let . If f(x) is continuous in then f (π/4) is
1
-1/2
-1
1/2
B.
-1/2
The differential equation representing the family of curves , where c > 0, is a parameter, is of order and degree as follows:
order 1, degree 2
order 1, degree 1
order 1, degree 3
order 2, degree 2
C.
order 1, degree 3
The differential equation whose solution is Ax^{2} + By^{2} = 1, where A and B are arbitrary constants is of
second order and second degree
first order and second degree
first order and first degree
second order and first degree
D.
second order and first degree
The differential equation for the family of curves,x^{2} +y^{2} -2ay = 0 where a is an arbitrary constant is
2(x^{2}-y^{2})y' = xy
(x^{2}+y^{2})y' = xy
2(x^{2}+y^{2})y' = xy
(x^{2}- y^{2})y' = xy
A.
2(x^{2}-y^{2})y' = xy
2x + 2yy′ - 2ay′ = 0
If f is a real-valued differentiable function satisfying |f(x) – f(y)| ≤ (x – y)^{2} , x, y ∈ R and f(0) = 0, then f(1) equals
-1
0
2
1
B.
0
The solution of the differential equation ydx + (x + x^{2}y) dy = 0 is
-1/ XY =C
-1/XY + log y = C
1/XY + log y = C
log y =Cx
B.
-1/XY + log y = C
y dx + x dy + x^{2}y dy = 0
A spherical iron ball 10 cm in radius is coated with a layer of ice of uniform thickness than melts at a rate of 50 cm^{3} /min. When the thickness of ice is 5 cm, then the rate at which the thickness of ice decreases, is
B.
A function y = f(x) has a second order derivative f″(x) = 6(x – 1). If its graph passes through the point (2, 1) and at that point the tangent to the graph is y = 3x – 5, then the function is
(x-1)^{2}
(x-1)^{3}
(x+1)^{3}
(x+1)^{2}
B.
(x-1)^{3}
f''(x) = 6(x - 1) f'(x) = 3(x - 1)2 + c ........ (i)
At the point (2, 1) the tangent to graph is y = 3x - 5 Slope of tangent = 3
∴ f'(2) = 3(2 - 1)2 + c = 3
3 + c = 3
⇒ c = 0
∴ From equation (i) f'(x) = 3(x - 1)^{2}
f'(x) = 3(x - 1)2 f(x) = (x - 1)3 + k ...... (ii)
Since graph passes through (2, 1)
∴ 1 = (2 - 1)^{2} + k k = 0
∴ Equation of function is f(x) = (x - 1)^{3}
Let f be differentiable for all x. If f(1) = - 2 and f′(x) ≥ 2 for x ∈ [1, 6] , then
f(6) ≥ 8
f(6) < 8
f(6) < 5
f(6) = 5
A.
f(6) ≥ 8
As f(1) = - 2 & f′(x) ≥ 2 ∀ x ∈ [1, 6]
Applying Lagrange’s mean value theorem