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If f(r)(x),g(r)(x)h,(r)(x),r=1,2,3 are p...

If `f_(r)(x),g_(r)(x)h,_(r)(x),r=1,2,3` are polynomials in x such that `f_(r)(a)=g_(r)(a)-h_(r)(a)r=1,2,3` and `F(x)=|(f_(1)(x), f_(2)(x), f_(3)(x)),(g_(1)(x),g_(2)(x),g_(3)(x)),(h_(1)(x),h_(2)(x),h_(3)(x))|`, then `F'(x)` at `x=a` is ……

A

0

B

`f_(1)(a)g_(2)(a)h_(3)(a)`

C

1

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
A
We have,
`F(x)=|{:(f_(1)(x),f_(2)(x),f_(3)(x)),(g_(1)(x),g_(2)(x),g_(3)(x)),(h_(1)(x),h_(2)(x),h_(3)(x)):}|`
`:." "F'(x)=|{:(f_(1)'(x),f_(2)'(x),f_(3)'(x)),(g_(1)(x),g_(2)(x),g_(3)(x)),(h_(1)(x),h_(2)(x),h_(3)(x)):}|+|{:(f_(1)(x),f_(2)(x),f_(3)(x)),(g_(1)'(x),g_(2)'(x),g_(3)'(x)),(h_(1)(x),h_(2)(x),h_(3)(x)):}|+|{:(f_(1)(x),f_(2)(x),f_(3)(x)),(g_(1)(x),g_(2)(x),g_(3)(x)),(h_(1)'(x),h_(2)'(x),h_(3)'(x)):}|`
Since `f_(n)(a)=g_(n)(a)=h_(n)(a),n=1,2,3`. Therefore, two rows in each determinant become identical on putting `x=a.`
Hence, `F'(a)=0`.
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