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Question: Let f(x) = \({x^2} + \lambda x + \mu \cos x\), \(\lambda \) is a positive integer \(\mu \) is a real...

Let f(x) = x2+λx+μcos⁡x{x^2} + \lambda x + \mu \cos x, λ\lambda is a positive integer μ\mu is a real number. The number of ordered pairs (λ,μ\lambda ,\mu ) for which f(x) = 0 and f (f(c)) = 0 have the same set of real roots.
(A)\left( A \right) 0
(B)\left( B \right) 1
(C)\left( C \right) 2
(D)\left( D \right) 3

Explanation

Solution

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(x)=x2+λx+μcos⁡xf\left( x \right) = {x^2} + \lambda x + \mu \cos x............. (1)
λ\lambda is a positive integer μ\mu is a real number.
Now we have to find out the number of ordered pairs (λ,μ\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,
⇒f(f(c))=[f(c)]2+λf(c)+μcos⁡f(c)\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,
⇒0=[0]2+λ(0)+μcos⁡(0)\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,
⇒0=0+0+μ(1)\Rightarrow 0 = 0 + 0 + \mu \left( 1 \right)
⇒μ=0\Rightarrow \mu = 0
So the value of μ\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,
⇒f(x)=x2+λx+(0)cos⁡x\Rightarrow f\left( x \right) = {x^2} + \lambda x + \left( 0 \right)\cos x
⇒f(x)=x2+λx\Rightarrow f\left( x \right) = {x^2} + \lambda x
Now from equation (2) we have,
⇒0=x2+λx\Rightarrow 0 = {x^2} + \lambda x
⇒x(x+λ)=0\Rightarrow x\left( {x + \lambda } \right) = 0
⇒x=0,−λ\Rightarrow x = 0, - \lambda
As it is given that λ\lambda is a positive integer and the value of μ\mu is fixed which is zero.
So for the fixed value of μ\mu the number of ordered pairs (λ,μ\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 μ\mu as above then substitute this value in equation (1) and solve for x, so as the value of μ\mu is fixed so the number of ordered pairs is only 1.