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Class 10 Math Unit 2 Chapter Wise Class Test 3 New Syllabus 2027 | PDF free download
Class 10 Math Chapter 2 Class Test 3 is based on Quadratic Equations and Inequalities and is part of the 786Times Chapter Wise Test Series. This Class 10 Mathematics Chapter 2 Test includes important quadratic equations, inequalities, roots of quadratic equations, discriminant, standard form, and related mathematical concepts. Students can use this Class 10 Math Test for chapter-wise preparation, revision, practice, and Class 10 board exam preparation. The test contains both objective MCQs and subjective questions to help students prepare effectively for their Mathematics examination.
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Unit 02:
Total Time: 35 min
Chapter Wise Class Test
(Test # 3)
Mathematics
Total Marks: 25
ABCD
1ABCD
2ABCD
3ABCD
ABCD
4ABCD
5ABCD
6ABCD
ABCD
7ABCD
8ABCD
NOTE: Four possible answers A, B, C and D to each question are given. The choice which you think is correct, fill that circle in front of that question with Marker or Pen ink. Cutting or filling two or more circles will result in zero mark in that question.
TIME: 10 Min
OBJECTIVE
MARKS: 8
Q.1:(8)
1.Sum of the roots of 3x2 + 5x − 12 = 0 is:
(A) 53(B) −53(C) 123(D) 35
2.What are the roots of (x − 3) (x + 3) = 0?
(A) 3, −3(B) 3, 3(C) −3, −3(D) 9, 0
3.If b2 − 4ac = 0, then the roots of ax2 + bx + c = 0 are:
(A) unequal(B) irrational(C) imaginary(D) equal
4.Subject “c” of x − 2c = b is:
(A) x + b(B) b − x(C) x − b2(D) b − x2
5.When b2 − 4ac = 0, the roots are:
(A) imaginary(B) irrational and unequal(C) real and equal(D) real and unequal
6.The discriminant of 4x2 + 4x + 1 = 0 is:
(A) 0(B) 4(C) 17(D) 16
7.The corresponding equation to the inequality: 3x2 − 5x + 7 ≤ 0 is:
(A) 3x2 + 5x = −7(B) 3x − 5x2 − 7 = 0(C) 3x2 − 5x + 7 = 0(D) 3x2 = − 5x + 7
8.If I = PRT, then T is:
(A) IPR(B) IRP(C) PRT(D) IPR
Answer Key: 1-(B) 2-(A) 3-(D) 4-(C) 5-(C) 6-(A) 7-(C) 8-(D)
TIME: 25 Min
SUBJECTIVE
MARKS: 17
(PART – I)
Q.2 Write short answers to following questions.(6×2=12)
(i)
Form a quadratic equation whose roots are; 6, 32 .
Sum of roots = 6 + 32 = 152; Product = 6 × 32 = 9.
Equation: x2 − (sum)x + product = 0 ⇒ x2 − 152x + 9 = 0 ⇒ 2x2 − 15x + 18 = 0
Equation: x2 − (sum)x + product = 0 ⇒ x2 − 152x + 9 = 0 ⇒ 2x2 − 15x + 18 = 0
(ii)
If the equation x2 + 2(1+k)x + k2 = 0 has equal roots, then find the value of k.
Here a = 1, b = 2(1+k), c = k2. For equal roots b2 − 4ac = 0:
4(1+k)2 − 4k2 = 0 ⇒ 4(1 + 2k + k2) − 4k2 = 0 ⇒ 4 + 8k = 0 ⇒ k = −12
4(1+k)2 − 4k2 = 0 ⇒ 4(1 + 2k + k2) − 4k2 = 0 ⇒ 4 + 8k = 0 ⇒ k = −12
(iii)
Write the quadratic equation 5(2x + 3) = −9x2 in standard form.
10x + 15 = −9x2 ⇒ 9x2 + 10x + 15 = 0 (standard form ax2 + bx + c = 0)
(iv)
Solve x2 − 4x − 5 = 0 by completing square method.
x2 − 4x = 5. Add (42)2 = 4 on both sides: x2 − 4x + 4 = 9 ⇒ (x − 2)2 = 9 ⇒ x − 2 = ±3 ⇒ x = 5 or x = −1.
Solution set = {5, −1}
Solution set = {5, −1}
(v)
Write the formula for the product of the roots of a quadratic equation.
For ax2 + bx + c = 0, product of roots = ca
(vi)
Form the equation whose roots are 5 + √ 7 and 5 − √ 7 .
Sum = (5 + √7) + (5 − √7) = 10; Product = 52 − (√7)2 = 25 − 7 = 18.
Equation: x2 − 10x + 18 = 0
Equation: x2 − 10x + 18 = 0
(PART – II)
Note: Attempt the following question.(5)
Q.3:
A ball is thrown upward with an initial velocity of 40ms−1. Calculate the maximum height it reaches above ground level.
Given: initial velocity vi = 40 ms−1, final velocity at maximum height vf = 0, acceleration a = −g = −10 ms−2.
Using vf2 = vi2 + 2ah:
0 = (40)2 − 2(10)h ⇒ 20h = 1600 ⇒ h = 80 m
(If g = 9.8 ms−2 is used, h = 160019.6 ≈ 81.6 m.)
Using vf2 = vi2 + 2ah:
0 = (40)2 − 2(10)h ⇒ 20h = 1600 ⇒ h = 80 m
(If g = 9.8 ms−2 is used, h = 160019.6 ≈ 81.6 m.)
