Test: Application of Derivatives - 1 - 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 .

In the case of strictly decreasing functions, the slope of the tangent line is always negative. This implies that the derivative of a strictly decreasing function is always negative. The derivative represents the rate of change of the function, and if the function is strictly decreasing, the rate of change is negative.

therefore , slopes of the tangents at (1,1) and (- 1 , -1)are equal. Hence, the two tangents in reference are parallel.

f (x) = 2 – 3x ⇒ f ‘ (x) = - 3 < 0 for all x ∈ R . so, f is strictly decreasing function.

Since g is continuous at a , therefore , if g (a) > 0 , then there is a nhd.of a, say (a-e , a+ e) in which g (x) is positive .This means that f ‘ (x)>0 in this nhd of a and hence f (x) is increasing at a.

The correct option is C 
a=c=1/4, b=1/2
y=x is a tangent
∴ slopes are equal.
dy/dx=2ax+b
⇒1=2a+b at (1,1)⋯(1)
Also, the parabola passes through (1,1)
⇒a+b+c=1⋯(2)
The parabola passes through (−1,0)
⇒0=a−b+c⋯(3)
Solving (1),(2),(3), we get -
∴a=c=1/4
and b=1/2

The modulus function has V-shaped graph,which means that it has only one minima.