Q114 / 155 · WBJEE 2019 · Mathematics
Let $f$ and $g$ be differentiable on the interval $I$ and let $a, b \in I, a < b$. Then,
- Correct answerAIf $f(a)=0=f(b)$, the equation $f^{\prime}(x)+f(x) g^{\prime}(x)=0$ is solvable in $(a, b)$
- BIf $f(a)=0=f(b)$, the equation $f^{\prime}(x)+f(x) g^{\prime}(x)=0$ may not be solvable in $(a, b)$
- Correct answerCIf $g(a)=0=g(b)$, the equation $g^{\prime}(x)+k g(x)=0$ is solvable in $(a, b), k \in \mathbb{R}$
- DIf $g(a)=0=g(b)$, the equation $g^{\prime}(x)+k g(x)=0$ may not be solvable in $(a, b), k \in \mathbb{R}$.
Solution
From option (a)
We have,
$f(a)=f(b)=0$
$\Rightarrow f^{\prime}(a) \cdot f^{\prime}(b) < 0$
Again, let $h(x)=f^{\prime}(x)+f(x) g^{\prime}(x)$
$\Rightarrow h(a)=f^{\prime}(a)+fg^{\prime}(a)=f^{\prime}(a)$
and $h(b)=f^{\prime}(b)+f(b) g^{\prime}(b)=f^{\prime}(b)$
$\therefore \quad h(a) \cdot h(b)=f^{\prime}(a) \cdot f^{\prime}(b) < 0$
$\therefore h(x)=0$ has root between $(a, b)$
From option (c)
We have, $g(a)=g(b)=0$
$\Rightarrow \quad g^{\prime}(a)\cdot g^{\prime}(b) < 0$
Again, let $m(x)=g^{\prime}(x)+k g(x)$
$\Rightarrow m(a)=g^{\prime}(a)+k g(a)=g^{\prime}(a)$
$\Rightarrow \quad m(b)=g^{\prime}(b)+k g(b)=g^{\prime}(b)$
$\therefore \quad m(a) \cdot m(b)=g^{\prime}\cdot g^{\prime}(b) < 0$
$\therefore \quad m(x)=0$ has root between $(a, b)$ So, option (a) and (c) are correct.
Hence, the correct answer is options: 'A', 'C'.