In this article you will get a collection of important 4 marks questions to prepare for the CBSE class 9 Mathematics annual exam 2019. All these questions have been prepared after carrying the thorough analysis of previous years' question papers and latest syllabus. All the questions are provided with proper solutions.

In CBSE Class 9 Mathematics Exam 2019, Section - D will comprise 8 questions of 4 marks each.

Students must practice the questions given here as it will help them to 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**

**Given below are some sample questions from CBSE Class 9 Mathematics: Important 4 Marks Questions:**

**Q.** The polynomials a*x*^{3} – 3*x*^{2} +4 and 2*x*^{3 }– 5*x* +*a* when divided by (*x* – 2) leave the remainders *p* and *q* respectively. If *p* – 2*q* = 4, find the value of *a*.

**Sol.**

Let, *f*(*x*) = a*x*^{3} – 3*x*^{2} +4

And *g*(*x*) = 2*x*^{3 }– 5*x* +*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 *× 2^{3} – 3 × 2^{2} + 4

**⟹**** ** *p* = 8*a* – 12 + 4

**⟹**** ** *p* = 8*a* – 8 ....(i)

And *g*(2)= 2* *× 2^{3} – 5 × 2 + *a*

**⟹**** ** *q* = 16 – 10 + *a*

**⟹**** ** *q* = 6 + *a *....(ii)

But *p* – 2*q* = 4 (Given)

**⟹**** **8*a* – 8 – 2(6 + *a*) = 4 (Using equations (i) and (ii))

**⟹**** **8*a* – 8 – 12 − 2*a* = 4

**⟹**** **6*a* – 20 = 4** **

**⟹**** **6*a* = 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 = 45^{o}and 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.

**CBSE Class 9 Mathematics Sample Paper**

**Q.** If *a* + *b* + *c* = 6 and *ab* + *bc* + *ca* = 11, find the value of *a*^{3 }+*b*^{3 }+*c*^{3 }− 3*abc*.

**Sol.**

(*a* + *b* + *c*)^{2} = *a*^{2 }+ *b*^{2} +*c*^{2} +2(*ab *+ *bc *+ *ca*)

(6)^{2 }=* a*^{2 }+ *b*^{2} +*c*^{2} + 2 × 11

*a*^{2 }+ *b*^{2} +*c*^{2} = 36 – 22 = 14

*a*^{3 }+*b*^{3 }+*c*^{3 }− 3*abc* =(* a* + *b* + *c*)[* a*^{2 }+ *b*^{2} +*c*^{2} −(*ab *+ *bc *+ *ca*)]

= 6 × (14 − 11)

= 6 × 3 = 18

**To get the complete set of questions, click on the following link:**

**Students may also check the following links to explore more stuff, important for CBSE Class 9 Annual exam preparations:**

**CBSE Class 9 Maths, Science, English, Hindi and Social Maths Question Papers****CBSE Class 9 Maths Chapter-wise Important Topics and Questions****CBSE Class 9 Exam 2019: Examination pattern and question paper design for all subjects****CBSE Class 9 Mathematics NCERT Exemplar Problems: All Chapters****CBSE Class 9 Mathematics NCERT Solutions**