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গণিত প্রশ্ন

8.

If F(x)=∣f(x)ϕ(x)g(x)ϕ(x)∣, then prove that, F(x+h)−F(x)=∣f(x+h)−f(x)ϕ(x+h)g(x+h)−g(x)ϕ(x+h)∣+∣f(x)ϕ(x+h)−ϕ(x)g(x)ϕ(x+h)−ϕ(x)∣ \begin{array}{l}\text { If } F(x)=\left|\begin{array}{ll}f(x) & \phi(x) \ g(x) & \phi(x)\end{array}\right| \text {, then prove that, } F(x+h)-F(x)= \ \left|\begin{array}{cc}f(x+h)-f(x) & \phi(x+h) \ g(x+h)-g(x) & \phi(x+h)\end{array}\right|+\left|\begin{array}{ll}f(x) & \phi(x+h)-\phi(x) \ g(x) & \phi(x+h)-\phi(x)\end{array}\right|\end{array}

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Proof:

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

Therefore,
F(x+h)=∣f(x+h)ϕ(x+h)g(x+h)ϕ(x+h)∣=ϕ(x+h)[f(x+h)−g(x+h)]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)=ϕ(x+h)[f(x+h)−g(x+h)]−ϕ(x)[f(x)−g(x)]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,
D1=∣f(x+h)−f(x)ϕ(x+h)g(x+h)−g(x)ϕ(x+h)∣=ϕ(x+h)[(f(x+h)−f(x))−(g(x+h)−g(x))]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))]
=ϕ(x+h)[f(x+h)−g(x+h)]−ϕ(x+h)[f(x)−g(x)]= \phi(x+h)[f(x+h) - g(x+h)] - \phi(x+h)[f(x) - g(x)]

and
D2=∣f(x)ϕ(x+h)−ϕ(x)g(x)ϕ(x+h)−ϕ(x)∣=[ϕ(x+h)−ϕ(x)][f(x)−g(x)]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)]
=ϕ(x+h)[f(x)−g(x)]−ϕ(x)[f(x)−g(x)]= \phi(x+h)[f(x) - g(x)] - \phi(x)[f(x) - g(x)]

Adding D1D_1 and D2D_2:
D1+D2=ϕ(x+h)[f(x+h)−g(x+h)]−ϕ(x+h)[f(x)−g(x)]+ϕ(x+h)[f(x)−g(x)]−ϕ(x)[f(x)−g(x)]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)]
=ϕ(x+h)[f(x+h)−g(x+h)]−ϕ(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)=LHS= F(x+h) - F(x) = \text{LHS}

Hence proved.

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