objective type problems in higher secondary mathematics for cbse aieee iitjee

Objective-type ProblemsinFundamentalsofHigher Secondary Mathematics

writtenby

S.Ganesan, M.Tech.

PREFACE

Students of Higher Secondary Mathematics course are to learn a plenty of new concepts like Determinants, Vectors, Complex Numbers, Conics, Derivatives and Integrals. It is the primary objective of this work to help them to get a thorough understanding of the fundamentals of these concepts early in their two-year course. The secondary objective is to help the students who are going to face competitive examinations like Board Exams, AIEEE and IITJEE in a couple of months to refresh their understanding of the fundamentals of these concepts.

Fifty problems have been carefully selected from IITJEE Screening Test papers from 1998 to 2005. All problems have been fully solved. I hope that students can learn the fundamentals of the above topics thoroughly by practising these problems a few times.

~ S.Ganesan.

Email: [email protected]

About the Author

The author is a graduate in Mechanical Engineering from College of Engineering, Guindy, Madras and a postgraduate in Mechanical Engineering from Indian Institute of Technology, Madras. He is a professional in utility and industrial boilers industry.

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1. If , and are linearly dependent vectors, and , then

(a) α = 1, β = -1(a) α = 1, β = ±1(a) α = -1, β = ±1(a) α = ±1, β = 1

Answer: (d)Solution:

[a b c] = 0⇒ = 0 ⇒ β = 1⇒ ⇒ 1 + α2 + β2 = 3⇒ α = ±1

2. For three vectors , which of the following expressions is not equal to any of the remaining three?(a) (b) (c) (d) (

Answer: (c)

Solution: =

== -

( =

3. If are unit coplanar vectors, then the scalar triple product is equal to

(a) 0(b) 1(c) -√3(d) √3

Answer: (a)Solution:

If are coplanar, then linear combinations of these vectors are also coplanar. Scalar triple product of coplanar vectors is zero.4. Let , and . Then

depends on(a) only x(b) only y(c) neither x nor y

(d) both x and yAnswer: (c)Solution:

= = [1(1+x-y) - x(1-x)] – [1(x2 – y2)]= 1 + x – y – x + x2 – x2 + y= 1

5. Let and . If is a unit vector, then the maximum value of the scalar triple product is

(a) -1(b) √10 + √6(c) √59(d) √60

Answer: (c)Solution:

= = = = 1

This will be maximum when

∴ Maximum value of = = √59

6. If are two unit vectors such that are perpendicular to each other, then the angle between If is

(a) 45°(b) 60°(c) cos-1 -1(d) cos-1 (2/7)

Answer: (b)Solution:

= 0⇒ ⇒ ⇒ ⇒ ⇒ 1 ∙ 1 ∙ cos θ = ½⇒ θ = cos-1 = 60°½

7. If , and , then is equal to(a)

(b) (c) 2(d)


Let b1 + b2 + b3 = 1

= But ⇒ b3 – b2 = 0 ; b1 – b3 = 1 ; b1 – b2 = 1⇒ b3 = b2 ; b3 = b1 – 1 ; b2 = b1 – 1∴ b1 + b2 + b3 = b1 + (b1 – 1) + (b1 – 1) = 1⇒ b1 = 1 ; b3 = 0 ; b2 = 0⇒

8. A unit vector coplanar with and and orthogonal to5 is

(a) (b) (c) (d)


[α( ) + β( )] ∙ [5 ] = 0⇒ [(2α + β) )] ∙ [5 ] = 0⇒ (2α + β)5 + (α – β)2 + (α + β)6 = 0⇒ β = -2α∴ Required vectors = (2α - 2α)

= Selecting α = 1, required vector = Unit vector along this vector =

=

9. If then f(100) is equal to

(a) 0(b) 1(c) 100(d) -100


C2 →C2 + C1

= 0⇒ f(100) = 0

10. If ω is an imaginary cube root of unity, then (1+ω-ω2)7 equals(a) 128ω(b) -128ω(c) 128ω2

(d) -128ω2Answer: (d)Solution:

(1+ω-ω2)7 = (-ω2-ω2)7 ∵ 1+ω+ω2 = 0= (-2)7 (ω2)7= -128ω14= -128ω2 ∵ ω14 = (ω3)4ω2 and ω3 = 1

11. If = x+iy, then(a) x = 3, y =1(a) x = 1, y =3(a) x = 0, y =3(a) x = 0, y =0


= -3i = 0

12. If i = √-1, then is equal to(a) 1 - i√3(b) -1 + i√3(c) i√3

(d) - i√3Answer: (c)Solution:

Let ω = 4 + 5ω334 + 3ω365 = 4 + 5ω333+1 + 3ω363+2

= 4 + 5ω + 3ω2= 1 + 2ω + (3 + 3ω + 3ω2)= 1 + (-1 + i√3) + 0= i√3

13. If arg(z) < 0, then arg(-z) – arg(z) =(a) π(b) –π(c) -π/2(d) π/2


Let arg(z) = -θarg(-z) = arg(i2z) = arg(i ∙ i ∙ z)

= -θ + π/2 + π/2(∵ each multiplication of z by i rotates z through π/2 anti-clockwise)= -θ + π

arg(z) – arg(-z) = -θ + π – (-θ)= π

14. Let ω = . Then the value of the determinant is

(a) 3ω(b) 3ω(ω-1)(c) 3ω2(d) 3ω(1-ω)


= (-1-ω2)ω4 – ω4 – (ω4-ω2) + (ω2+1+ω2)= -ω6 - 3ω4 + 3ω2 + 1= -1 - 3ω + 3ω2 + 1= 3ω(ω-1)

15. If ω (≠ 1) is a complex cube root of unity, the least value of n ∊ N for which(1 + ω2)n = (1 + ω4)n is(a) 6(b) 5(c) 3(d) 2


(1 + ω2)n = (1 + ω4)n⇒ (-ω)n = (1 + ω)n⇒ (-ω)n = (-ω2)n⇒ ωn = ω2nWhen n = 3, ω3 = ω6.

16. If in the expansion of (1+x)m(1-x)n, the coefficients of x and x2 are 3 and -6 respectively, then m is(a) 6(b) 9(c) 12(d) 24


(1+x)m(1-x)n ==

(m – n) = 6 and

⇒ m2 – m + n2 – n – 2mn = -12⇒ m + n = 12 + (m-n)2⇒ m + n = 21 (∵ m – n = 6)Solving (m –n) = 6 and (m + n) = 21, m = 12

17. For 2 ≤ r ≤ n, (a) (b) (c) (d)


= = =

18. In the binomial expansion of (a-b)n, n ≥ 5, the sum of the 5th and 6th terms is zero. Then a/b equals(a) (n-5)/6(b) (n-4)/5(c) 5/(n-4)(d) 6/(n-5)


= 0⇒ =⇒

= = = (n-4)/5

19. If P(x,y) is any point on the ellipse 16x2 + 25y2 = 400, and F1(3,0) and F2(-3,0) are its foci, then PF1+PF2 equals (a) 8(b) 6(c) 10(d) 12


⇒ ⇒ ( Dividing LHS and RHS by 400 )⇒ But for an ellipse, PF1+PF2 = 2a = 10

20.If x1 , x2 , x3 as well as y1 , y2 , y3 are in G.P. with the same common ratio, then the points (x1,y1), (x2,y2) and (x3,y3)(a) lie on a straight line(b) lie on an ellipse(c) lie on a circle(d) are vertices of a triangle


x2 = x1r & x3 = x1r2y2 = y1r & y3 = y1r2

=

= R2 → R2 – rR1 & R3 → r2R1

= R3 → R3 – rR2

= 0∴ Given 3 points should lie on a straight line.

21. In a ∆ PQR, ∠R = π/2. If tan (P/2) and tan (Q/2) are the roots of equation ax2+bx+c=0 (a≠0), then

(a) a+b = c(b) b+c = a(c) a+c = b(d) b = c


tan (P/2) + tan (Q/2) = -b/atan (P/2) ∙ tan (Q/2) = c/aP+Q = π/2 ⇒ (P+Q)/2 = π/4 ⇒ tan ((P+Q)/2) = 1

⇒ 1 = = ⇒ -b = a-c⇒ c = a+b

22. The number of integer values of m, for which the x-coordinate of the point of intersection of the lines 3x + 4y = 9 and y = mx + 1 is also an integer, is(a) 2(b) 0(c) 4(d) 1


3x + 4 (mx + 1) = 9

⇒ 3x + 4mx + 4 = 9⇒ (3 + 4m)x = 5⇒ x = 5 / (3 + 4m)As x is an integer, (3 + 4m) must be a divisor of 5.∴ 3 + 4m = -1, +1, +5, -5.But positive values for (3 + 4m) will result in fractional values of m.∴ 3 + 4m = -1 or -5⇒ m = -4 / 4 = -1 , or, -8 / 4 = -2

23. The equation of the directrix of the parabola y2 + 4y + 4x + 2 = 0 is(a) x = -1(b) x = 1(c) x = -3/2(d) x = 3/2


y2 + 4y + 4x + 2 = 0⇒ (y2 + 4y + 4) + 4x + 2 = 4⇒ (y + 2)2 = 2 – 4x⇒ (y + 2)2 = -4 (x – )½

⇒ Y2 = -4X, where Y = y + 2, and X = x – ½Axis of the parabola is parallel to X-axis and the parabola opens left.Vertex of the parabola is at (+ , -2) and a = 1.½

Directrix: x = + + 1½

⇒ x = 3/2

24. For the hyperbola which of the following remains constant with change in α

(a) abscissas of vertices(b) abscissas of foci(c) eccentricity(d) directrix


a = cos α and b = sin α with vertex at (0,0).c2 = a2 + b2 = cos2 α + sin2 α = 1⇒ c = ae = constant.∴ Abscissas of foci are independent of α.

25. The number of distinct real roots of = 0 in -π/4 ≤ x ≤ π/4 is

(a) 0(b) 2(c) 1(d) 3

Answer: (c)

Solution: = 0

⇒ = 0 (C1 → C1+C2+C3)

⇒ = 0⇒ (sin x + 2cos x)[(sin2 x – cos2 x)–(sin x cos x – cos2 x)+(cos2 x – sin x cos x)] =0⇒ (sin x + 2cos x) [sin2 x + cos2 x – sin x cos x] = 0⇒ (sin x + 2cos x) (sin x – cos x)2 = 0Either sin x + 2cos x = 0 or, sinx – cos x = 0Case1: sin x + 2cos x = 0

⇒ sin x = -2cos x⇒ tan x = -2⇒ x = tan-1 (-2)Rejected as x < -π/4.

Case2: sin x - cos x = 0⇒ sin x = cos x⇒ tan x = 1⇒ x = tan-1 1Accepted as x = π/4.

26. If α+β = π/2 and β+γ = α, then tan α equals(a) 2(tan β + tan γ)(b) tan β + tan γ(c) tan β + 2tan γ(d) 2tan β + tan γ


β+γ = α ⇒ γ = α – βtan γ = tan (α – β) =

= = =

⇒ tan α = tan β + 2tan γ27. If the angles of a ∆ are in the ratio 4:1:1, then the ratio of the largest side to the perimeter is

(a) √3 : (2+√3)(b) 1 : √3(c) 1 : (2+√3)(d) 2 : 3


Let A:B:C :: 4:1:1⇒ 4x + x + x = 180 ⇒ x = 30⇒ A = 120° , B = 30° , C = 30°Largest side is that side opposite to the largest angle : aUsing the Sine Rule,

⇒ a = √3 b and c = bs = a + b + c = √3 b + b + b = (2+√3)b⇒

28. If a, b, c are the sides of a ∆ such that a : b : c = 1 : √3 : 2, then the ratio A:B:C is equal to(a) 3:2:1(b) 3:1:2(c) 1:2:3(d) 1:3:2


Let a, b, c be x, √3x, 2xUsing the Cosine Rule,

cos A = ⇒ A = cos-1 (√3/2) = 30°Again using the Cosine Rule,cos B = ⇒ B = cos-1 (1/2) = 60°C = 180 – (30+60) = 90⇒ A:B:C = 30:60:90 = 1:2:3

29. Let h(x) = f(x) [f(x)]2 + [f(x)]3 for every real number x. Then,(a) h is increasing whenever f is increasing(b) h is increasing whenever f is decreasing(c) h is constant whenever f is decreasing(d) nothing can be said in general


h(x) = f(x) [f(x)]2 + [f(x)]3⇒ h’(x) = f(x) 2 f(x) f’(x) + [f(x)]2 f’(x) + 3 [f(x)]2 f’(x)

= 6 f’(x) [f(x)]2 ⇒ h’(x) > 0 whenever f’(x) > 0∴ h(x) increases whenever f(x) increases.

30. Let h(x) = min {x, x2}, for every real number x. Then(a) h is not continuous at two values of x(b) h is differentiable for all x

(c) h’(x) = 1, for 0 < x < 1(d) h is not differentiable at two values of x.


By inspection, we can conclude that h is continuous for all values of x h’(x) = 2x for 0 < x < 1 h’(x) is not differentiable at x = 0 and x =1 due to different slopes on either sides of these points.

31. If , then the value of f(1) is(a) 1/2 (b) 0(c) 1(d) -1/2 Answer: (a)Solution:Differentiating both sides wrt x,f(x) = 1 – x f(x)⇒ (1+x) f(x) = 1⇒ f(x) = 1 / (1+x)⇒ f(1) = 1/2

32. If f(x) = 3x – 5, then f-1(x)(a) is given by 1 / (3x – 5)(b) is given by (x + 5) / 3(c) does not exist because f is not one-one(d) does not exist because f is not onto.Answer: (b)Solution:Let y = 3x – 5x = (y + 5) / 3

∴ f-1(x) = (x + 5) / 333. Let f(x) = x – [x], for every real number x, where [x] is the integral part of x. Then, is

(a) 1(b) 2(c) 0(d) ½Answer: (a)Solution:

= 0 as x is an odd function.

34. The order of the differential equation whose general solution is given by , where C1 , C2 , C3 , C4, C5 are arbitrary constants, is(a) 5(b) 4(c) 3(d) 2Answer: (c)Solution:Let k1 = C1 + C2Let k2 = C3

Let k3 =

Now we have, y = k1cos (x + k2) – k3exThere are only three constants now. ∴ 3rd order.35. A solution of the differential equation y” – xy’ + y = 0 is (a) y = 2(b) y = 2x(c) y = 2x – 4(d) y = 2x2 – 4Answer: (c)Solution:Since the order of the DE is one, either y = 2x or y = 2x – 4 will qualify.Let us take up y = 2x – 4 for check-up. Substituting it in the DE,4 – x ∙ 2 + 2x – 4 = 036. is

(a) 2(b) -2(c) ½(d) -½Answer: (c)Solution:

= ( ∵ cos 2x = 1 – 2sin2 x)= = = =

= = ½

37. The function f(x) = sin4 x + cos4 x increases if(a) 0 < x < π/8(b) π/4 < x < 3π/8(c) 3π/8 < x < 5π/8(d) 5π/8 < x < 3π/4Answer: (b)Solution:f’(x) = 4sin3 x cos x - 4cos3 x sin x= 4sin x cos x (sin2 x – cos2 x)[∵ 2sin x cos x = sin 2x & cos2 x - sin2 x = cos 2x]= 2 ∙ sin 2x ∙ (-1) ∙ cos 2x= -sin 4xFor f’(x) > 0, sin 4x < 0sin θ < 0 in III and IV quadrants.Taking III quadrant, π < 4x < 3π/2⇒ π/4 < x < 3π/8Taking IV quadrant, 3π/2 < 4x < 2π⇒ 3π/8 < x < π/2

38. Let f: R→R be any function. Define g: R→R by g(x) = |f(x)| for all x. Then g is (a) onto if f is onto(b) one-one if f is one-one(c) continuous if f is continuous(d) differentiable if f is differentiableAnswer: (c)Solution:Let h(x) = |x|⇒ h[f(x)] = |f(x)| = g(x)A composite function of two continuous functions is also continuous. We know that h(x) is continuous. Therefore if f(x) is continuous, g(x) is also continuous.39. If x2 + y2 = 1, then

(a) yy” – 2(y’)2 + 1 = 0(b) yy” + (y’)2 + 1 = 0(c) yy” + (y’)2 - 1 = 0(d) yy” + 2(y’)2 + 1 = 0Answer: (b)Solution:x2 + y2 = 1Differentiating implicitly with respect to x,2x + 2yy’ = 0⇒ x + yy’ = 0Again differentiating,1 + yy” + y’y’ = 0⇒ yy” + (y’)2 + 1 = 040. If f(x) = , then =

(a) 0(b) 1(c) 2(d) 3Answer: (c)Solution:

f(x) = f(x) = ecos x sin x , -2 ≤ x ≤ 2f(-x) = ecos (-x) sin (-x)= -ecos x sin x= - f(x)∴ f(x) is odd for -2 ≤ x ≤ 2= 0 + = 2 ∙ = 2 ∙ 1 = 2

41. Let . Then f decreases in the interval

(a) (-∞, -2)(b) (-2, -1)(c) (1, 2)(d) (2, ∞)Answer: (c)Solution:f’(x) = ex (x-1) (x-2)ex > 0 for all x.(x-1) (x-2) < 0 if 1 < x < 2⇒ f’(x) < 0 if 1 < x < 242. If the normal to the curve y = f(x) at the point (3,4) makes an angle 3π/4 with the positive x-axis, then f’(3) =(a) -1(b) -3/4(c) 4/3(d) 1Answer: (d)Solution:Angle of normal = 3π/4⇒ Angle of tangent = 3π/4 – π/2 = π/4⇒ f’(3) = tan π/4 = 143. The value of the integral is

(a) 3/2(b) 5/2(c) 3(d) 5Answer: (b)Solution:Let us denote as ln x for convenience and consistent with engineering practice.Please note that ln x is defined only for x > 0, and ln x < 0, for 0 < x < 1.

We can rewrite as

We know that by putting ln x = u ⇒ dx/x = du

= = - [-(-1)½ 2] + [2½ 2]= 5/2

44. The domain of definition of is(a) R \ {-1, -2}(b) (-2, ∞)(c) R \ {-1, -2, -3}(d) (-3, ∞) \ {-1, -2}

Answer: (d)Solution:x + 3 > 0 ⇒ x > -3x2+3x+2 ≠ 0⇒ (x+1) (x+2) ≠ 0⇒ x ≠ -1 and x ≠ -2∴ Domain of f(x) is (-3, ∞) \ {-1, -2}45. Let g(x) = 1 + x – [x] and then, for all values of x, f[g(x)] is equal to(a) x(b) 1(c) f(x)(d) g(x)

Answer: (b)Solution:For all x, [x] ≤ x⇒ x – [x] ≥ 0⇒ 1 + x – [x] ≥ 1⇒ g(x) ≥ 1∴ g(x) > 0 for all values of xBut for all values of x > 0, f(x) = 1∴ f[g(x)] = 1 for all x.46. Let f: (0, ∞) → R and F(x) = . If F(x2) = x2(1 + x), then f(4) equals(a) 5/4(b) 7(c) 4(d) 2Answer: (c)Solution:

F(x2) = ∴ F(x2) = x2 + x3Differentiating both sides with respect to x,F’(x2) ∙ 2x = 2x + 3x2⇒ f(x2) = (2 + 3x) / 2 [∵ F’(x) = f(x)]⇒ f(4) = f(22) = (2 + 3 ∙ 2) / 2 = 4

47. The length of the longest interval in which the function 3sin x – 4sin3 x is increasing, is(a) π/3(b) π/2(c) 3π/2(d) πAnswer: (a)Solution:3sin x – 4sin3 x = sin 3xFor θ > 0, sin θ increases in the interval 3π/2 ≤ θ ≤ 2π + π/2⇒ 3π/2 ≤ 3x ≤ 2π + π/2⇒ 3π/2 ≤ 3x ≤ 5π/2

⇒ π/2 ≤ x ≤ 5π/6∴ Length of the interval= (5π/6) – (π/2) = π/348. Let the function f: R→R be defined by f(x) = 2x + sin x for all x ∈ R. Then f is(a) one-to-one and onto(b) one-to-one but not onto(c) onto but not one-to-one(d) neither one-to-one nor ontoAnswer: (a)Solution:f(x) = 2x + sin x⇒ f’(x) = 2 + cos x⇒ f’(x) > 0 for all x ∈ R ∵ cos x ≥ -1 for all x ∈ R⇒ f(x) is a strictly increasing function for all x ∈ R⇒ f(x) is one-to-oneSince f(x) takes all values of (-∞, ∞), it is onto as well.49. If f: [0, ∞) → [0, ∞) and f(x) = x / (1 + x) then the function f is(a) one-to-one and onto(b) one-to-one but not onto(c) onto but not one-to-one(d) neither one-to-one nor ontoAnswer: (b)Solution:As x ≥ 0, x / (x + 1) ≤ 1 ⇒ f(x) is not onto.

⇒ Since f(x) is strictly increasing for all x, f(x) is one-to-one.50. Domain of the definition of function is

(a) [-1/4, 1/2](b) [-1/2, 1/9](c) [-1/2, 1/2](d) [-1/4, 1/4]Answer: (a)Solution:Let θ = sin-1 (2x).

From the definition of inverse sine function, θ is defined only in the interval,-π/2 ≤ θ ≤ π/2But for f(x) to be in R, θ + π/6 ≥ 0 ⇒ θ ≥ -π/6. This requirement redefines the interval of θ such that-π/6 ≤ θ ≤ π/2⇒ -π/6 ≤ sin-1 (2x) ≤ π/2Taking sine throughout the inequality, we have-1/2 ≤ 2x ≤ 1⇒ -1/4 ≤ x ≤ 1/2

objective type problems in higher secondary mathematics for cbse aieee iitjee

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c solution

b c d answer

b solution

d solution

b1 b1

y answer

b1 b2 b3

unit vectors