Solved Examples on Application of Derivatives

Illustration 1:The maximum value of

(cos a₁). (cos a₂)…. (cos a_n), under the restrictions 0 = a₁, a₂,… , a_n≤π/2

(cot a₁). (cot a₂)…. (cot a_n) = 1 is

1. 1/2^n/2                                                            2. 1/2ⁿ

3. 1/2n                                               4. 1

Solution: Given cot a₁. cot a₂….....cot a_n = 1

Then, cos a₁/sin a₁ .cos a₂/sin a₂ . cos a₃/sin a₃ ….. cos a_n/sin a_n = 1.

Hence, cos a₁.cos a₂…. cos a_n = k …….. (1)

and sin a₁. sin a₂……. sin a_n = k …….. (2)

by multiplying equations (1) and (2) we get,

(cos a₁. cos a₂…… cos a_n) x (sin a₁. sin a₂……. sin a_n) = k²

Then k² = 1/ 2.2.2…….. n times . (2 sin a₁. cos a₁) (2 sin a₂. cos a₂) (2 sin a₃. cos a₃)…………  (2 sin a_n. cos a_n)

Hence, k² = 1/2ⁿ (sin 2a₁).(sin 2a₂)…….. (sin 2a_n)

               = 1/2ⁿ sin 2a_i = 1 for all 1≤ i ≤ n

Hence, k = 1/2^n/2. Hence the correct option is (1).

Illustration 2:If ax² + bx + c = 0, a, b, c ∈ R. Find the condition that this equation would have at least one root in (0, 1).

Solution: Let f’(x) = ax² + bx + c

Integrating both sides,

=>f(x) = ax³ / 3 + bx² / 2 + cx + d

=>f(0) = d and f(1) = a/3 + b/2 + c + d

Since, Rolle’s theorem is applicable

=>f(0) = f(1)

=> d = a/3 + b/2 + c + d

=> 2a + 3b + 6c = 0

Hence required condition is 2a + 3b + 6c = 0

Illustration 3: If at each point of the curve y = x³ – ax² + x + 1 the tangents is inclined at an acute angle with the positive direction of the x-axis, then find the interval in which a lies.

Solution:Since the tangent is always inclined at an acute angle with the x-axis, hence and the use the concept of quadratic equation that ax² + bx + c > 0 for all x ∈R if a > 0 and D < 0.

y = x³ – ax² + x + 1 and the tangent is inclined at an acute angle with the positive direction of x-axis,

Hence,dx/dy> 0, So, 3x² – 2ax + 1 >0, for all x ∈ R

=> (2a)² – 4 (3)(1) < 0

=> 4(a²– 3)< 0

=> (a – √3) (a + √3) < 0

So, – √3 < a <√3.