Question

If f(\theta ) = \left| {\begin{array}{*{20}{c}} 1&{\cos \theta }&1 \\\ { - \sin \theta }&1&{ - \cos \theta } \\\ { - 1}&{\sin \theta }&1 \end{array}} \right| and A and B are respectively the maximum and minimum values of$f(\theta )$, then $(A,B)$ is equal to
A. $(3, - 1)$
B. $(4,2 - \sqrt 2 )$
C. $(2 + \sqrt 2 ,2 - \sqrt 2 )$
D. $(2 + \sqrt 2 , - 1)$

Answer 1

We first find the determinant of the given matrix and then using the trigonometric identities find the simplest form of the determinant. In the end we find the maximum and minimum values by comparing them with suitable identity.

Formula used:
a) We use the formula to solve a determinant

a&b;&c; \\\ d&e;&f; \\\ g&h;&i; \end{array}} \right| = a\left| {\begin{array}{*{20}{c}} e&f; \\\ h&i; \end{array}} \right| - b\left| {\begin{array}{*{20}{c}} d&f; \\\ g&i; \end{array}} \right| + c\left| {\begin{array}{*{20}{c}} d&e; \\\ g&h; \end{array}} \right|.$$ Then use the following formula to calculate the determinant of order $2 \times 2.$ $$\left| {\begin{array}{*{20}{c}} a&b; \\\ c&d; \end{array}} \right| = ad - bc.$$ b) $${\cos ^2}\theta - {\sin ^2}\theta = 2\cos \theta $$ c) For any angle x, $$\sqrt 2 \leqslant \sin x + \cos x \leqslant \sqrt 2 $$ **Complete step-by-step answer:** We have a determinant of order $$3 \times 3$$. We have the determinant $$f(\theta ) = \left| {\begin{array}{*{20}{c}} 1&{\cos \theta }&1 \\\ { - \sin \theta }&1&{ - \cos \theta } \\\ { - 1}&{\sin \theta }&1 \end{array}} \right|$$ Use the method to solve the determinant. Here we multiply the terms in the first row to their respective $2 \times 2$determinants and then solve the $2 \times 2$ determinants. $$ \Rightarrow f(\theta ) = 1\left| {\begin{array}{*{20}{c}} 1&{ - \cos \theta } \\\ {\sin \theta }&1 \end{array}} \right| - \cos \theta \left| {\begin{array}{*{20}{c}} { - \sin \theta }&{ - \cos \theta } \\\ { - 1}&1 \end{array}} \right| + 1\left| {\begin{array}{*{20}{c}} { - \sin \theta }&1 \\\ { - 1}&{\sin \theta } \end{array}} \right|$$ Now from the method of solving determinant of order $2 \times 2$ We use the following formula to calculate the determinant of order $2 \times 2.$ $$\left| {\begin{array}{*{20}{c}} a&b; \\\ c&d; \end{array}} \right| = ad - bc.$$ $$ \Rightarrow f(\theta ) = 1[1 - ( - \sin \theta \cos \theta )] - \cos \theta [ - \sin \theta - (\cos \theta )] + 1[ - {\sin ^2}\theta - ( - 1)]$$ Open the brackets and multiply the terms

\Rightarrow f(\theta ) = 1[1 + \sin \theta \cos \theta ] - \cos \theta [ - \sin \theta - \cos \theta ] + 1[ - {\sin ^2}\theta + 1] \\
\Rightarrow f(\theta ) = 1 + \sin \theta \cos \theta + \cos \theta \sin \theta + ({\cos ^2}\theta - {\sin ^2}\theta ) + 1 \\

Use the formula $${\cos ^2}\theta - {\sin ^2}\theta = 2\cos \theta $$ on the right hand side of the equation. $$ \Rightarrow f(\theta ) = 2 + 2\sin \theta \cos \theta + \cos 2\theta $$ Now write $$2\cos \theta \sin \theta = \sin 2\theta $$ on the right hand side of the equation. $$ \Rightarrow f(\theta ) = 2 + \sin 2\theta + \cos 2\theta $$ Now we will find the minimum and maximum values of $$f(\theta ) = 2 + \sin 2\theta + \cos 2\theta $$. We know that $$\sqrt 2 \leqslant \sin x + \cos x \leqslant \sqrt 2 $$ So, x be any angle, $$\sqrt 2 \leqslant \sin 2\theta + \cos 2\theta \leqslant \sqrt 2 $$ Therefore, minimum value of $$f(\theta ) = 2 + \sin 2\theta + \cos 2\theta $$ is $$f(\theta ) = 2 - \sqrt 2 $$ And, maximum value of $$f(\theta ) = 2 + \sin 2\theta + \cos 2\theta $$ is $$f(\theta ) = 2 + \sqrt 2 $$ So, $$B = 2 - \sqrt 2 ,A = 2 + \sqrt 2 $$ So, $$(A,B) = (2 + \sqrt 2 ,2 - \sqrt 2 )$$ **So, the correct answer is “Option C”.** **Note:** Students many times make mistakes when calculating the determinant as they sometimes repeat the negative sign which we put while calculating the second determinant. Also, for minimum and maximum values always use previous results which you are familiar with.Students should remember the method for calculating the determinant value and should know the range of trigonometric functions i.e For any angle x, $$\sqrt 2 \leqslant \sin x + \cos x \leqslant \sqrt 2 $$ for solving these types of questions.