SATHEE: Limit Continuity and Differentiability 7 Question 30

Limit Continuity and Differentiability 7 Question 30

30. Let $f : - \frac{1}{2}, 2 \to R$ and $g : - \frac{1}{2}, 2 \to R$ be functions defined by $f (x) = [x^{2} - 3]$ and $g (x) \neq x | f (x) + | 4 x - 7 ∣ f (x)$ , 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 $- \frac{1}{2}, 2$

(b) $f$ is discontinuous exactly at four points in $- \frac{1}{2}, 2$

(d) $g$ is not differentiable exactly at five points in $- \frac{1}{2}, 2$

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Answer:

Correct Answer: 30. $(A) \to p$ ; (B) $\to r$

Solution:

Given, $x = \sec θ - \cos θ$ and $y = \sec^{n} θ - \cos^{n} θ$ On differentiating w.r.t. $θ$ respectively, we get

$\frac{d x}{d θ} = \sec θ \tan θ + \sin θ$

$\begin{aligned} and \frac{d y}{d θ} = n \sec^{n - 1} θ \cdot \sec θ \tan θ - n \cos^{n - 1} θ \cdot (- \sin θ) \\ \Rightarrow \frac{d x}{d θ} = \tan θ (\sec θ + \cos θ) \\ and \frac{d y}{d θ} = n \tan θ (\sec^{n} θ + \cos^{n} θ) \\ \Rightarrow \frac{d y}{d x} = \frac{n (\sec^{n} θ + \cos^{n} θ)}{\sec θ + \cos θ} \\ ∴ \frac{d y}{d x}^{2} = \frac{n^{2} {(\sec^{n} θ + \cos^{n} θ)}^{2}}{(\sec θ + \cos θ)^{2}} \\ = \frac{n^{2} {(\sec^{n} θ - \cos^{n} θ)}^{2} + 4}{(\sec θ - \cos θ)^{2} + 4} = \frac{n^{2} (y^{2} + 4)}{(x^{2} + 4)} \\ \Rightarrow (x^{2} + 4) \frac{d y}{d x}^{2} = n^{2} (y^{2} + 4) \\ 31. Let φ (x) = | \begin{array}{ccc} A (x) & B (x) & C (x) \\ A (α) & B (α) & C (α) \\ A^{'} (α) & B^{'} (α) & C^{'} (α) \end{array} | \end{aligned}$

Given that, $α$ is repeated root of quadratic equation $f (x) = 0$ .

$∴$ We must have $f (x) = (x - α)^{2} \cdot g (x)$

$\begin{array}{ll} ∴ & φ^{'} (x) = | \begin{array}{ccc} A^{'} (x) & B^{'} (x) & C^{'} (x) \\ A (α) & B (α) & C^{'} (α) \\ A^{'} (α) & B^{'} (α) & C^{'} (α) \end{array} | \\ \Rightarrow & φ^{'} (α) = | \begin{array}{ccc} A^{'} (α) & B^{'} (α) & C^{'} (α) \\ A (α) & B (α) & C (α) \\ A^{'} (α) & B^{'} (α) & C^{'} (α) \end{array} | = 0 \end{array}$

$\Rightarrow x = α$ is root of $φ^{'} (x)$ .

$\Rightarrow (x - α) is a factor of φ^{'} (x) also.$

or we can say $(x - α)^{2}$ is a factor of $f (x)$ .

$\Rightarrow φ (x) is divisible by f (x) .$

Limit Continuity and Differentiability 7 Question 30

30. Let $f : - \frac{1}{2}, 2 \to R$ and $g : - \frac{1}{2}, 2 \to R$ be functions defined by $f (x) = [x^{2} - 3]$ and $g (x) \neq x | f (x) + | 4 x - 7 ∣ f (x)$ , where $[y]$ denotes the greatest integer less than or equal to $y$ for $y \in R$ . Then,

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Limit Continuity and Differentiability 7 Question 30

30. Let f:−12,2→R and g:−12,2→R be functions defined by f(x)=[x2−3] and g(x)≠x|f(x)+|4x−7∣f(x), where [y] denotes the greatest integer less than or equal to y for y∈R. Then,

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30. Let $f : - \frac{1}{2}, 2 \to R$ and $g : - \frac{1}{2}, 2 \to R$ be functions defined by $f (x) = [x^{2} - 3]$ and $g (x) \neq x | f (x) + | 4 x - 7 ∣ f (x)$ , where $[y]$ denotes the greatest integer less than or equal to $y$ for $y \in R$ . Then,