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
\[\begin{align}
  & {{(k+1)}^{2}}\le {{(\sqrt{{{k}^{2}}+9})}^{2}} \\
 & {{k}^{2}}+1+2k\le {{k}^{2}}+9 \\
 & 2k\le 9-1 \\
 & 2k\le 8 \\
 & k\le 4 \\
\end{align}\]
Also,
\[\begin{align}
  & {{(k+1)}^{2}}\ge {{(-\sqrt{{{k}^{2}}+9})}^{2}} \\
 & {{k}^{2}}+1+2k\le {{k}^{2}}+9 \\
 & 2k\le 9-1 \\
 & 2k\le 8 \\
 & k\le 4 \\
\end{align}\]
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 \[{{k}^{2}}+1+2k\le {{k}^{2}}+9\] .