উচ্চতর গণিত ১ম পত্র প্রশ্ন

AI-তৈরি নমুনা উত্তর

Proof:

Given,
$F(x) = \begin{vmatrix} f(x) & \phi(x) \\ g(x) & \phi(x) \end{vmatrix} = \phi(x)[f(x) - g(x)]$

Therefore,
$F(x+h) = \begin{vmatrix} f(x+h) & \phi(x+h) \\ g(x+h) & \phi(x+h) \end{vmatrix} = \phi(x+h)[f(x+h) - g(x+h)]$

Now, the LHS is:
$F(x+h) - F(x) = \phi(x+h)[f(x+h) - g(x+h)] - \phi(x)[f(x) - g(x)]$

Now, evaluating the RHS:
Let,
$D_1 = \begin{vmatrix} f(x+h)-f(x) & \phi(x+h) \\ g(x+h)-g(x) & \phi(x+h) \end{vmatrix} = \phi(x+h)[(f(x+h) - f(x)) - (g(x+h) - g(x))]$
$= \phi(x+h)[f(x+h) - g(x+h)] - \phi(x+h)[f(x) - g(x)]$

and
$D_2 = \begin{vmatrix} f(x) & \phi(x+h)-\phi(x) \\ g(x) & \phi(x+h)-\phi(x) \end{vmatrix} = [\phi(x+h) - \phi(x)][f(x) - g(x)]$
$= \phi(x+h)[f(x) - g(x)] - \phi(x)[f(x) - g(x)]$

Adding $D_1$ and $D_2$:
$D_1 + D_2 = \phi(x+h)[f(x+h) - g(x+h)] - \phi(x+h)[f(x) - g(x)] + \phi(x+h)[f(x) - g(x)] - \phi(x)[f(x) - g(x)]$
$= \phi(x+h)[f(x+h) - g(x+h)] - \phi(x)[f(x) - g(x)]$
$= F(x+h) - F(x) = \text{LHS}$

Hence proved.