SATHEE: Functions 2 Question 16

Functions 2 Question 16

17.

If $f (x) = \sin^{2} x + \sin^{2} x + \frac{π}{3} + \cos x \cos x + \frac{π}{3}$ and g $\frac{5}{4} = 1$ , then $(g \circ f) (x) = \dots$ .

(1996, 2M)

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Answer:

Correct Answer: 17. (1)

Solution:

$f (x) = \sin^{2} x + \sin^{2} (x + π / 3) + \cos x \cos (x + π / 3)$

$\Rightarrow f (x) = \sin^{2} x + (\sin x \cos π / 3 + \cos x \sin π / 3)^{2}$ $+ \cos x \cos (x + π / 3)$

$+ \cos x (\cos x \cos π / 3 - \sin x \sin π / 3)$

$\Rightarrow f (x) = \sin^{2} x + \frac{\sin^{2} x}{4} + \frac{3 \cos^{2} x}{4} + \frac{2 \cdot \sqrt{3}}{4} \sin x \cos x$ $+ \frac{\cos^{2} x}{2} - \cos x \sin x \cdot \frac{\sqrt{3}}{2}$

$= \sin^{2} x + \frac{\sin^{2} x}{4} + \frac{3 \cos^{2} x}{4} + \frac{\cos^{2} x}{2}$

$= \frac{5}{4} \sin^{2} x + \frac{5}{4} \cos^{2} x = \frac{5}{4}$

and $g \circ f (x) = g f (x) = g (5 / 4) = 1$

Alternate Solution

$f (x) = \sin^{2} x + \sin^{2} (x + π / 3) + \cos x \cos (x + π / 3)$

$\Rightarrow f^{'} (x) = 2 \sin x \cos x + 2 \sin (x + π / 3) \cos (x + π / 3)$

$- \sin x \cos (x + π / 3) - \cos x \sin (x + π / 3)$

$= \sin 2 x + \sin (2 x + 2 π / 3) - [\sin (x + x + π / 3)]$

$= 2 \sin \frac{2 x + 2 x + 2 π / 3}{2} \cdot \cos \frac{2 x - 2 x - 2 π / 3}{2}$

$- \sin (2 x + π / 3)$

$= 2 [\sin (2 x + π / 3) \cdot \cos π / 3] - \sin (2 x + π / 3)$

$= 2 \sin (2 x + π / 3) \cdot \frac{1}{2} - \sin 2 x + \frac{π}{3} = 0$

$\Rightarrow f (x) = c$ , where $c$ is a constant.

But $f (0) = \sin^{2} 0 + \sin^{2} (π / 3) + \cos 0 \cos π / 3$

Therefore, $(g \circ f) (x) = g [f (x)] = g (5 / 4) = 1$

Functions 2 Question 16

17.

Answer:

Alternate Solution

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Functions 2 Question 16

17.

Answer:

Alternate Solution

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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