SATHEE: Functions 1 Question 2

Functions 1 Question 2

2. Let $f (x) = a^{x} (a > 0)$ be written as $f (x) = f_{1} (x) + f_{2} (x)$ , where $f_{1} (x)$ is an even function and $f_{2} (x)$ is an odd function. Then $f_{1} (x + y) + f_{1} (x - y)$ equals

(2019 Main, 8 April II)

(a) $2 f_{1} (x + y) \cdot f_{2} (x - y)$

(b) $2 f_{1} (x + y) \cdot f_{1} (x - y)$

(d) $2 f_{1} (x) \cdot f_{1} (y)$

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Answer:

Correct Answer: 2. (d)

Solution:

Given, function $f (x) = a^{x}, a > 0$ is written as sum of an even and odd functions $f_{1} (x)$ and $f_{2} (x)$ respectively.

Clearly, $f_{1} (x) = \frac{a^{x} + a^{- x}}{2}$ and $f_{2} (x) = \frac{a^{x} - a^{- x}}{2}$

So, $f_{1} (x + y) + f_{1} (x - y)$

$= \frac{1}{2} [a^{x + y} + a^{- (x + y)}] + \frac{1}{2} [a^{x - y} + a^{- (x - y)}]$

$= \frac{1}{2} a^{x} a^{y} + \frac{1}{a^{x} a^{y}} + \frac{a^{x}}{a^{y}} + \frac{a^{y}}{a^{x}}$

$= \frac{1}{2} a^{x} a^{y} + \frac{1}{a^{y}} + \frac{1}{a^{x}} \frac{1}{a^{y}} + a^{y}$

$\begin{matrix} = \frac{1}{2} a^{x} + \frac{1}{a^{x}} a^{y} + \frac{1}{a^{y}} \\ = 2 \frac{a^{x} + a^{- x}}{2} \frac{a^{y} + a^{- y}}{2} = 2 f_{1} (x) \cdot f_{1} (y) \end{matrix}$

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Functions 1 Question 2

2. Let $f (x) = a^{x} (a > 0)$ be written as $f (x) = f_{1} (x) + f_{2} (x)$ , where $f_{1} (x)$ is an even function and $f_{2} (x)$ is an odd function. Then $f_{1} (x + y) + f_{1} (x - y)$ equals

Answer:

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Functions 1 Question 2

2. Let f(x)=ax(a>0) be written as f(x)=f1(x)+f2(x), where f1(x) is an even function and f2(x) is an odd function. Then f1(x+y)+f1(x−y) equals

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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2. Let $f (x) = a^{x} (a > 0)$ be written as $f (x) = f_{1} (x) + f_{2} (x)$ , where $f_{1} (x)$ is an even function and $f_{2} (x)$ is an odd function. Then $f_{1} (x + y) + f_{1} (x - y)$ equals