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```Solved Examples on Application of Derivatives

Illustration 1:The maximum value of

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

(cot a1). (cot a2)…. (cot an) = 1 is

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

3. 1/2n                                               4. 1

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

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

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

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

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

(cos a1. cos a2…… cos an) x (sin a1. sin a2……. sin an) = 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 an. cos an)

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

= 1/2n sin 2ai = 1 for all 1≤ i ≤ n

Hence, k = 1/2n/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.
```
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