Derivatives

The correct answer is (B) : (1, 0)



Now at x = ,
= 1 =
∴ f(x) is continuous in [0, 2π]
Now, at x = π
L.H.D = = 0
R.H.D = =

∴ f(x) is not differentiable at x = π
∴ (m, n) = (1, 0)

The correct answer is 4
Suppose h(x) = f(x).g′(x)
As f(x) is even


and g(x) is even ⇒ g′(x) is odd
and g(1) = 2 ensures one root of g′(x) is 0.
So , h(x) = f(x).g′(x) has minimum fives zeroes
Therefore h′(x) = f′(x).g′(x) + f(x).g′(x)=0,
has minimum 4 zeroes

Given, .


Substituting :


Now, calculate :

Simplify:




Final Result: .

To determine the number of local maximum and minimum points for the function , we need to analyze its derivative. Using the Leibniz rule for differentiation under the integral sign, we have:

Setting , we solve:
This gives:\

We then perform a sign test to identify intervals of increase and decrease. Consider points before 0, between 0 and 3, 3 and 5, and after 5 in the expression . Substituting a test value from each interval, we determine:

• : (increasing)
• : (decreasing)
• : (increasing)
• : (decreasing)

Local extrema occur at changes from increasing to decreasing (local maximum) or from decreasing to increasing (local minimum). Hence, we find:

• Local maxima at
• Local minimum at

As both local maxima and minima occur twice considering points beyond 5, the respective counts of local maximum and minimum points are: 2 and 2.