Answer
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Hint – In this particular question use the concept that if f(x) = 0 then there exists a point c such that f (c) = 0, later on use the concepts that the ordered pairs are the required solution of the given equation so use these concepts to reach the solution of the question.
Complete step-by-step answer:
Given equation, $f\left( x \right) = {x^2} + \lambda x + \mu \cos x$............. (1)
$\lambda $ is a positive integer $\mu $ is a real number.
Now we have to find out the number of ordered pairs ($\lambda ,\mu $) such that f(x) = 0 and f (f(c)) = 0 have the same set of real roots.
So, f (x) = 0............... (2)
Now in place of x put c in equation (2) we have,
f (c) = 0 ............. (3)
And it is also given that f (f(c)) = 0................... (4)
Now in place of x put f (c) in equation (1) we have,
$ \Rightarrow f\left( {f\left( c \right)} \right) = {\left[ {f\left( c \right)} \right]^2} + \lambda f\left( c \right) + \mu \cos f\left( c \right)$
Now substitute the values from equation (3) and (4) in the above equation we have,
$ \Rightarrow 0 = {\left[ 0 \right]^2} + \lambda \left( 0 \right) + \mu \cos \left( 0 \right)$
Now as we know that the value of cos0 is 1 so we have,
$ \Rightarrow 0 = 0 + 0 + \mu \left( 1 \right)$
$ \Rightarrow \mu = 0$
So the value of $\mu $ is fixed.
Now it is given that f(x) = 0 and f (f(c)) = 0 have the same set of real roots.
So from equation (1) we have,
$ \Rightarrow f\left( x \right) = {x^2} + \lambda x + \left( 0 \right)\cos x$
$ \Rightarrow f\left( x \right) = {x^2} + \lambda x$
Now from equation (2) we have,
$ \Rightarrow 0 = {x^2} + \lambda x$
$ \Rightarrow x\left( {x + \lambda } \right) = 0$
$ \Rightarrow x = 0, - \lambda $
As it is given that $\lambda $ is a positive integer and the value of $\mu $ is fixed which is zero.
So for the fixed value of $\mu $ the number of ordered pairs ($\lambda ,\mu $) = 1
So there is only one ordered pair.
So this is the required answer.
Hence option (B) is the correct answer.
Note – Whenever we face such types of question the key concept we have to remember is that the value of cos 0 is 1, so first find out the simplified equation using the given properties as above then use cos 0 =1 as above and find out the value of $\mu $ as above then substitute this value in equation (1) and solve for x, so as the value of $\mu $ is fixed so the number of ordered pairs is only 1.
Complete step-by-step answer:
Given equation, $f\left( x \right) = {x^2} + \lambda x + \mu \cos x$............. (1)
$\lambda $ is a positive integer $\mu $ is a real number.
Now we have to find out the number of ordered pairs ($\lambda ,\mu $) such that f(x) = 0 and f (f(c)) = 0 have the same set of real roots.
So, f (x) = 0............... (2)
Now in place of x put c in equation (2) we have,
f (c) = 0 ............. (3)
And it is also given that f (f(c)) = 0................... (4)
Now in place of x put f (c) in equation (1) we have,
$ \Rightarrow f\left( {f\left( c \right)} \right) = {\left[ {f\left( c \right)} \right]^2} + \lambda f\left( c \right) + \mu \cos f\left( c \right)$
Now substitute the values from equation (3) and (4) in the above equation we have,
$ \Rightarrow 0 = {\left[ 0 \right]^2} + \lambda \left( 0 \right) + \mu \cos \left( 0 \right)$
Now as we know that the value of cos0 is 1 so we have,
$ \Rightarrow 0 = 0 + 0 + \mu \left( 1 \right)$
$ \Rightarrow \mu = 0$
So the value of $\mu $ is fixed.
Now it is given that f(x) = 0 and f (f(c)) = 0 have the same set of real roots.
So from equation (1) we have,
$ \Rightarrow f\left( x \right) = {x^2} + \lambda x + \left( 0 \right)\cos x$
$ \Rightarrow f\left( x \right) = {x^2} + \lambda x$
Now from equation (2) we have,
$ \Rightarrow 0 = {x^2} + \lambda x$
$ \Rightarrow x\left( {x + \lambda } \right) = 0$
$ \Rightarrow x = 0, - \lambda $
As it is given that $\lambda $ is a positive integer and the value of $\mu $ is fixed which is zero.
So for the fixed value of $\mu $ the number of ordered pairs ($\lambda ,\mu $) = 1
So there is only one ordered pair.
So this is the required answer.
Hence option (B) is the correct answer.
Note – Whenever we face such types of question the key concept we have to remember is that the value of cos 0 is 1, so first find out the simplified equation using the given properties as above then use cos 0 =1 as above and find out the value of $\mu $ as above then substitute this value in equation (1) and solve for x, so as the value of $\mu $ is fixed so the number of ordered pairs is only 1.
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