Question

Let $f,g:[-1,2]\to R$ 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:$\begin{array}{|llll|}\hline e& x=-1& x=0& x=2\\ f(x)& 3& 6& 0\\ g(x)& 0& 1& -1\\ \hline\end{array}$In each of the intervals (-1,0) and (0,2) the function $(f-3g)$ never vanishes. Then the correct statement(s) is (are):

• (A) $f^{\prime }(x)-3g^{\prime }(x)=0$ has exactly three solutions in (-1,0)$\cup (0,2)$
• (B) $f^{\prime }(x)-3g^{\prime }(x)=0$ has exactly one solution in (-1,0) and exactly one solutions in (0,2)
• (C) $f^{\prime }(x)-3g^{\prime }(x)=0$ has exactly two solutions in (-1,0) and exactly two solutions in (0,2)
• (D) None of the above

Answer 1

The Correct Option is B

Explanation:
$f,g:[-1,2]\to R$$f(x)$ is twice differentiable on (-1,2)$f(-1)=3,g(-1)=0$$f(0)=6,g(0)=1$$f(2)=0,g(2)=-1$$(f-3g{)}^{\prime \prime }\ne 0$ on (-1,0) and (0,2)Number of solutions of $f^{\prime }(x)-3g^{\prime }(x)=0$ in $(-1,0)\cup (0,2)=?$Let $h(x)=f(x)-3g(x)$. Then $h(-1)=f(-1)-3g(-1)=3$$h(0)=f(0)-3g(0)=6-3(1)=3$Therefore, by Rolle's theorem, $h^{\prime }(x)$ that is, $F^{\prime }(x)-3g^{\prime }(x)=0$ has at least one root in (-1,0).Also $h(2)=f(2)-3g(2)=0-3(-1)=3$Hence, again by Rolle's theorem, $f^{\prime }(x)-3g^{\prime }(x)=0$ has at least one root in (0,2).That is, $f^{\prime }(x)-3g^{\prime }(x)=0$ has at least 2 roots in (-1,2).Since $(f-3g{)}^{\prime \prime }\ne 0$ for (-1,0) and (0,2)So, $f(x)$ has no point of inflexion in (-1,0) and (0,2) .Therefore, $(f^{\prime }-3g^{\prime })(x)\ne 0$ in (-1,0) and (0,2) , that is, $(f^{\prime }-3g^{\prime })(x)\ne 0$ exactly once in (-1,0) and exactly once in (0,2).Hence, the correct option is (B).