In this article, you will get a collection of important 4 marks questions to prepare for the CBSE Class 9 Mathematics Annual Exam 2020. All these questions have been prepared after carrying the thorough analysis of previous years' question papers and the latest syllabus. All the questions are provided with proper solutions.
In CBSE Class 9 Mathematics Exam 2020, Section - D will comprise 6 questions of 4 marks each.
Students must practice the questions given here as it will help them not only assess their preparation level but also know the important topics which need to be prepared for the annual exam with more concentration.
CBSE Class 9 Mathematics Syllabus 2019-2020
Given below are some sample questions from CBSE Class 9 Mathematics: Important 4 Marks Questions:
Q. The polynomials ax3 – 3x2 +4 and 2x3 – 5x +a when divided by (x – 2) leave the remainders p and q respectively. If p – 2q = 4, find the value of a.
Sol.
Let, f(x) = ax3 – 3x2 +4
And g(x) = 2x3 – 5x +a
When f(x) and g(x) are divided by (x – 2) the remainders are p and q respectively.
⟹ f(2) = p and g(2) = q
⟹ f(2) = a × 23 – 3 × 22 + 4
⟹ p = 8a – 12 + 4
⟹ p = 8a – 8 ....(i)
And g(2)= 2 × 23 – 5 × 2 + a
⟹ q = 16 – 10 + a
⟹ q = 6 + a ....(ii)
But p – 2q = 4 (Given)
⟹ 8a – 8 – 2(6 + a) = 4 (Using equations (i) and (ii))
⟹ 8a – 8 – 12 − 2a = 4
⟹ 6a – 20 = 4
⟹ 6a = 24
⟹ a = 24/6
⟹ a = 4
CBSE Class 9 Mathematics Exam 2018: Important 3 Marks Questions
Q. Construct a ΔABC in which BC = 3.8 cm, ∠B = 45oand AB + AC = 6.8cm.
Sol.
Steps of Construction
1. Draw BC = 38 cm.
2. Draw a ray BX making an ∠CBX = 45°.
3. From BX, cut off line segment BD equal to AB + BC i.e., 6.8 cm.
4. Join CD.
5. Draw the perpendicular bisector of CD meeting BD at A.
6. Join CA to obtain the required
Justification:
Clearly, A lies on the perpendicular bisector of CD.
∴ AC = AD
Now, BD = 6.8 cm
⟹ BA + AD = 6.8 cm
⟹ AB + AC = 6.8 cm
Hence, is the required triangle.
Q. If a + b + c = 6 and ab + bc + ca = 11, find the value of a3 +b3 +c3 − 3abc.
Sol.
(a + b + c)2 = a2 + b2 +c2 +2(ab + bc + ca)
(6)2 = a2 + b2 +c2 + 2 × 11
a2 + b2 +c2 = 36 – 22 = 14
a3 +b3 +c3 − 3abc =( a + b + c)[ a2 + b2 +c2 −(ab + bc + ca)]
= 6 × (14 − 11)
= 6 × 3 = 18
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