NEW SYLLABUS NEW Board of Intermediate … 4 + 1 + 1 = 6 AB = BC = CA⇒Equilateral triangle formed ...

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Board of Intermediate EducationMathematics - I(B)

Model Paper (English Version) Time: 3 Hours (Given at the end of the Text Book) Max. Marks: 75

Note : The Question Paper consists of three sections A, B and C

SECTION-A

I. (i) Very short answer type questions.(ii) Answer ALL questions.(iii) Each question carries TWO marks. 10 ×× 2 = 20

1. Find the value of x, if the slope of the line passing through (2, 5) and (x, 3) is 2

2. Transform the equation x + y + 1 = 0 into the normal form.

3. Show that the points (1, 2, 3), (2, 3, 1) and (3, 1, 2) form an equilateral triangle.

4. Find the angle between the planes 2x − y + z = 6 and x + y + 2z = 7

Lt 2 x5. Show that { + x + 1} = 3

x → 0+ x

Lt ex + 3− e36. Find

x → 0 x

7. If f(x) = axex2find f'(x) (where a > 0, a ≠ 1).

dy8. If y = log [sin (logx)], find .dx

9. Find the approximate value of 3√65

10. Find the value of 'C' in Rolle's theorem for the function f(x) = x2 + 4 on [−3, 3]

SECTION - B 5 ×× 4 = 20

II. Short answer questions

(i) Answer any FIVE questions.

(ii) Each question carries FOUR marks.

11. A(2, 3) and B(−3, 4) are two given points. Find the equation of the locus of P, so

that the area of the triangle PAB is 8.5 sq.units.

12. When the axes are rotated through an angle π6

, find the transformed equation of



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SYLLABUS NEW SYLLABUS

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x2 + 2√3 xy − y2 = 2 a2.

13. Find the points on the line 3x − 4y − 1 = 0 which are at a distance of 5 units from

the point (3, 2)

cos ax − cos bx if x ≠ 0

x214. Show that f(x) = { 1

2(b2 − a2) if x = 0

where a and b are real constants is continuous at '0'

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

16. A particle is moving in a straight line so that after t seconds its distance s (in cms)

from a fixed point on the line is given by s = f(t) = 8t + t3. Find (i) the velocity

at time t = 2 sec (ii) the intitial velocity (iii) acceleration at t = 2 sec.

17. Show that the tangent at any point θ on the curve x = c sec θ, y = c tan θ is

y sin θ = x − c cos θ.

SECTION - C 5 ×× 7 = 35

III. Long answer questions

(i) Answer any FIVE questions.

(ii) Each question carries SEVEN marks.

18. Find the equation of straight lines passing through (1, 2) and making an angle of

60° with the line √3 x + y + 2 = 0

19. Show that the area of the triangle formed by the lines ax2 + 2hxy + by2 = 0 and

√n2 h2 − ab lx + my + n = 0 is .

am2 − 2hlm + bl2

20. Find the value of k, if the lines joining the origin to the points of intersection of

the curve 2x2 − 2xy + 3y2 + 2x − y − 1 = 0 and the line x + 2y = k are mutually

perpendicular.

21. If a ray with d.c's l, m, n makes angles α, β, γ and δ with four diagonals of a cube,

4then show that cos2 α + cos2 β + cos2 γ + cos2 δ =

3

3at 3at2 dy22. If x = , y = then find .

1 + t3 1 + t3 dx

23. At any point t on the curve x = a (t + sin t); y = a (1 − cos t), find lengths of

tangent and normal




24. A wire of length l is cut into two parts which are bent respectively in the form of

a square and a circle. Find the lengths of the pieces of the wire, so that the sum

of the areas is the least.

SOLUTIONS

5 − 31. Slope of the line passing through (2, 5) and (x, 3) is

2 − xBut given slope is 2.

5 − 3∴ We have = 2 (or) 2 = 2(2 − x) ∴ x = 22 − x

2. x + y + 1 = 0

x + y = −1

√ √Divide both sides by 1 + 1 = 2

1 1 −1( )x + ( )y = (or) √

2 √2 √

2

1 1 1(− )x + (− )y = √

2 √2 √

2

⇒ x cos α + y sin α = p form where

−1 5πcos α = = cos

√2 4

−1 5πsin α = = sin

√2 4

1p =

√2

5π 5π 1∴ Normal form: x cos + y sin =

4 4 √2

3. A = (1, 2, 3); B = (2, 3, 1); C = (3, 1, 2)

√

√

√

AB = (2 − 1)2 +(3 − 2)2 + (1 − 3)2 = 1 + 1 + 4 = 6

√

√

√

BC = (3 − 2)2 + (1 − 3)2 + (2 − 1)2 = 1 + 4 + 1 = 6

√

√

√

CA = (1 − 3)2 + (2 − 1)2 + (3 − 2)2 = 4 + 1 + 1 = 6

AB = BC = CA ⇒ Equilateral triangle formed




a1a2 + b1b2 + c1c24. θ = cos−1 ( )√

(a1

2 + b12 + c1

2)(a22 + b2

2 + c22)

Here a1 = 2; b1 = −1; c1 = 1

a2 = 1; b2 = 1; c2 = 2

(2)(1) + (-1)(1) + (1)(2) ∴ θ = cos−1 ( )√

√

(2)2 + (−1)2 + (1)2 (1)2 + (1)2 + (2)2

3 1 π= cos−1( ) = cos−1 ( ) =

6 2 3

2 x5. Lt ( + x + 1)x → 0+ x

x= 2(1) + 0 + 1 (... Lt = Lt (1) = 1)x → 0+ x x → 0

= 3

ex + 3 − e36. Lt

x → 0 x

e3 (ex − 1)= Lt

x → 0 x

ex − 1= e3. Lt ( )x → 0 x

= e3(1)

= e3

7. f(x) = ax. ex2

f'(x) = ax.(ex2.2x) + ex2

(ax log a)

= axex2( 2x + loga)

8. y = log [sin (logx)]

dy 1 d = . [sin (logx)]dx sin (logx) dx

1 d= . cos (logx) . (logx)

sin (logx) dx

1 cot (logx)= cot (logx) . = x x




9. 3√65 =

3√ 64 + 1

x = 64; δx = 1; x + δx = 65

f(x) = x1/3; f(x + δx) = (x + δx)

1/3

1 1f'(x) = . x

−2/3 = 3 3x

2/3

f(x + δx) _∼ f(x) + f'(x). δx

1(x + δx)

1/3 _∼ x1/3 + . δx

3x2/3

1 ⇒ (65)

1/3 _∼ (64)1/3 + . 1

3 . (64)2/3

1= 4 +

48

= 4.0208

10. f(x) = x2 + 4

f(−3) = (−3)2 + 4 = 13

f(3) = (3)2 + 4 = 13

∴ f is is differentiable on [−3, 3]

By Rolle's theorem, there exists c ∈ [−3, 3] such that f'(c) = 0

f'(x) = 2x and f'(c) = 2c

f'(c) = 0 ⇒ c = 0 and c ∈ [−3, 3]

11. Given A = (2, 3); B = (−3, 4) are two points

Let P = (x1, y1) be a variable point

Also given that ∆ PAB = 8.5

1 2 −3 x1 2⇒ = 8.5

2 3 4 y1 3

⇒ 8 + 9 − 3y1 − 4x1 + 3x1 − 2y1 = 17

⇒ − x1 − 5y1 + 17 = 17

⇒ (x1 + 5y1 − 17)2 = 289

⇒ x12 + 25 y1

2 + 289 + 10 x1y1 − 170 y1 − 34 x1 = 289




⇒ x12 + 10 x1y1 + 25y1

2 − 34x1 − 170y1 = 0

∴ Equation of Locus of P is given as:

x2 + 10 xy + 25 y2 − 34 x − 170 y = 0

12. Given equation: x2 + 2√3 xy − y2 = 2 a2

θ = π6

is the angle of rotation of axes

X√3 − Y

∴ x = X cos θ − Y sinθ = X cos π6

− Y sin π6

= 2

X + Y√3

y = X sin θ + Y cos θ = X sin π6

+ Y cos π6

= 2

Now, the transformed equatiion is given as follows:

X√3 − Y 2 X√

3 − Y X + Y√

3 X + Y√

3 2( ) + 2√

3 ( ) ( ) − ( ) = 2 a2

2 2 2 2

⇒ 3X2 − 2√3 XY + Y2 + 2√

3 (√

3 X2 + 3XY − XY− √

3 Y2)

− (X2 + 2√3 XY + 3Y2) = 8 a2

⇒ 3X2 − 2√3 XY + Y2 + 6 X2 + 4√

3 XY − 6 Y2 − X2 − 2√

3 XY− 3 Y2

= 8 a2

⇒ 8 X2 − 8 Y2 = 8 a2 (or) X2 − Y2 = a2

13. Let P(h, k) be any point on the line 3x − 4y − 1 = 0 which is at a distance of '5'

units from the point (3, 2) = A (say).

3h − 1∴ We have 3h − 4k − 1 = 0 (or) = k → (1)

4

Now AP = 5 ⇒ AP2 = 25

⇒ (h − 3)2 + (k − 2)2 = 25

3h − 1 2

⇒ (h − 3)2 + ( − 2) = 25 [from (1)]4

⇒ 16 (h − 3)2 + (3h − 9)2 = 400

⇒ 25 h2 − 150 h + 225 = 400

⇒ 25 h2 − 150 h − 175 = 0

⇒ h2 − 6h − 7 = 0

⇒ (h − 7) (h + 1) = 0

∴ h = 7 (or) −1




h = 7 ⇒ k = 5 and h = −1 ⇒ k = −1 [from (1)]

∴ (7, 5) and (−1, −1) are the points required

cos ax − cos bx14. Consider Lt f(x) = Lt [ ] x ≠ 0

x → 0 x → 0 x2

(1 − cos ax) − (1 − cos bx)= Lt [ − ]x → 0 x2

1 − cos ax 1 − cos bx= − Lt [ − ]x → 0 x2 x2

a2 (1 − cos ax) b2 (1 − cos bx)= − Lt [ − ]x → 0 a2 x2 b2 x2

1 − cos ax 1 − cos bx= −a2. Lt + b2 . Lt

ax → 0 (ax)2 bx → 0 (bx)2

1 1= − a2 ( ) + b2( )2 2

1= (b2 − a2) = f(0) (given)

2

∴ f(x) is continuous at x = 0

15. f(x) = sin 2x

f(x + h) = sin [2(x + h)] − sin (2x + 2h)

f(x + h) − f(x) = sin (2x + 2h) − sin 2x

2x + 2h + 2x 2x + 2h − 2x= 2 cos sin

2 2= 2 cos (2x + h). sin h

f(x + h) − f(x) sin h = 2 . cos (2x + h).

h h

f(x + h) − f(x) sin hLt [ ] = Lt [2 cos (2x + h). ]h→0 h h→0 h

sin h⇒ f'(x) = 2. Lt cos (2x + h) Lt

h→0 h→0 h

sin θ= 2 . cos (2x + 0) . 1 [... Lt = 1]θ→0 θ

∴ f'(x) = 2 cos 2x




16. s = f(t) = 8t + t3

dsVelocity = = f'(t) = 8 + 3t2

dt

(i) Velocity at t = 2 sec. is 8 + 3(2)2 = 20 cms/sec

(ii) Initial velocity is obtained at t = 0

hence, initial velocity = 8 + 3(0)2 = 8 cm/sec

d ds(iii) Acceleration = ( ) = f"(t) = 6t

dt dt

∴ Acceleration at t = 2 sec. is 6(2) = 12 cm/sec2

dx17. x = c. sec θ ⇒ = c. sec θ. tan θ

dθ

dyy = c . tan θ ⇒ = c. sec2 θ

dθ

dy dy/dθ c. sec2 θ sec θ 1 = = = = dx dx/dθ c. sec θ tan θ tan θ sin θ

1Now, equation of tangent at 'θ' is y − c tan θ = (x − c sec θ)

sin θ

⇒ y sin θ − c tan θ sin θ = x − c sec θ

c sin2 θ c⇒ y sin θ − = x −

cos θ cos θ

c c sin2 θ⇒ y sin θ = x − +

cos θ cos θ

c= x − (1 − sin2 θ)

cos θ

c cos2 θ= x −

cos θ

= x − c cos θ

∴ Tangent at 'θ' is given by the equation: y sin θ = x − c cos θ

18. Slope of the line √3x + y + 2 = 0 is −√

3 = m1 (say)

Slope of the required line is m2 (say), 60° is the angle between these two lines.

m1 − m2∴ tan θ = 1 + m1 m2




−√3 − m2⇒ tan 60° =

1 − √3 m2

−(√3 + m2)

⇒ √3 =

1 − √3 m2

⇒ √3 (1 − √

3 m2) = −(√

3 + m2)

Squaring both sides, we get

3(1 − √3 m2)2 = (√

3 + m2)2

⇒ 3(1 + 3m22 − 2√

3 m2) = 3 + 2√

3 m2 + m2

2

⇒ 8 m22 − 8√

3 m2 = 0

⇒ 8m2 (m2 − √3) = 0

⇒ m2 = 0 (or) m2 = √3

The equation of the line passing through (1, 2) and with m2 = 0 is

y − 2 = 0 (x − 1) ie, y − 2 = 0

The equation of the line passing through (1, 2) and with m2 = √3 is

y − 2 = √3(x − 1) ie., √

3x − y + 2 - √

3 = 0

∴ The required equations are y = 2 and y = √3x + 2 - √

3 = 0

19. Let ax2 + 2hxy + b y2 ≡ (l1x + m1y) (l2x + m2y) = 0

Multiplication and comparison of like terms will give a = l1l2; b = m1m2;

2h = l1m2 + l2m1

Also, the lines forming the triangle are:

l1x + m1y = 0 → (1)

l2x + m2y = 0 → (2)

and the 3rd line given is lx + my + n = 0 → (3)

From (1) and (2) , O = (0, 0) is the first vertex.

From (1) and (3) we get the 2nd vertex as follows:

x y 1

m1 0 l1 m1

m n l m

x y l = = m1n − 0 0 − nl1 l1m − lm1




m1n −nl1∴ A = ( , ) is the 2nd vertexl1m − lm1 l1m − lm1

Similarly, solving (2), (3) the 3rd vertex can be obtained as

m2n −nl2B = ( , )l2m − lm2 l2m − lm2

Now area of ∆ OAB is given by

1 −n2l2m1 + n2l1m2∆ = 2 (l1m − lm1) (l2m − lm2)

1 n2 (l1m2 − l2m1)= 2 l1l2m2 − (l1m2 + l2m1) ml + m1m2 l2

1 n2 √(l1m2 +

l2m1)2 −

4l1m2l2

m1= 2 am2 − 2hml + bl2

1 n2 √ 4h2 − 4ab

= 2 am2 − 2hml + bl2

1 2n2 √h2 − ab

= 2 am2 − 2hml + bl2

n2 √h2 − ab

∴ ∆ = is the required area.am2 − 2hml + bl2

x + 2y20. Line: x + 2y = k ∴ = 1 → (1)

k

Curve: 2x2 − 2xy + 3y2 + 2x − y − 1 = 0

(2x2 − 2xy + 3y2) + (2x − y)(1) − 1(1)2 = 0

(x + 2y) x + 2y 2⇒ (2x2 − 2xy + 3y2) + (2x − y) − ( ) = 0

k k

⇒ k2 (2x2 − 2xy + 3y2) + k (2x2 + 4xy − xy − 2y2) − (x + 2y)2 = 0

⇒ 2 k2x2 − 2 k2 xy + 3 k2y2 + 2 kx2 + 3 kxy − 2 ky2 − x2 − 4xy − 4y2 = 0

⇒ x2 (2k2 + 2k − 1) + xy (−2k2 + 3k −4) + y2 (3k2 − 2k − 4) = 0

These lines are given mutually perpendicular.

∴ x2 coeff + y2 coeff = 0

⇒ 2k2 + 2k − 1 + 3k2 − 2k − 4 = 0

⇒ 5k2 − 5 = 0 (or) k2 − 1 = 0 ⇒ k = ± 1




21. Let the direction cosines

of the given ray be (l, m, n)

of α, β, γ, δ are the angles made

by the ray with the four diagonals

of the cube, then

l + m + n l + m − n − l +m+n l − m + ncos α = , cos β = , cos γ = , cos δ =

√3 √

3 √

3 √

3

1∴ cos2α + cos2β + cos2γ + cos2δ = [(l+m+n)2 (l+m−n)2 + (−l +m+n)2 + (l−m+n)2]3

1= [4(l2 + m2 + n2)]

34

= ( ... l2 + m2 + n2 = 1)3

3at dx (1+t3)(3a) − (3at) (3t2)22. x = ⇒ =

1+t3 dt (1 + t3)2

3a + 3at3 − 9at3=

(1 + t3)2

3a(1 − 2t3)=

(1 + t3)2

3at2 dy (1 + t3)(6at) − (3at2)(3t2)y = ⇒ =

1 + t3 dt (1 + t3)2

6at + 6at4 − 9at4=

(1 + t3)2

3at (2 − t3)=

(1 + t3)2

3at(2 − t3)

dy dy/dt

(1 + t3)2

∴ = = dx dx/dt 3a (1 − 2t3)

(1 + t3)2



Y

B (0, a, 0)

P (a, a, a)(0, a, a) E

C (0, 0, a)

O (0, 0, 0)

F(a, 0, a)

A (a, 0, 0)

D (a, a, 0)

X

Z


dy t(2 − t3)⇒ =

dx 1 − 2t3

23. x = a(t + sin t)

dx = a (1 + cos t)dt

y = a(1 − cos t)

dy = a(sin t)dt

dy dy/dt a(sin t)∴ = =

dx dx/dt a (1 + cos t)

2 sin t/2 cos t/2= 2 cos2 t/2

sin t/2= = tan t2 = m(say)

cos t/2y √

1+m2

Length of tangant = m

a (1 − cos t) √1 + tan2

(t/2)

= tan t/2

a. 2 sin2 t2 sec ( t2 )

= sin

t2 sec ( t2 )

= 2a sin ( t2 )

Length of normal = y√1 + m2

= a (1 − cos t) √1+tan2 t/2

t t t sin t/2 t t= 2a sin2 sec = 2a sin . = 2a sin . tan

2 2 2 cos t/2 2 2

24. Let 'x' be the side of the square and 'r' be the radius of the circle

Now l = perimeter of the square + circumference of the circle

⇒ l = 4x + 2Πr

l − 4x⇒ = r → (1)

2Π

S = area of square + area of circle




⇒ S = x2 + Πr2

l − 4x 2⇒ S = x2 + Π ( )2Π

(l − 4x)2⇒ S(x) = x2 +

4Π

2(l − 4x)(−4)Now, S'(x) = 2x +

4Π

2(l − 4x)⇒ S'(x) = 2x −

Π

2(l − 4x)(−4)S"(x) = 2 −

Π

8(l − 4x)S"(x) = 2 +

Π

2 (l − 4x)S'(x) = 0 ⇒ 2x − = 0

Π

l − 4x⇒ x = (or) Πx + 4x = l

Π

l∴ l = (Π + 4)x (or) x = → (2)

Π + 4

lFor this value of x = , S"(x) is observed to be positive and hence 'S' takes

Π + 4minimum value.

4l l −

Π+4 lΠ l(2), (1) ⇒ r = = =

2Π 2Π (Π+4) 2(Π+4)

4l∴ Length of the square piece = 4x =

Π + 4

2Πl ΠlLength of the circular piece = 2Πr = =

2(Π + 4) Π+4

(Prepared by C. Sadasiva Sastry)



NEW SYLLABUS NEW Board of Intermediate … 4 + 1 + 1 = 6 AB = BC = CA⇒Equilateral triangle formed ...

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Transcript of NEW SYLLABUS NEW Board of Intermediate … 4 + 1 + 1 = 6 AB = BC = CA⇒Equilateral triangle formed ...