CBSE Class 11th Maths, NCERT & R.D Sharma Solutions & Sample Papers

Sample Papers

MCQs

Sets Relations And Functions Trigonometric Functions Principle Of Mathematical Induction Complex Numbers And Quadratic Equations Linear Inequalities Permutations and Combinations Binomial Theorem Sequence and Series Straight Lines Conic Sections Introduction To Three Dimensional Geometry Limits and Derivatives Mathematical Reasoning Statistics […]

Limits and Derivatives, Class 11 Mathematics R.D Sharma Solutions

LIMITS Page 29.11 Ex 29.1 Q1. Answer : limx→0 xx Left hand limit: limx→0- xx Let x=0-h, where h→0.⇒limh→0 0-h0-h=limh→0 -hh=-1 Right hand limit: limx→0+ xxLet x=0+h, where h→0.limh→0 0+h0+h=limh→0 hh=1 Left hand limit ≠ Right hand limit Thus,lim x→0 xx does not exist. Q2. Answer : fx=2x+3,x≤2x+k,x>2Left hand limit:limx→2- fx=limx→2- 2x+3Let x=2-h, where h→0.limh→0 […]

Introduction to Three Dimensional Geometry, Class 11 Mathematics R.D Sharma Solutions

Page 28.6 Ex 28.1 Q1. Answer : (i) The x-coordinate, the y-coordinate and the z-coordinate of the point (5, 2, 3) are all positive. Therefore, this point lies in XOYZ octant. (ii) The x-coordinate, the y-coordinate and the z-coordinate of the point (−5, 4, 3) are negative, positive and positive, respectively. Therefore, this point lies […]

Conic Sections, Class 11 Mathematics R.D Sharma Solutions

CIRCLES Page 24.16 Ex 24.1 Q1. Answer : Let (h, k) be the centre of a circle with radius a. Thus, its equation will be x-h2+y-k2=a2. (i) Here, h = −2, k = 3 and a = 4 ∴ Required equation of the circle: x+22+y-32=42 ⇒x+22+y-32=16 (ii) Here, h = a, k = b and radius […]

Binomial Theorem, Class 11 Mathematics R.D Sharma Solutions

Page 18.11 Ex 18.1 Q1. Answer : (i) (2x + 3y)5 =C05(2x)5(3y)0+C15(2x)4(3y)1+C25(2x)3(3y)2+C35(2x)2(3y)3+C45(2x)1(3y)4+C55(2x)0(3y)5 =32×5+5×16×4×3y+10×8×3×9y2+10×4×2×27y3+5×2x×81y4+243y5=32×5+240x4y+720x3y2+1080x2y3+810xy4+243y5 (ii) (2x − 3y)4 =C04(2x)4(3y)0-C14(2x)3(3y)1+C24(2x)2(3y)2-C34(2x)1(3y)3+C44(2x)0(3y)4=16×4-4×8×3×3y+6×4×2×9y2-4×2x×27y3+81y4=16×4-96x3y+216x2y2-216xy3+81y4 (iii) x-1×6=C06 x61x0-C16 x51x1+C26 x41x2-C36 x31x3+C46 x21x4-6C5 x11x5+C66 x01x6=x6-6 x5×1x+15 x4×1×2-20×3×1×3+15×2×1×4-6 x×1×5+1×6=x6-6×4+15×2-20+15×2-6×4+1×6 (iv) (1 − 3x)7 =C07(3x)0-C17(3x)1+C27(3x)2-C37(3x)3+C47(3x)4-C57(3x)5+C67(3x)6-C77(3x)7=1-7×3x+21×9×2-35×27×3+35×81×4-21×243×5+7×729×6-2187×7=1-21x+189×2-945×3+2835×4-5103×5+5103×6-2187×7 (v) (ax-bx)6=C06(ax)6(bx)0-C16(ax)5(bx)1+C26(ax)4(bx)2-C36(ax)3(bx)3+C46(ax)2(bx)4-C56(ax)1(bx)5+C66(ax)0(bx)6 =a6x6-6a5x5×bx+15a4x4×b2x2-20a3b3×b3x3+15a2x2×b4x4-6ax×b5x5+b6x6=a6x6-6a5x4b+15a4x2b2-20a3b3+15a2b4x2-6ab5x4+b6x6 (vi) xa-ax6=C06xa6ax0-C16xa5ax1+C26xa4ax2-C36xa3ax3+C46xa2ax4-C56xa1ax5+C66xa0ax6=x3a3-6x2a2+15xa-20+15ax-6a2x2+a3x3 (vii) x3-a36=C06(x3)6(a3)0-C16(x3)5(a3)1+C26(x3)4(a3)2-C36(x3)3(a3)3+C46(x3)2(a3)4-C56(x3)1(a3)5+C66(x3)0(a3)6=x2-6×5/3a1/3+15×4/3a2/3-20xa+15×2/3a4/3-6×1/3a5/3+a2 (viii) (1+2x-3×2)5Consider 1-2x and 3×2 as two separate entities and apply the binomial […]

Probability, Class 11 Mathematics R.D Sharma Solutions

Probability Page No.33.6 Ex.33.1 Q1. Answer : If a coin is tossed once, the possible outcomes are a head (H) and tail (T). Therefore, the sample space of this experiment is S = { H, T}. Q2. Answer : If a coin is tossed twice, the possible outcomes are HH, HT, TH, TT. Therefore, the […]

Mathematical Reasoning, Class 11 Mathematics R.D Sharma Solutions

Mathematical Reasoning Page No.31.4 Ex.31.1 Q1. Answer : (i) Listen to me, Ravi! It is an exclamatory sentence. Therefore, it is not a statement. (ii) Every set is a finite set. It is a false assertive sentence because there are some sets that are infinite like the set of all real numbers. Therefore, it is […]

Straight Lines, Class 11 Mathematics R.D Sharma Solutions

Page 23.13 Ex 23.1 Q1. Answer : (i) θ=-π4 ∴ Slope of the line=m=tanθ⇒Slope of the line=tan-π4=-tanπ4=-1 Hence, the slope of the line is -1. (ii) θ=2π3 ∴ Slope of the line=m=tanθ⇒Slope of the line=tan2π3=-tanπ3=-3 Hence, the slope of the line is -3. (iii) θ=3π4 ∴ Slope of the line=m=tanθ⇒Slope of the line=tan3π4=-tanπ4=-1 Hence, the […]

Sequence and Series, Class 11 Mathematics R.D Sharma Solutions

ARITHMETIC PROGRESSION Page 19.4 Ex 19.1 Q1. Answer : Given : an = n2-n+1For n=1, a1 = 12-1+1 = 1For n=2, a2 = 22-2+1 = 3For n=3, a3 = 32-3+1 = 7For n=4, a4 = 42-4+1 =13For n=5, a5 = 52-5+1 = 21 Thus, the first five terms of the sequence are 1, 3, 7, […]

Permutations & Combinations, Class 11 Mathematics R.D Sharma Solutions

PERMUTATIONS Page. 16.4 Ex 16.1 Q1. Answer : (i) 30!28! = 30×29×28!28! ∵ n!= n(n-1)! =30×29 = 870 (ii) 11!-10!9!=11×10×9! – 10×9!9! ∵ n! =n(n-1)! =9!(110-10)9! =100 (iii) LCM of (6!,7! and 8!): n! = n(n-1)! Therefore, (6!,7! and 8!) can be rewritten as: 8! = 8×7×6! 7! = 7×6! 6! = 6! ∴ LCM […]

Linear Inqualities, Class 11 Mathematics R.D Sharma Solutions

Page 15.10 Ex 15.1 Q1. Answer : We have, 12x<50⇒x<5012 Dividing both the sides by 12⇒x<256(i) x∈RThen, the solution of the given inequation is -∞,256.(ii) x∈ZThen, the solution of the given inequation is ……….-3, -2, -1, 0, 1, 2, 3, 4.(iii) x∈NThen, the solution of the given inequation is 1, 2, 3, 4. Q2. Answer […]

Complex Numbers & Quadratic Equations, Class 11 Mathematics R.D Sharma Solutions

COMPLEX NUMBERS Page 13.1 Ex – 13.1 Q1. Answer : i i457 =i4×114+1 =i4114×i =i ∵i4=1ii i528 =i4×132 =i4132 =1 ∵i4=1iii 1i58=1i4×14+2 =1i414×i2 =1i2 ∵i4=1 =-1 ∵i2=-1iv i37+1i67=i4×9+1+1i4×16+3 =i49×i+1i416×i3 =i-1i ∵i3=-i =i-1i×ii =i-ii2 =i–i ∵i2=-1 =2i v i41+1i2579 =i4×10+1+1i4×64+19 =i410×i+1i464×i9 =i+1i9 ∵ i4=1 =i+ii29 =i-i9 ∵i2=-1=0 vi i77+i70+i87+i4143=i4×19+1+i4×17+2+i4×21+3+i4×103+23 =i419×i+i417×i2+i421×i3+i4103×i2 =i-1-i-13 ∵i4=1,i3=-i and i2=-1 =-23 =-8vii i30+i40+i60=i4×7+2+i4×10+i4×15 […]

Mathematical Induction, Class 11 Mathematics R.D Sharma Solutions

Page 12.2 Ex – 12.1 Q1. Answer : We have: P(n): n(n + 1) is even. Now, P(3) = 3(3 + 1) = 12 (Even) Therefore, P(3) is even. Page 12.3 Ex – 12.1 Q2. Answer : We have: P(n): n3+n is divisible by 3.Thus, we have:P(3) =33+3 =27+3=30 It is divisible by 3.Hence, P(3) is […]

Trigonometric Functions, Class 11 Mathematics R.D Sharma Solutions

1. MEASUREMENT OF ANGLES Page 4.12 Ex – 4.1 Q1. Answer : We have:π rad =180°∴1 rad=180π° i 18π5=180π×9π5° =36×9° =324° ii-5π6=180π×-5π6° =-30×5° =-150° iii 18π5c=180π×18π5° =36×18° =648° iv -3c=180π×-3° =18022×7×-3° =-378022° =-1711822° =-171°1822×60′ =-171°49111′ =-171°49’111×60” =-171°49’5.45” ≈-171°49’5” (v) 11c=180π×11° =18022×7×11° =630° vi 1c=180π×1° =18022×7×1° =63011° =57311° =57°311×60′ =57°16411′ =57°16’411×60” =57°16’21.81”≈57°16’22” Q2. Answer : We have:180°=π rad∴1°=π180 […]

Relations & Functions, Class 11 Mathematics R.D Sharma Solutions

Page 2.8 Ex – 2.1 Q1. Answer : (i) a3+1, b-23=53, 13 By the definition of equality of ordered pairs, we have: a3+1, b-23=53, 13 ⇒a3+1 = 53 and b-23=13⇒a3= 53-1 and b = 13+23⇒a3=23 and b = 1⇒a = 2 and b = 1 (ii) (x + 1, 1) = (3, y − 2) By the […]

Sets, Class 11 Mathematics R.D Sharma Solutions

Page 1.2 Ex – 1.1 Q1. Answer : Well-defined collections are sets. Example: The collection of good teachers in a school is not a set, It is a collection. Thus, we can say that every set is a collection, but every collection is not necessarily a set. The collection of vowels in English alphabets is a set. […]

Axiomatic Probability, Class 11 Mathematics NCERT Solutions

1. There will be total 10 MCQ in this test. 2. Please keep a pen and paper ready for rough work but keep your books away. 3. The test will consist of only objective type multiple choice questions requiring students to mouse-click their correct choice of the options against the related question number. 4. No negative […]