Please give some imp questions of class 9 maths tomorrow is my exam
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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
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
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.
THE FIGURE OF THE CONSTRUCTED TRIANGLE IS IN THE ATTACHMENTY PLEASE CHECK IT.............
HOPE IT HELPS
PLEASE MARK MY ANSWER AS BRAINLIEST....
PLEASE...............
I REQUEST YOU............
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
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
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.
THE FIGURE OF THE CONSTRUCTED TRIANGLE IS IN THE ATTACHMENTY PLEASE CHECK IT.............
HOPE IT HELPS
PLEASE MARK MY ANSWER AS BRAINLIEST....
PLEASE...............
I REQUEST YOU............
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