Answer
Verified
424.2k+ views
Hint- We will use the basic definitions of inverse cosine to solve this question.
As we know, if $y = {\cos^{ - 1}}x$ then \[x\] must be in the range of (-1, 1) and y must be in the range of $(0,\pi ).$
Complete step-by-step solution -
Given equation is ${\cos ^{ - 1}}p + {\cos ^{ - 1}}q + {\cos ^{ - 1}}r = 3\pi $
Let ${y_1} = {\cos ^{ - 1}}p,{y_2} = {\cos ^{ - 1}}q,{y_3} = {\cos ^{ - 1}}r$
Now, we have to evaluate the values of p, q, r.
As we know if $y = {\cos^{ - 1}}x,{\text{ then - 1}} \leqslant {\text{x}} \leqslant {\text{1 and 0}} \leqslant {\text{y}} \leqslant \pi $
Hence, the given equation will hold only when each has the highest value of y.
So, ${y_1} = {y_2} = {y_3} = \pi $
$
\Rightarrow {\cos ^{ - 1}}p = {\cos ^{ - 1}}q = {\cos ^{ - 1}}r = \pi \\
\Rightarrow p = q = r = \cos \pi \\
\Rightarrow p = q = r = - 1 \\
$
Now, we have to find the value of ${p^2} + {q^2} + {r^2} + 2pqr$
By substituting the value $p = q = r = - 1$ we get
$
\Rightarrow {( - 1)^2} + {( - 1)^2} + {( - 1)^2} + 2( - 1)( - 1)( - 1) \\
\Rightarrow 3 - 2 \\
\Rightarrow 1 \\
$
Hence, the correct option is B.
Note- To solve inverse trigonometric equations remember the basic concept of solving the algebraic equations and remember the definitions of inverse trigonometric functions such as about their domain and range. The identities of inverse trigonometric functions must be remembered with their domain and range.
As we know, if $y = {\cos^{ - 1}}x$ then \[x\] must be in the range of (-1, 1) and y must be in the range of $(0,\pi ).$
Complete step-by-step solution -
Given equation is ${\cos ^{ - 1}}p + {\cos ^{ - 1}}q + {\cos ^{ - 1}}r = 3\pi $
Let ${y_1} = {\cos ^{ - 1}}p,{y_2} = {\cos ^{ - 1}}q,{y_3} = {\cos ^{ - 1}}r$
Now, we have to evaluate the values of p, q, r.
As we know if $y = {\cos^{ - 1}}x,{\text{ then - 1}} \leqslant {\text{x}} \leqslant {\text{1 and 0}} \leqslant {\text{y}} \leqslant \pi $
Hence, the given equation will hold only when each has the highest value of y.
So, ${y_1} = {y_2} = {y_3} = \pi $
$
\Rightarrow {\cos ^{ - 1}}p = {\cos ^{ - 1}}q = {\cos ^{ - 1}}r = \pi \\
\Rightarrow p = q = r = \cos \pi \\
\Rightarrow p = q = r = - 1 \\
$
Now, we have to find the value of ${p^2} + {q^2} + {r^2} + 2pqr$
By substituting the value $p = q = r = - 1$ we get
$
\Rightarrow {( - 1)^2} + {( - 1)^2} + {( - 1)^2} + 2( - 1)( - 1)( - 1) \\
\Rightarrow 3 - 2 \\
\Rightarrow 1 \\
$
Hence, the correct option is B.
Note- To solve inverse trigonometric equations remember the basic concept of solving the algebraic equations and remember the definitions of inverse trigonometric functions such as about their domain and range. The identities of inverse trigonometric functions must be remembered with their domain and range.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Using the following information to help you answer class 12 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Summary of the poem Where the Mind is Without Fear class 8 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Write an application to the principal requesting five class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE