SATHEE: Functions 1 Question 13

Functions 1 Question 13

14.

The values of $f (x) = 3 \sin \sqrt{\frac{π^{2}}{16} - x^{2}}$ lie in the interval…

(1983, 2M)

True/False

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Answer:

Correct Answer: 14. $0, \frac{3}{\sqrt{2}}$

Solution:

Given, $f (x) = 3 \sin \sqrt{\frac{π^{2}}{16} - x^{2}}$

$\Rightarrow$ Domain $\in - \frac{π}{4}, \frac{π}{4}$

$∴$ For range, $f^{'} (x) = 3 \cos \sqrt{\frac{π^{2}}{16} - x^{2}} \cdot \frac{1 (- 2 x)}{2 \sqrt{\frac{π^{2}}{16} - x^{2}}} = 0$

Where, $\cos \sqrt{\frac{π^{2}}{16} - x^{2}} = 0$ or $x = 0$

neglecting $\cos \sqrt{\frac{π^{2}}{16} - x^{2}} = 0 \Rightarrow \frac{π^{2}}{16} - x^{2} = \frac{π^{2}}{4}$

$\Rightarrow x^{2} = - \frac{3 π^{2}}{16}$ , never possible

$\Rightarrow x = 0$

Thus,

$f (0) = 3 \sin \frac{π}{4} = \frac{3}{\sqrt{2}}$

and

$f - \frac{π}{4} = f \frac{π}{4} = 0$

Hence, range $\in 0, \frac{3}{\sqrt{2}}$

Functions 1 Question 13

14.

True/False

Answer:

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Functions 1 Question 13

14.

True/False

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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