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Q. 33. Determine the value of (1997  5 Marks)
Ans. π^{2}
Solution.
.....(1)
Putting cos x = t, – sin x dx = dt
When x → 0,t →1 and when x → p,t → 1
Q. 34. Let f(x) = Maximum {x^{2}, (1 – x)^{2}, 2x(1 – x)}, where 0 < x < 1. Determine the area of the region bounded by the curves y = f(x), xaxis, x = 0 and x = 1.
Ans.
Solution. We draw the graph of y = x^{2}, y = (1– x)^{2} and y = 2x (1– x) in figure.
Let us find the point of intersection of y = x2 and y = 2x (1– x)
The x – coordinate of the point of intersection satisfies the equation x^{2} = 2x (1– x), ⇒ 3x^{2} = 2x ⇒ 0 or x = 2 /3
∴ At B, x = 2/3
Similarly, we find the x coordinate of the points of intersection of y = (1 – x)^{2} and y = 2x (1– x) are x = 1/3 and x = 1
∴ At A, x = 1/3 and at C x = 1
From the figure it is clear that
The required area A is given by
Q. 35. Prove that Hence or otherwise, evaluate the integral
Ans. log 2
Solution.
...... (1)
Q. 36. Let C_{1} and C_{2} be the graphs of the functions y = x^{2} and y = 2x, 0 < x < 1 respectively. Let C_{3} be the graph of a function y = f(x), 0 < x < 1, f(0) = 0. For a point P on C_{1}, let the lines through P, parallel to the axes, meet C_{2} and C_{3} at Q and R respectively (see figure.) If for every position of P (on C_{1}), the areas of the shaded regions OPQ and ORP are equal, determine the function f(x).
Ans. f (x) = x^{3}x^{2}
Solution. f (x) = x^{3}x^{2}
Let P be on C_{1}, y = x^{2} be (t, t^{2})
∴ ordinate of Q is also t^{2}.
Now Q lies on y = 2x, and y = t2
∴ x = t^{2}/2
For point R, x = t and it is on y = f (x)
∴ R is [t, f (t)]
...(1)
Equating (1) and (2), we get,
Differentiating both sides,we get,
t^{2}  t^{3} = f (t)
∴ f (t) = x^{3}  x^{2}.
Q. 37.
Ans. π/2
Solution.
Q. 38. Let f(x) be a continuous function given by
Ans.
Solution.
∵ f (x) is continuous at x = – 1 and x = 1
∴ (–1)^{2} + a (–1) + b = – 2 and 2 = (1)^{2} + a . 1 + b i.e. a – b = 3 and a + b = 1
On solving we get a = 2, b = –1
Given curves are y = f (x), x = – 2y^{2} and 8x + 1 = 0
Solving x = – 2 y^{2} , y = x^{2} + 2x –1 (x < –1) we get x = – 2
Also y = 2x, x = – 2 y^{2} meet at (0, 0)
The required area is the shaded region in the figure.
∴ Required area
NOTE THIS STEP :
Q. 39. For x > 0, let Find the function
Here, lnt = log_{e}t.
Solution.
Hence Proved.
Q. 40. Let b ≠ 0 and for j = 0, 1, 2, …, n, let Sj be the area of the region bounded by the yaxis and the curve xe^{a}y = sin by, Show th at S_{0}, S_{1}, S_{2}, …, S_{n} are in geometric progression. Also, find their sum for a = 1 and b = π.
Ans.
Solution. Given that x = sin by. e^{ay} ⇒ – e–ay < x < e–^{ay}
The figure is drawn taking a and b both +ve. The given curve oscillates between x = e^{–ay} and x = – e^{–ay}
Integrating by parts,
Q. 41. Find the area of the region bounded by the curves y = x^{2}, y = 2 – x^{2} and y = 2, which lies to the right of the line x = 1.
Ans.
Solution. The given curves are y = x^{2} which is an upward parabola with vertex at (0, 0)
.......(2)
a downward parabola with vertex at (0, 2)
..........(3)
An upward parabola with vertex at (0, – 2)
y = 2 .........(4)
A straight line parallel to x – axis
x = 1 ..........(5)
A straight line parallel to y – axis
The graph of these curves is as follows.
∴ Required area = BCDEB
Q. 42. If f is an even function then prove that
Solution. Given that f (x) is an even function, then to prove
.......(2)
[As f is an even function] Adding two values of I in (1) and (2) we get
Let x  π /4 = t so that dx = dt
as x → 0, t → π /4 and as x → π/4, t → π/2π/4 = π/4
R.H.S. Hence proved.
Q. 43.
Ans. 2π
Solution. We have,
[∵ cos x is independent of θ]
(Using Leibnitz thm.)
Q. 44. Find the value of
Ans.
Solution.
The second integral becomes zero integrand being an odd function of x.
{using the prop. of even function and also x = x for 0 < x < π /3}
Let x + π /3 = y ⇒ dx = dy
also as x → 0,y → π /3 as x → π /3 , y → 2π /3
∴ The given integral becomes
Q. 45.
Ans.
Solution. Let
= I_{1}+I_{2}
Now using the property that
Integrating by parts, we get
Q. 46. Find the area bounded by the curves x^{2} = y, x^{2} = –y and y^{2} = 4x – 3.
Ans.
Solution. The given curves are, x^{2} = y .........(i)
x^{2} = – y .........(ii)
y^{2} = 4 x –3 .........(iii)
Clearly point of intersection of (i) and (ii) is (0, 0). For point of intersection of (i) and (iii), solving them as follows
x^{4} 4x+3 = 0 (x1)(x^{3} + x^{2} +x3) = 0
or ( x  1)2 ( x2 + 2x + 3)= 0 ; ⇒ x = 1 and then y = 1
∴ Req. point is ( 1, 1). Similarly point of intersection of (ii) and (iii) is (1, – 1). The graph of three curves is as follows:
We also observe that at x = 1 and y = 1
for (i) and (iii) is same and hence the two curves touch each other at (1, 1).
Same is the case with (ii) and (iii) at (1, –1).
Required area = Shaded region in figure = 2 (Ar OPA)
Q. 47. f(x) is a differ en tiable function an d g(x) is a dou ble differentiable function such that f(x) < 1 and f '(x) = g(x). If f^{2}(0) + g^{2}(0) = 9. Prove that there exists some c∈ (3, 3) such that g (c).g ''(c) < 0 .
Solution. Given that f (x) is a differentiable function such that f’(x) = g (x), then
Similarly
First let us consider g (0) >
Let us suppose that g'' (x) be positive for all x ∈ (–3, 3).
Then g” (x) > 0 ⇒ the curve y = g (x) is open upwards.
Now one of the two situations are possible. (i) g(x) is increasing
a contradiction as
∴ at least at one of the point c ∈ (–3,3), g'' (x) < 0.
But g (x) > 0 on (– 3, 3)
Hence g(x) g''(x) < 0 at some x ∈ (– 3, 3).
(ii) g (x) is decreasing
a contradiction as
∴ at least at one of point c ∈ (3, 3) g "(x) should be – ve. But g(x) > 0 on (–3 , 3).
Hence g (x) g'' (x) < 0 at some x ∈ (–3 , 3).
Secondly let us consider g (0) <
Let us suppose that g'' (x) be – ve on (– 3 , 3). then g'' (x) < 0 ⇒ the curve y = g(x) is open downward.
Again one of the two situations are possible (i) g (x) is decreasing then
a contradiction as
∴ At least at one of the point c ∈ (– 3, 3), g'' (x) is + ve. But g (x) < 0 on (– 3, 3).
Hence g(x) g'' (x) < 0 for some x ∈ (– 3, 3).
(ii) g (x) is increasing then
a contradiction as as
∴ At least at one of the point c ∈ (– 3, 3) g'' (x) is + ve.
But g (x) < 0 on ( –3, 3).
Hence g (x) g'' (x) < 0 for some x ∈ (– 3,3).
Combining all the cases, discussed aboe, we can conclude that at least at one point in (– 3, 3), g (x) g”(x) < 0.
Q. 48. is a quadratic function and its maximum value occurs at a point V. A is a point of intersection of y = f(x) with xaxis and point B is such that chord AB subtends a right angle at V. Find the area enclosed by f(x) and chord AB.
Ans.
Solution.
⇒ 4a^{2} f (1) + 4af(1) + f (2) =3a^{2}+ 3a
4b^{2} f (1) + 4bf (1) + f (2) = 3b^{2}+3a
4c^{2} f (1) + 4cf (1) + f (2) = 3c^{2}+3c
Consider the equation
4 x^{2} f (1) + 4 xf (1) + f (2) = 3x^{2}+3x or
[4 f (1)  3]x^{2} + [4 f (1)  3]x +f (2)= 0
Then clearly this eqn. is satisfied by x = a,b,c
A quadratic eqn. satisfied by more than two values of x means it is an identity and hence
Let f (x) = px^{2} +qx +r [f (x) being a quadratic eqn.]
Solving the above we get
It’s maximum value occur at f’ (x) = 0 i.e., x = 0 then f (x) = 1, ∴ V ( 0, 1)
Let A (–2, 0) be the point where curve meet x –axis.
Equation of chord AB is
Required area is the area of shaded region given by
Q. 49.
Solution.
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