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Standard XII

Mathematics

Proof by mathematical induction

If x may ha...

Question

If $x$ may have any real value find which is the greater $x^{3} + 1$ or $x^{2} + x$ .

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Solution

$x^{3} + 1 - (x^{2} + x) = x^{3} - x^{2} - (x - 1) = (x^{2} - 1) (x - 1) = {(x - 1)}^{2} (x + 1)$
Now ${(x - 1)}^{2}$ is positive, hence
$x^{3} + 1 < x^{2} + x$ if $x + 1 < 0$ or $x^{3} + 1 > x^{2} + x$ if $x + 1 > 0$
According as $x + 1$ is positive or negative; that is, according as $x > - 1$ or $x < - 1$ .
If $x = - 1$ , the inequality becomes an equality.

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(iii). h(x) = sinx + cos, 0 < (iv). f(x) = sinx − cos x, 0 < x < 2π

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(vii).

(viii).

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$f (x) = s i n x - c o s x, 0 < x < 2 π$

$f (x) = x^{3} - 6 x^{2} + 9 x + 15$

$g (x) = \frac{x}{2} + \frac{2}{x}, x > 0$

$g (x) = \frac{1}{x^{2} + 2}$

$f (x) = x \sqrt{1 - x}, x > 0$

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If x may have any real value find which is the greater x3+1 or x2+x.

x3+1−(x2+x)=x3−x2−(x−1)=(x2−1)(x−1)=(x−1)2(x+1)Now (x−1)2 is positive, hencex3+1<x2+x if x+1<0 or x3+1>x2+x if x+1>0 According as x+1 is positive or negative; that is, according as x>−1 or x<−1.If x=−1, the inequality becomes an equality.

If $x$ may have any real value find which is the greater $x^{3} + 1$ or $x^{2} + x$ .