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# Subjective Type Questions: Definite Integrals and Applications of Integrals - 3 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

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Q. 33. Determine the value of         (1997 - 5 Marks)

Ans. π2

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 {x2, (1 – x)2, 2x(1 – x)}, where 0 < x < 1. Determine the area of the region bounded by the curves y = f(x), x-axis, x = 0 and x = 1.

Ans.

Solution. We draw the graph of y = x2, 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 x2 = 2x (1– x), ⇒ 3x2 = 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 C1 and C2 be the graphs of the functions y = x2 and y = 2x, 0 < x < 1 respectively. Let C3 be the graph of a function y = f(x), 0 < x < 1, f(0) = 0. For a point P on C1, let the lines through P, parallel to the axes, meet C2 and C3 at Q and R respectively (see figure.) If for every position of P (on C1), the areas of the shaded regions OPQ and ORP are equal, determine the function f(x).

Ans. f (x) = x3-x2

Solution. f (x) = x3-x2

Let P be on C1, y = x2 be (t, t2)

∴ ordinate of Q is also t2.
Now Q lies on y = 2x, and y = t2

∴ x = t2/2

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,

t2 - t3 =- f (t)

∴ f (t) = x3 - x2.

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 = – 2y2 and 8x + 1 = 0

Solving x = – 2 y2 , y = x2 + 2x –1 (x < –1) we get x = – 2

Also y = 2x,  x = – 2 y2 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 = loget.

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 y-axis and the curve xeay = sin by, Show th at S0, S1, S2, …, Sn 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 = x2, y = |2 – x2| and y = 2, which lies to the right of the line x = 1.

Ans.

Solution. The given curves are y = x2 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/2-π/4 = π/4

R.H.S. Hence proved.

Q. 43.

Ans.

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

= I1+I2

Now using the property that

Integrating by parts, we get

Q. 46. Find the area bounded by the curves x2 = y, x2 = –y and y2 = 4x – 3.

Ans.

Solution. The given curves are, x2 = y .........(i)

x2 = – y .........(ii)

y2 = 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

x4 -4x+3 = 0 (x-1)(x3 + x2 +x-3) = 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 f2(0) + g2(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

∴ 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

∴ 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

∴ 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 x-axis 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.

⇒ 4a2 f (-1) + 4af(1) + f (2) =3a2+ 3a
4b2 f (-1) + 4bf (1) + f (2) = 3b2+3a
4c2 f (-1) + 4cf (1) + f (2) = 3c2+3c

Consider the equation

4 x2 f (-1) + 4 xf (1) + f (2) = 3x2+3x or

[4 f (-1) - 3]x2 + [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) = px2 +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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