Let f: be defined as f (t) =
f(t )dt
A minimum value of is
Correct Answer :- d
Explanation : f(x) = ∫(0 to x) te-t^2 dt
f'(x) = xe-x^2 = 0
Therefore, x = 0
f"(x) = e-x^2(1 - 2x2)
f"(0) = 1 > 0
Therefore, the minimum value is f(0) = 0
Let f(x) = ∫ex(x−1)(x−2)dx then f decrease in the interval
f(x) = ∫ex(x−1)(x−2)dx
For decreasing function ,f′(x)<0,f′(x)<0
⇒ex(x−1)(x−2) < 0
⇒(x−1)(x−2) < 0
⇒ ex(x−1)(x−2) < 0
⇒ (x−1)(x−2) < 0
⇒1 < x < 2,
∵ ex > 0 ∀ x ∈ R
The value of x1 and x2 with x1 < x2 such that has the largest value are
Let f(x) = Then the real roots of tlie equation x2 — f'(x) = 0 are
is equal to
ANSWER :- A
Solution :- limx→0 1/x[∫y→a e^sin2tdt− ∫x+y→a e^sin2tdt]
limx→0 [∫y→a e^sin2tdt− ∫x+y→a e^sin2tdt]/x
Hence it is 0/0 form, Apply L-hospital rule.
lim x-->0 0 - {e^(sin^2a).0 - e^(sin^2(x+y).1}/1
lim x-->0 e^(sin^2(x+y))
= e^(sin^2y)
If f(x) is differentiable and equals
If
The area of the region between the curves y = and bounded by the lines x = 0 and
Let f(x) = where g is a real valued continuous function on. Then f′ (x) is equal to
Here f(x) = where g is a real valued continuous function on R.
Apply Lebnitz's rule, we obtain
f′ (x) = 0
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