Test: Applications of Derivatives(16 Sep) - JEE MCQ

Hence equation of tangent to the given curve at (0 , 1) is :

(y−1) = 2(x−0),i.e..y−1 = 2x 

x³ - 18x² + 96x = x(x² - 18x + 96) = x[(x-9)² + 15]
 =x(x-9)² +15 ≥ 0

Also, for x < 0 (slightly) , f ‘(x) < 0 and for x > 0 (slightly) f ‘(x) >0. Hence f has a local minima at x = 0 .

Since the graph cuts the lines y = -1 and y = 1 , therefore ,it must cut y = 0 atleast once as the graph is a continuous curve in this case.

, which does not exist at x = 2 . However , we find that  , at x = 2 . Hence , there is a vertical tangent to the given curve at x = 2 .The point on the curve corresponding to x = 2 is (2 , 0). Hence , the equation of the tangent at x = 2 is x = 2

f ‘(0) = 0 , f ‘’ (0) = 0 and f ‘’’(0) = 6 . So, f has a point of inflexion at 0.

f ‘ (x) = 0 ⇒ f (x)is constant in (0 , 1)and also in (2, 4). But this does not mean that f (x) has the same value in both the intervals . However , if f (c) = f (d) , where c ∈ (0 , 1) and d ∈ (2, 4) then f (x) assumes the same value at all x ∈ (0 ,1) U (2, 4) and hence f is a constant function.

 has a local maximum value = - 2 at x = - 1 and a local minimum value = 2 at x = 1.

Sinx cosx = 1/2 (sin2x) and minimum value of sin2x is – 1 .