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The number of polynomials $p: R \rightarrow R$ satisfying $p(0)=0, p(x)>x^2$ for all $x \neq 0$ and $p^{\prime \prime}(0)=\frac{1}{2}$ is
$0$
$1$
more than $1,$ but finite
infinite
Solution
(a)
We have,
$p(x) > x^2, p(0)=0, p^{\prime \prime}(0)=\frac{1}{2}$
Let $g(x)=p(x)-x^2$
$g(x) > 0, \forall x \neq 0$
$\text { and } \quad g(0)=p(0)-0=0$
$\Rightarrow x=0 \text { should be minima. }$
$\because g^{\prime \prime}(x) \text { should be } \geq 0 \text { at } x=0$
$\text { Now, } \quad g^{\prime}(x)=p^{\prime}(x)-2 x$
$\quad g^{\prime \prime}(x)^2=p^{\prime \prime}(x)-2$
$g^{\prime \prime}(0)=p^{\prime \prime}(0)-2=\frac{1}{2}-2=-\frac{3}{2}$
But $g^{\prime \prime}(0) \geq 0$
$\because$ No polynomial exists.
Similar Questions
Let $f, g:[-1,2] \rightarrow R$ be continuous functions which are twice differentiable on the interval $(-1,2)$. Let the values of $f$ and $g$ at the points $-1.0$ and $2$ be as given in the following table:
$x=-1$ | $x=0$ | $x=2$ | |
$f(x)$ | $3$ | $6$ | $0$ |
$g(x)$ | $0$ | $1$ | $-1$ |
In each of the intervals $(-1,0)$ and $(0,2)$ the function $(f-3 g)^{\prime \prime}$ never vanishes. Then the correct statement(s) is(are)
$(A)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly three solutions in $(-1,0) \cup(0,2)$
$(B)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly one solution in $(-1,0)$
$(C)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly one solution in $(0,2)$
$(D)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly two solutions in $(-1,0)$ and exactly two solutions in $(0,2)$