CBSE Paper 2020 Math Delhi (Set 2)

General Instructions:

Read the following instructions very carefully and strictly follow them:

1. This question paper comprises four sections – A, B, C and D.

This question paper carries 36 questions. All questions are compulsory.

1. Section A – Question no. 1 to 20 comprises of 20 questions of one mark each.
2. Section B – Question no. 21 to 26 comprises of 6 questions of two marks each.
3. Section C – Question no. 27 to 32 comprises of 6 questions of four marks each.
4. Section D – Question no. 33 to 36 comprises of 4 questions of six marks each.

1. There is no overall choice in the question paper. However, an internal choice has

been provided in 3 questions of one mark, 2 questions of two marks, 2 questions of four marks and 2 questions of six marks. Only one of the choices in such questions have to be attempted.

1. In addition to this, separate instructions are given with each section and question, wherever necessary.
2. Use of calculators is not permitted.

Question 1

If <math>\begin{bmatrix} x & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -2 & 0 \end{bmatrix} = 0</math>, then x equals

1. 0 (b) -2 (c) -1
2. 2

Solution:

Given: <math>\begin{bmatrix} x & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -2 & 0 \end{bmatrix} = \mathbf{O}</math> <math>\Rightarrow [x-2 \quad 0] = [0 \quad 0]</math> <math>\Rightarrow x = 2</math>

Hence, the correct answer is option (d).

Question 2

<math>\int 4^x 3^x dx</math> equals

1. <math>\frac{12^x}{\log 12} + C</math>

1. <math>\frac{4^x}{\log 4} + C</math>
2. <math>\left(\frac{4^x \cdot 3^x}{\log 4 \cdot \log 3}\right) + C</math>
3. <math>\frac{3^x}{\log 3} + C</math>

Solution:

Consider the given integral <math>\int 4^x 3^x dx</math> <math>=\int 12^x dx</math> <math>=\frac{1}{\ln 12}\int (12^x \ln 12) dx</math> <math>=\frac{12^x}{\ln 12} + C</math> Hence, the correct answer is option (a).

Question 3

A number is chosen randomly from numbers 1 to 60. The probability that the chosen number is a multiple of 2 or 5 is

1. 2/5
2. 3/5
3. 7/10
4. 9/10

Solution:

The number of multiples of 2 from 1 to 60 are given by, <math>60 = 2 + (n_1 - 1)2</math> <math>\Rightarrow 2(n_1-1)=58</math> <math>\Rightarrow n_1 = 30</math> The number of multiples of 5 from 1 to 60 are given by, <math display="block">60 = 5 + (n_2 - 1)5</math> <math>\Rightarrow</math> 5 <math>(n_2-1)=55</math> <math>\Rightarrow n_1 = 12</math> The number of multiples of both 2 and 5 from 1 to 60 are given by, <math>60 = 10 + (n_3 - 1)10</math> <math>\Rightarrow 10(n_3-1)=50</math> <math>\Rightarrow n_3 = 6</math>