SATHEE: Application of Derivatives 2 Question 2

Application of Derivatives 2 Question 2

2. Let $f (x) = e^{x} - x$ and $g (x) = x^{2} - x, \forall x \in R$ . Then, the set of all $x \in R$ , where the function $h (x) = (f \circ g) (x)$ is increasing, is

(2019 Main, 10 April II)

(a) $0, \frac{1}{2} \cup [1, \infty)$

(b) $- 1, \frac{- 1}{2} \cup \frac{1}{2}, \infty$

(d) $\frac{- 1}{2}, 0 \cup [1, \infty)$

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Solution:

The given functions are

$and g (x) = x^{2} - x, \forall x \in R$

Then, $h (x) = (f \circ g) (x) = f (g (x))$

Now, $h^{'} (x) = f^{'} (g (x)) \cdot g^{'} (x)$

$\begin{aligned} = (e^{g (x)} - 1) \cdot (2 x - 1) = (e^{(x^{2} - x)} - 1) (2 x - 1) \\ = (e^{x (x - 1)} - 1) (2 x - 1) \end{aligned}$

$∵$ It is given that $h (x)$ is an increasing function, so $h^{'} (x) \geq 0$

$\Rightarrow (e^{x (x - 1)} - 1) (2 x - 1) \geq 0$

Case I $(2 x - 1) \geq 0$ and $(e^{x (x - 1)} - 1) \geq 0$

$\Rightarrow x \geq \frac{1}{2}$ and $x (x - 1) \geq 0$

$\Rightarrow x \in [1 / 2, \infty)$ and $x \in (- \infty, 0] \cup [1, \infty)$ , so $x \in [1, \infty)$

Case II $(2 x - 1) \leq 0$ and $[e^{x (x - 1)} - 1] \leq 0$

$\Rightarrow x \leq \frac{1}{2}$ and $x (x - 1) \leq 0 \Rightarrow x \in - \infty, \frac{1}{2}$ and $x \in [0, 1]$

So, $x \in 0, \frac{1}{2}$

From, the above cases, $x \in 0, \frac{1}{2} \cup [1, \infty)$ .

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Application of Derivatives 2 Question 2

2. Let $f (x) = e^{x} - x$ and $g (x) = x^{2} - x, \forall x \in R$ . Then, the set of all $x \in R$ , where the function $h (x) = (f \circ g) (x)$ is increasing, is

Solution:

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Application of Derivatives 2 Question 2

2. Let f(x)=ex−x and g(x)=x2−x,∀x∈R. Then, the set of all x∈R, where the function h(x)=(f∘g)(x) is increasing, is

Solution:

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एनसीईआरटी पुस्तकें और समाधान

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अन्य संसाधन

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नमूना मॉक टेस्ट

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2. Let $f (x) = e^{x} - x$ and $g (x) = x^{2} - x, \forall x \in R$ . Then, the set of all $x \in R$ , where the function $h (x) = (f \circ g) (x)$ is increasing, is