Question:

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

Updated On: May 12, 2024
  • 4
  • 8
  • 10
  • 12
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The Correct Option is B

Solution and Explanation

Given $7 \cos x + 5 \sin x = 2k + 1$
We know that $-r \le a \cos\theta +b \sin\theta \le r$ where $r=\sqrt{a^{2} +b^{2}}$
$ \therefore \, \, - \sqrt{49 +25} \le 7\cos x +5 \sin x \le\sqrt{49+25}$
$ \Rightarrow - \sqrt{74} \le 2k +1 \le \sqrt{74}\Rightarrow -8.6 \le2k + 1 \le 8.6$
$ \Rightarrow -9.6 \le2k \le 7.6 \Rightarrow -4.8 \le k \le 3.8 $
$ \Rightarrow \, \, k $ can take only 8 integral values.
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Concepts Used:

General and Middle Terms

In the expansion of a binomial term (a + b) raised to the power of n, on the basis of value ‘n’, we can write the general and middle terms. Prior to getting into the general and middle terms in binomial expansion,

Expansion of Binomial Term

Here,

First-term = nC0 anb0

Second-term = nC1 an–1b1

Third-term = nC2 an–2b2

Read More: Binomial Expansion Formula

Middle Term of the Binomial Expansion

As it is known, the expansion of (a + b)n contains an (n + 1) number of terms. On the basis of value ‘n’, we can write the middle term or terms of (a + b)n.

That means, that if ‘n’ is even, there will be only one middle term and if ‘n’ is odd, there will be two middle terms.

When ‘n’ is Even

In case ‘n’ is even, then the number of terms in the expansion will be n + 1. Since n is even so n + 1 is odd. Therefore, the ((n+1+1)/2)th