Free Booklet1

..............................................................................................................Mathematics – IB

Practice Paper – 3

SECTION – A

I. 1. Find the equation of the straight line passing through the

point (–2, 4) and making non-zero intercepts on the axis of

coordinates whose sum is zero.

Sol. Let x – intercept = a

y – intercept = b

Given that a + b = 0 ⇒ b = –a

Intercept formxa

yb

+ = 1

⇒ xa

ya

+− = 1

⇒ x – y = a

If this line passing through the point (–2, 4) then

–2 – 4 = a

⇒ a = –6

Hence required straight line equation is

x – y = –6

⇒ x – y + 6 = 0.

2. Find the value of P if the straight lines x + P = 0, y + 2 = 0,

3x + 2y + 5 = 0 are concurrent.

Sol. Given straight line equations are

x + P = 0 ——— (1)

y + 2 = 0 ⇒ y = –2 ——— (2)

3x + 2y + 5 = 0 ——— (3)

Solving (2) & (3)

3x + 2(–2) + 5 = 0 ⇒ 3x + 1 = 0

⇒ x = −13

SOLUTIONS FOR PRACTICE PAPER - 3

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..............................................................................................................Mathematics – IB

Practice Paper – 3

Point of intersection of (2) & (3) is −

−

13

2,

Since (1), (2), (3) are concurrent

This point lies on (1)

⇒ −13

+ P = 0

⇒ P = 13

3. Find the ratio in which the XZ-plane divides the line joining

A(–2, 3, 4) and B(1, 2, 3).

Sol. Given A = (–2, 3, 4)

B = (1, 2, 3)

XZ - Plane divides the line joining AB in the ratio

= –y1 : y2

= –3 : 2

= 3 : 2 externally

4. Find the direction cosines of the normal to the plane

x + 2y + 2z – 4 = 0.

Sol. Given plane equation is

x + 2y + 2z – 4 = 0

⇒ x + 2y + 2z = 4

⇒ x

y Z1 2 2

2

1 2 2

2

1 2 22 2 2 2 2 2 2 2 2+ ++

+ ++

+ +

4

1 2 22 2 2=

+ +

⇒13

23

23

43

x y z+ + =

∴ The direction cosines of the normal to the plane are

13

23

23

, ,

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..............................................................................................................Mathematics – IB

Practice Paper – 3

5. Compute lime 1

xx 0

3x

→

−

.

Sol. lim limx

x

x

xex

ex→ →

−=

−0

3

0

313

13

= 3(1)

= 3

∴ limx

xex→

−0

3 1 = 3

6. Evaluate lim1 cos 2mx

sin nxx 0 2→

_ (m. n ∈∈∈∈∈ Z).

Sol. limcos

sinx

mx

nx→

−0 2

1 2 = lim

sin

sinx

mx

nx→ 0

2

2

2

= 2 lim

sin

sinx

mx

xnx

x

→ 0

2

2

2

2

= 2

limsin

limsin

x

x

mxx

nxx

m

n

→

→

=0

2

0

2

2

2

2

7. Find the derivative of tan–1 (log x).

Sol. Let y = Tan–1 (log x)

differentating w.r. to 'x' on bothsides, we have

dydx x

ddx

=+

1

1 2(log ). (log x)

= 1

1

12+ (log )

.x x

= 1

1 2x x+ ( )

log

Free Booklet4

..............................................................................................................Mathematics – IB

Practice Paper – 3

8. Find the derivative of sin–12x

1+ x2

.

Sol. Let y = sin−

+

12

2

1

x

x

put x = Tan θ ⇒ θ = Tan–1x

= sin−

+

12

2

1

Tan

Tan

θθ

= sin–1(sin 2θ)

= 2θ

= 2 Tan–1xDifferentiating w.r.to 'x' on bothsides, we have

dydx

ddx

= 2 (Tan–1x)

= 2.1

1 2+ x =

2

1 2+ x

9. Find dy and ∆∆∆∆∆y of y = f(x) = x2 + x at x = 10 when ∆∆∆∆∆x = 0.1

Sol. Given f(x) = x2 + x, x = 10 and ∆x = 0.1

dy = f'(x) ∆x

= (2x + 1) ∆x

= [2(10) + 1] (0.1)

= (21) (0.1)

= 2.1

∆y = f(x + ∆x) – f(x)

= (x + ∆x)2 + (x + ∆x) – (x2 + x)

= x2 + 2x ∆x + (∆x)2 + x + ∆x – x2 – x

= 2x ∆x + (∆x)2 + ∆x

= 2(10) (0.1) + (0.1)2 + 0.1

= 2 + 0.01 + 0.1

= 2.11

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..............................................................................................................Mathematics – IB

Practice Paper – 3

10. Verify Rolle's theorem for the function f(x) = x2 + 4 in

[–3, 3].

Sol. Given f(x) = x2 + 4

Since f is a second degree polynomial

∴ f is continuous on [–3, 3] and f is desivable on (–3, 3)

Also f(–3) = 9 + 4 = 13

f(3) = 9 + 4 = 13

∴f(–3) = f(3)

∴ f satisfies all the conditions of Role's theorum.

∴ There exists c∈(–3, 3) such that f1 (c) = 0

But f1(x) = 2x

f1 (c) = 2c

0 = 2c

⇒ c = 0 ∈ (–3, 3)

Hence Rolle's theorum is verified.

SECTION – B

II. 11.If the distance from the Point P to the Points (2, 3) and (2, –3)

are in the ratio 2 : 3, then find the equation of the Locus of P.

Sol. Let A = (2, 3) and B = (2, –3)

Let p(x1, y1) be any point on the locus

Given geometric condition is PA : PB = 2 : 3

⇒ PAPB

=23

⇒ 3PA = 2PB

⇒ 9PA2 = 4PB2

⇒ 9[(x1 – 2)2 + (y1 – 3)2] = 4 [(x1 – 2)2 + (y1 + 3)2]

⇒ 9 x x y y12

1 12

14 4 6 9− + + − +

= 4 x x y y12

1 12

14 4 6 9− + + + +

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..............................................................................................................Mathematics – IB

Practice Paper – 3

⇒ 9 912

12x y+ – 36x1 – 54y1 + 117 = 4 41

212x y+ – 16x1 + 24y1 + 52

⇒ 5 512

12x y+ – 20x1 – 78y1 + 65 = 0

∴ Locus of P is 5x2 + 5y2 – 20x – 78y + 65 = 0.

12. When the axes are rotated through an angle 45°, the

transformed equation of a curve is 17x2 – 16xy + 17y2 =

225. Find the original equation of the curve.

Sol. Angle of rotation θ = 45°

x = x cos θ + y sin θ y = –x sin θ + y cos θ

= 2 cos 45° + y sin 45° = – x sin 45° + y cos 45°

= x1

2

+ y

1

2

= –x

1

2

+ y

1

2

= x y+

2=

− +x y

2

The original equation of 17x2 – 16xy + 17y2 = 225 is

17x y x y x y+

−

+

− +

216

2 2

2

+ 17− +

x y

2

2

= 225

⇒ 17

x xy y y x x xy y2 2 2 2 2 222

162

1722

+ +

−

−

+

− +

= 225

⇒ 17x2 + 34xy + 17y2 – 16y2 + 16x2 + 17x2 – 34xy + 17y2

= 450

⇒ 50x2 + 18y2 = 450

⇒ 25x2 + 9y2 = 225

13. Find the equation of the straight line passing through the

points (–1, 2) and (5, –1) and also find the area of the triangle

formed by it with the axes of coordinates.

Sol. Let A = (–1, 2)

B = (5, –1)

Free Booklet7

..............................................................................................................Mathematics – IB

Practice Paper – 3

Equation of the straight line passing through the points A and B

is

Y – Y1 = y yx x

2 1

2 1

−− (x – x1)

⇒ y – 2 = − −

+1 25 1 (x + 1)

⇒ 6y – 12 = –3x – 3

⇒ 3x + 6y – 9 = 0

⇒ x + 2y – 3 = 0

⇒ x + 2y = 3

⇒ x y3 3 2

+/

= 1

x – intercept = 3, y – intercept = 3/2

Area of ∆OAB = 12

( int ) ( int )x ercept y ercept− −

= 12

332

( )

= 94

sq.units.

14. Check the continuity of the following function at 2 :

f(x) =

12

(x 4) if 0 < x <2

0 if x =2

2 8x if x >2

2

3

−

−

−

Sol. Given f(x) =

12

4 0 2

0 2

2 8 2

2

3

( )x if x

if x

x if x

− < <

=

− >

−

Free Booklet8

..............................................................................................................Mathematics – IB

Practice Paper – 3

limx → −2 f(x) = lim

x → −2 12

(x2 – 4)

= 12 0

limh →

[(2 – h)2 – 4]

= 12

(4 – 4)

= 0

limx → +2 f(x) = lim

x → +22

8

23−

= lim( )h h→

−+

0 3

28

2

= 2 – 88

= 2 – 1= 1

∴ limx → −2 f(x) ≠ lim

x → +2 f(x)

∴ f is discontinuous at x = 2

15. Find the derivative of the function sin 2x from the first principle.

Sol. Let f(x) = sin 2xBy first principle

f1(x) = lim( ) ( )

h

f x h f xh→

+ −0

= limsin ( ) sin

h

x h xh→

+ −0

2 2

= limcos sin

h

x h x x h x

h→

+ +

+ −

0

22 2 2

22 2 2

2

= limh → 0 2 cos (2x + h) lim

sinh

hh→ 0

= 2 cos (2x + 0).1

= 2 cos 2x

∴ f1x = 2cos 2x.

Free Booklet9

..............................................................................................................Mathematics – IB

Practice Paper – 3

16. Find the equation of tangent and normal to the curve y = x3 + 4x2 at (–1, 3).

Sol. Given curve equation is y = x3 + 4x2 ——— (1)

dydx

= 3x2 + 8x

m = dydx

(–1, 3) = 3(–1)2 + 8(–1)

= 3 – 8

= –5

The equation of the tangent to the curve (1) at (–1, 3) is

y – y1 = m(x – x1)

y – 3 = –5(x + 1)

y – 3 = –5x – 5

⇒ 5x + y + 2 = 0

The equation of the normal to the curve (1) at (–1, 3) is

y – y1 = −1m

(x – x1)

⇒ y – 3 = −−

15

(x + 1)

⇒ 5y – 15 = x + 1

⇒ x – 5y + 16 = 0

17. The volume of a cube is increasing at a rate of 9 (centimetres)3

per second. How fast is the surface area increasing when

the length of the edge is 10 centimeters ?

Sol. Let x be the length of the edge of the cube, v be its volume ands be its surface area.

Given dvdt

= 9 cm3/sec

Since v = x3

dvdt

= 3x2 dxdt

Free Booklet10

..............................................................................................................Mathematics – IB

Practice Paper – 3

9 = 3x2 dxdt

⇒ dxdt

= 32x

Since s = 6x2

dsdt

= 12x dxdt

= 12x 32x

= 36x

When x = 10

dsdt

= 3610

= 3.6 cm2/sec

SECTION – C

III.18. Find the circumcenter of the triangle whose vertices are (–2, 3),(2, –1) and (4, 0).

Sol. Let A = (–2, 3)B = (2, –1)C = (4, 0)

Let s (α, β) be the circum center of the ∆ABCThen SA = SB = SCSA = SB⇒ SA2 = SB2

⇒ (α + 2)2 + (β – 3)2 = (α – 2)2 + (β + 1)2

⇒ α2 – 4α + 4 + β2 + 6β + 9 = α2 – 4α + 4 + β2 + 2β + 1⇒ 8α – 8β + 8 = 0⇒ α – β + 1 = 0 ——— (1)SB = SC⇒ SB2 = SC2

⇒ (α – 2)2 + (β + 1)2 = (α – 4)2 + (β – 0)2

Free Booklet11

..............................................................................................................Mathematics – IB

Practice Paper – 3

Y

A

BX

OX'

Y'

⇒ α2 – 4α + 4 + β2 + 2β + 1 = α2 – 8α + 16 + β2

⇒ 4α + 2β – 11 = 0 ——— (2)

Solving (1) & (2)

α β 1–1 1 1 –1

2 –11 4 2

α β11 2 4 11

12 4−

=+

=+

α β9 15

16

= =

⇒ α = 32

, β = 52

∴ Circumcenter, S = 32

52

,

19. Show that the area of the triangle formed by the lines ax2 +

2hxy + by2 = 0 and lx + my + n = 0 is

n h ab

am 2h/m+b

2 2

2 2

_

_ l .

Sol. Let ��

OBandOA be the pair of straight lines represented by theequation

ax2 + 2hxy + by2 = 0 (see figure)

and AB be the line lx + my + n = 0

Let ax2 + 2hxy + by2

≡ (l1x + m1y) (l2x + m2y),

and OBandOA be the lines.

l1x + m1y = 0 and

l2x + m2y = 0 respectively.

Free Booklet12

..............................................................................................................Mathematics – IB

Practice Paper – 3

Y

O

B2x + 3y = k

A

X

Let A = (x1, y1) and B = (x2, y2).

Then l1x1 + m1y1 = 0 and lx1 + my1 + n = 0.

So, by the rule of cross–multiplication, we obtain

111

1

1

1

mm1

ny

nmx

��� and therefore

x1 = 11

1

mmnm��

; y1 = 11

1

mmn��

�

Similarly x2 = 22

2

mmnm��

; y2 = 22

2

mmn��

�

∴ Area of ∆ OAB = 21

|x1y2 – x2y1|

= )mm()mm(

)mm(n21

2211

12212

����

��

= |mm)mmm(m|

mm4)mm(n

21

2211221

221

21212

12212

������

����

= |bmh2am|

ab4h4n21

22

22

��

(Since l1l2 = a, m1m2 = b and l1m2 + l2m1 = 2h)

= |bmh2am|

abhn22

22

��.

20. Show that the lines joining the origin to the points of

intersection of the curve x2 – xy + y2 + 3x + 3y – 2 = 0 and

the line x – y – 2 = 0 are mutually perpendicular.

Sol. Equation of the curve is

x2 – xy + y2 + 3x + 3y – 2 = 0 —— (1)

Equation of AB is x – y – 2 = 0

x – y = 2x y−

2 = 1 —— (2)

Free Booklet13

..............................................................................................................Mathematics – IB

Practice Paper – 3

Homogenising, (1) with the help of (2) combined equation ofOA, OB is

x2 – xy + y2 + 3x.1 + 3y.1 – 2.12 = 0

x2 – xy + y2 + 3(x + y)x y−

2 – 2

( )x y− 2

2 = 0

x2 – xy + y2 + 3

2 (x2 – y2) – (x2 – 2xy + y2) = 0

x2 – xy + y2 + 3

2x2 –

3

2y2 – x2 + 2xy – y2 = 0

3

2x2 + xy –

3

2y2 = 0

a + b = 3

2 –

3

2 = 0

∴ OA, OB are perpendicular.

21. Find the direction cosines of two lines which are connected

by the relations l + m + n = 0 and mn – 2nl – 2lm = 0.

Sol. Given l + m + n = 0 ——— (1)

mn – 2nl – 2lm = 0 ——— (2)

From (1), l = – (m + n)

Substituting in (2),

mn + 2n (m + n) + 2m (m + n) = 0

mn + 2mn + 2n2 +2m2 + 2mn = 0

2m2 + 5mn + 2n2 = 0

(2m + n) (m + 2n) = 0

2m = –n or m = –2n

Case (i) : 2m1 = –n1

From l1 = –m1 – n1

= –m1 + 2m1 = m1

l1 1 1

1 1 2= =

−m x

Free Booklet14

..............................................................................................................Mathematics – IB

Practice Paper – 3

D.Rs of the first line are 1, 1, –2

D.Cs of this line are 1

6

1

6

2

6, , −

Case (ii) : m2 = –2n2

From (1) l2 = –m2 – n2 = +2n2 – n2 = n2

l2 2 2

1 2 1=

−=

m n

d.cs of the second line are 1, –2, 1

d.cs of this line are 1

6

2

6

1

6, ,

−

22. If y = tan–1 1+ x + 1_ x

1+ x _ 1_ x

2 2

2 2

0 < |x| < 1, then find

dydx

.

Sol. Put x2 = cos 2θ

y = tan–1 1 2 1 2

1 2 1 2

+ + −

+ − −

cos cos

cos cos

θ θ

θ θ

= tan–1 2 2

2 2

2 2

2 2

cos sin

cos sin

θ θ

θ θ

+

−

= tan–1 cos sincos sin

θ θθ θ

+−

= tan–1

11

+−

tantan

θθ

= tan–1 tanπ

θ4

+

= π4

+ θ

= π4

+ 12

cos–1 (x2)

dydx x

xx

x=

−

−× =

−

−

12

1

12

14 4

( )

Free Booklet15

..............................................................................................................Mathematics – IB

Practice Paper – 3

x x

x x

x x

x x

80 – 2x

30 – 2x

23. Find the lengths of subtangent, subnormal at a point t on

the curve y = a (sin t – t cos t), x = a(cos t + t sin t).

Sol. Equations of the curve are x = a (cos t + t sin t)

x = a (cos t + t sin t)

dxdt

= a (–sin t + t cos t) = at cos t

y = a (sin t – t cos t)

dydt

= a (cos t – cos t + t sin t)

= a t sin t

dydt

dydt

dxdt

at tat t

=

=sincos

= tan t

Length of the sub-tangent = y

f x

a t tt

1

1’( )

(sin t cos )tan

=−

= |a cot t (sin t – t cos t)|

Length of the sub-normal = |y1. f'(x1)|

= |a(sin t – t cos t) tan t|

= |a tan t (sin t – t cos t|

24. From a rectangular sheet of dimensions 30 cm × 80 cm fourequal squares of side x cms are removed at the corners andthe sides are taken turned up so as to form an openrectangular box. Find the value of x, so that the volume ofthe box is the greatest.

Sol. Length of the box = 80 – 2x = l

Breadth of the box = 30 – 2x = b

Height of the box = x = h

Volume = lbh

= (80 – 2x) (30 – 2x). x

= x (2400 – 220x + 4x2)

Free Booklet16

..............................................................................................................Mathematics – IB

Practice Paper – 3

f(x) = 4x3 – 220x2 + 2400x

f'(x) = 12x2 – 440x + 2400

= 4[3x2 – 110x + 600]

f'(x) = 0 ⇒ 3x2 – 110x + 600 = 0

x = 110 12100 7200

6

± −

= 110 70

6180

6406

303

203

±= =or or

If x = 30,

b = 30 – 2x

= 30 – 2(30)

= –30 < 0

⇒ x ≠ 30

∴ x = 203

f"(x) = 24x – 440

When x = 203

, f"(x) = 24. 203

– 440

= 160 – 440

= –280 < 0

f(x) is maximum when x = 203

Volume of the box is maximum when x = 203

cm.

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