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Q. 17. If f and g are continuous function on [0, a] satisfying f(x) = f(a – x) and g(x) + g(a – x) = 2, then show that (1989  4 Marks)
Solution.
Hence the result.
Q. 18. Show that (1990  4 Marks)
Solution.
...(1)
...(2)
Adding (1) and (2), we get
[Using the property,
Hence Proved.
Q. 19. Prove that for any positive integer k, 2[cos x + cos 3x + .............. + cos (2k – 1) x] (1990  4 Marks)
Hence prove that
Solution. To prove :
It is equivalent to prove that
sin 2kx = 2 sin x cos x + 2 cos 3x sinx+ ... + 2 cos(2k 1) x sinx
Now, R.H.S. = (sin2 x) + (sin4 x sin2x) + (sin6 x sin4x) +.... +(sin 2kx  sin(2k 2)x)
= sin2kx = L.H.S. Hence Proved.
[Using the identity proved above]
+2 cos(2k 1) x cos x]dx
+(cos 6 x + cos 4x) + (cos 2kx) + cos(2k 2) x ]dx
cos(2k 2)x] + cos 2 k x dx
= π /2
Hence Proved
Q. 20. Compute the area of the region bounded by the curves y = ex In x and (1990  4 Marks)
Ans.
Solution. The given curves are
y = ex logex … (1)
and … (2)
The two curves intersect where ex log x
At x = 1/e or ex = 1, log x =  log e =1,y =1
So that is one point of intersection and at x = 1,
log 1 = 0 ∴ y = 0
∴ (1, 0) is the other common point of intersection of the curves. Now in between these two points,
i.e. log x is – ve, throughout
Clearly under the condition stated above y_{1} < y_{2} both being –ve in the interval
The rough sketch of the two curves is as shown in fig. and shaded area is the required area.
∴ The required area = shaded area
Q. 21. Sketch the curves and identify the region bounded by x = 1/2, x = 2, y = In x and y = 2^{x}. Find the area of this region. (1991  4 Marks)
Ans.
Solution. The given curves are x = 1/2 ....(1), x = 2...(2), y = ln x...(3),y = 2^{x}....(4)
Clearly (1) and (2) represent straight lines parallel to y  axis at distances 1/2 and 2 units from it, respectively. Line x = 1/2 meets meets (3) at (2, ln 2) and (4) at (2, 4).
The graph of curves are as shown in the figure.
Required area = ABCDA
Q. 22. If ‘f’ is a continous function with then show that every line y = mx
intersects the curve (1991  4 Marks)
Solution. We are given that f is a continuous function and
To sh ow that ever y line y = mx intersects the curve
If possible, let y = mx intersects the given curve, then Substituting y = mx in the equation of the curve we get
Then F (x) is a continuous function as f (x) is given to be continuous.
Also F ( x) → ∞ as x→ ∞
But F (0) = –2
Thus F (0) = –ve and F(b) = +ve where b is some value of x, and F (x) is continuous.
Therefore F (x) = 0 for some value of x Î (0,b) or eq. (1) is solvable for x.
Hence y = mx intersects the given curve.
Q. 23. (1991  4 Marks)
Ans.
Solution.
Consider, 2x  p = y n ⇒
When x → 0,y → π when x → π ,y → π
∴ We get
⇒ cos y / 2 dy = 2du
Also as y → 0,u → 0 and as y → π,u → 1
Q. 24. Sketch the region bounded by the curves y = x^{2} and Find the area. (1992  4 Marks)
Ans.
Solution. The given curves are y = x^{2} and Here y = x^{2 }is upward parabola with vertex at origin.
Also, is a curve symm. with respect to yaxis.
At x = 0,y = 2
∴ Curve is decreasing on (0, ∞)
⇒ At (0,2) tangent to curve is parallel to x – axis.
As x → ∞ , y → 0
∴ y = 0 is asymptote of the given curve.
For the given curves, point of intersection : solving their equations we get x = 1, y = 1, i.e., (1,1).
Thus the graph of two curves is as follows:
Q. 25. Determine a positive integer n < 5,such that (1992  4 Marks)
Ans. n = 3
Solution. Given that
where n ∈ N and n < 5
To find the value of n.
.........(1)
Q. 26. (1993  5 Marks)
Ans.
Solution.
Put x^{2} + 1 = t ,2 x dx = dt
when x → 2,t → 5 , x → 3,t → 10
Q. 27. Show that where n is a positive integer and 0 < n < π. (1994  4 Marks)
Ans. 2n + 1  cosγ
Solution. To prove that
Now we know that  sin x is a periodic function of period π , So using the property..
where n ∈ I and f (x) is a periodic function of period T
Q. 28. In what ratio does the xaxis divide the area of the region bounded by the parabolas y = 4x – x^{2} and y = x^{2} – x ? (1994  5 Marks)
Ans. 121 : 4
Solution. The given equations of parabola are
y = 4 x  x^{2} or (x  2)^{2} = (y 4) ..... (1)
...... (2)
Solving the equations of two parabolas we get their points of intersection as
Here the area below x  axis,
Area above x  axis,
∴ Ratio of areas above x – axis and below x – axis.
Q. 29. Use mathematical induction to prove that I_{m} = m π , m = 0, 1, 2, ....... (1995  5 Marks)
Solution. Given
To prove: I_{m} = mπ,m= 0, 1, 2, ......
For m = 0
∵ Result is true for m = 0
For m = 1,
∴ Result is true for m = 1
Let the result be true for
Consider
Now, 1  cos(k+ 1)x
= 1 – cos kx cosx + sin kx sin x
= 1 + cos kx cos x + sin kx sin x – 2 cos kx cos x
= 1 + cos (k – 1) x – 2 cos kx cos x
= 2 – (1 – cos (k – 1) x) –2 cos kx cos x
= 2 (1 – cos kx cos x) – (1– cos (k – 1) x)
= 2 – 2 cos kx + 2 cos kx – 2 cos kx cos x – [1– cos ( k – 1) x]
= 2 (1 – cos kx) + 2 cos kx (1– cos x) – (1 – cos (k – 1) x)
= 2 (kπ) + 2(0)  (k  1)π [Using (i)]
= 2kπ kπ+π = (k +1)π
Thus result is true for m=k + 1 as well. Therefore by the principle of mathematical induction, given statement is true for all m = 0, 1, 2, ...............
Q. 30. Evaluate the definite integral : (1995  5 Marks)
Ans.
Solution.
Q. 31. Consider a square with vertices at (1, 1), (– 1, 1), (– 1, – 1) and (1, – 1). Let S be the region consisting of all points inside the square which are nearer to the origin than to any edge. Sketch the region S and find its area. (1995  5 Marks)
Ans.
Solution. Let us consider any point P (x, y) inside the square such that its distance from origin < its distance from any of the edges say AD
........ (1)
Above represents all points within and on the parabola 1. If we consider the edges BC then OP < PN will imply
........ (2)
Similarly if we consider the edges AB and CD, we will have
........ (3)
........ (4)
Hence S consists of the region bounded by four parabolas meeting the axes at
The point L is intersection of P_{1} and P_{3} given by (1) and (3).
y^{2}  x^{2} = 2( x  y) = 2( yx) 0
∴ y  x = ∴ y = x ∴ x^{2} + 2x  1 = 0 ⇒ (x + 1)^{2} = 2
Q. 32. Let A_{n} be the area bounded by the curve y = (tan x)n and the lines x = 0, y = 0 and x = π/4. Prove that for n > 2, and deduce (1996  3 Marks)
Solution.
Since 0 < tan x < 1, when 0 < x < π /4 , we have
0 < (tan x)^{n +1} < (tanx)n for each n ∈ N
Now, for n > 2
........(2)
Combining (1) and (2) we get
Hence Proved.
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