Class X (CBSE 2019) Mathematics Abroad (Set-3)

General Instructions:

1. All questions are compulsory.
2. The question paper consists of 30 questions divided into four sections - A, B, C and D.
3. Section A comprises 6 questions of 1 mark each. Section B contains 6 questions of 2 marks each. Section C contains 10 questions of 3 marks each. Section D contains 8 questions of 4 marks each.

1. There is no overall choice. However, an internal choice has been provided in two questions

of 1 mark, two questions of 2 marks, four questions of 3 marks each and three questions of 4 marks each. You have to attempt only one of the alternative in all such questions.

1. Use of calculators is not permitted.

Question 1

Which term of the A.P. -4, -1, 2, ... is 101?

SOLUTION:

We have been given an arithmetic progression where

<math>a = -4, d = -1 - (-4) = 3</math> and <math>a_n = 101</math>

We need to find which term of the given AP is 101 so, we need to find n.

Using <math>a_n = a + (n-1)d</math>

Substituting the values in the formula we get

<math>101 = -4 + (n-1)3</math>

<math>101 + 7 = 3n</math>

<math>3n = 108</math> <math>n = 36</math>

Therefore, 36th term of given A.P is 101.

Question 2 Evaluate: tan 65°<br>cot 25° OR

Express (sin 67° + cos 75°) in terms of trigonometric ratios of the angle between 0° and 45°.

SOLUTION:

tan 65°<br>cot 25° <math display="block">= \frac{\tan(90^\circ - 25^\circ)}{\cot 25^\circ} \quad (\because \tan(90^\circ - \theta) = \cot\theta)</math> <math>=\frac{\cot 25^{\circ}}{\cot 25^{\circ}}</math> <math>=1</math> OR <math>(\sin 67^{\circ} + \cos 75^{\circ})</math>

<math>=(\sin(90^{\circ}-23^{\circ}) + \cos(90^{\circ}-25^{\circ}))</math> (: <math>\sin(90^{\circ}-\theta) = \cos\theta</math> and <math>\cos(90^{\circ}-\theta) =</math> <math>\sin\theta</math>) = (<math>\cos 23^{\circ} + \sin 25^{\circ}</math>) Question 3

Find the value of k for which the quadratic equation <math>kx(x-2) + 6 = 0</math> has two equal roots.

SOLUTION:

Given quadratic equation is: <math display="block">kx\left( x-2\right) +6=0</math> <math>\Rightarrow kx^2 - 2kx + 6 = 0</math> For a quadratic equation to have equal roots, <math display="block">b^2 - 4ac = 0</math> Comparing the given equation with general equation <math>ax^2+bx+c=0</math> We get <math>a = k</math>, <math>b = -2k</math> and <math>c = 6</math> <math>(-2k)^2 - 4(k)(6) = 0</math> <math>\Rightarrow 4k^2 - 24k = 0</math> <math>\Rightarrow 4k(k-6)=0</math> Therefore, <math>k=0</math> and <math>k=6</math>

Question 4