Question

If the equation $kcosx-3sinx=k+1$ has a solution for x then

A.$[4,\infty )$

B.$[-4,4]$

C.$(-4,4)$

D.$(-\infty ,4]$

Answer 1

Hint: We have one property that the maximum value and minimum value of the expression, $a\cos x+b\sin x$ is $\sqrt{{{a}^{2}}+{{b}^{2}}}$ and $-\sqrt{{{a}^{2}}+{{b}^{2}}}$ . Using this property find the maximum and minimum value of $kcosx-3sinx$ . Now, $\left( k+1 \right)$ should lie between the maximum and minimum value of the expression $kcosx-3sinx$ and solve it further.

Complete step-by-step answer:
According to the question, we have the expression,
$kcosx-3sinx=k+1$ ……………….(1)
We know the property that the maximum and minimum of the expression $a\cos x+b\sin x$ is
$\sqrt{{{a}^{2}}+{{b}^{2}}}$ and $-\sqrt{{{a}^{2}}+{{b}^{2}}}$ .
$a\cos x+b\sin x$ ……………….(2)
Maximum value = $\sqrt{{{a}^{2}}+{{b}^{2}}}$ ………………..(3)
Minimum value = $-\sqrt{{{a}^{2}}+{{b}^{2}}}$ …………………..(4)
Replacing a by k and b by -3 in equation (2), equation (3), and equation (4), we get
$kcosx-3sinx$
Maximum value = $\sqrt{{{k}^{2}}+{{(-3)}^{2}}}=\sqrt{{{k}^{2}}+9}$ ………………..(5)
Minimum value = $-\sqrt{{{k}^{2}}+{{(-3)}^{2}}}=-\sqrt{{{k}^{2}}+9}$ …………………..(6)
From equation (1), we have $kcosx-3sinx=k+1$ , and from equation (5) and equation (6) we have the maximum and minimum value of the expression $kcosx-3sinx$ . So, $\left( k+1 \right)$ should lie between the maximum and minimum value of the expression $kcosx-3sinx$ .
$-\sqrt{{{k}^{2}}+9}\le k+1\le \sqrt{{{k}^{2}}+9}$
$k+1\le \sqrt{{{k}^{2}}+9}$ or $k+1\ge -\sqrt{{{k}^{2}}+9}$ ………………..(7)
Squaring equation (7), we get

& {{(k+1)}^{2}}\le {{(\sqrt{{{k}^{2}}+9})}^{2}} \\\ & {{k}^{2}}+1+2k\le {{k}^{2}}+9 \\\ & 2k\le 9-1 \\\ & 2k\le 8 \\\ & k\le 4 \\\ \end{aligned}$$ Also, $$\begin{aligned} & {{(k+1)}^{2}}\ge {{(-\sqrt{{{k}^{2}}+9})}^{2}} \\\ & {{k}^{2}}+1+2k\le {{k}^{2}}+9 \\\ & 2k\le 9-1 \\\ & 2k\le 8 \\\ & k\le 4 \\\ \end{aligned}$$ Solving both inequalities we get $$k\le 4$$ . Hence, the correct option is D. Note: In this question, one might make a mistake in solving inequality $$k+1\ge -\sqrt{{{k}^{2}}+9}$$ . After squaring both sides one might write it as $${{k}^{2}}+1+2k\ge {{k}^{2}}+9$$ which is wrong. We should remember that we sign is reversed then inequality sign changes. So, when we square both sides, then the negative sign will change into a positive sign. Therefore, we have to change the inequality sign and after squaring the equation should be written as $${{k}^{2}}+1+2k\le {{k}^{2}}+9$$ .