Cbse Sample Papers For Class 11 Maths Mock Paper 1
CBSE Class 11 Math Sample Papers Mock Paper1 (20212022)
Entering the Senior Secondary level education is a huge step for students. The syllabus to be covered in a year seems tremendous. Thus with a huge bulk to cover, they find very little time for practice and revision. To make it easier, Extramarks comes up with the best CBSE Sample Papers For Class 11 Maths Mock Paper 1. These CBSE Sample Papers For Class 11 Maths Mock Paper 1 are formulated as per the CBSE guidelines keeping the latest exam patterns in mind. Students can download these CBSE Sample Papers For Class 11 Maths Mock Paper 1 in a PDF file and solve it at their convenience.
CBSE Sample Paper for Class 11 Math with Solutions – Mock Paper1
The maximum utilisation from any Mock paper can be derived only if after attempting the same can be assessed as well. This is possible with Extramarks. It provides CBSE Sample Papers For Class 11 Maths Mock Paper 1 with solutions for the betterment of students. With this, it becomes possible for students to evaluate their answer sheets and assign them marks as well. This boosts their morale and they become confident by attempting CBSE Sample Papers For Class 11 Maths Mock Paper 1 and improving every time. Thus they can succeed in the final exams with flying colours. Apart from solutions, another advantage of these CBSE Sample Papers For Class 11 Maths Mock Paper 1 is their availability in the form of a PDF, too, with a download option. By downloading, students can avail of this pdf in offline mode as well. This is a boon for students living in remote areas where there is a connectivity issue. By downloading these CBSE Sample Papers For Class 11 Maths Mock Paper 1, they don’t need to be interrupted in between their revision due to any technical glitches.
How Can Solving Math Sample Paper Class 11 Help?
Practice makes a man perfect. Mathematics is a subject where skills need to be whetted with practice. It also helps students to learn basic answering skills to score high in the final exams. This is where Extramarks proves to be very useful. The CBSE Sample Papers For Class 11 Maths Mock Paper 1 provided by Extramarks are very useful in revision. These CBSE Sample Papers For Class 11 Maths Mock Paper 1 are perfectly formatted to add advantages to the revision schedule of students. Below is given what can be gained from solving these CBSE Sample Papers For Class 11 Maths Mock Paper 1:
 Standard CBSE BoardLevel Questions
After completing the syllabus, there is a dire need for a platform to judge the capabilities and knowledge students have learnt so far. These CBSE Sample Papers For Class 11 Maths Mock Paper 1 serve this function and students assess their preparation levels. These CBSE Sample Papers For Class 11 Maths Mock Paper 1 are prepared on the blueprint of question papers followed by the Central Board of Secondary Education authorities. It means solving CBSE Sample Papers For Class 11 Maths Mock Paper 1 makes students accustomed to the standard questions as set by CBSE. Thus students will be motivated to prepare a strategy to solve questions in the final exam.
 Identify weak areas
When we revise, we feel ready to attempt the exam. But sometimes we are not able to filter which topics we are thorough with and for which topics we are not well acquainted. But when we attempt any Mock exam, it not only helps us judge our flaws and mistakes, but also we can improve those very topics before the exam. For this students can obtain CBSE Sample Papers For Class 11 Maths Mock Paper 1 from Extramarks. It will no doubt give them an excellent platform to practice. They can also ameliorate their problemsolving skills by identifying their weaknesses. These CBSE Sample Papers For Class 11 Maths Mock Paper 1 can be used as a brilliant tool to find out the level of preparation and hence can be made better.
 Time Efficiency
Efficiency and time management are a must to solve an elaborative question paper. This requires a lot of practice by attempting mocks. CBSE Sample Papers For Class 11 Maths Mock Paper 1 are designed on the pattern of actual exams. Thus by solving these CBSE Sample Papers For Class 11 Maths Mock Paper 1, students can easily learn how to manage time and can become more efficient. By sitting in their comfort zone, they can develop a flow and efficiency to attempt the final exam by practising through CBSE Sample Papers For Class 11 Maths Mock Paper 1.
Thus students should visit the Extramarks website to access and download these CBSE Sample Papers For Class 11 Maths Mock Paper 1 to pace up their preparation.
How to Solve Mock Papers for Class 11 Mathematics?
Solving CBSE Sample Papers For Class 11 Maths Mock Paper 1 before the examination is very useful to students. The advantages of these CBSE Sample Papers For Class 11 Maths Mock Paper 1 are that students can familiarise themselves with the pattern of the question paper; they learn to manage time to finish the exam in time; and also they can recall formulas from different chapters and topics as questions from various chapters are integrated into these CBSE Sample Papers For Class 11 Maths Mock Paper 1. While solving CBSE Sample Papers For Class 11 Maths Mock Paper 1 students should follow these steps :
Step 1: To Download Mock Papers
The first step towards the process is to download a couple of sample papers available under CBSE Sample Papers For Class 11 Maths Mock Paper 1 on the website of Extramarks. Extramarks provides these CBSE Sample Papers For Class 11 Maths Mock Paper 1 for downloading. This makes it more accessible for students as they will be free from the issues of buffering and low internet connectivity. Not only this, they can take the printouts of these CBSE Sample Papers For Class 11 Maths Mock Paper 1.
Step 2: Going Through the Papers
The second and very expected step is to go through the entire CBSE Sample Papers For Class 11 Maths Mock Paper 1. This gives a fair idea regarding what sort of questions are being asked and students can judge a pattern being followed in the questions.
Step 3: Solve Maximum Questions
Once done with reading, the next and very obvious step is to attempt as many questions as possible. Continue the process for all the downloaded mock exams and solve the maximum number of questions. Mark the questions which are not familiar.
Step 4: Revisit the Wrong Answers
Evaluation and Amendment are the next steps. Make a fair assessment and pick up the questions which were answered wrongly. Analyse the solutions of the same through Extramarks CBSE Sample Papers For Class 11 Maths Mock Paper 1 under which the solutions to the questions are given too.
The topics which were unfamiliar in step 3 should be revisited and relearnt in more detail.
Step 5: Repeat the Process
Practice is the key to success. Thus, the last and ultimate step is to repeat the process until students master every concept and topic. Do not panic if the mistake Count is high because the higher the number of mistakes during practice, the fewer will be in the final exams. Following these steps, one can easily score good grades in Mathematics exams.
Class 11 Math Syllabus – Overview
The Curriculum of Mathematics for Class 11 covers a variety of topics. These topics can be studied from the NCERT books for a clear understanding of concepts. In totality, there are 5 units in the syllabus of Mathematics, each unit comprising three or four subtopics. All these topics are included in CBSE Sample Papers For Class 11 Maths Mock Paper 1 available on the Extramarks website under the CBSE Sample Papers column.
Chapter 1 deals with Sets and their properties. Chapter second is about Relations and Functions is an important chapter as it is repeated in Class 12 as well. Thus its importance increases as having a firm grip over the chapter in Class 11 itself can save time in Class 12 board preparation.
Chapter 3 is the one which the majority of students are scared of, I.e. Trigonometric Functions. However, this daunting chapter, if covered in a detailed method in this Class, can be of immense help in Class 12 as students will be wellversed with the formulas beforehand.
Chapter 4 is the easiest and scoring chapter. It is the Principle of Mathematical Induction. With a proper understanding, students can never get the answers from this chapter wrong.
Chapter 5 deals with Complex Numbers and Quadratic Equations. Linear Inequalities are elaborated on in Chapter 6. This is very easy to understand and hence can fetch marks.
If seen from a competition point of view, Chapter 7 maintains its significance. Permutations and Combinations are a part of almost every important competitive exam under the Quants section.
Binomial Theorem is the keynote of Chapter 8. Again Chapter 9, dealing with Sequence and Series, is vital for school exams and competitive exams.
3 chapters are dedicated to Geometry.
Limits and Derivatives in Chapter 13 is a newer concept for students. And the curriculum ends with the topics of Mathematical Reasoning and Probability.
For comprehensive knowledge, the details of these chapters and their marks distribution can be accessed from the Extramarks CBSE syllabus.
Q1. The equation of the line which makes equal intercepts on the axes and passes through the point (4, 5) is
Opt: x + y = 3. x + y = 5. x + y = 7. x + y = 9.
Ans: x + y = 9.
Q2. Different words have been formed with all the letters of the word EQUATION (use each letter exactly once) so that the words begin and end with a consonant. The number of such words, is
Opt:
40320.
4320.
1440.
720.
Ans: 4320.
Q3. If ‘P(n) = n^{2} – n + 41’ is prime, then it is not true for
Opt:
P(41).
P(40).
P(39).
P(38).
Ans: P(41).
Q4.
Opt: a purely imaginary number. a purely real number.
Ans: a purely real number.
Q5.
Opt:
Ans:
Q6. The foot of the perpendicular from the origin to a straight line lies at the point (7, 5). Then the equation of the line is
Opt:
7x – 5y + 74 = 0. 7x + 5y + 74 = 0. 7x + 5y – 74 = 0. 7x – 5y – 74 = 0.
Ans: 7x – 5y – 74 = 0.
Q7. The ratio in which xyplane divides the join of (1, 2, 3) and (4, 5, 6) is
Opt:
3:1 internally. 3:1 externally. 2:1 internally. 1:2 externally.
Ans: 1:2 externally.
Q8.
Opt:
1. 3. 6. 9.
Ans: 6.
Q9. The number of real solutions of the equation 2sin^{2} x + 7 sin x + 3 = 0 in the interval [0, 4π] is
Opt:
0.
1.
2.
3.
Ans: 2.
Q10. The set (A ∪ B ∪ C) ∩ (A ∩ B^{C} ∩ C^{C})^{C} ∩ C^{C }is equal to
Opt:
A ∩ C^{C.}
A ∩ C.
B ∩ C^{C.}
B ∩ C.
Ans: B ∩ C^{C.}
Q11. The expression
$\sqrt{\frac{1\u2013\mathrm{sin\alpha}}{1+\mathrm{sin\alpha}}}+\sqrt{\frac{1+\mathrm{sin\alpha}}{1\u2013\mathrm{sin\alpha}}}$is equal to
Opt:
1/cos α.
cos α.
2/cos α.
–1/cos α.
Ans: 2/cos α.
Q12. The probability that a student will receive A, B, C or D grade is 0.40, 0.35, 0.15 and 0.10 respectively. The probability that a student will receive C or D grade is
Opt:
0.10.
0.25.
0.40.
0.50.
Ans: 0.25.
Q13.
Opt:
2.
0.
1/2.
∞.
Ans: 2.
Q14.
$\begin{array}{}\text{The\hspace{0.33em}domain\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}real\hspace{0.33em}function\hspace{0.33em}f\hspace{0.33em}defined\hspace{0.33em}by}\\ \text{f(x)}=\frac{{{x}}^{2}+2{x}+1}{{{x}}^{3}+6{{x}}^{2}+12{x}+8}\text{\hspace{0.33em}is}\end{array}$Opt:
{2}.
( ∞. ∞).
R{2}.
R{2}.
Ans: R{2}.
Q15. A set of lines ax + by + c = 0, where 5a + 7b + 4c = 0 is
concurrent at the point
Opt:
(5/4, 3/4). (5/4, 1/2). (5/4, 7/4). (3/4, 1/2).
Ans: (5/4, 7/4).
Q16.
Opt:
^{18}C_{6}.2^{6}.
^{18}C_{6}.2^{5}.
^{18}C_{6}.2^{4}.
^{18}C_{6}.2^{2}.
Ans: ^{18}C_{6}.2^{6}.
Q17.
Opt:
isosceles rightangled triangle.
equilateral triangle.
acuteangled scalene triangle.
obtuseangled triangle.
Ans: equilateral triangle.
Q18. If the arithmetic mean of a distribution is 20 and the standard deviation is 7, then the coefficient of variation is
Opt:
35.
30.
27.
14.
Ans: 35.
Q19. ASSERTION – REASON BASED QUESTIONS
In the following questions, a statement of assertion (A) is followed by a statement of Reason (R). Choose the correct answer out of the following choices.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Assertion (A): If (12, x), (13, y), (14, z) are in A × B and n(A) = 3 and n(B) = 3, then A = {12, 13, 14} and B = {x, y, z}.
Reason (R): If n(A) = p and n(B) = q then n(A × B) = pq.
Ans: (b) Both A and R are true, but R is not the correct explanation of A.
Explanation:
We know that if n(A) = p and n(B) = q then n(A × B) = pq.
If (12, x), (13, y), (14, z) are in A × B, this means the first element of each ordered pair belongs to A and the second element belongs to B.
Thus, A = {12, 13, 14} and B = {x, y, z}.
Hence, n(A) = 3 and n(B) = 3.
Q20. If 2x + i (x – y) = 5, where x and y are real numbers, find the values of x and y.
Ans:
We have 2x+ i(x – y) = 5
or 2x+ i(x – y) = 5 + 0.i
Comparing the real and imaginary parts, we get
2x = 5 and x – y = 0
⇒ x = 5/2 and x = y
Thus x = y = 5/2.
Q21. Write the negation of each of the following statements:
(i) For every natural number x, x + 0 = x = 0 + x.
(ii) There exists a capital for every state in India.
Ans:
(i) There exists a natural number x such that x + 0 ≠ x = 0 + x.
(ii) There exists a state in India which does not have its capital.
Q22. Find the equation of the parabola with focus (2, 0) and directrix x = –2.
Ans: Since the focus lies on the xaxis, thus xaxis it self is the axis of the parabola.
Since the directrix is x = –2 and the focus is (2, 0), the parabola is to be of the form
y^{2} = 4ax with a = 2.
Hence the required equation is : y^{2} = 4(2)x = 8x.
Q23.
${\text{Find the integral value of}}{x}{:}{\text{\hspace{0.33em}}}{7}{x}{}{5}{\le}{3}{x}{+}{7}$Ans:
$\begin{array}{}\text{We\hspace{0.33em}have,}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}7{x}5\le 3{x}+7\\ \Rightarrow \text{\hspace{0.33em}}7{x}3{x}\le 7+5\\ \Rightarrow \text{\hspace{0.33em}}4{x}\le 12\\ \Rightarrow \text{\hspace{0.33em}}{x}\le 3\\ \Rightarrow \text{\hspace{0.33em}}{x}=\infty \dots 3,2,1,0,1,2,3\end{array}$Q24. Find the equation of the straight line which makes an angle of 60° with the positive direction of xaxis and cuts an intercept of 6 units on the y – axis.
Ans: We have, m = tan 60° = (1/√3) and c = + 6.
The equation of the line is given by : y = mx + c Putting values , y = (1/√3)x + 6
(√3)y = x + 6√3
x – (√3)y + 6√3 = 0
Q25. Find the term independent of x in the expansion of
${\left({x}+\frac{1}{{x}}\right)}^{6}.$Ans:
$\begin{array}{}\text{Suppose}({r}+1)\text{thtermisindependentof}{x}\text{inthe\hspace{0.33em}expansionof}{\left({x}+\frac{1}{{x}}\right)}^{6}\text{.}\\ {{T}}_{{r}+1}={\text{\hspace{0.33em}}}^{6}{{C}}_{{r}}\text{\hspace{0.33em}}{{x}}^{6{r}}{\left(\frac{1}{{x}}\right)}^{{r}}\\ {{T}}_{{r}+1}={\text{\hspace{0.33em}}}^{6}{{C}}_{{r}}\text{\hspace{0.33em}}{{x}}^{6{r}{r}}\\ \text{Tobeindependentof}{x},\text{wehave}\\ 62{r}=0\Rightarrow {r}=3\\ \therefore \text{\hspace{0.33em}}{{T}}_{3+1}{=}^{6}{{C}}_{3}\text{\hspace{0.33em}}{{x}}^{63}{\left(\frac{1}{{x}}\right)}^{3}=20\\ \text{Hence,}{4}^{\text{th}}\text{\hspace{0.33em}termisindependentof}{{x}}_{1}\text{i.e.,}20.\end{array}$Q26.
${\text{Show that}}{r}{\text{does not have integral value for}}{6}{{\text{\hspace{0.33em}}}}^{{5}}{{P}}_{{r}}{=}{7}{{\text{\hspace{0.33em}}}}^{{6}}{{P}}_{{r}{}{1}}{.}$Ans:
$\begin{array}{}\text{Wehave}6{\text{\hspace{0.33em}}}^{5}{{P}}_{{r}}=7{\text{\hspace{0.33em}}}^{6}{{P}}_{{r}1}\\ \text{or\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}6\times \frac{5!}{(5{r})!}=7\times \frac{6!}{(6{r}+1)!}\\ \mathrm{or}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\frac{6!}{(5{r})!}=\frac{7\times 6!}{(6{r}+1)(6{r})(6{r}1)!}\\ \text{or}(7{r})(6{r})=7\\ \text{or\hspace{0.33em}}{{r}}^{2}13{r}+35=0,\text{whichdoesnotgiveintegralvalueof}{r}\text{.}\end{array}$Q27. A coin is tossed and a die is thrown. Find the probability that the outcome will be a head and a number greater than 4.
Ans:
Probability (head) = 1/2
Probability (number greater than 4) = 2/6 = 1/3 ( since there are two numbers greater than 4 in a die i.e., 5 and 6)
Required probability = 1/2 × 1/3 = 1/6.
Q28. A, B and C are mutually exclusive and exhaustive events associated with a random experiment.
Find P(A) if P(B) = (5/2)P(A) and P(C) = (3/2)P(A).
Ans:
Let P(A) = p
P(B) = (5/2) P(A)
= (5/2)P
and
P(C) = (3/2)P(A)
= (3/2)P
Since A, B and C are mutually exclusive and exhaustive events associated with a random experiment, thus A ∪ B ∪ C = S
P( A ∪ B ∪ C) = P(S) = 1
P(A) + P(B) + P(C) = 1
p + (5/2)p + (3/2)p = 1
(10/2)p =1
p = 1/5
P(A) = p
= 1/5.
Q29. If three angles A, B, C, are in A.P. Prove that:
$\mathrm{cotB}=\frac{\mathrm{sinA}\u2013\mathrm{sinC}}{\mathrm{cosC}\u2013\mathrm{cosA}}$
Ans:
${R}.{H}.{S}=\frac{\mathrm{sinA}\u2013\mathrm{sinC}}{\mathrm{cosC}\u2013\mathrm{cosA}}\phantom{\rule{0ex}{0ex}}=\frac{2\mathrm{sin}\frac{{A}\u2013{C}}{2}\mathrm{cos}\frac{{A}+{C}}{2}}{2\mathrm{sin}\frac{{A}+{C}}{2}\mathrm{sin}\frac{{A}\u2013{C}}{2}}\phantom{\rule{0ex}{0ex}}=\mathrm{cot}\left(\frac{{A}+{C}}{2}\right)=\mathrm{cotB}={L}.{H}.{S}\phantom{\rule{0ex}{0ex}}\begin{array}{c}\because {A},{B},{C}\text{are in A.P}\\ \therefore 2{B}={A}+{C}\end{array}$Q30. Prove that :
$\mathrm{tan}3{A}\xb7\mathrm{tan}2{A}\xb7\mathrm{tan}{A}=\mathrm{tan}3{A}\u2013\mathrm{tan}2{A}\u2013\mathrm{tan}{A}$Ans:
$\mathrm{We}\mathrm{have},\phantom{\rule{0ex}{0ex}}3{A}=2{A}+{A}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}3{A}=\mathrm{tan}\left(2{A}+{A}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}3{A}=\frac{\mathrm{tan}2{A}+\mathrm{tan}{A}}{1\u2013\mathrm{tan}2{A}\xb7\mathrm{tan}{A}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}3{A}\left(1\u2013\mathrm{tan}2{A}\xb7\mathrm{tan}{A}\right)=\mathrm{tan}2{A}+\mathrm{tan}{A}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}3{A}\u2013\mathrm{tan}3{A}\xb7\mathrm{tan}2{A}\xb7\mathrm{tan}{A}=\mathrm{tan}2{A}+\mathrm{tan}{A}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}3{A}\u2013\mathrm{tan}2{A}\u2013\mathrm{tan}{A}=\mathrm{tan}3{A}\xb7\mathrm{tan}2{A}\xb7\mathrm{tan}{A}$Q31. Find the value of p so that the three lines 3x + y – 2 = 0, px + 2y – 3 = 0 and 2x – y – 3 = 0 may intersect at one point.
Ans: The intersection point of two lines 2x – y – 3 = 0 and 3x + y – 2 = 0
can be obtained by solving these equations :
3x + y – 2 = 0
2x – y – 3 = 0
———————–
5x = 5
x = 1 and y = 2 – 3x = 2 – 3 = –1.
Since the point (1, –1) lies on the line px + 2y – 3 = 0,
it will satisfy the equation of the line.
p(1) + 2(–1) – 3 = 0
p – 2 – 3 = 0
p = 5
Q32. Find the equation of a circle which is concentric to the given circle x^{2} + y^{2} – 4x – 6y – 3 = 0 and which touches the x axis.
Ans: The centre of the given circle is given by
(–g, –f) = (–1/2 coff. of x, –1/2 coff. of y) = (2,3)
The centre of the given circle is (2,3). The centre of concentric circles are same , therefore the centre of the required circle is (2,3).
Since the circle touches the xaxis therefore the radius of the required circle is 3 units.
The equation of the required circle is
(x – 2)^{2} + (y – 3)^{2} = 3^{2}
⇒ x^{2} + y^{2} – 4x – 6y + 4 = 0
Q33.
${\text{Solve}}\frac{{(}{2}{x}{}{1}{)}}{{3}}{\ge}\frac{{(}{3}{x}{}{2}{)}}{{4}}{}\frac{{(}{2}{}{x}{)}}{{5}}{.}$Ans:
$\begin{array}{}\text{Wehave,\hspace{0.33em}}\frac{(2{x}1)}{3}\ge \frac{(3{x}2)}{4}\frac{(2{x})}{5}\text{\hspace{0.33em}or}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\frac{(2{x}1)}{3}\ge \frac{\left\{5\right(2{x}2)4(2{x}\left)\right\}}{20}\\ \Rightarrow \text{\hspace{0.33em}}20(2{x}1)\ge 3(19{x}18)\\ \Rightarrow \text{\hspace{0.33em}}34\ge 17{x}\\ \text{Therefore,}2\ge {x}\text{or}{x}\le 2\\ \text{Hence,thesolutionsetis}(2,\infty )\text{.}\end{array}$Q34. Prove that: 1 + 2 + 3 +…+ n = n(n + 1)/2 for all n ∈ N.
Ans:
$\text{Let}{P}\left({n}\right)=1+2+3+\dots \dots +{n}=\frac{{n}\left({n}+1\right)}{2}\phantom{\rule{0ex}{0ex}}{P}\left(1\right)=1=\frac{1\left(1+1\right)}{2}\Rightarrow 1=1\Rightarrow {P}\left(1\right)\text{is True.}\phantom{\rule{0ex}{0ex}}\text{Assuming}{P}\left({k}\right)\text{is True.}\phantom{\rule{0ex}{0ex}}{P}\left({k}\right)=1+2+3+\dots \dots +{k}=\frac{{k}\left({k}+1\right)}{2}\phantom{\rule{0ex}{0ex}}\text{Now}{P}\left({k}+1\right)=1+2+3+\dots \dots +{k}+\left({k}+1\right)=\frac{\left({k}+1\right)\left({k}+2\right)}{2}\phantom{\rule{0ex}{0ex}}=\frac{{k}\left({k}+1\right)}{2}+\left({k}+1\right)\phantom{\rule{0ex}{0ex}}=\frac{{{k}}^{2}+{k}+2{k}+2}{2}\phantom{\rule{0ex}{0ex}}=\frac{{k}\left({k}+1\right)+2\left({k}+1\right)}{2}\phantom{\rule{0ex}{0ex}}=\frac{\left({k}+1\right)\left({k}+2\right)}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {P}\left({k}+1\right)\text{is True}\phantom{\rule{0ex}{0ex}}\text{Hence by principle of mathematical induction}{P}\left({n}\right)\text{is true for all}{n}\in {N}\text{.}$Q35. Find the domain of the function
${f}\left({x}\right)=\frac{{{x}}^{2}+2{x}+1}{{{x}}^{2}\u20137{x}+12}.$
Ans:
Since, x^{2} – 7x +12 = (x – 4)(x – 3), the function is defined for all real numbers
except x = 4 and x = 3. Hence, the domain of f is R – {3, 4}.
Q36. The following is the record of goals scored by team A in a football session:
No. of goals scored

0

1

2

3

4

No. of matched

1

9

7

5

3

For the team B, mean number of goals scored per match was 2 with a standard deviation 1.25 goals. Find which team may be considered more consistent?
Ans:
$Mean=\frac{{\sum}_{i=1}^{n}{\left(f\right)}_{i}{\left(x\right)}_{i}}{n}\phantom{\rule{0ex}{0ex}}=\frac{50}{25}\phantom{\rule{0ex}{0ex}}=2\phantom{\rule{0ex}{0ex}}Variation\left({\left(\sigma \right)}^{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{{\sum}_{i=1}^{n}{\left(f\right)}_{i}{\left(({\left(x\right)}_{i}\stackrel{\u02c9}{x})\right)}^{2}}{n}\phantom{\rule{0ex}{0ex}}=\frac{30}{25}\phantom{\rule{0ex}{0ex}}=1.2\phantom{\rule{0ex}{0ex}}Standard\text{}deviation\phantom{\rule{0ex}{0ex}}=\sqrt{1.2}\phantom{\rule{0ex}{0ex}}=1.09\phantom{\rule{0ex}{0ex}}Standard\text{}deviation\text{}of\text{}team\text{}B\phantom{\rule{0ex}{0ex}}=1.25\text{}goals\phantom{\rule{0ex}{0ex}}Mean\text{}of\text{}team\text{}B\phantom{\rule{0ex}{0ex}}=2\phantom{\rule{0ex}{0ex}}The\text{}team\text{}with\text{}lower\text{}standarddeviation\text{}is\text{}more\text{}consistent{.}\phantom{\rule{0ex}{0ex}}Therefore,\text{}team\text{}A\text{}is\text{}more\text{}consistent{.}$Mean=n∑i=1n(f)i(x)i=2550=2Variation((σ)2)=n∑i=1n(f)i(((x)i−xˉ))2=2530=1.2Standard deviation=1.2=1.09Standard deviation of team B=1.25 goalsMean of team B=2The team with lower standarddeviation is more consistent.Therefore, team A is more consistent.
Q37. Find the number of 4digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even?
Ans:
Number of given digits = 5
Number of digits in each number = 4
Number of 4digit numbers = ^{5}P_{4}
= 5!
= 120
Now, we find 4 digit even numbers.
Only 2 numbers are possible at units place (2, 4) as we need even number.
If no digit is repeated, then number of ways to fill remaining three places of 4digit numbers = ^{4}P_{3}
= 4! /(4 – 3)!
= 4! / 1!
= 24
Number of ways to arrange 2 even numbers at one’s place = 2
Total formed 4digit even numbers = 24 x 2
= 48
Thus, there are 48, 4 digit even numbers.
Q38. The English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonants can be formed from the alphabet?
Ans:
$\begin{array}{l}\mathrm{Number}{ of vowels in the English alphabets}=5\\ \mathrm{Number}{ of consonants in the English alphabets}=21\\ \mathrm{Number}{ of words formed by 2 vowels and 2 consonants}\\ ={{\hspace{0.17em}}}^{5}{\mathrm{C}}_{2}\times {{\hspace{0.17em}}}^{21}{\mathrm{C}}_{2}\\ ={\hspace{0.17em}}\frac{5!}{2!(52)!}\\ \times \frac{21!}{2!(212)!}\\ =\frac{5!}{2!3!}\times \frac{21!}{2!{\hspace{0.17em}}19!}\\ =10\times 210\\ =2100\\ \mathrm{These}{ 2100 words has 4 letters only. }\\ {Number of ways to arrange 4 letters}={{\hspace{0.17em}}}^{4}{\mathrm{P}}_{4}\\ =\frac{4!}{(44)!}\\ =4!=24\\ \mathrm{So},{ the number of total words formed by 2 vowels and }\\ {2 consonants}=2100\times 24\\ {\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=50400\end{array}$
Q39. How many litres of water will have to be added to 1125 litres of the 45% solution of acid so that the resulting mixture will contain more than 25% but less than 30% acid content?
Ans:
$\begin{array}{}\mathrm{Let}\text{Volumeofaddedwatertoacid}=\text{xlitres}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}Volumeof45\%solutionofacid}=\text{1125litres}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}Totalsolutionofacid}=\left(1125+\text{x}\right)\text{\hspace{0.33em}litres}\\ \mathrm{Therefore},\\ 45\text{\%\hspace{0.33em}}\mathrm{of}\text{\hspace{0.33em}}112525\text{\%}\left(1125+\text{x}\right)\\ \mathrm{and}\text{\hspace{0.33em}}45\text{\%\hspace{0.33em}}\mathrm{of}\text{\hspace{0.33em}}112530\text{\%}\left(1125+\text{x}\right)\\ \mathrm{or}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\frac{45}{100}\text{\hspace{0.33em}}\times 1125\frac{25}{100}\text{\hspace{0.33em}}\times \left(1125+\text{x}\right)\\ \mathrm{and}\text{\hspace{0.33em}}\frac{45}{100}\text{\hspace{0.33em}}\times \text{\hspace{0.33em}}1125\frac{30}{100}\text{\hspace{0.33em}}\times \left(1125+\text{x}\right)\\ \mathrm{or}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}45\text{\hspace{0.33em}}\times 112525\text{\hspace{0.33em}}\times \left(1125+\text{x}\right)\\ \mathrm{and}\text{\hspace{0.33em}}445\text{\hspace{0.33em}}\times \text{\hspace{0.33em}}112530\text{\hspace{0.33em}}\times \left(1125+\text{x}\right)\\ \mathrm{or}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}5062528125+25\text{x}\\ \mathrm{and}\text{\hspace{0.33em}}5062533750+30\text{x}\\ \mathrm{or}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}506252812525\text{x}\\ \mathrm{and}\text{\hspace{0.33em}}506253375030\text{x}\\ \mathrm{or}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}2250025\text{x}\\ \mathrm{and}\text{\hspace{0.33em}}1687530\text{x}\\ \mathrm{or}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}900\text{x}\\ \mathrm{and}\text{\hspace{0.33em}}562.5\text{x}\\ \Rightarrow \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}x}\in \left(562.5,900\right)\\ \mathrm{Thus},\text{addedwatershouldbemorethan}562.5\text{litres\hspace{0.33em}butlessthan900litres.}\end{array}$Q40.
Ans:
Q41. In a school, selection of 11 players cricket team is to be done for the upcoming interschool cricket championship from the existing 17 players in the school cricket team. Rohit is the captain and fast bowler of the team so his place in team is reserved.
Based on this information answer the following questions:
(i) In how many ways team selection can be done?
(ii) If Soham is an allrounder player, so he must be in the team, then find the number of ways team selection can be done?
(iii) If in team 6 players can bowl and the team must include exactly 4 bowlers, then find the number of ways team selection can be done?
Ans:
(i) Given Rohit’s place in the team is reserved.
⸫ The remaining 10 players are selected from 17 – 1 = 16 players.
⸫ The required number of ways = ^{16}C_{10 }= 8008
(ii) Given Rohit is the captain and his place in the team is reserved and Soham must be in the team,
⸫ The remaining 9 players are selected from 17 – 2 = 15 players.
⸫ The required number of ways = ^{15}C_{9 }= 5005
(iii)
Given Rohit is captain and fast bowler and his place in the team is reserved.
Also, it is given that out of 17 players, 6 players can bowl.
A cricket team of 11 players is to be selected in such a way that there are exactly 4 bowlers.
∴ 3 bowlers can be selected out of remaining 5 bowlers in ^{5}C_{3} ways.
And, the remaining 7 players are selected out of 11 players in ^{11}C_{7} ways.
Thus, by multiplication principle, the required number of ways of
selecting cricket team is
=5C3×11C7=3!(5−3)5!×7!(11−7)!11!=3!2!5!×7!4!11!=3300
Q42. In a school, during health checkup of class XI students, age, weight, and height of 5 students were represented by sets A, B and C respectively as follows:
A = {16 yr, 16.25 yr, 16.5 yr, 16.75 yr, 17 yr}, B = {171 cm, 171.5 cm. 172 cm, 172.5 cm, 173 cm} and C = {53 kg, 53.25 kg, 53.5 kg, 53.75 kg, 54 kg},
Based on the above information answer the following questions:
(i) Elements from set B are related to the elements of set C as shown in arrow diagram. Identify that, is the given relation formed between weights and heights is a function? If yes, identify the function.
(ii) What are the numbers of nonempty relations from set A to set B?
Ans:
(i) Since, in the given arrow diagram for each element of set A there is a unique image in set C,
Hence, the given relation is a function.
Let the function be, y = f(x).
The ordered pairs of the function are, (53, 173), (53.25, 172.5), (53.5, 172), (53.75, 171.5), (54, 171).
The above ordered pairs have same difference between the elements of elements. So, corresponding to linear change in the weight, the height varies linearly. Hence it is a linear function. Let the function, y = mx + c. So, satisfying any two ordered pairs to find m and c.
Using, (53, 173) : 173 = 53m + c …(1)
Using, (54, 171) : 171 = 54m + c …(2)
Subtracting (1) from (2) we get m = – 2 and by substituting m = 2 in (1) we get c = 279.
So, the function f(x) = – 2x + 279.
(ii)
Since,A={16,16.25,16.5,16.75,17},B={171cm,171.5cm,172cm,172.5cm,173cm},⇒n(A)=5,n(B)=5,∴NumberofsubsetsofA×B=25×5=225⇒NumberofrelationsfromsetAtosetB=225,As,nullsetϕisanemptysubsetofA×BaswellemptyrelationfromsetAtosetB,∴Numberofnon−emptyrelationsfromsetAtosetB=225−1.
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