Test: Application Of Derivatives - 2 - JEE MCQ

Since f ‘ (x) = a , therefore , f (x) is strict increasing on R iff a > 0.

 therefore , slope of tangent at (1,1) = - 1 and the slope of tangent at (- 1 ,1) = 1. Hence , the two tangents in reference are at right angles.

And y has a relative minimum at x = 0 if f ‘(0)+a = 0 . a =0.

Since f ‘(x) = 2x > 0 for x > 0,and f ‘ (x) = 2x < 0 for x < 0 ,therefore on R , f is neither increasing nor decreasing. Infact , f is strict increasing on [0 , ∞) and strict decreasing on (- ∞,0].

Since f‘(x) = a , therefore , f (x) is strict increasing on R if f a > 0.

Hence ,slope of tangent at (a , b) = -b/a . Therefore , the equation of tangent at (a , b ) is ; 

 = 0, at x = -1 and x = 2 i.e a + 2b = 1 and 4b + a/2 = -1. ⇒ a = 2 , b = -1/2 

Conditions of Rolle’s Theorem are satisfied by f(x) in [2,3].Hence there exist atleast one real c in (2, 3) s.t. f ‘(c) = 0 . Therefore , the set S contains atleast one element

f (x) = mx + c is strict decreasing on R 
if f ‘ (x) < 0 i.e. if m < 0 .

f ‘ (x) = 2x – 2 = 2 (x - 1) < 0 if x < 1 i.e. x x∈ (−∞,1) x∈ (−∞,1). Hence f is strict decreasing in left decreasing in (−∞,1)

Only when x = 0 and when x = 0 , y = 1. So, only at (0,1) tangent is parallel to x- axis.

i.e. x = e. Note that , f ‘(x) changes sign from positive to negative as we move from left to right through e. So, f (e) is maximum i.e. maximum value of f (x) is f (e)

Obviously, every differentiable function is continuous but every continuous function isn't differentiable.

 Therefore , f is strictly increasing on R.

 which does not exist at x = 0 ∈ (-2 , 2). So , Rolle’s theorem is not applicable.

 therefore , slope of tangent at (0 , 0) = cos 0 = 1 and hence slope of normal at (0 , 0) is - 1 .

 so,at (0 ,0) , the curve y = x1/5 has a vertical tangent.

So, f has a local minima at 2 and a local maxima at - 2 .