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$: QUINTESSENCE: - Infomathsinfomathsonline.com/Que_papers/vectors … · Web view · 2017-03-01Ans. (c) z = x + y {as . For max. / min . x2 = 1 ( x = 1, – 1 as x > 0 . y > 0$
INFOMATHS12.DIFFERENTIATION (WORKSHEET)

1. Ans. (d) y = logex

In general,

2. Ans. (c) F(x) = xn F' (x) = nxn-1 F "(x) = n(n – 1) xn-2 ………………Fn(x) = n(n – 1) ……… 5.4.3.2.1 hence, gives expansion as

Put x = 1

= C0 – C1 + C2 – C3 + …… = 0

3. Ans. (b)

Taking log on both sides:-

So, at But y = e at

at

4. Ans. (a) log(x + y) – 2xy = 0 On differentiating both sides

….(*) Eq. at x = 0, log (0 + y) – 0 = 0 log y = 0 y = 1 at x = 0, y = 1 (*) Eq. gives.

Put x = 0, y = 1

5. Ans. (d) h(x) = f(x)2 + g(x)2

As f ' = g and f" = - f

f" = g' and f" = - f y' = - f Differentiating both sides:- h'(x) = 2f(x) . f '(x) + 2g (x) g'(x) = 2f(x) . g(x) – 2g(x) f(x) = 0So, h(x) is a constant function h(x) = K = constant Also, h (1) = 8, h(0) = 2 Not possible

6. Ans. (a) f(x) = x|x| and g(x) = sinx gof(x) = g(f(x))= g(x |x|) = sin (x|x|)

LHD at x = 0 = - 2(0) cos 0 = 0 RHD at x = 0 =2(0) cos 0 = 0 LHD = RHD g(f(x)) is difficult at x = 0 As g'(f(0)) = 0 g’(f(x)) is also constant at x = 0 Hence, gof'(x) is continuous at x = 0 as LHL = RHL = f(0)

7. Ans. (c) We know, if f(a + b) = f(a). f(b)f(5) = 2 f '(0) = 3 Using 1st principle method:-

….* Eq. form By L’Hospital rule

Since f(a + b) = f(a) . f(b) Take a = b = 0

= f(a) . f’(0) f(0) = f(0)2 f '(a) = f(a) . f '(0) f(0) [f(0) – 1] = 0 f' (5) = f(5) . f '(0) f(0) = 0, or 1 = 2 . 3 f(0) = 1 as f(0) 0 = 6 otherwise f(0 + 5)

= f(0) f(5) = 0.f(5) = 0 but f(5) = 2

* EQ. becomes form

8. Ans. (a) Derivative of wrto

Derivative of

wrto

Derivative of wrto 2tan-1 x = – 1

9. Ans. (b)

1 INFOMATHS/MCA/MATHS/

INFOMATHS

As f(x) is a constant func.

10. Ans. (b)

11. Ans. (c)

12. Ans. (d)

13. Ans. (d)

Where 2 and 3 vanishes.as in 2 we have all zero elements in R2 and in 3 we have, all 'O' elements in R3.

lly

=0which is(independent of p)

14. Ans. (c) y = xx taking log both sides:- logy = x log x. differentiating both sides:-

15. Ans. (c)

is type as f(a) = g(a)

k = 1

16. Ans. (d)

17. Ans. (b) at x = – 3 as |x| = – x if x < 0. f(x) = – 3 (2 + x) f ' (x) = – 3

18. Ans. (d) x2x – 2xx coty – 1 = 0 …* Eq. at x = 1, 1 – 2cot y – 1 = 0 cot y = 0

Now, Diff. *Eq. wrtox :- 2(1 + logx) . x2x – 2 [–xx.cosec2y.y’]–2xx (1 + logx) . cot y = 0 At x = 1

2 + 2y’(1) = 0y'(1) = –1

19. Ans. (d)

Differentiation at x = 0.

LHD

Take x = 0 – h, h 0

(which does not exist).

20. Ans. (a) f " (x) = tan2x = sec2x – 1 Integrating both sides:- f'(x) = tanx – x + cas f'(0) = 0 0 = cHence f'(x) = tan x – x Integrating both sides again:-

And f(0) = 0 0 = log(1) + c c = 0

Hence


INFOMATHS21. Ans. (c) y = (x2 + 1)sinx

Taking log both sides:- logy = sinx . log (x2 + 1) differentiating both sides:-

x

At x = 0 y' (0) = 0 {since y (0) = 1}

22. Ans. (c) y = xlogx Taking log both sides:-logy = (logx)2 differentiating both sides:-

23. Ans. (b) xy = ex-y Taking log both sides:- y.logx = (x – y) on differentiating both sides wrtox:-

24. Ans. (c) On squaring both sides:- x2 = y + x on differentiating:-

25. Ans. (c) y = 4x3 – 3x2 + 2x – 1

At , = 3 – 3 + 2 = 2

26. Ans. (d)

Divide Nr / Dr by

Take x = cos

27. Ans. (d) y = xy Taking log both sides:- Log y = ylogx. Diff. both sides:-

28. Ans. (b) = tan-11 – tan-1x

at x = 2,

29. Ans. (c) Take x = tan

30. Ans. (b) f(x) = |x|f(x) is a modulus function, which is continuous at x = 0 but not differentiable at x = 0. As there can’t be a tangent drawn at x = 0 on the fnc. y = (x)

31. Ans. (c) as f(x) is an odd function f(–x) = – f(x)So, – f'(–x) = – f'(x)f'(–x) = f'(x)Put x = 3. f'(–3) = f'(3) = – 2


INFOMATHS13. APPLICATION OF DERIVATIVES (WORKSHEET)

1. Ans. (b) u(t) = t2 – 2t

dx = (t2 – 2t)dt Integrating both sides:-

As the particle starts from rest So, at t = 0, x = 0 C = 0 Now, at t = 3 seconds. x = 9 – 9 = 0 mt. Means, that the body has not moved for the first 3 seconds.

2. Ans. (d) As the function f(x) has a local minima at x = – 1 So, f(x) is continuous at x = – 1 LHL = RHL = f(–1 )K – 2x = 2x + 3 At x = - 1 K + 2 = 1 K = - 1.

3. Ans. (c) f(x) = x2 – x + 1 f'(x) = 2x – 1 f(x) is strictly increasing if f'(x) > 0 2x – 1 > 0

And f(x) is strictly decreasing for x < Hence, on [0, 1] f(x) is not monotonic i.e. Neither st. increasing nor st. decreasing

4. Ans. (c) f(x) = x3 – 3x + 2 f'(x) = 3x2 – 3 at (2, 4) f'(x) = 9

So, slope of normal Eq. of normal will be

9y – 36 = – x + 2 x + 9y – 38 = 0

5. Ans. (c) f(x) = xx f' (x) = xx (1 + logx) f(x) is decreasing, if f'(x) < 0 xx. (1 + logx) < 0 …* Eq. here, x > 0 as logx is defined if x > 0 only xx > 0 * Eq. gives 1 + logx < 0 log x < – 1 x < e-1

f(x) decreases for

6. Ans. (d) ax2 + by2 = 1

…(i)Also subtract given equation ___(a – a')x2 + (b – b')y2 = 0

6. Ans. (d) condition of orthogonally is m1m2 = – 1 Solving both equations and using the condition of

orthogonality. We have

7. Ans. (c) z = x + y

{as

For max. / min

x2 = 1 x = 1, – 1 as x > 0

y > 0 x = 1, y = 1

DDT at x = 1

at x = 1, y = 1 Minimum value of z = 2.

8. Ans. (d)

for max / mn.

logx = 1 x = e


INFOMATHS

Maximum value of

9. Ans. (a) z = px + qy

For max. / min.

Put

and also for ,

Hence

10. Ans. (c) f(x) = x5 – 20x3 + 240x f'(x) = 5x4 – 60x2 + 240 f(x) is monotonically decreasing if f ' (x) < 0 5x4 – 60x2 + 240 < 0 x4 – 12x2 + 48 < 0 (x2 – 6)2 + 12 < 0 Which is never possible Hence, f'(x) > 0 for all x R f(x) is monotonically increasing function on R.

11. Ans. (d)

So, f(x) is minimum, if is max. x2 + 1 is Minimum. Which is possible, only if x = 0

So,

And is attained, when x

12. Ans. (a) f(x) = x2 + x + (i) As f(x) is a polynomial func. f(x) is continuous on [a, b](ii) f(x) is differentiable on (a, b) as f'(x) = 2x + Hence, LMV holds. at least one real value

c (a, b) s. that

2c + = (b + a) +

13. Ans. (b) The volume of a right circular cylinder is r2h

14. Ans. (b, d) f(x) = 8x5 – 15x4 + 10x2

f ' (x) = 40x4 – 60x3 + 20x = 20x [2x3 – 3x2 + 1] = 20x (x – 1)2 (2x + 1) For extreme points. f'(x) = 0

extreme values are attained at x = 0, 1, only

15. Ans. (d) Slope of tangent

Slope of normal

Equation of normal: - y – a(sin – cos ) = – cot [x – a (cos + sin]sin [y – a(sin – cos)] = – cos [x – a (cos + sin]sin.y – asin2 + asincos = – cos.x + acos2 + asincos x cos + y sin = a Distance of the normal line from origin is ‘a’ unit always

16. Ans. (a) Length of normal Here y2 = 8x

At (2, 4)

So, LON units.

17. Ans. (b) y2 = 2x3 – x2 + 3.

At (1, 4) Eq. of tangent:-

2y – 8 = x – 1 2y – x – 7 = 0

18. Ans. (b) As f(b) = f(a) Since, condition of Rolle’s theorem are valid.

= f(2) – f(1) = 0 {As f(a) = f(b)}


INFOMATHS

19. Ans. (d) f ' (x) = 0

x2 = 4 x = 2for local max. / local min at x = 2

x x < 2 x > 2f'(x) –ve +ve

As f '(x) changes sign from –ve to +ve at x = 2 f(x) has a local minima at x = 2.

20. Ans. (a) f(x) = x2 – 5x + 6 f'(x) = 2x – 5 at (2, 0) f ' (x) = – 1 = m1 and at (3, 0) f ' (x) = 1 = m2 m1m2 = - 1

21. Ans. (b) (i) f(x) is continuous on [0, b]

(ii) f(x) is differentiable on (0, b) as exists for (0, b)

LNV holds at least one real value of c (a, b) s that

As c = 1 b = 4

22. Ans. (d) f(x) = x.(x – 1)2 f ' (x) = (x – 1)2 + 2x(x – 1) = (x – 1) [x – 1 + 2x] = (x – 1) (3x – 1) For max. / min f'(x) = 0 (x – 1) (3x – 1) = 0

x = 1, f(1) = 0. and

is the max value.

23. Ans. (c) Let the no. be 'a and b'

We know,

24. Ans. (c) f(x) = alog|x| + bx2 + x

Since f(x) has extremum at x = 1 and x = 3 f '(1) = 0 and f '(3) = 0

a + 2b + 1 = 0 and a + 18b + 3 = 0

Subtracting both equations:-

–16b – 2 = 0

also, a + 2b + 1 = 0

So, and

25. Ans. (a) f(x) = 2sinx + sin2x f'(x) = 2cosx + 2cos2x for max / minima Put f'(x) = 0 2cosx + 2cos2x = 0 cosx + cos2x = 0 cosx + 2cos2x – 1 = 0 2cos2x + cosx – 1 = 0 (2cosx – 1) (cosx + 1) = 0

or cosx = – 1

, , f(0) = 0 f(2) = 0

f() = 0

Absolute mini. At

Absolute maxi. At

Maxi. at

26. Ans. (a)

Eq. (I)Point where the curve crosses the y-axis has its abscissa as '0' * Eq. gives y = b Hence the point on Y-axis is (0, b)

from Eq. (1)Eq. of tangent:-

ay – ab = - bx bx + ay – ab = 0

27. Ans. (b) for the curve x2 + y2 – 2x – 4y + 1 = 0 Differentiating both sides:-

Tangent is parallel to X-axis


INFOMATHS

If

x + y – 1 = 0 Substituting value of y = 1 – x in the Eq. of circle. x2 + (1 – x)2 – 2x – 4 (1 – x) + 1 = 0 2x2 – 2x + 1 – 2x – 4 + 4x + 1 = 0 2x2 – 2 = 0 x2 = 1 x = 1, y = 2, 0 points are (1, 0) (–1, 2)

28. Ans. (b) y2 = 4ax

at (a, 2a)

Eq. of normal. y – 2a = – 1 (x – a) x + y = 3a is the eq. of normal.

29. Ans. (c) Z = sinx (1 + cosx)

= – sin2x + cosx + cos2x = cos2x – 1 + cosx + cos2x = 2cos2x + cosx – 1 = (2cosx – 1) (cos x + 1)

For max / min. (2cosx – 1) (cosx + 1) = 0

, ,

At

is a point of maximum

30. Ans. (b) f(x) = xx f'(x) = xx. (1 + logx) for max / min f'(x) = 0 xx(1 + logx) = 0 1 + log x = 0 logx = – 1 x = e-1

Max. value is obtained at

max. value

31. Ans. (c) As the curve is x2 = 2y Let the point nearest to (0, 5) be (a, b) a2 = 2b

Now,

For max. min. 2b – 8 = 0 b = 4 and

hence the points are

32. Ans. (d) P(x) = (x – 2)2 . (x – )

= (x – 2) [x – 2 + 2(x – )] = (x – 2) (3x – 2 – 2)

For max / min Put (x – 2) (3x – 2 – 2) = 0 3x – 2 – 2 = 0 As x = 1 is the extremum point 3 – 2 – 2 = 0 2 = 1

is the other root

33. Ans. (d) y2 = 8(x + 2)

At (– 1, 3)

Eq. of tangent

3y – 9 = 4x + 4 4x – 3y + 13 = 0

34. Ans. (d) f(x) = ax3 + bx2 + cx + d f(–2) = – 8a + 4b – 2c + d = 0 … Eq. (1)f'(x) = 3ax2 + 2bx + c as f(x) has max. and min. value.

At x = – 1 and f'(–1) = 3a – 2b + c = 0 … Eq. (II)

and a + 2b + 3c = 0 … Eq. (III)

as odd


INFOMATHS

b + 3d = 7 E. IVSolving all the equations simultaneously We have a = 1, b = 1, c = – 1, d = 2

35. Ans. (b) f(x) = cosx + 10x + 3x2 + x3 f'(x) = – sinx + 10 + 6x + 3x2 3(x2 + 2x + 1) + 7 - sinx 3(x + 1)2 + 7 - sinx > 0 f(x) is increase f(x) is inc. in [-2, 3] f(-2) is min.f(-2) = cos2 - 20 + 12 – 8 = 1 – 16 = - 15 for max. mina f'(x) = 0 –sinx + 10 + 6x + 3x2 = 0

36. Ans. (a) f"(x) = 6 (x – 1)

Eq. (1)As eq. of tangent at point (2, 1) is y = 3x – 5.

Eq. (2)Hence, Eq. (1) and (2) gives:- 3x2 – 6x + c = 3 At (2, 1) 12 – 12 + c = 3 c = 3 Also, f(x) = x3 – 3x2 + cx + d As the where passes through (2, 1) 1 = 8 – 12 + 2c + d 2c + d = 5 Hence d = – 1 as c = 3 Eq. of the function f(x) = x3 – 3x2 + 3x – 1 = (x – 1)3.

14. INDEFINITE INTEGRATION(WORKSHEET)

1. Ans. (a) by adjustant.


INFOMATHS

As = f(x)ex + c

2. Ans. (b)

Using LATE RULE in first integral

3. Ans. (b)

= logx – x + c

4. Ans. (d)

5. Ans. (c)

As

6. Ans. (a) Or prove the result by substituting x = a tan dx = a sec2 d

7. Ans. (d)

on comparison with RHS.

We have = 1,

15. DEFINITE INTEGRATION(WORKSHEET)

1. Ans. (c) …Eq. (1)

Using property

… Eq. (ii)Adding Eq. (i) and (ii)


INFOMATHS

2. Ans. (c)

3. Ans. (c)

4. Ans. (d) Solved -assignment Q. No. 24

5. Ans. (a) Area between curves y = 2 – x2 and y = x2 firstly, the POI of curves are x2 = 2 – x2 2x2 = 2 x2 = 1 x = 1 y = 1Hence, the points are (1, 1) and (–1, 1).

Area

sq. units

sq. units

sq. Units

6. Ans. (d)

= (Remaining function being odd function

will vanish)

As , if f(–x) = f(x)

7. Ans. (b)

Using

We have

8. Ans. (d) The give Q. Eq. is homogenous in x and y. So, take y = vx

Integrating both sides:-v = logx + c

y = x [log x + c]as we have y (1) = 1 1 = c c = 1 Hence, the curve is y = x log x + x

9. Ans. (b) … Eq. (i)

Using

Adding both equations:-

Put cosx = t – sinx dx = dtAs x = 0, t = 1

x = t = – 1

= (tan-11 – tan-10)

10. Ans. (b) For the POI of the curves y = 2 – x and x2 + y2 = 4


INFOMATHS

x2 + (2 – x)2 = 4 2x2 – 4x + 4 = 4 2x2 – 4x = 0 2x (x – 2) = 0 x = 0, 2 y = 2, 0 respectively. Poi’s are (0, 2) and (2, 0) Area bounded by curves

11. Ans. (c)

12. Ans. (c) As So, the in local I = 0

13. Ans. (a) f(x) Being an odd function as f(–x) = f(x) I = 0

14. Ans. (d) POI of the curves y = x and y2 = 16x x2 = 16xx(x – 16) = 0 x = 0, 16 y = 0, 16 Hence POI’s are (0, 0) and (16, 16)

Area

sq. units.

15. Ans. (d)

Divide num / Der by cos2 x

16. Ans. (c) =

As f() is an even function.

Take d = 2.dx As = 0, x = 0

,

= 0

17. Ans. (c)

2t3et – 6t2et + 12tet – 12et 2e – 6e + 12e – 12e + 12 - 4e + 12 = 4 (3 – e)

18. Ans. (c)

19. Ans. (b) Take x2 =

2x dx = d

= e1 – [e – 1] = 1

20. Ans. (b)


INFOMATHS

Downward parabolic with vertex (1, 5)

Required Area from figure

21. Ans. (a)

= I1 + I2 Put x = - x in I1

2I1 = 0 I1 = 0 For I2 as x – [x] = [x] = fractional part So 0 [x] < 1

Also

as 0 < [x] < 1

22. Ans. (b)

= 2p'(2) – (p(2) – p(0)) = 2(-1) – (3 – 3) = - 2

23. Ans. (d) (by result)

24. Ans. (d) It’s rectangle with vertices (0, 0), (0, 3), (2, 1) (-1, 2)

So, length , breadth Area

25. Ans. (a) We have two triangles required Area = Area OAB + Area OAC

= 1 sq. unit

26. Ans. (a) Area between curves y = f(x) and y = 0 is

Differentiating both sides wrtox:-

27. Ans. (c)

= cos0 – cos = 1 + 1 = 2

28. Ans. (d) -assignment (solved) Booklet)

29. Ans. (d) … Eq. (1)

Using


INFOMATHS

… Eq. (2)

Adding Eq. (1) and (2):-

I = 0

30. Ans. (a) …* Eq. Differentiating both sides using Leibnitz rule:- f'(x) = 2sinx cosx. x – 2cosx.sinx.x f'(x) = 0 f(x) = K [constant function]

Also, substituting in * Eq.

So

31. Ans. (a)

By using

I1 = - I1 2I1 = 0 I1 = 0

32. Ans. (c)

Let

Put

So ans. (c) as as a = 2

33. Ans. (c)

34. Ans. (b) y = ex

35. Ans. (d)


INFOMATHS

a2 + 49 – 77 0 (a + 11) (a – 7) ≤ 0- 11 < a < 7 As a I+

a = 1, 2, 3, 4, 5, 6 sin values of a.

36. Ans. (b) By Leibnitz rule f '(x) = x sin x – 0 = x sin x .

37. Ans. ()

38. Ans. (a) y = x2, y = x Intersection points x2 = x x = 0, 1 y = 0, 1

Intersection points are (0, 0) and (1, 1) area bounded between y = x2, y = x

Also area bounded between y = x2, y = 1 Intersection points are 1 = x2 x = 1

Area bounded between

39. Ans. (d)

Case I: a < b < 0

Case II: a < 0, b > 0

Case III : a < a < b

From I, II, III ans. is |b| - |a|

40. Ans. (c)

By result

41. Ans. (c)

42. Ans. (a)

16.DIFFERENTIAL EQUATIONS(WORKSHEET)1. Ans. (a)


INFOMATHS(separating the variables) Now, integrating both sides:-

(x + a) (1 – ay) = ayc Hence, solution is (x + a) (1 – ay) = yc

2. Ans. (d)

Integrating both sides:-

…*Eq.

as x = 1, y = 1 * Eq. gives

At x = – 1

y = – 1

3. Ans. (b) Take 4x + y + 1 = z


z = 2tan (2x + c) 4x + y + 1 = 2tan (2x + c)

4. Ans. (b)

Comparing the Eq. with LD Eq.

There P = 1, Q = t

IF Solutions is given as

s (IF)

S = t – 1 + ce-t

5. Ans. (a)

Comparing with LD Eq.

We have , Q = 2y2

I. Factor Solution is given as

x. IF

x = y (y2 + c)


INFOMATHS

6. Ans. (b)

Comparing with

Here, Q = 2x

I factor

7. Ans. (c) y(2x + y2)dx + x(x + 3y2)dy = 0 …(*)Mdx + Ndy = 0 M = y(2x + y2), N = x(x + 3y2) My = 2x + 3y2, Nn = 2x + 3y2 My = Nn D.E. is exact Solution of *

(By taking y cos tan) (by taking tan s not containing x w.r.t. y)

x2y + xy3 = c

8. Ans. (c)

Put

Integral

9. Ans. (b)

…(*)

Leibnitz D.E. w.r.t. y

IF =

Solutions of * is

9. Ans. (b)

Let tan-1y z

xez = z + c

10. Ans. (a)

Multiplying both sides by 2y:-

Take y2 = z

On comparing with

We have Q = 1

I. factor Solution is given as

z = x(logx + c)y2 = x(logx + c)

11. Ans. () Eq. of a circle having centre (h, k) and radius = r units is (x – h)2 + (y – k)2 = r2 … Eq. (1)


INFOMATHSDifferentiating both sides:-

…Eq. (2)Diff. again:-

Differentiating again wrtox:-

y' . (y")2 = – y" – 2y' (y")2 + x.y"' + (y')2 . y"'

14. Ans. (c) 9yy' + 4x = 0

9y dy = – 4x dx. Integrating both sides:-

9y2 = – 4x2 + c 4x2 + 9y2 = c

15. Ans. (c) x dy – y dx = 0 x dy = y dx


logy = logx + logc logy = log(xc) y = xc which represents an equation of a straight line.

16. Ans. (c)

On integrating both sides:-

Squaring both sides t = x2 + c2 +2xcx2 + y2 + 2cx + c2 – 1 = 0 is the required equation which represents a circle with centre (–c, 0) variable an x axis and radius = 1 unit.

Part of ques. no. 75. 3y' . (y")2 + y" = xy"' + (y')2 . y"' = y"' [x + (y')2]

12. Ans. (a)

and Q = sin x

I. factor Solutions is given as

y . IF

xy + x cosx = sinx + c

13. Ans. (b) x.y'= 2y

On integrating both sides:-

logy = log (xc)2

y = (xc)2

as the curve passes through (1, 2) 2 = c2

Here, the Eq. of curve will be y = 2x2 Which passes through (4, 32)

17. Ans. ()

18. Ans. (c)

19. Ans. (c)

20. Ans. repeat Q.9.

21. degree = 6


INFOMATHS

17. MATRICES DETERMINANTS (WORKSHEET)1. Ans. () As |kA| = kn |A|

Where n is order of A But n 3So, |kA| k|A|, for any value of k. Hence |aA|=a|A| does not hold true for any value of a.

2. Ans. ()

3. Ans. () As, A is a skew symmetric matrix. Then all diagonal elements will be 'O'Hence a11 = a22 = a33 = ……. ann = 0 So, all the choices are true.

4. Ans. () As.

So Eq. has a unique solution. 18 INFOMATHS/MCA/MATHS/

INFOMATHS

5. Ans. (b) Consider

b + 2c = 1 e + 2f = 0 Eq. *h + 2i = 0

Eq. **

To evaluate :-

Eq. (I)

2 Eq. ** – Eq. * gives (6a + 8b + 10c) – (b + 2c) = 0 – 1 = – 1 (6d + 8e + 10f) – (e + 2f) = 2 – 0 = 2 (6g + 8h + 10i) – (h + 2i) = 0 So, Eq. I given

Required matrix as

6. Ans. (c) As system of equation is homogenous.

and

given b – 5 = 0 b = 5 Infinitely many So Eq. has solution for just are value of b.

7. Ans. (b)

(2 – y) [y2 – 15y + 50 – 24] – 2 [20 – 2y – 18] + 3[8 – 15 +3y] = 0 (2 – y) [y2 – 15 + 260 – 2[–2y + 2] + 3[3y – 7] = 0 2y2–30y+52 – y3 + 15y2 – 26y + 4y – 4 + 9y – 21 =0 – y3 + 17y2 – 43y + 27 = 0 y3 – 17y2 + 43y – 27 = 0 clearly y = 1 is the root of the equations.

8. Ans. (a) Case (i) Diagonal element is include only 1 zero. No. of symmetric matrices. = 3C1 3 ways {A remaining 4 zeros may be set in just 3 ways}} = 9 matrices. Case (iii) Diagonal elements include 3 zeroes: - No. of matrices = 3C3 3 ways {As remaining 2 zeroes may be set in just 3 ways} = 3 matrices. T. no. of matrices = 12.

9. Ans. (d) (M – N)2 = (M – N) (M – N) = M. (M – N) – M. (M – N)= M2 – M.N – N.M – N2.

10. Ans. (c) (MN – NM)T = (MN)T – (NM)T = NT.MT – MT.NT

= N.M – M.N. {As MT = M, NT = N} = – (MN – NM)Hence MN – NM is skew symmetric.

11. Ans. (b)

12. Ans. (b) As the matrix is singular |A| = 0

13. Ans. (b)

14. Ans. (b) (b) x + w2y + wz = 0


INFOMATHSwx + y + w2z = 0w2x + wy + z = 0

As = 0 using C1 C1 + C2 + C3 So, So Eq. has Infinitely many consistent.

3(–4k – 9) – k (–4 – 6) – 2(3 – 2k) = 0 – 12k – 27 + 10 k – 6 + 4k = 0

2k – 33 = 0

15. Ans. (d)

16. Ans. (b)

17. Ans. () = 0n using C1 C1 + C2 + C3 since 1 + w + w2 = 0 & = 0 if all the entries of a row or a column are '0'

18 Ans. (b)

19. Ans. ()

20. Ans. ()

21. Ans. ()

22. Ans. ()

23. Ans. (c) As B = – 4-1BA (A + B)2 = (A + B) . (A + B) = A2 + A.B + BA + B2 …* Eq. Also B = –A-1BA Pre-multiplying both sides by A:- AB = – (AA-1) BA = – IBA AB = – BA So * gives:- (A + B)2 = A2 + B2 {As AB = – BA}

24. Ans. (d) Considering the equations

using R1 R1 – 3R2

and R3 R3 – 2R2

= 0

and

using R2 R2 – R1 and R3 R3 – 4R1

= – 25 + 25 = 0

Similarly 2 = 3 = 0 So, Sol Eq. has infinitely many solutions!

using C3 C3 + 3C1 and C2 C2 + 2C1

= 50 – 32 = 18 0

Hence A is non singular matrix.

25. Ans. (d)

26. Ans. (b) (b)

A.B

27. Ans. (c)

(1 + xyz) (x – y) (y – z) (z – x) = 0 1 + xyz = 0 xyz = – 1

28. Ans. (d) So Eq. has no. solutions if |A| = 0


INFOMATHS

Using C1 C1 + C2 + C3

R3 R3 – R1 R2 R2 – R1

(a + 2) (a –)2 = 0 a = – 2, 1 Also, At least are determinant should be non-zero i.e. |1| = 0

29. Ans. (c) A is invertible, if |A| 0

Expanding along R1 : - 1 – K (0 – K) 01 + k2 0Which is true for k R.

30. Ans. (c)

31. Ans. (d)


Using R3 R3 – R1 And R2 R2 – R1

x(x + w2 – w) (x + w – w2) = 0 x[x + w(w – 1)] [x – w(w – 1)] = 0 x[x2 – w2 (w – 1)2] = 0 x[x2 – w2 (w2 + 1 – 2w)] = 0 x[x2 – w2[–w – 2w]] = 0 x[x2 – w2(–3w)] = 0 x[x2 + 3w3] = 0 x(x2 + 3) = 0 x = 0

32. Ans. (b) As A is singular matrix |A| = 0 A . adjA is always a zero matrix.

33. Ans. (c)

= A

A4 = A2.A2 = I So, A4 + A3 – A2 = I + A – I = A

34. Ans. (d) = 0 as the elements of determinants are consecutive.

35. Ans. (c)

= A

36. Ans. (d)

A2 – 5A + 7I = ?Characteristics eq. of A is |A – I| = 0

(3 – ) (2 – ) + 1 = 0 2 – 5 + 6 + 1 = 0 2 – 5 + 7 = 0 As, according to cayley Hamiltonian theorem Each matrix satisfies its characteristics equation. A2 – 5A + 7I = 0

37. Ans. (c)

38. Ans. ()

39. Ans. (b) If |A| = 3 |adjA| = |A|n-1

|adjA| |3|2 = 9

40. Ans. (b) As A2 – 2A + I = B


INFOMATHS-2 + 2b = 4 (b – 1)2 = 4 2b = 6 b – 1 = 2b = 3 b = 3, – 1

41. Ans. (d)

= 1(5 – 1) –10 (5 – 10) + 14 (5 – 50) = 4 + 50 – 630 = – 630 = – 576.

42. Ans. (c)

= x(x2 – 12) – 2(3x – 15) + 5 (12 – 5x) = x3 – 12x – 6x + 30 + 30 – 25x = x3 – 43x + 60

43. Ans. (b)

|A| = 1

A-1 = adjA = AT

44. Ans. (b)

= 1

45. Ans. () As A is orthogonal AT . A = I Then |A| cannot be determined.

46. Ans. (b) as

47. Ans. (c) As a, b, c are the roots of x3 + px2 + q = 0Then a + b + c = – p ab + ac + bc = 0abc – qq


= (a +b + c) [– (c – b)2 – (a – b) (a – c)] – (a + b + c)2 [(c – b)2 + (a – b) (a – c)] – (a + b + c) [c2 + b2 – 2bc + a2 – ac – ba + bc] – (a + b + c) [a2 + b2 + c2 – ab – bc – ca] – (–p) [a2 + b2 + c2 – (0)] + p(a2 + b2 + c2) + p(a + b + c)2 – 2(ab + bc + ca)] + [(–p)2 – 2(0) + p p2 = + p3

48. Ans. (c)

49. Ans. (d)

= 6(16 – 18)–5(24 – 24) + 4(27 – 24) = – 12 – 0 + 12 = 0 So, So Eq. has Infinitely Many solutions since it is a Homogenous set of Equations

50. Ans. (a)

= a(a2.

51. Ans. (b) Given Homogenous equations have a non-trivial solutions if = 0

a(3 – 2) – 4(b – c) + 1(2b – 3c ) = 0 a – 4b + 4c + 2b – 3c = 0 a – 2b + c = 0 a + c = 2b a, b, c are in AP.

52. Ans. (d)

53. Ans. (c) and

A. B is not possible. But B.A is possible Also (A + B) does not exist.

54. Ans. (b) A-1 exist, if |A| 0

(– + 6) + 4(–2 + 3) + 1 (4 – ) 0 – + 6 + 4 + 4 – 0


INFOMATHS–2 + 14 0 7Or R, 87

55. Ans. (c)

A2 = I

56. Ans. (a)

4x – x + y

On comparing :- 2x + y = 3 and 3x + y = 2 Subtracting both equations – x = + 1 x = – 1 and y = 5 x. y = – 5

57. Ans. (d)

C1 C1 + C2 + C3

Using R3 R3 – R1 and R2 R2 – R1

(a + b + c) [– (b – c)2 – (a – b) (a – c)] – (a + b + c) [(b – c)2 + (a2 – ac – ab + bc)] – (a + b + c) [(b2 + c2 – 2bc + a2 – ac – ab + bc)] – (+a + b + c) [(a2 + b2 + c2 – ab – bc – ca] – (a + b + c) [(a – b)2 + (b – c)2 + (c – a)2]So is negative, if a + b + c is positive.

58. Ans. (d)

R1 R1 + R2+ R3

C3 C3 – 2C2

– (a + b + c) [– (a – c)2 – (a – b) (c – b)] (a + b + c) [(a – c)2 + ac – ab – bc + b2] (a + b + c) [a2 + b2 + c2 – ab – bc – ca] = a3 + b3 + c3 – 3abc

59. Ans. (a)

C2 C2 + C3

= 0 as C1 and C2 are identical columns.

60. Ans. (b)

x3 – 67x + 126 = 0 As – a is one of the roots. Let the other roots be and + – 9 = 0 Eq. I – 9 – 9 = 67 Eq. II r = – 126 Eq. IIISo, 2 and 7 are the roots satisfying the above equations.

61. Ans. (b) 1. (17a + 14b) = 3 + 2 17a + 14b = 5 Here are no such values of a and b that satisfies the above equations.

62. Ans. (c) |A . B| = |A| . |B| = 8 . 2 = 16 {As determinant of a diagonal matrix is the product of diagonal entries only.}

63. Ans. (c) as

|M| = 6

64. Ans. (a)

R4 R4 – R1 R3 R3 – R1 R2 R2 – R1


INFOMATHS

= 1 {as is a diagonal matrix}

65. Ans. (c)

Similarly,

66. Ans. (d) The matrix Is invertible if |A| 020 – 4k = 0 k = 5

67. Ans. (a) Characteristic equation will be |A – I| = 0

(2 – )2 – 3 = 0 2 – 4 + 4 – 3 = 0 2 – 4 + 1 = 0 And as we know, every characteristic equation satisfies the matrix. A2 – 4A + I = 0

68. Ans. (a) |A + AT| = |A| + |AT| is not true.

69. Ans. (b)

= unit matrx.

70. Ans. (c) AB = 0 |A . B| = |0|Either, |A| = 0 or |B| = 0 A = 0 or B = 0

71. Ans. (d) |A| = an integer, if all the entries of the matrix are integers.

72. Ans. (c) AB = A and BA = B. Then B2 = B.B = (BA) (BA) {As BA = B}= B(AB) . A using associative law = (BA) A As BA = B = BA = B again, BA = B

73. Ans. (b)

using C3 C3 – 3C1 C2 C2 – 2C1

= 10 – 12 – 2

74. Ans. (c)

On comparison:- a = 2, b = 1 – a + c = 0 c = 2, and – b + d = 1d = 1 + b = 2 a = 2, b = 1, c = 2, d = 2

Hence

75. Ans. (a) Matrix multiplication is non commutative

76. Ans. (a)

So,

AAT = I Hence, A is an orthogonal matrix.

77. Ans. (d)

= 0

78. Ans. (d)

=(a+b+c)2 . as determinant of a diagonal matrix is 1.

79. Ans. (a) AB = 0 |AB| = 0 |B| = 0


INFOMATHSas |A| 0

80. Ans. (b)

C1 C1 + C2 + C3

C3 C3 – C1 and C2 C2 – C1

= (3 + p) p2

p = 0, p = 0, p = – 3

81. Ans. (c)

C1 C1 + C2 + C3

R3 R3 – R1 R2 R2 – R1

= x(–x2) =–x3

x = 0

82. Ans. (b) Characteristic Equation of A is

|A – I| = 0

(4 – ) (1 – ) + 2 = 0 2 – 5 + 4 + 2 = 0 2 – 5 + 6 = 0 As A will satisfy the chara Eq. A2 – 5A + 6I = 0 Hence A2 – 5A + 6I is a null matrix.

83. Ans. (b)

84. Ans. (a)

85. Ans. (b)

86. Ans. (d)

87. Ans. (d)

18.VECTORS(WORKSHEET)1. Ans. (b)

2. Ans. (d)

3. Ans. (d) = P.V of B – P.V of A

of D – P.V of C

cos = 1 = 0

4. Ans. () If a point (x, y) is rotated through an angle in anticlockwise direction. Then new co-ordinates of point are (x, cos – y, sin, x,sin + y, cos) Now, as has components 2p and 1 and the vector is rotated through an angle . Then new coordinates will be (2 pcos , 2psin+ cos) comparing it with (given new coordinates (p + 1, 1) We have 2p cos – sin = p + 1 and 2 psin + cos = 1 squaring and adding both equations 4p2 + 1 = (p + 1)2 + 1 4p2 + 1 = p2 + 2p + 2 3p2 = 2p + 1 3p2 – 2p – 1 = 0 3p2 – 3p + p – 1 = 0 3p (p – 1) + (p – 1) = 0 (3p + 1) (p – 1) = 0


INFOMATHS

5. Ans. (c) Given system of equations is homogeneous and has a non-zero solution. = 0

C3 C3 – C1 and C2 C2 – C1

Expanding determinant along R3:- 2[(a+1)2+a2+a(a+1)]–2[(a+2)2+a2+a(a+2)] = 0 (a + 1)2 + a2 + a(a + 1) – (a + 2)2 – a2 – a(a + 2) = 0 (a + 1)2 – (a + 2)2 + a [(a + 1) – (a + 2)] = 0 (2a + 3) (–1) + a(–1) = 0 2a + 3 + a = 0 3a + 3 = 0 a = – 1

6. Ans. (c) and consider

Also, as vectors makes equal angles with each other.

y + z = 1 and x + y = 1 z – x = 0 Z = x

hence as

Squaring both sides:- 3x2 – 2x + 1 = 2 3x2 – 2x – 1 = 0 3x2 – 3x + x – 1 = 0 3x (x – 1) + (x – 1) = 0 (3x + 1) (x – 1) = 0

x = 1,

if x = 1, then y = 0, z = 1

vector

Now, does not make an obtuse angle of with vector i.

So, , and

Now,

7. Ans. (d) No. of students = 30Average = 45 sum of marks of students = 30 45 = 1350As marks were observed wrongly and new marks were increased by 24 and 34. New sum of marks = 1350 + 24 + 34 = 1408

New corrected average = 46.93

= 47 (approximately)

8. Ans. ()

& (Comparing Both Sides)

{as and are unit vectors}

x 1

9. Ans. (d)

10. Ans. (c) As Here R is the resultant force of P and Q. inclined at angle with each other

25P2 = 25P2 + 24P2cos 24P2cos = 0 cos = 0 = 90

11. Ans. (c) We know, if and are the adjacent sides of a parallelogram.

Then Area

sq. units.


INFOMATHS

12. Ans. (d) Let vector parallel to be

Now take vector be perpendicular to

then

x + y = 0 Eq. * Now, as

Comparing both sides:- + x = 3 + y = 0 z = 4 x = 3 – , y = – , z = 4 Substituting values of x and y in * eq. (3 – ) – = 0 2 = 3

13. Ans. (c) As are mutually perpendicular

= 6

14. Ans. (b) Question 8 Repeated

15. Ans. (b)

16. Ans. (b) We know that value of a parallelepiped

with edges is equal to

Volume

= 1(1 – 0) – x (0 – x ) + (–x) v = 1 + x3 – x

For max / min put

3x2 – 1 = 0

Now,

at we have gives minima.

Hence value is minimize at

17. Ans. (d)

18. Ans. (c) Value of parallelepiped with edges and

is

Volume = 1 – (–)2 + 1(–)

V = 3 – + 1

For max / min

32 – 1 = 0

Now,

At gives minima

Volume is minimized at

19. Ans. (b)

20. Ans. (c)

Taking dot product with and :-

… Eq. (1)

… Eq. (2)

… Eq. (3)


INFOMATHS

Adding 1 and 2 :-

= 0 … Eq. *

Subtracting Eq. (3) from Eq. *

as , ,

Then

or

21. Ans. (b) as = 42

= 16

22. Ans. (b) As if A, B, C, D are the vertices of a tetrahedron Then volume of tetrahedron

Where, = (k + 1, k + 6, k + 36)

– (k, k, k) = (1, 6, 36)

= (k, k + 2, k + 5) – (k, k, k)

= (0, 2, 5)

= (k, k, k + 6) – (k, k, k) (0, 0, 6)

hence = 12

value of tetrahedron

cubic units.

23. Ans. (c) If vectors and are orthogonal then

– 2 – 2 + 12 = 0 2 = 10 = 5

24. Ans. (c) As

Area of parallelogram

Here,

= 49 sq. units

25. Ans. (a)

Squaring both sides:-

Parallelogram is a rectangle as two adjacent sides are

26. Ans. (b) …

Eq. *

and

Sol. Taking dot product with on both sides of Eq. *

and

As if any of the two vectors are

identical

Adding all the 3 equations:-

27. Ans. (a) and


INFOMATHS… Eq. (1)

and

giving … Eq. (2)

From equation 1 and 2

… Eq. *

But as vectors are non coplanar. for some scalars , ,

where = = Hence in Eq. *:- 1 + 2 = 0 and + b = 0

and = – 6

Since the vectors and are non collinear.

Eq. 1 gives 28. Ans. (b)

29. Ans. (a) Let the required vectors

be

Projection of an

3 = x

Projection of on

4 = y

Projection of on

12 = z

Vector

Units.

30. Ans. (d)

31. Ans. (c) Pv of B – Pv of A

As points A, B, C, D are coplanar.

32. Ans. (a)

Expanding the determinant along R1:- x (0– z) + y [–x + z (–1+x)] = 0 –xz – yx + yz (–1 + x) = 0 –xz – yx – yz + xyz = 0 xyz = xz + yx + yz

32. Ans. (a) As are unit vectors

as,

on comparing both sides:-

and

cos = 0

34. Ans. () as

and

So, is the odd one.

35. Ans. () As is true.

and

statement (iii) and (iv) are true.

36. Ans. () See 3-D assignment–a question 6 answer.

37. Ans. ()

38. Ans. ()

Squaring both sides and adding both equations

and


INFOMATHS

and

on adding

Hence

39. Ans. () As vectors and are perpendicular

x + 3x + 8 = 0 4x + 8 = 0 x = – 2

40. Ans. () As, the vector equation of a plane passing through a fixed point having and normal as is

3(x – 1) – (y – 2) + 2(z – 3) = 0 3x – y + 2z – 7 = 0

40. Ans. () Repeated question 20.

42. Ans. () = tan

43. Ans. () Taking dot product both sides with

and

= 0

Adding all the equations:-

44. Ans. () Let the given vectors be coplanar. Scalar triple product = 0

(2 – 1) = 0

= 0,

Hence, the vectors are non coplanar for all except

2 values of i.e. = 0,

45. Ans. ()

. {as = 0

As are coplanar vectors.

46. Ans. () as and between and is acute then


INFOMATHS2x2 + 3x + 1 > 0 (2x + 1) (x + 1) > 0 x < – 1 or x > – 1 …* Eq. (i)

Now as cos and

then cos < 0

–x < 0 x > 0 … Eq. (2)combining Eq. (1) and (2)x > 0

48. Ans. () As is angle between vectors and .

sin = cos . tan = 1

49. Ans. () Let and be the adjacent sides of the parallelogram.

According to the triangle law of vector addition, and in ABD,

50. Ans. () As, if then

2 – 4 – = 0 = – 2

51. Ans. () = 0 {Since

52. Ans. ()

53. Ans. () 1 = Pv of B – Pv of A

= Pv of C – Pv of A

Hence points A, B, C are collinear.

54. Ans. () as

Hence, vectors and are parallel. Since two vectors are parallel if one of the vectors can be represented in terms of other.

55. Ans. ()

sin = cos tan = 1

56. Ans. ()

57. Ans. () Angle between two vectors and is

= cos-10

58. Ans. ()

Either , , cos = 0

, , = 90

59. Ans. ()

and is angle between and

as is a unit vector.

v

2cos = – 1


INFOMATHS

60. Ans. () If vectors and are parallel and are of equal magnitudes. Then they are equal and have same sense of direction.

61. Ans. ()As and are adjacent sides and are equal vectors

62. Ans. ()In ABC, triangle Acc. To triangle law of vector addition

and in ABDAcc. To triangle law of vector addition.

So, the diagonal vectors are and

63. Ans. () As we know, if and are adjacent sides of a triangle.

Then area

Here, P, Q, R are vertices of the triangle.

Area PQR

square units.

19. 3-DIMENSIONAL GEOMETRY(WORKSHEET) 1. Ans. (b)

Let image of point P be P’ (a, b, c) wrto plane x – 2y = 0 using shortcut method:-

c – 4 = 0

c = 4

Hence image of point P (–1, 3, 4) is


P(-1, 3, 4)

x – 2y = 0

P(a, b, c)

INFOMATHS2. Ans. (a) Length of the perpendicular drawn from (1,

0, 2) to pi

(assumed)

x + 1 = 3 x = 3 – 1 y = – 2 + 2 z = – – 1

M(3 – 1, – 2+ 2, – – 1) Now, DRs of the vertical line will be <3 – 2, – 2+ 2, – – 3>D.R’s of horizontal line are <3, – 2, –1>

As Pn AB 3(3 – 2) – 2 (–2 + 2) –1(_– 3) = 0 9 – 6 + 4 – 4 + + 3 = 0 14 – 7 = 0

Point M will be (3 – 1, – 2 + 2, – – 1)

Length of line PM will be

units.

3. Ans. (d)

= 7 units.

= 7 units

= 7 units = 7 units

Now for diagonals.

Hence ABCD is a rhombus.

4. Ans. (b) Similarly as question 3.

5. Ans. (d) As A (1, –1, 0) 3(–2, 1, 8) and C(–1, 2, 7) are consecutive vertices of a parallelogram. Hence D.R.’s of AB = D.R’s of DC As parallel lines will have same D.R.’s. <–3, 2, 8> = <–1 – a, 2 – b, 7 – c>Where the vertices of D are assumed as (a, b, c) on comparison: – (a + 1) = – 3 2 – b = 2 7 – c = 8

a + 1 = 3 b = 0 c = – 1 a = 2hence vertex D will be (2, 0, – 1)

6. Ans. (b) The points (0, 0, 0) (1, 0, 0) (0, 2, 0) and (0, 0, 4) lines as a sphere.

7. Ans. () Let the point at which the line join A (3, 1, –2) and B(–2, 7, –4) intersects the XY plane be M(a, b, 0) and assumes that the point M divides AB in ratio : 1

Using section formula

as – 4 – 2 = 0

M will be (8, – 5, 0)

8. Ans. () As x + y + z = 0 x + y = – z Cubing both sides:-(x + y)3 = z3 x3 + y3 + 3xy (x + y) = – z3 x3 + y3 + z3 – 3xyz = 0 k = 3.

10. Ans. ()

Using image formula

On comparison:-

and


P(1, 0, 2)

A M B

P(p, q, r)

2x + y + z – 6 = 0

INFOMATHS

12. Ans. ()

Value of frustum

Volume

Using similarity condition:

Volume of frustum

16. Ans. () Equation of line through (–1, 3, 3) is

(assumed)

x + 1 = x = – 1 y – 3 = 2 y = 2 + 3z – 3 = 3 z = 3 + 3 Now, as the point hits the XY plane z = 0 Hence 3 + 3 = 0 = – 1

Point at which it hits the xy plane is (–2, 1, 0).

9. Ans. () x = a, z + b, y = c, z + d,

and x = a2z + b2, y = c2z + d2

are the two straight lines with

DR’s As < a, c, 1 > and < a2, c2, 1 >Since the lines are . a1a2 + c1c2 + 1 = 0

20. COMPLEX NUMBERS (WORKSHEET)

1. Ans. ()

R2 R2 – R1 – R3

= 0 {as determined value is “0” if all elements of a row are ‘0’

2. Ans. () Using Homers’ method

1 9 35 – 1 4 - 5 + 4i –5 + 4i – 36 – 4i – 21 +16i – 164

1 4 + 4i – 1 –4i 20 + 16i 160

Hence, the remainder is 160.

3. Ans. () and

is the equation of right bisector of a

line join 0 + 0i and

4. Ans. () Consider z1 = x1 + iy, and z2 = x2 + iy2


INFOMATHS

On squaring both sides:-

Squaring both sides again

x1x2 + y1y2 = +2x1x2y1y2

x1x2 + y1y2 + 2x1x2y1y2

arg z2 = arg z1 arg z2 – arg z1 = 0

5. Ans. () |z + 4| 3 represents a circle centred at 4 + 0i with radius as 3 units. Now |z + 1| = |–6| = 6

6. Ans. ()

On rationalizing

Conjugate of z is Now, rationalizing

7. Ans. () As z2 + z + 1 = 0Then roots of the Q. Eq. are w and w2.

Now,

… Eq. *

as z = w, or z = w2 and z12 = w12 = 0

Eq. * gives ‘0’ as answer.

8. Ans. ()

9. Ans. () No. of solution of as z = x + iy (x + iy)2 + x – iy = 0 x2 – y2 + i2xy + x – iy = 0 x2 – y2 + x + i(2x – 1)y = 0 + 0i comparing the real and imaginary parts. x2 – y2 + x = 0

(2x – 1)y = 0 , or y = 0

when

x2– y2 + x = 0 gives

Solutions are

and also, as if y = 0 x2 + x = 0 x(x + 1) = 0 x = 0, – 1 solutions are z = 0 + 0j and z = – 1 + 0i Hence, there are 4 solutions of the equation.

10. Ans. () We know, that triangle formed by cube roots of unity is an equilateral triangle with length of each side as units. Also, area of an equilateral triangle

(side)2 sq. units.

11. Ans. () as ‘z’ has constant modulus z = x + iy

x2 + y2 = k2 Now, as z2 is purely imaginary. (x + iy)2 = x2 – y2 + i2xy Gives x2 – y2 = 0 {since a purely imaginary complex no. has no real part} x2 = y2 Also, as x2 + y2 = k2 2x2 = k2

If

And if There are 4 solutions as

and

12. Ans. () (1 + w – w2)7 = ?as 1 + w + w2 = 0 1 + w = – w2 (1 + w – w2)7 = (–w2 – w2)7 = (–2w2)7 = 128w14 = – 128w2.


INFOMATHS

13. Ans. () As an rationalizing

= i the smallest natural no. for which in = 1 is n = 4.

14. Ans. () As and are the roots of x2 + x + 1 = 0 = w and = w2. 4 = w2 = w and 7 = (w2)7 = w14 = w2 Then Q. Eq. will remain same as the roots are uncharged. Required Q. Eq. will be x2 + x + 1 = 0

15. Ans. () If P is a point an Hyperbola. Then |PF2| – |PF1| = 2a {where F1 and F2 are the foci of hyperbola & 2a is length of transverse axis.}So, |z + i| – |z – i| = k represents a hyperbola, if 0 < k < 2

16. Ans. ()

On rationalizing As, the complex no. is purely real. Imaginary part = 0 (1 – 2x2)x + 3x = 0 1 – 2x2 + 3 = 0 2x2 = 4 x2 = 2

17. Ans. () As we know that 4th roots of unity are i, – i, 1 and – 1

= i2 + i2 + 1 + 1 = 0

18. Ans. ()

C1 C1 – C3 and C2 C2 – C3

Expanding the determinant along R1:- 0 – (ei – 1) (1 – e-i) = 0 (ei – 1) (1 – e-i) = 0 cos + isin– 2 + cos – isin = 0 2cos – 2 = 0 cos = 1 = 0

19. Ans. () 4th roots of unity are 1, –1, i, – i

20. Ans. () (x2 + y2 + (4 – 3i) (x + iy) + (4 + 3i) (x – iy) = 0 Assuming that z = x + iy x2+y2+4x + 3y–3xi + 4yi + 4x + 3y+3xi – 4yi = 0 x2 + y2 + 8x + 6y = 0 centre of circle will be (–4, – 3) and radius units.

23. Ans. () as , , , are the fourth roots of unity , , and are i, –i, 1 and – 1

Using C1 C1 + C2 + C3 + C4 We have all ‘0’ elements in C1 = 0 {as if all elements of a column are ‘0’ the value becomes ‘0’}

27. Ans. () As 1, w, w2 are the cube roots of unity w3 = 1 and 1 + w + w2 = 0 (1 + w – w2)7 + (1 – w + w2)7 = (–w2 – w2)7 + (–w – w)7

= (– 2w2)7 + (–2w)7 = – 128w14 – 128w7 = –128w2 – 128w = – 128 (w2 + w) = 128

28. Ans. () similar to question 9.

29. Ans. ()

= cis 2 + cis (–2) = 1 + 1 = 2

30. Ans. () 1, w, w2, w3, ………. wn-1 are the ‘n’ nth roots of unity. xn – 1 = (x – 1) (x – w) (x – w2) ……. (x – wn-1) Differentiating both sides wrtox:- n.xn-1 = 1.(x – w) (x – w2) ……. (x – wn-1) + (x – 1) . 1 (x – w2) ….. (x – wn-1) + (x – 1) (x – w). 1 ……. (x – wn-1) ……………………………………………………….……………………………………………………….+(x – 1) (x – w) (x – w2) ……. (x – wn-1) Put x = 1 in above equation. n = (1 – w) (1 – w2) (1 – w3) …. (1 – wn-1)

31. Ans. ()


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