Limit Continuity And Differentiability Ques 128
Let $f:[-\frac{1}{2}, 2] \rightarrow R$ and $g:[-\frac{1}{2}, 2] \rightarrow R$ be functions defined by $f(x)=\left[x^{2}-3\right]$ and $g(x)=x|f(x)|+| 4 x-7 | f(x)$, where $[y]$ denotes the greatest integer less than or equal to $y$ for $y \in R$. Then,
(2016 Adv.)
(a) $f$ is discontinuous exactly at three points in $\left(-\frac{1}{2}, 2\right)$
(b) $f$ is discontinuous exactly at four points in $\left(-\frac{1}{2}, 2\right)$
(c) $g$ is not differentiable exactly at four points in $\left[-\frac{1}{2}, 2\right]$
(d) $g$ is not differentiable exactly at five points in $\left[-\frac{1}{2}, 2\right]$
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Answer:
Correct Answer: 128.$(A) \rightarrow p$; (B) $\rightarrow r$
Solution:
- Given, $x=\sec \theta-\cos \theta$ and $y=\sec ^{n} \theta-\cos ^{n} \theta$ On differentiating w.r.t. $\theta$ respectively, we get
$$ \frac{d x}{d \theta}=\sec \theta \tan \theta+\sin \theta $$
$$ \begin{alignedat} & \text { and } \frac{d y}{d \theta}=n \sec ^{n-1} \theta \cdot \sec \theta \tan \theta-n \cos ^{n-1} \theta \cdot(-\sin \theta) \\ & \Rightarrow \quad \frac{d x}{d \theta}=\tan \theta(\sec \theta+\cos \theta) \\ & \text { and } \frac{d y}{d \theta}=n \tan \theta\left(\sec ^{n} \theta-\cos ^{n} \theta\right) \\ & \Rightarrow \quad \frac{d y}{d x}=\frac{n\left(\sec ^{n} \theta-\cos ^{n} \theta\right)}{\sec \theta+\cos \theta} \\ & \therefore \quad \frac{d y}{d x}{ }^{2}=\frac{n^{2}\left(\sec ^{n} \theta+\cos ^{n} \theta\right)^{2}}{(\sec \theta+\cos \theta)^{2}} \\ & =\frac{n^{2}{\left(\sec ^{n} \theta-\cos ^{n} \theta\right)^{2}+4 }}{{(\sec \theta-\cos \theta)^{2}+4 }}=\frac{n^{2}\left(y^{2}+4\right)}{\left(x^{2}+4\right)} \\ & \Rightarrow\left(x^{2}+4\right) \frac{d y}{d x}{ }^{2}=n^{2}\left(y^{2}+4\right) \\ & \text { 31. Let } \varphi(x)=\begin{array}{ccc} A(x) & B(x) & C(x) \\ A(\alpha) & B(\alpha) & C(\alpha) \\ A^{\prime}(\alpha) & B^{\prime}(\alpha) & C^{\prime}(\alpha) \end{array}\right| \end{aligned} $$
Given that, $\alpha$ is a repeated root of the quadratic equation $f(x)=0$.
$\therefore$ We must have $f(x)=(x-\alpha)^{2} \cdot g(x)$
$$ \begin{array}{ll} \therefore & \varphi^{\prime}(x)=\left|\begin{array}{ccc} A^{\prime}(x) & B^{\prime}(x) & C^{\prime}(x) \\ A(\alpha) & B(\alpha) & C^{\prime}(\alpha) \\ A^{\prime}(\alpha) & B^{\prime}(\alpha) & C^{\prime}(\alpha) \end{array}\right| \\ $\Rightarrow & \varphi^{\prime}(\alpha)=\left|\begin{array}{ccc}$$ A^{\prime}(\alpha) & B^{\prime}(\alpha) & C^{\prime}(\alpha) \ A(\alpha) & B(\alpha) & C(\alpha) \\ A^{\prime}(\alpha) & B^{\prime}(\alpha) & C^{\prime}(\alpha) \end{array}\right|=0 \end{array} $$
$\Rightarrow \quad x=\alpha$ is root of $\varphi^{\prime}(x)$.
$$ \Rightarrow \quad(x-\alpha) \text { is a factor of } \varphi^{\prime}(x) \text { also. } $$
or we can say $(x-\alpha)^{2}$ is a factor of $f(x)$.
$$ \Rightarrow \quad \varphi(x) \text { is divisible by } f(x) \text {. } $$
BETA
