Question

The value \[y\] for which the equation \[4\sin x + 3\cos x = {y^2} - 6y + 14\] has a real solution, is/are:
(A) \[3\]
(B) \[5\]
(C) \[ - 3\]
(D) None of these

Answer 1

Hint: In order to solve this question, first of all we will write the expression \[4\sin x + 3\cos x\] in the inequality form with its maximum and minimum values or we can say in its range using max and min values. Then we will write \[{y^2} - 6y + 14\] instead of \[4\sin x + 3\cos x\] .After that we will take two cases individually and solve it for the value of \[y\] . And hence we will get the required result.
Here, maximum value of the expression \[4\sin x + 3\cos x\] is \[5\] and the minimum value of the expression \[4\sin x + 3\cos x\] is \[ - 5\]

Complete answer:
The given equation is:
\[4\sin x + 3\cos x = {y^2} - 6y + 14 - - - \left( i \right)\]
And we have to find the value of \[y\] for which the given line has a real solution.
Now take the left part of the equation i.e., \[4\sin x + 3\cos x\]
Here, the maximum value of the above expression is \[5\]
And the minimum value is \[ - 5\]
Therefore, we can write
\[ \Rightarrow - 5 \leqslant 4\sin x + 3\cos x \leqslant 5\]
Using equation \[\left( i \right)\] we can write
\[ \Rightarrow - 5 \leqslant {y^2} - 6y + 14 \leqslant 5\]
Now case 1:
\[ - 5 \leqslant {y^2} - 6y + 14\]
Adding \[5\] both sides, we get
\[{y^2} - 6y + 19 \geqslant 0\]
Here, this is true for all real values of \[y\]

Now case 2:
\[ \Rightarrow {y^2} - 6y + 14 \leqslant 5\]
Subtracting \[5\] from both sides, we get
\[{y^2} - 6y + 9 \leqslant 0\]
This equation can never be less than \[0\] as the coefficient of \[y\] is positive.
Therefore, this is equal to zero.
i.e., \[{y^2} - 6y + 9 = 0\]
Solving using middle term split method, we get
\[ \Rightarrow {y^2} - 3y - 3y + 9 = 0\]
\[ \Rightarrow y\left( {y - 3} \right) - 3\left( {y - 3} \right) = 0\]
On simplifying, we get
\[ \Rightarrow {\left( {y - 3} \right)^2} = 0\]
\[ \Rightarrow y = 3\]
Hence, there exist one value of \[y\] that is, \[3\] for which the equation \[4\sin x + 3\cos x = {y^2} - 6y + 14\] has a real solution.

Hence, option (A) is the correct answer.

Note:
The main concept used here is the maximum and minimum value. So, students should be careful about how to calculate maximum and minimum values of the given expression. Also, student must know the technique of writing the range of the given expression in order to tackle such problems. Also, while solving these questions, take care of inequality concept as well to get the required result.