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
Q.10. If a < 0, & f(x) = eax + e-ax is monotonically decreasing. Find interval to which x belongs.
Ans: Given a < o &
f(x) = eax + e-ax is decreasing
f'(x) < 0 aeax - ae-xx < 0
But a < 0
(e2ax - 1) > 0 e2ax > 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
is increasing function.