application-of-derivatives-7

Dumb Question: How y = |x² - 3| = - (x² - 3) for neighbour hood of x = .

Ans: y = |x² - 3| since |x² - 3| is + ve quantity.
If we put x = , x² - 3 = - 1 but is + ve quantity so, we put - ve to make it + ve.
So, y = |x² - 3| = - (x² - 3)




Q.6. If ax² + bx + c = 0, a,b,c R. Find condition that this eq. would have at least one root in (0, 1).

Ans: Let f’(x) = ax² + bx + c
Integrating both sides,
f(x) = + cx + d ………………………………….. (i)
f(0) = d   &   f(1) = + c + d
Since, Rolle’s theorem is applicable.
[Dumb Question: Why Rolle’s theorem is applicable ?
 Ans: Rolle’s theorem is applicable b/c f(x) is polynomial function which is continuous & differentiable everywhare.]
f(0) = f(1)

2a + 3b + 6c = 0

required condition.


Q.7. Let f(x) & g(x) be differentiable for 0 x 2 such that f(0) = 2, g(0) = 1 & f(2) = 8. Let there exists a real no. c in [0, 2] such that f'(c) = 3g'(c) find value of g(2).

Ans: As f(x) & g(x) are cobt. & differentiable in ... then there exists atleast one value 'c' such that
f'(c) =
g'(c) =
But   = 3
        
     g(2) = 3


Q.8. f(x) is a polynomial of degree 4 with real coeff. such that f(x) = 0 is satisfied by x = 1, 2, 3 only, them and f'(1) f'(2) f'(3) ?

Ans: f(x) = 0 has only roots 1, 2, 3 only
But it is degree 4 eq.
Any one of 1, 2, or 3 is 4 repeated root of f(x) = 0
f'(1) or f'(2) or f'(3) any one of them must be zero.
f'(1) f'(2) f'(3) = 0


Q.9. Find interval for which f(x) = x - cosx is increasing or dectreasing.

Ans:    f(x) = x - cosx
Differentiating w.r.t. x
          f'(x) = 1 - sinx
we know
         -1 sinx |   for all x R
.... sin 0   for x v R.


Q.10. If a < 0, & f(x) = e^ax + e^-ax is monotonically decreasing. Find interval to which x belongs.

Ans: Given a < o &
    f(x) = e^ax + e^-ax is decreasing
f'(x) < 0   ae^ax - ae^-xx < 0

But a < 0
(e^2ax - 1) > 0   e^2ax > 1
2ax > 0   ax > 1
x < 0   as   a < 0


Medium Type:

Q.1. Find values of 'k' for which point min. of function f(x) = 1 + k²x - x³ satisfy inequality < 0.

Ans:  
Since it always +ve
(x + 2)(x + 3) < 0
- 3 < x < - 2 ........................................................ (i)
    f(x) = 1 + k²x - x³
    f'(x) = k² - 3x²
    f''(x) = - 6x
For max/min, f(x) = 0
x = ±
Let x₁ =   &   x₂ = -
f''(x₁) < 0   &   f''(x₂) > 0
f(x) is max. at x = x₁ & min. at x = x₂
But x₂ is min. which lies b/w
    - 3 < x₂ < - 2    (from relation (i))
- 3 < - < - 2
3 > |k| > 2
k (- 3, - 2) U (2, 3)


Q.2. If f(x) = , x > 0 determine bigger of two no. & ^e.

Ans. y = f(x) = , x > 0
Taking log on both sides
ln y = ln x
Differentiating both side
   
f'(x) = [1 - ln x]
Let f'(x) = 0
log x = 0   or   x = e
    f''(x) =
  f''(e) = + 0
f''(e) < 0
f(x) has maxima at x = e
f(e) > f() for all x > 0



Q.3. If f(x) = ax³ + bx² + cx + d where a, b, c, d are real no.s & 3b² < c², is an increasing cubic function & g(x) = af'(x) + bf''(x) + c², then prove that is an increasing function.

Ans: f(x) = ax³ + bx² + cx + d
       f'(x) = 3ax² + 2bx + c
Since f(x) is increasing
    f'(x) > 0    3ax² + 2bx + c > 0
3a > 0   &   D < 0
3a > 0   &   b² - 3ac < 0
a > 0   &   b² < 3ac ........................................ (i)
g(x) = af'(x) + bf''(x) + c²
g(x) = 3a²x² + 2abx + ac + 6abx + 2b² + c²
g(x) = 3a²x² + 8abx + (2b² + c² + ac)
D = 64a²b² - 4.3a²(2b² + c² + ac)
   = 4a²(10b² - 3c² - 3ac)
Since 3ac > b² from (i)
- 3ac < - b²
4a²(10b² - 3c² - 3ac) < 4a²(10b² - 3c² - b²)
=  4a²(9b² - 3c²)
Since 3b² < c²
=  12a²(3b² - c²) = -ve
 D < 0
g(x) > 0 v x R
  is increasing function.