Applications Of Derivatives

Material used:

For fixed volume ,

Minimizing gives .

Step 1: f'(x)=3x²\ge0 for all x. Hence, f(x) is increasing on [-1,1].
Step 2: Root of f(x)=0 is x=1, which is not in (-1,1). Thus, only statement I is correct.

At an extreme point, f'(x)=0 Hence, the slope of the tangent is zero, so it is parallel to the x-axis.

Critical points occur when f'(x)=0. Differentiating, the trigonometric part vanishes for all x only when: a²-3a+2=0 ⟹ a=1,2 Excluding these values gives the interval: (0,1)∪(1,4)

Let radius =r, height =h. Step 1: Volume fixed: V=π r²h Step 2: Material used: M=2π rh + (5)/(4)π r² Step 3: Using h=(V)/(π r²) and minimizing M w.r.t. r, we obtain: (r)/(h)=(2)/(3) Hence, required ratio is (2)/(3).

The slope of the tangent is given by the derivative of : At , the slope is .
Using the point-slope form at point (1,1): Thus, option (1) is correct.