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If `|x^n x^(n+2)x^(2n)1x^a a x^(n+5)x^(a+6)x^(2n+5)|=0,AAx in R ,w h e r en in N ,`then value of `a`is`n`b. `n-1`c. `n+1`d. none of theseA. nB. n-1C. n+1D. none of these |
Answer» Correct Answer - C Taking `x^(5)` common from last row we bet `x^(5) |{:(x^(n),,x^(n+2),,x^(2n)),(1,,x^(a),,a),(x^(n),,x^(a+1),,x^(2n)):}|=0 ,AA x in R` `rArr a +1 = n+2 " or " a = n+1` (as it will make first and third row is identical ) |
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