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5. Continuity and Differentiation
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If the equation
${a_n}{x^{n - 1}} + \,{a_{n - 1}}{x^{n - 1}} + \,......\, + \,{a_1}x = 0,\,{a_1} \ne 0,n\, \geqslant \,2,$
has a positive root $x= \alpha ,$ then the equation
$n{a_n}{x^{n - 1}} + \,(n - 1){a_{n - 1}}{x^{n - 1}} + \,......\, + \,{a_1} = 0$
has a positive root which is
A
Equal to $\alpha$
B
Greater than or equal to $\alpha$
C
Smaller than $\alpha$
D
Greater than $\alpha$
Solution
Let $f(x)=a_0 x^n+a_1 x^{n-1}+\ldots+a_n x$
Now for $x=0$
$f(0)=0$
And as $\alpha$ is one root
$f (\alpha)=0$
Hence $f^{\prime}( x )$ will have one root in $(0, \alpha)$
As $f^{\prime}(x)=n a_n x^{n-1}+(n-1) a_{n-1} x^{n-2}+\ldots+a_1=0$
Then root of $n a_n x^{n-1}+(n-1) a_{n-1} x^{n-2}+\ldots+a_1=0$ is less than $\alpha$
Standard 12
Mathematics
