SATHEE: Trigonometrical Equations 3 Question 3

Trigonometrical Equations 3 Question 3

3. The number of integral values of $k$ for which the equation $7 \cos x + 5 \sin x = 2 k + 1$ has a solution, is

(2002, 1M)

(a) 4

(b) 8

(d) 12

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Answer:

Correct Answer: 3. (b)

Solution:

We know that,

$\begin{aligned} - \sqrt{a^{2} + b^{2}} \leq a \sin x + b \cos x \leq \sqrt{a^{2} + b^{2}} \\ ∴ - \sqrt{74} \leq 7 \cos x + 5 \sin x \leq \sqrt{74} \\ i.e. - \sqrt{74} \leq 2 k + 1 \leq \sqrt{74} \end{aligned}$

Since, $k$ is integer, $- 9 < 2 k + 1 < 9$

$\Rightarrow - 10 < 2 k < 8 \Rightarrow - 5 < k < 4$

$\Rightarrow$ Number of possible integer values of $k = 8$ .

Trigonometrical Equations 3 Question 3

3. The number of integral values of $k$ for which the equation $7 \cos x + 5 \sin x = 2 k + 1$ has a solution, is

Answer:

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Trigonometrical Equations 3 Question 3

3. The number of integral values of k for which the equation 7cos⁡x+5sin⁡x=2k+1 has a solution, is

Answer:

Engineering Preparation

Medical Preparation

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Other Resources

Syllabus

Test Platform

Sample Mock Test

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NCERT Exercise Video Solution

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3. The number of integral values of $k$ for which the equation $7 \cos x + 5 \sin x = 2 k + 1$ has a solution, is