \\g(a) & g(b) \end {vmatrix}\) = f(a)g(x) – f(x)g(a)…..(1) => F’(x) = f’(a)g’(x) – f’(x)g(a) Since, f(x) and g(x) are continuous in [a, b] and differentiable in (a, b), So, F(x) is continuous in [a, b] and differentiable in (a, b) Also from (1), F(a) = f(a)g(a) – f(a)g(a) = 0 And F(b) = f(a)g(b) – f(b)g(a) Now, by the mean value theorem, there exists at least one point c, a < c < b, such that, F’(c) = (F’(b) – F’(a)) / (b – a) => f(a) g’(c) – g(a) f’(c) = (f(a)g(b) – f(b)g(a) – 0)/b – a Or, f(a)g(b) – f(b)g(a) = (b – a)( f(a) g’(c) – g(a) f’(c)) =>\(\begin{vmatrix}f(a) & f(b) \\g(a) & g(b) \end {vmatrix}\) = (b – a)\(\begin{vmatrix}f(a) & f'(c) \\g(a) & g'(c) \end {vmatrix}\)

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