UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem

UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem (द्विपद प्रमेय)

These Solutions are part of UP Board Solutions for Class 11 Maths. Here we have given UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem (द्विपद प्रमेय).

प्रश्नावली 8.1

प्रश्न 1 से 5 तक प्रत्येक व्यंजक को प्रसार ज्ञात कीजिए:
प्रश्न 1.
(1 – 2x)5
UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 8.1 1

UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 8.1 2

UP Board Solutions

प्रश्न 3.
(2x – 3)6
UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 8.1 3

UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 8.1 4

UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 8.1 5
UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 8.1 5.1

UP Board Solutions

प्रश्न : द्विपद प्रमेय का प्रयोग करके निम्नलिखित का मान ज्ञात कीजिए (प्रश्न 6 से 9 तक)

प्रश्न 6.
(96)3
UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 8.1 6

प्रश्न 7.
(102)5
UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 8.1 7

प्रश्न 8.
(101)4
UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 8.1 8

प्रश्न 9.
(99)5
UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 8.1 9

UP Board Solutions

प्रश्न 10.
द्विपद प्रमेय का प्रयोग करते हुए बताइए कौन-सी संख्या बड़ी है-
(1.1)10000 या 1000
UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 8.1 10

प्रश्न 11.
(a + b)4 – (a – b)4 का विस्तार कीजिए। इसका प्रयोग करके (√3 + √2)4 – (√3 – √2)4 का मान ज्ञात कीजिए।
UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 8.1 11

प्रश्न 12.
(x + 1)6 + (x – 1)6 का मान ज्ञात कीजिए। इसका प्रयोग करके या अन्यथा (√2 + 1)6 + (√2 – 1)6 का मान ज्ञात कीजिए।
UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 8.1 12

UP Board Solutions

प्रश्न 13.
दिखाइए कि 9n+1 – 8n – 9, 64 से विभाज्य है जहाँ n एक धन पूर्णाक है।
UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 8.1 13

UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 8.1 14

UP Board Solutions

प्रश्नावली 8.2

प्रश्न 1 और 2 में गुणांक ज्ञात कीजिए:

प्रश्न 1.
(x + 3)8 में x5 का।
UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 8.2 1

प्रश्न 2.
(a – 2b)12 में a5b7 का।
UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 8.2 2
UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 8.2 2.1

UP Board Solutions

प्रश्न 3 व 4 के प्रसार में व्यापक पद लिखिए।

प्रश्न 3.
(x2 – y)6.
UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 8.2 3

प्रश्न 4.
(x2 + yx)12, x ≠ 0.
UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 8.2 4

प्रश्न 5.
(x – 2y)12 के प्रसार में चौथा पद ज्ञात कीजिए।
UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 8.2 5

UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 8.2 6

UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 8.2 7

UP Board Solutions
UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 8.2 7.1

UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 8.2 8
UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 8.2 8.1

प्रश्न 9.
(1 + a)m+n के प्रसार में सिद्ध कीजिए कि am तथा an के गुणांक बराबर हैं।

UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 8.2 9

UP Board Solutions

प्रश्न 10.
(x + 1)n के प्रसार में (r – 1) वाँ, r वाँ और (r + 1) वें पदों के गुणांक में 1 : 3 : 5 का अनुपात हो तो n तथा r का मान ज्ञात करो।
UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 8.2 10
UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 8.2 10.1

UP Board Solutions

प्रश्न 11.
सिद्ध कीजिए कि (1 + x)2n के प्रसार में xn का गुणांक, (1 + x)2n-1 के प्रसार में xn के गुणांक का दुगुना होता है।
UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 8.2 11

प्रश्न 12.
m का धनात्मक मान ज्ञात वीजिए जिसके लिए (1 + x)mके प्रसार में x2 का गुणांक 6 हो।
UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 8.2 12

अभ्यास 8 पर विविध प्रश्नावली

प्रश्न 1.
यदि (a + b)n के प्रसार में प्रथम तीन पद क्रमशः 729, 7290 तथा 30375 हों तो a, b तथा n ज्ञात कीजिए।
UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 1
UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 1.1

UP Board Solutions

प्रश्न 2.
यदि (3 + ax)9 के प्रसार में x2 और x3 के गुणांक समान हों, तो a का मान ज्ञात कीजिए।
UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 2

प्रश्न 3.
द्विपद प्रमेय का प्रयोग करते हुए गुणनफल (1 + 2x)6 (1 – x)7 में x5 का गुणांक ज्ञात कीजिए।
UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 3
UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 3.1

प्रश्न 4.
यदि a और b भिन्न-भिन्न पूर्णाक हों, तो सिद्ध कीजिए कि an – bn का एक गुणनखंड (a – b) है, जबकि n एक धन पूर्णाक है।
UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 4

UP Board Solutions

प्रश्न 5.
(√3 + √2)6 – (√3 – √2)6 का मान ज्ञात कीजिए।
UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 5

UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 6

UP Board Solutions

प्रश्न 7.
(0.99)5 प्रसार के पहले 3 पदों का प्रयोग करते हुए इसका निकटतम मान ज्ञात कीजिए।
हल:
UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 7

UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 8
UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 8.1

UP Board Solutions

UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 9
UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 9.1

UP Board Solutions

प्रश्न 10.
(3x2 – 2ax + 3a2)3 का द्विपद प्रमेय से प्रसार ज्ञात कीजिए।
UP Board Solutions for Class 11 Maths Chapter 8 Binomial Theorem 10

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CBSE Sample Papers for Class 10 Maths Paper 4

These Sample papers are part of CBSE Sample Papers for Class 10 Maths. Here we have given CBSE Sample Papers for Class 10 Maths Paper 4.

CBSE Sample Papers for Class 10 Maths Paper 4

Board CBSE
Class X
Subject Maths
Sample Paper Set Paper 4
Category CBSE Sample Papers

Students who are going to appear for CBSE Class 10 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 4 of Solved CBSE Sample Paper for Class 10 Maths is given below with free pdf download solutions.

Time allowed: 3 Hours
Maximum Marks: 80

General Instructions

 

  • All questions are compulsory.
  •  The question paper consists of 30 questions divided into four sections A, B, C andD.
  • Section A contains 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,
  • There is no overall choice. However, an internal choice has been provided in four questions of 3 marks each and three questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.
  • Use of calculators is not permitted.

Section-A

This Calculator computes the Degree and Leading Coefficient Calculator term of a given Polynomial.

Question 1.
The values of the remainder r, when a positive integer a is divided by 3 are 0 and 1 only. Justify your answer.

Question 2.
Find the altitude of an equilateral triangle when each of its side is ‘a’ cm.

Question 3.
If x =[latex s=2]\frac { 2 }{ 3 } [/latex] and x = – 3 are roots of the quadratic equation ax2 + 7x + b = 0, find the values of a and b.

Question 4.
If A + B = 90° and sec A =[latex s=2]\frac { 5 }{ 3 } [/latex] , then find the value of cosec B.

Question 5.
The first three terms of an AP respectively are 3y – 1, 3y + 5 and 5y + 1. Then find y.

Find the Value of x is used to consider unknown value.

Question 6.
Find the value of x such that PQ = QR where co-ordinates of P, Q, R are (6, -1), (1, 3), and (x, 8) respectively.

Section-B

Question 7.
Find the LCM of 66 & 486 by the Prime factorisation method. Hence find their HCF.

Question 8.
The sum of the 5th and the 9th terms of an AP is 30. If its 25th term is three times its 8th term, find the AP.

Question 9.
A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, then find the number of blue balls in the bag.

Question 10.
Find the area of the triangle ABC with A (1, – 4) and mid-points of sides through A being (2, -1) and (0,-1).

Question 11.
Find the value of a so that the point (3, a) lies on the line represented by 2x – 3y = 5.

Question 12.
Two dice are thrown simultaneously. What is the probability that the sum of the numbers appearing on the dice is a prime number?

Section-C

Question 13.
Find the HCF of 81 and 237 and express it as a linear combination of 81 and 237.

Question 14.
If a and (1 are the zeroes of the quadratic polynomial p(s) = 3 s2 – 6s + 4, find the value of
[latex s=2]\frac { \alpha }{ \beta } +\frac { \beta }{ \alpha } +2\left( \frac { 1 }{ \alpha } +\frac { 1 }{ \beta } \right) +3\alpha \beta [/latex].

Question 15.
In fig., PSR, RTQ and PAQ are three semicircles of diameters 10 cm, 3 cm and 7 cm respectively. Find the perimeter ofthe shaded region.
[Use π = 3.14]
CBSE Sample Papers for Class 10 Maths Paper 4 img 1

Question 16.
150 spherical marbles, each of diameter 1.4 cm, are dropped in a cylindrical vessel of diameter 7 cm containing some water, which are completely immersed in water. Find the rise in the level of water in the vessel.
OR
Volume and surface area of a solid hemisphere are numerically equal. What is the diameter ofhemisphere?

Question 17.
The three vertices ofaparallelogram ABCD are A(3,^l), B(-l,-3)andC(-6,2). Find the coordinates of vertex D and find the area of ABCD.

Question 18.
If sec θ + tan θ = p, then prove that [latex s=2]\frac { { p }^{ 2 }-1 }{ { p }^{ 2 }+1 } [/latex] = sin θ
OR
If α + β = 90° and α = 2β , then find the value of cos2 α + sin2 β

Question 19.
If the median for the following frequency distribution is 28.5, find the values ofx and y:
CBSE Sample Papers for Class 10 Maths Paper 4 img 2
OR
The mean of marks scored by 100 students was found to be 40. Later on it was discovered that a score of 53 was misread as 83. Find the correct mean.

Question 20.
In the adjoining figure, PA and PB are tangents to a circle with centre O. If OP is equal to the diameter of the circle, prove that ∆ABP is an equilateral triangle.
CBSE Sample Papers for Class 10 Maths Paper 4 img 3

Question 21.
Solve: 2x2 +3y2 = 35; [latex s=2]\frac { { x }^{ 2 } }{ 2 } +\frac { { y }^{ 2 } }{ 3 } [/latex] = 5

Question 22.
Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of triangle PQR. Prove that ∆ABC ~ ∆PQR
OR
In the given figure, ∆ABC and ∆DBC are on the same base BC. AD and BC intersect at O. Prove that
[latex]\frac { ar(\triangle ABC) }{ ar(\triangle DBC) } =\frac { AO }{ DO } [/latex].
CBSE Sample Papers for Class 10 Maths Paper 4 img 4

Section-D

Question 23.
On a straight line passing through the foot of a tower, two points C and D are at distances of 4 m and 16m from the foot respectively. If the angles of elevation from C and D of the top of the tower are complementary, then find the height of the tower.
OR
The angles of elevation and depression of the top and the bottom of a tower from the top of a building, 60 m high, are 30° and 60° respectively. Find the difference between the heights of the building and the tower and the distance between them.

Question 24.
An iron pole consisting of a cylindrical portion 110 cm. high and of base diameter 12cm. is surmounted by a cone 9 cm. high. Find the mass of the pole, given that 1 cm3 of iron has 8 gram mass (approx.).
[use 71 = 355/113].

Question 25.
If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then prove that the two triangles are similar.
OR
If a line divides any two sides of a triangle in the same ratio, then prove that the line is parallel to the third side.

Question 26.
CBSE Sample Papers for Class 10 Maths Paper 4 img 5

Question 27.
Draw a circle of radius 4 cm. Take a point P outside the circle. Without using the centre of the circle, draw two tangents to the circle from point P.

Question 28.
The frequency distribution of scores obtained by 230 candidates in a medical entrance test is as follows:
CBSE Sample Papers for Class 10 Maths Paper 4 img 6
Draw cumulative frequency curve or ogive by more than method.

Question 29.
If the equation (1 + m2) x2 + 2mcx + c2 – a2 = 0 has coincident roots show that c2 = a2 (1 + m2)
or c = ±a[latex]\sqrt { 1+{ m }^{ 2 } } [/latex] .
OR
If x = 4 and x = -5 are roots of3x2-2mx + 2n = 0, find the values of ‘m’ and ‘n’.

Question 30.
If four numbers in A.P. are such that their sum is 50 and the greatest number is 4 times the least, then find the numbers.

Solutions
Section-A

Solution 1.
No. According to Euclid’s division lemma, a=3q + r, where 0 ≤ r < 3 and r is an integer. Therefore, the values of r can be 0, 1 or 2. (1)

Solution 2.
CBSE Sample Papers for Class 10 Maths Paper 4 img 7

Solution 3.
CBSE Sample Papers for Class 10 Maths Paper 4 img 8

Solution 4.
Given, A + B = 90° and sec A = [latex s=2]\frac { 5 }{ 3} [/latex]
⇒ sec(90°-B) =[latex s=2]\frac { 5 }{ 3} [/latex] (∵ A + B = 90°)
∴ cosec B = [latex s=2]\frac { 5 }{ 3} [/latex] (1)

Solution 5.
a1 = 3y- 1, a2 = 3y+ 5, a3 = 5y+ 1
∴ a2 – a1 = 3 – a1
⇒(3y + 5) – (3y- 1) = (5y + 1) – (3y +5) ⇒6 = 2y-4 (1)
⇒ 2y=10 ⇒ y=5

Solution 6.
Since, PQ = QR ⇒ Q is mid-point of PR.
∴ Using mid-point formula,
1 = [latex s=2]\frac { 6+x }{ 2} [/latex] ⇒ 6 + x = 2 ⇒x = -4. (1)

Section-B

Solution 7.
The Prime factorisation of 66 & 486 gives
66 = 2 × 3 × 11
486 = 2 × 3 ×3 ×3 × 3 × 3= 2 × 35 (1/2)
∴The LCM of these two integer is
2 × 35 × 11 = 5346 (1/2)
HCF (66,486) = [latex]\frac { 66\times 486 }{ LCM(66,486) } =\frac { 66\times 486 }{ 5346 } [/latex] = 6 (1)

Solution 8.
Given : a5 + a9 = 30
a25 = 3a8
Now, a + 4d + a+8d = 30
⇒ 2a+ 12d = 30
⇒ a + 6d = 15 …(i) (1/2)
and, a + 24d=3a + 21d ⇒2a-3d = 0 …(ii) (1/2)
From eqs. (i) and (ii)
CBSE Sample Papers for Class 10 Maths Paper 4 img 9

Now, put d = 2 in eq. (i)
a+ 12=15 ⇒ a = 3
Required A.P. = 3,5,7,………….. (1/2)

Solution 9.
Let the number of blue balls = x
∴ Total number ofballs = 5 + x
P (blue ball) = [latex s=2]\frac { x }{ 5+x } [/latex] (1/2)
P (red ball) = [latex s=2]\frac { 5 }{ 5+x } [/latex] (1/2)
Given that P (blue) = 2 × p (red)
[latex s=2]\frac { x }{ 5+x } [/latex] = 2 × [latex s=2]\frac { 5 }{ 5+x } [/latex]
⇒ [latex s=2]\frac { x }{ 5+x } [/latex] = [latex s=2]\frac { 10 }{ 5+x } [/latex]
On solving we get x = 10 (1)

Solution 10.
CBSE Sample Papers for Class 10 Maths Paper 4 img 10
P is the mid-point ofAB
∴ x+1=4⇒x = 3 [By Mid-Point formula]
y-4 = -2 ⇒ y = 2
⇒ B(3,2) (1/2)
Similarly,
z + 1 = 0 ⇒z = -1
and t – 4 = -2 ⇒t = 2
⇒ C(—1,2)
∴ Area ∆ABC (1/2)
= [latex s=2]\frac { 1 }{ 2 } [/latex][1(2 – 2)+3(2+4)-1(-4-2)] ⇒[latex s=2]\frac { 1 }{ 2 } [/latex] ×24 = 12 sq units (1)

Solution 11.
Since, (3, a) lies on the line 2x – 3y = 5
So, 2 × 3 – 3a = 5 (1)
⇒ 6 —3a = 5 ⇒ a= [latex s=2]\frac { 1 }{ 3 } [/latex] (1)

Solution 12.
Total number of possible outcomes when two dice are thrown simultaneously =36 (1/2)
Sum of the numbers appearing on the dice
is a prime number i.e., 2,3,5,7 and 11
So, the possible outcomes are (1,1), (1,2), (2, 1), (1,4), (2,3), (3,2), (4,1), (1,6), (2,5),
(3,4), (4,3), (5,2), (6,1), (5,6) and (6,5).
Number of possible outcomes = 15 (1)
∴ Required probability = [latex s=2]\frac { 15 }{ 36 } [/latex] = [latex s=2]\frac { 5 }{ 12 } [/latex] (1/2)

Section-C

Solution 13.
Given integers are 81 and 237 such that 81 < 237.
Applying Euclid’s division lemma to 81 and 237, we get
CBSE Sample Papers for Class 10 Maths Paper 4 img 11
Since the remainder 75 ≠ 0. So, consider the divisor 81 and the remainder 75 and apply division lemma to get
CBSE Sample Papers for Class 10 Maths Paper 4 img 12
The remainder at this stage is zero. So, the last divisor i.e. 3 is the HCF of 81 and 237.
To represent the HCF as a linear combination of the given two numbers, we start from the last but one step and successively eliminate the previous remainders as follows :
From (iii), we have
3 = 75 – 6 × 12
⇒3 = 75-(81 – 75 × 1) ×12
[Substituting 6 = 81 -75 × 1 obtained from (ii)]
⇒ 3 = 75 – 12 × 81 + 12 × 75
⇒ 3 = 13 × 75 – 12 × 81
⇒ 3 = 13 × (237-81 × 2)-12 × 81 [Substituting 75 = 237 – 81 × 2 obtained from (i)] (1)
⇒ 3 = 13 × 237 – 26 × 81 – 12 × 81
⇒ 3 = 13 × 237 – 26 × 81 – 12 × 81
⇒ 3 = 13 × 237-38 × 81
⇒ 3 = 237 × + 81 y, where x = 13 and
y = -38 ….v (1)

Solution 14.
CBSE Sample Papers for Class 10 Maths Paper 4 img 13
CBSE Sample Papers for Class 10 Maths Paper 4 img 14

Solution 15.
Perimeter of shaded region
= Perimeter (QTR+ QAP + PSR) (1)
= π[latex]\left[ 5+\frac { 3 }{ 2 } \frac { 7 }{ 2 } \right] =\pi \left[ \frac { 20 }{ 2 } \right] [/latex] =10π = 31.4cm (2)

Solution 16.
Let the radius of spherical marble = 0.7 cm (1/2)
Volumeofl marble =[latex s=2]\frac { 4 }{ 3 } [/latex]πr3 = [latex s=2]\frac { 4 }{ 3 } [/latex]π(0.7)3 cm3 (1/2)
Volume of 150 marble = 200π(0.7)3 cm3 (1/2)
Let h be the rise in the height of water
∴ Volume of water raised = Volume of 150 marbles (1/2)
So, π × 72 × h = 200π(0.7)3 ⇒ h = [latex]\frac { 200\times 7\times 7\times 7 }{ 7\times 7\times 10\times 10\times 10 } [/latex]
⇒ h = 1.4 cm (1)
OR
Let the radius ofhemisphere = r
Now, volume ofhemisphere = [latex s=2]\frac { 2 }{ 3 } [/latex] πr3 (1/2)
Surface area ofhemisphere = 3πr2 (1/2)
A.T.Q, volume ofhemisphere = surface area ofhemisphere (1/2)
⇒ [latex s=2]\frac { 2 }{ 3 } [/latex] πr3 = 3πr2 ⇒r= [latex s=2]\frac { 9 }{ 2 } [/latex]units (1)

Solution 17.
Suppose the co-ordinates of vertex D are (x, y), then
Mid-point of AC = Mid-point of BD (For parallelogram ABCD) (1/2)
CBSE Sample Papers for Class 10 Maths Paper 4 img 15

Solution 18.
CBSE Sample Papers for Class 10 Maths Paper 4 img 16
OR
CBSE Sample Papers for Class 10 Maths Paper 4 img 17

Solution 19.
CBSE Sample Papers for Class 10 Maths Paper 4 img 18
CBSE Sample Papers for Class 10 Maths Paper 4 img 19
OR
CBSE Sample Papers for Class 10 Maths Paper 4 img 20

Solution 20.
Let OP meet the circle at Q. Join AQ. As OP is equal to the diameter of the circle and OQ is radius, so OQ = QP i.e. Q is mid-point of OP. Since PA is tangent to the circle at A and OA is its radius, OA ⊥L AP i.e. ∠OAP = 90°.
In right triangle OAP, Q is mid-point of hypotenuse,
CBSE Sample Papers for Class 10 Maths Paper 4 img 21
∴AQ = OQ = QP
Also OA = OQ (radii of same circle)
⇒ OA=OQ = AQ ⇒ ∆OAQ is equilateral
⇒ ∠AOQ = 60° ⇒ ∠AOP = 60°. (1)
In ∆OAP, ∠OPA + ∠AOP + ∠OAP =180°
⇒ ∠OPA+60°+ 90° = 180°
⇒ ∠OPA= 30°
⇒ ∠APB = 60° (∴OP is bisector of ZAPB) (1)
Also PA = PB ⇒ ∠PAB = ∠PBA.
In ∆PAB, ∠PAB + ∠PBA+ ∠APB = 180°
⇒ 2 ∠PAB+ 60° =180°
⇒ ∠PAB=60°
⇒ Triangle ABP is equilateral. (1)

Solution 21.
∴Letx2 = u, y2 = v
⇒ 2u + 3v=35 and [latex]\frac { u }{ 2 } +\frac { v }{ 3 } [/latex] = 5 (1/2)
⇒ 2u + 3v = 35 …(i)
⇒ 3u + 2v = 30 …(ii) (1/2)
Multiply (i) by 3 and (ii) by 2 and subtracting (ii) from (i), we have
⇒ 6u – 6u + 9v – 4v= 105 – 60
⇒ 5v = 45 ⇒v = 9
Substituting v = 9 in (1), we get 2u + 2 7 = 3 5
2u = 8 => u = 4 ⇒x2 = 4, y2 = 9
∴ x = ± 2,y = ± 3 is the required solution. (1)

Solution 22.
CBSE Sample Papers for Class 10 Maths Paper 4 img 22
OR
CBSE Sample Papers for Class 10 Maths Paper 4 img 23

Section-D

Solution 23.
Suppose AB be a tower and there are two points C and D at the distances of 4 m and 16 m from the foot of the tower respectively
CBSE Sample Papers for Class 10 Maths Paper 4 img 24
Since, the angles of elevation from C and D of the top of the tower are complementary.
So,e,+e2 = 90° …(i) (1/2)
Let the height of the tower be h.
Then, from equation (i), tan (θ1 + θ2) = tan 90° (1/2)
⇒ [latex]\frac { \tan { { \theta }_{ 1 } } +\tan { { \theta }_{ 2 } } }{ 1-\tan { { \theta }_{ 1 } } \tan { { \theta }_{ 2 } } } =\frac { 1 }{ 0 } [/latex]
⇒ 1- tan θ1 tan θ2 =0 ⇒ tan θ1 tan θ2 = 1 (1)
⇒ [latex]\frac { h }{ 4 } \times \frac { h }{ 16 } [/latex] = 1 ⇒ h<2 = 64 ⇒ h = 8 m (∵Height cannot be negative) (1)
Hence, the height of the tower is 8 m.
OR
CBSE Sample Papers for Class 10 Maths Paper 4 img 25

Solution 24.
CBSE Sample Papers for Class 10 Maths Paper 4 img 26

Solution 25.
CBSE Sample Papers for Class 10 Maths Paper 4 img 27
CBSE Sample Papers for Class 10 Maths Paper 4 img 28
OR
CBSE Sample Papers for Class 10 Maths Paper 4 img 29
CBSE Sample Papers for Class 10 Maths Paper 4 img 30

Solution 26.
CBSE Sample Papers for Class 10 Maths Paper 4 img 31

Solution 27.
CBSE Sample Papers for Class 10 Maths Paper 4 img 32
(i) Draw a line segment 4 cm.
(ii) Take a point P outside the circle and draw a secant PAB, intersecting the circle at A and B.
(iii) Produce AP to C such that AP = CP.
(iv) Draw a semi-circle with CB as diameter.
(v) Draw PD⊥L CB, intersecting the semi-circle at D.
(vi) With P as centre and PD as radius draw arcs to intersect the given circle at T and T’.
(vii) Join PTand PT’. Then, PTand PT’ are the required tangents.

Solution 28.
First convert the given frequency distribution table to More Than Type frequency distribution table.
CBSE Sample Papers for Class 10 Maths Paper 4 img 33
Now mark the lower limits along X-axis and cumulative frequencies along F-axis, and plot the points (400,230), (450,210), (500,175), (550,135), (600,103), (650,79), (700,52), (750,34). Join the points listed above by smooth free hand curve to obtain the more than type ogive.
CBSE Sample Papers for Class 10 Maths Paper 4 img 34

Solution 29.
The equation (1 + m2) x2 + 2mcx + c2 – a2 = 0
For coincident (Repeated roots) D = 0 (1/2)
⇒ (2mc)2 -4(1+ m2) (c2 – a2) = 0 (1/2)
⇒ 4m2c2 – 4(c2 – a2 + m2c2 – m2a2) = 0 (1/2)
⇒ m2c2 – c2 + a2 – m2c2 + m2a2 = 0 (1/2)
⇒ m2a2 – c2 + a2 = 0 (1/2)
⇒ m2a2 + a2 = c2 ⇒ a2 (1 +m2) = c2 (1)
⇒ c = ±a[latex]\sqrt { 1+{ m }^{ 2 } } [/latex] Hence proved. (1/2)
OR
Put x = 4, we get 3(4)2 – 2m (4) + 2n = 0 (1/2)
⇒ 48-8m + 2n = 0 =>2n-8m = -48 ⇒ n-4m = -24 ….(i) (1)
Put x = -5, we get 3 (-5)2 – 2m (-5) + 2n = 0 (1/2)
⇒ 75 + 10m + 2n = 0 ⇒2n + 10m=-75 ….(ii) (1)
Solving (i) and (ii) we get, m = – [latex s=2]\frac { 3 }{ 2 } [/latex]and n = -30 (1)

Solution 30.
Let (a-3d),(a-d),(a + d),(a + 3d) are the four numbers
∴ Sum = 50
⇒ (a-3d) + (a-d) + (a + d) + (a-3d) = 50
⇒ a= [latex s=2]\frac { 25 }{ 2 } [/latex] (1)
also, (a + 3d) = 4(a-3d) (1)
⇒ 5 d=a
⇒ d = [latex s=2]\frac {5 }{ 2 } [/latex] (1/2)
5,10, 15 and 20 are the required numbers ofA.P. (1)

We hope the CBSE Sample Papers for Class 10 Maths paper 4 help you. If you have any query regarding CBSE Sample Papers for Class 10 Maths paper 4, drop a comment below and we will get back to you at the earliest.

Balaji Class 9 Maths Solutions Chapter 6 Remainder Theorem and Factor Theorem Ex 6.2

Balaji Class 9 Maths Solutions Chapter 6 Remainder Theorem and Factor Theorem Ex 6.2 शेषफल प्रमेय तथा गुणनखण्ड प्रमेय

Ex 6.2 Remainder Theorem and Factor Theorem अतिलघु उत्तरीय प्रश्न (Very Short Answer Type Questions)

प्रश्न 1.
बहुपद (25x2 – 1) + (1 – 5x) का एक गुणनखण्ड ज्ञात कीजिए।
हलः
(25x2 – 1) + (1 – 5x)2 = 25x2 – 1 + 1 + 25x2 – 10x = 50x2 – 10x = 10x(5x – 1)
x बहुपद का एक गुणनखण्ड है।

Balaji Class 9 Maths Solutions Chapter 6 Remainder Theorem and Factor Theorem Ex 6.2

प्रश्न 2.
बहुपद x3 – 6x2 + 11x – 6 के गुणनखण्ड ज्ञात कीजिए।
हल:
x3 – 6x2 + 11x – 6 में x = 1 रखने पर
शेषफल = (1)3 – 6(1)2 + 11(1) – 6 = 1 – 6 + 11 – 6 = 0
अतः (x – 1) इसका एक गुणनखण्ड है।
इसी प्रकार (x – 2) व (x – 3) भी इसके गुणनखण्ड हैं।

प्रश्न 3.
यदि (x + 1) बहुपद f (x) = 2x2 + kx, का एक गुणनखण्ड है तो k का मान ज्ञात कीजिए। (NCERT Exemplar)
हलः
यदि (x + 1), f(x) = 2x2 + kx का एक गुणनखण्ड है तो x + 1 = 0
∴ x = 0 – 1 = -1 रखने पर f(-1) = 0
f(-1) = 2(-1)2 + k(-1)
0 = 2 – k ⇒ k = 2

प्रश्न 4.
यदि (x – 2) बहुपद 4x3 + 3x2 – 4x + k का एक गुणनखण्ड है तो k का मान ज्ञात कीजिए।
हलः
यदि (x – 2), f(x) = 4x3 + 3x2 – 4x + k का एक गुणनखण्ड है तो x – 2 = 0 या x = 2 रखने पर
∴ f(2) = 0
4(2)3 + 3(2)2 – 4(2) + k = 0
32 + 12 – 8 + k=0
36 + k = 0 ⇒ k = -36

Ex 6.2 Remainder Theorem and Factor Theorem लघु उत्तरीय प्रश्न – I (Short Answer Type Questions – I)

प्रश्न 5.
a का मान ज्ञात कीजिए यदि (x + 1) बहुपद 2x3 – ax2 -(2a – 3)x + 2 का एक गुणनखण्ड है।
हलः
यदि (x + 1) बहुपद 2x3– ax2 – (2a – 3)x + 2 का एक गुणनखण्ड है तो
x + 1 = 0 या x = -1 रखने पर।
शेषफल = 0
2(-1)3 – a(-1)2 – (2a – 3)(-1) + 2 = 0
-2 – a + 2a – 3 + 2 = 0
a – 3 = 0 ⇒ a = 3

Balaji Class 9 Maths Solutions Chapter 6 Remainder Theorem and Factor Theorem Ex 6.2

प्रश्न 6.
k का मान ज्ञात कीजिए, यदि (x – 3) बहुपद k2x2 – kx – 2 का गुणनखण्ड है।
हलः
यदि (x – 3), k2x2 – kx – 2 का एक गुणनखण्ड है तो x – 3 = 0 या x = 3 रखने पर
शेषफल = 0
k2. (3)2 – k. 3 – 2 = 0
9k2 – 3k – 2 = 0
9k2 –(6 – 3)k – 2 = 0
9k2 – 6k + 3k – 2 = 0
3k(3k – 2) + 1(k – 2) = 0
(3k – 2)(3k + 1) = 0
यदि 3k – 2 = 0 ∴ k = [latex]\frac{2}{3}[/latex]
यदि 3k + 1 = 0 ∴ k = [latex]\frac{1}{3}[/latex]

Ex 6.2 Remainder Theorem and Factor Theorem लघु उत्तरीय प्रश्न – II (Short Answer Type Questions – II)

Remainder Theorem Calculator is a free online tool that displays the quotient and remainder of division for the given polynomial expressions.

प्रश्न 7.
यदि (x – 1) बहुपद x4 – 3x3 + bx2 + 8x – 4 का एक गुणनखण्ड है, तो b का मान ज्ञात कीजिए।
हलः
यदि (x – 1), x4 – 3x3 + bx2 + 8x – 4 का एक गुणनखण्ड है तो x – 1 = 0 या x = 1 रखने पर
शेषफल = 0
(1)4 – 3(1)3 + b(1)2 + 8(1) – 4 = 0
1 – 3 + b + 8 – 4 = 0
b + 2 = 0
b = 0 – 2 ⇒ b = -2

प्रश्न 8.
सिद्ध कीजिए कि (x – 3) व (x + 4) बहुपद x2 + x – 12 के गुणनखण्ड हैं।
हलः
बहुपद x2 + x – 12 के गुणनखण्ड (x – 3) तथा (x + 4) होंगे।
यदि x – 3 = 0 या x = 3 रखने पर शेषफल = (3)2 + 3 – 12 = 9 + 3 – 12 = 0
यदि x + 4 = 0 या x = -4 रखने पर शेषफल = (-4)2 – 4 – 12 = 16 – 16 = 0
∴ (x – 3) व (x + 4) बहुपद x2 + x – 12 के गुणनखण्ड हैं।

Balaji Class 9 Maths Solutions Chapter 6 Remainder Theorem and Factor Theorem Ex 6.2

प्रश्न 9.
गुणनखण्ड प्रमेय का प्रयोग करके जाँचिये कि g(x), बहुपद f(x) का गुणनखण्ड है या नहीं।
(i) f(x) = x3 – 6x2 -19x + 84 तथा g(x) = x – 7
(ii) f(x) = x3 – 3x2 +4x – 4 तथा g(x) = x – 2
(iii) f(x) = 3x4 + 17x3 + 9x2 – 7x – 10 तथा g(x) = x + 5
(iv) f(x) = 2x3 + 4x + 6 तथा g(x) = x + 1
हल:
(i) f(x) = x3 – 6x2 – 19x + 84 तथा g(x) = x – 7
g(x) = x – 7 = 0 या x = 7 का मान f(x) में रखने पर
f(7) = (7)3 – 6(7)2 – 19(7) + 84
= 343 – 294 – 133 + 84
=427 – 427 = 0
अतः g(x), f(x) का एक गुणनखण्ड है।

(ii) f(x) = x3 – 3x2 + 4x – 4 तथा g(x) = x – 2
g(x) = 0 या x – 2 = 0 या x = 2 रखने पर
f(2) = (2)3 – 3(2)2 + 4(2) – 4
= 8 – 12 + 8 – 4
= 16 – 16 = 0
अतः g(x), f(x) का एक गुणनखण्ड है।

(iii) f(x) = 3x4 + 17x3 + 9x2 – 7x – 10 तथा g(x) = x + 5
g(x) = 0 या x + 5 = 0 या x = -5 रखने पर
f(-5) = 3(-5)4 + 17(-5)3 + 9(-5)2 – 7(-5) – 10
= 3 × 625 – 17 × 125 + 9 × 25 + 35 – 10
= 1875 – 2125 + 225 + 25
= 2125 – 2125 = 0
अतः g(x), f(x) का एक गुणनखण्ड है।

(iv) f(x) = 2x3 + 4x + 6 तथा g(x) = x +1
g(x) = 0 या x + 1 = 0 या x = -1 रखने पर
f(-1) = 2(-1)3 + 4(-1) + 6
= -2 – 4 + 6 = 0
अतः g(x), f (x) का एक गुणनखण्ड है।

Balaji Class 9 Maths Solutions Chapter 6 Remainder Theorem and Factor Theorem Ex 6.2

प्रश्न 10.
सिद्ध कीजिए कि 2x4 – 6x3 + 3x2 + 3x – 2; x2 – 3x + 2 से पूर्णतया विभाजित है।
हलः
2x4 – 6x3 + 3x2 + 3x – 2 को x2 – 3x + 2 से भाग करने पर
∵ x2 – 3x + 2 = (x – 2)(x -1) यदि x – 2 = 0 या x = 2 रखने पर
शेषफल = 2(2)4 – 6(2)3 + 3(2)2 + 3(2) – 2
= 32 – 48 + 12 + 6 – 2
= 50 – 50 = 0
∴ (x – 2) से पूर्णतया विभाजित है।
यदि x – 1 = 0 या x = 1 रखने पर
शेषफल = 2(1)4 – 6(1)3 + 3(1)2 + 3(1) – 2
= 2 – 6 + 3 + 3 – 2
= 8 – 8 = 0
∴ (x – 1) से पूर्णतया विभाजित है।

प्रश्न 11.
सिद्ध कीजिए कि (x – 1), बहुपद x10 – 1 तथा x11 – 1 का गुणनखण्ड है।
हलः
(x – 1), बहुपद x10 – 1 का गुणनखण्ड होगा। यदि x – 1 = 0 या x = 1 रखने पर
x10 – 1 का शेषफल = (1)10 – 1 = 1 – 1 = 0
∴ (x – 1), x10 – 1 का गुणनखण्ड है।
x11 – 1 का शेषफल = (1)11 – 1 = 1 – 1 = 0
∴ (x – 1), x11 – 1 का गुणनखण्ड है।

प्रश्न 12.
बहुपद 4x3 + 16x2 – x + 5 से क्या घटाया जाये कि ऐसा बहुपद प्राप्त हो जो (x + 5) से पूर्णतया विभाजित हो?
हलः
यदि (x + 5) से 4x3 + 16x2 – x + 5 को पूर्णतया विभाजित किया जाए तो
x + 5 = 0 या x = 0 – 5 = -5 रखने पर
शेषफल = 4(-5)3 + 16(-5)2 – (-5) + 5
= 4(-125) + 16(25) + 5 + 5
= -500 + 400 + 10 = -90

Balaji Class 9 Maths Solutions Chapter 6 Remainder Theorem and Factor Theorem Ex 6.2

प्रश्न 13.
गुणनखण्ड प्रमेय के प्रयोग से k का मान ज्ञात कीजिए यदि (x + 2), बहुपद (x + 1)7 + (2x + k)3 का एक गुणनखण्ड है।
हलः
यदि (x + 2), बहुपद (x + 1)7 + (2x + k)3 का एक गुणनखण्ड है तो
x + 2 = 0 या x = 0 – 2 = -2 रखने पर
शेषफल = 0
(-2 + 1)7 + (2 × -2 + k)3 = 0
(-1)7 + (-4 + k)3 = 0
-1 + (-4 + k)3 = 0
(-4 + k)3 = 1
(-4 + k)3 = (1)3
-4 + k = 1
k = 1 + 4 = 5

प्रश्न 14.
m व n के मान ज्ञात कीजिए यदि (x – 1) तथा (x + 2), बहुपद 2x3 + mx2 + nx – 14 के गुणनखण्ड हैं।
हलः
यदि (x – 1), बहुपद 2x3 + mx2 + nx – 14 का एक गुणनखण्ड है तो x – 1 = 0 या x = 1 रखने पर
2(1)3 + m(1)2 + n(1) – 14 = 0
2 + m + n – 14 = 0
m + n = 12 ………….(1)
यदि (x + 2), बहुपद 2x3 + mx2 + nx – 14 का एक गुणनखण्ड है तो x + 2 = 0 या x = -2 रखने पर
2(-2) + m(-2)2 + n(-2) – 14 = 0
-16 + 4m – 2n – 14 = 0
Balaji Class 9 Maths Solutions Chapter 6 Remainder Theorem and Factor Theorem Ex 6.2
समीकरण (1) में m का मान रखने पर 9 + n = 12
n = 12 – 9 = 3

Balaji Class 9 Maths Solutions Chapter 6 Remainder Theorem and Factor Theorem Ex 6.2

प्रश्न 15.
α व β के मान ज्ञात कीजिए यदि (x + 1) तथा (x + 2), बहुपद x3 + 3x2 – 2αx + β के गुणनखण्ड हैं।
हलः
यदि (x + 1), x3 + 3x2 – 2αx + β का एक गुणनखण्ड है तो
x + 1 = 0 या x = 0 – 1 = -1 रखने पर
शेषफल = (-1)3 + 3(-1)2 – 2α . (-1) + β = 0
-1 + 3 + 2α + β = 0
2 + 2α + β = 0
2α + β = -2 ………………….(1)
यदि (x + 2), x3 + 3x2 – 2α.x + β का एक गुणनखण्ड है। तो .
x + 2 = 0 या x = 0 – 2 = -2 रखने पर
शेषफल ⇒ (-2)3 + 3(-2)2 – 2α (-2) + β = 0
Balaji Class 9 Maths Solutions Chapter 6 Remainder Theorem and Factor Theorem Ex 6.2

प्रश्न 16.
गणनखण्ड प्रमेय का प्रयोग करके सिद्ध कीजिए कि a + b, b + c, c + a बहुपद (a + b + c)3 – (a3 + b3 + c3) के गुणनखण्ड हैं।
हल:
∵ a + b एक गुणनखण्ड होगा (a + b + c)3 – (a3 + b3 + c3) का
यदि a + b = 0 या a = – b रखने पर,
शेषफल = (-b + b + c)3 – (-b3 + b3 + c3) = c3 – c3 = 0
∴ (a + b) इसका एक गुणनखण्ड है। .
∵ b + c एक गुणनखण्ड है (a + b + c)3 – (a3 + b3 + c3) का
यदि b + c = 0 या b = -c रखने पर,
शेषफल = (a – c + c)3 – (a3 – c3 + c3) = a3 – a3 = 0
∴ (b + c) इसका एक गुणनखण्ड है।
∵ c + a एक गुणनखण्ड है (a + b + c)3 – (a3 + b3 + c3) का।
यदि c + a = 0 या c = -a रखने पर ,
शेषफल = (a + b – a)3 – (a3 + b3 – a) = b3 – b3 = 0
∴ c + a इसका एक गुणनखण्ड है।

Balaji Publications Mathematics Class 9 Solutions

UP Board Solutions for Class 9 Maths Chapter 2 Polynomials

UP Board Solutions for Class 9 Maths Chapter 2 Polynomials (बहुपद)

These Solutions are part of UP Board Solutions for Class 9 Maths. Here we have given UP Board Solutions for Class 9 Maths Chapter 2 Polynomials (बहुपद).

प्रश्नावली 2.1

प्रश्न 1.
निम्नलिखित व्यंजकों में कौन-कौन एक चर में बहुपद हैं और कौन-कौन नहीं हैं? कारण के साथ अपने उत्तर दीजिए :
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-7

प्रश्न 2.
निम्नलिखित में से प्रत्येक में x² का गुणांक लिखिए :
(i) 2 + x² + x
(ii) 2 – x² + x3
(iii) [latex]\frac { \pi }{ 2 }[/latex] x² + x
(iv) √2 x – 1
हल :
(i) 2 + x² + x में x² का गुणांक = 1
(ii) 2 – x² + x3 में x² का गुणांक = -1
(iii) [latex]\frac { \pi }{ 2 }[/latex] x² + x में x² का गुणांक = [latex]\frac { \pi }{ 2 }[/latex]
(iv) √2 x – 1 अर्थात 0.x2 + √2 x – 1 में x² का गुणांक = 0

UP Board Solutions

प्रश्न 3.
35 घात के द्विपद का और 100 घात के एकपदी का एक-एक उदाहरण दीजिए।
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-8

प्रश्न 4.
निम्नलिखित बहुपदों में से प्रत्येक बहुपद की घात लिखिए :
(i) 5x3 + 4x² + 7x
(ii) 4 – y²
(iii) 5t – √7
(iv) 3
हल :
(i) 5x3 + 4x² + 7x में चर x की अधिकतम घात = 3
दिए हुए बहुपद की घात= 3
(ii) 4 – y² में चर y की अधिकतम घात = 2
दिए हुए बहुपद की घात = 2
(iii) 5t – √7 में चर है की अधिकतम घात = 1
दिए हुए बहुपद की घात = 1
(iv) 3 एक अचर पद है अर्थात 3.x0
दिए हुए बहुपद की घात = 0

Synthetic Division Calculator, Calculator will divide polynomial by binomial using synthetic divsion.

प्रश्न 5.
बताइए कि निम्नलिखित बहुपदों में कौन-कौन बहुपद रैखिक है, कौन-कौन द्विघाती हैं और कौन-कौन त्रिघाती हैं :
(i) x² + x
(ii) x – x3
(iii) y + y² + 4
(iv) 1 + x
(v) 3t
(vi) r²
(vii) 7x3
हल :
(i) बहुपद x² + x में चर x की अधिकतम घात = 2
यह बहुपद द्विघाती है।
(ii) बहुपद x – x3 में चर x की अधिकतम घात = 3
यह बहुपद त्रिघाती है।
(iii) बहुपद y + y² + 4 में चर y की अधिकतम घात = 2
यह बहुपद द्विघाती है।
(iv) बहुपद 1 + x में चर x की अधिकतम घात 1 है।
यह बहुपद रैखिक है।
(v) बहुपद 3t में चर है की अधिकतम घात 1 है।
यह बहुपद रैखिक है।
(vi) बहुपद r² में चर r की अधिकतम घात 2 है।
यह बहुपद द्विघाती है।
(vii) बहुपद 7x3 में चर x की अधिकतम घात 3 है।
यह बहुपद त्रिघाती है।

प्रटनावली 2.2

प्रश्न 1.
निम्नलिखित पर बहुपद 5x – 4x² + 3 के मान ज्ञात कीजिए।
(i) x = 0
(ii) x = – 1
(iii) x = 2
हल :
माना बहुपद p (x) = 5 – 4x² + 3
(i) x = 0 पर बहुपद p (x) का मान
p(0)= 5 (0) – 4 (0)² + 3 = 3
(ii) x = -1 पर बहुपद p (x) का मान
p(-1) = 5 (-1) – 4 (-1)² + 3 = – 5 – 4 + 3 = -6
(iii) x = 2 पर बहुपद p (x) का मान
p(2) = 5 (2) – 4 (2)2 + 3 = 10 – 16 + 3 = -3

UP Board Solutions

प्रश्न 2.
निम्नलिखित बहुपदों में से प्रत्येक बहुपद के लिए p (0), p (1) और p (2) ज्ञात कीजिए :
(i) p(y) = y² – y + 1
(ii) p(t) = 2 + t + 2t² – t3
(iii) p(x) = x3
(iv) p(x) = (x – 1)(x + 1)
हल :
(i) p(y) = y² – y + 1
p (0) = 0² – 0 + 1 = 0 – 0 + 1 = 1
p (1) = 1² – 1 + 1 = 1 – 1 + 1 = 1
p(2) = 2² – 2 + 1 = 4 – 2 + 1 = 3
(ii) p(t) = 2 + t + 2t² – t3
p(0) = 2 + 0 + 2 (0)² – (0)3 = 2
p (1) = 2 + 1 + 2 (1)² – (1)3 = 2 + 1 + 2 – 1 = 4
p (2) = 2 + 2 + 2 (2)² – (2)3 = 2 + 2 + 8 – 8 = 4
(iii) p (x) = x3
p(0) = (0)3 = 0
p (1) = (1)3 = 1
p (2) = (2)3 = 8
(iv) p (x) = (x – 1) (x + 1)
p(0) = (0 – 1) (0 + 1) = (-1) (1) = -1
p (1) = (1 – 1) (1 + 1) = (0) (2) = 0
p (3) = (2 – 1) (2 + 1) = (1) (3) = 3

प्रश्न 3.
सत्यापित कीजिए कि दिखाए गए मान निम्नलिखित स्थितियों में संगत बहुपद के शून्यक हैं :
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-9
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-10
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-1

प्रश्न 4.
निम्नलिखित स्थितियों में से प्रत्येक स्थिति में बहुपद को शून्यक ज्ञात कीजिए :
(i) p(x) = x + 5
(ii) p(x) = x – 5
(iii) p(x) = 2x + 5
(iv) p(x) = 3x – 2
(v) p(x) = 3x
(vi) p(x) = ax; a ≠ 0
(vii) p (x) = cx + d; c ≠ 0, c, d वास्तविक संख्याएँ हैं।
हल :
(i) बहुपद p (x) = x + 5 का शून्यक ज्ञात करने के लिए इसे शून्य के बराबर रखते हैं।
p(x) = 0
⇒ x + 5 = 0
⇒ x = – 5
p(3) को शून्यक = – 5
(ii) बहुपद p (x) = x – 5 को शून्यक ज्ञात करने के लिए इसे शून्य के बराबर रखते हैं।
p (x) = 0
⇒ x – 5 = 0
⇒ x = 5
p(x) का शून्यक = 5
(iii) बहुपद p (x) = 2x + 5 का शून्यक ज्ञात करने के लिए इसे शून्य के बराबर रखते हैं।
p(3) = 0
⇒ 2x + 5 = 0
⇒ 2x = – 5
⇒ x = [latex]\frac { -5 }{ 2 }[/latex]
p (x) का शून्यके = [latex]\frac { -5 }{ 2 }[/latex]
(iv) बहुपद p (x) = 3x – 2 का शून्यक ज्ञात करने के लिए इसे शून्य के बराबर रखते हैं।
p (5) = 0
⇒ 3x – 2 = 0
⇒ 3x = 2
⇒ x = [latex]\frac { 2 }{ 3 }[/latex]
p (x) का शून्यक = [latex]\frac { 2 }{ 3 }[/latex]
(v) बहुपद p (x) = 3x का शून्यक ज्ञात करने के लिए इसे शून्य के बराबर रखते हैं।
p (x) = 0
⇒ 3x = 0
⇒ x = 0
p (x) का शून्यक = 0
(vi) बहुपद p(x) = ax; a ≠ 0 का शून्यक ज्ञात करने के लिए इसे शून्य के बराबर रखते हैं।
p(x) = 0
⇒ ax = 0
⇒ x = 0 (a ≠ 0)
p(x) का शून्यक = 0
(vii) बहुपद p (x) = cx + d, c ≠ 0 का शून्यक ज्ञात करने के लिए इसे शून्य के बराबर रखते हैं।
p(x) = 0
cx + d = 0
cx = -d
x = [latex]\frac { -d }{ c }[/latex] (c ≠ 0)
p (x) का शून्यक = [latex]\frac { -d }{ c }[/latex]

UP Board Solutions

प्रश्नावली 2.3

प्रश्न 1.
x3 + 3x² + 3x + 1 को निम्नलिखित से भाग देने पर शेषफल ज्ञात कीजिए :
(i) x + 1
(ii) x – [latex]\frac { 1 }{ 2 }[/latex]
(iii) x
(iv) x + π
(v) 5 + 2x
हल :
माना p (x) = x3 + 3x² + 3x + 1
(i) माना x + 1 = 0 ⇒ x = -1
p (x) को + 1 से भाग देने पर शेषफल
p(- 1) = (-1)3 + 3(-1)² + 3(-1) + 1 = -1 + 3 – 3 + 1 = 0
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-11
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-12

प्रश्न 2.
x3 – ax² + 6x – a को x – a से भाग देने पर शेषफल ज्ञात कीजिए।
हल :
माना p (x) = x3 – ax² + 6x – a तथा x – a = 0
p (x) को x – a से भाग देने पर शेषफल = (a)3 – a(a)² + 6(a) – a = a3 – a3 + 6a – a = 5a

प्रश्न 3.
जाँच कीजिए कि 7 + 3x, 3x3 + 7x का एक गुणनखण्ड है या नहीं।
हल :
माना p (x) = 3x + 7x
यदि 7 + 3x, p (x) का एक गुणनखण्ड है तो p (x) को 7 + 3x से भाग देने पर शेषफल शून्य होना चाहिए।
माना 7 + 3x = 0 ⇒ 3x = – 7 ⇒ x = [latex]\frac { -3 }{ 7 }[/latex]
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-13

प्रश्नावली 2.4

प्रश्न 1.
बताइए कि निम्नलिखित बहुपदों में से किस बहुपद का एक गुणनखण्ड (x + 1) है।
(i) x3 + x2 + x + 1
(ii) x4 + x3 + x2 + x + 1
(iii) x4 + 3x3 + 3x2 + x + 1
(iv) x3 – x2 – (2 + √2) x + √2
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-14

UP Board Solutions

प्रश्न 2.
गुणनखण्ड प्रमेय लागू करके बताइए कि निम्नलिखित स्थितियों में से प्रत्येक स्थिति में g (x), p (x) का एक गुणनखण्ड है या नहीं :
(i) p(x) = 2x3 + x2 – 2x – 1, g (x) = x + 1
(ii) p(x) = x3 + 3x2 + 3x + 1, g (3) = x + 2
(iii) p(x) = x3 – 4x2 + x + 6, g (x) = x – 3
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-15

प्रश्न 3.
k का मान ज्ञात कीजिए जबकि निम्नलिखित स्थितियों में से प्रत्येक स्थिति में (x – 1), p (x) का एक गुणनखण्ड हो :
(i) p(3) = x2 + x + k
(ii) p(x) = 2x2 + kx + √2
(iii) p(x) = kx2 – √2 x + 1
(iv) p(x) = kx2 – 3x + k
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-16
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-2

प्रश्न 4.
गुणनखण्ड ज्ञात कीजिए :
(i) 12x2 – 7x + 1
(ii) 2x2 + 7x + 3
(iii) 6x2 + 5x – 6
(iv) 3x2 – x – 4
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-17

प्रश्न 5.
गुणनखण्ड ज्ञात कीजिए :
(i) x3 – 2x2 – x + 2
(ii) x3 – 3x2 – 9x – 5
(iii) x3 + 13x2 + 32x + 20
(iv) 2y3 + y2 – 2y – 1
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-18
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-19

प्रश्नावली 2.5

प्रश्न 1.
उपयुक्त सर्वसमिकाओं को प्रयोग करके निम्नलिखित गुणनफल ज्ञात कीजिए :
(i) (x + 4) (x + 10)
(ii) (x + 8) (x – 10)
(iii) (3x + 4) (3x – 5)
(iv) (y2 + [latex]\frac { 3 }{ 2 }[/latex]) (y2 – [latex]\frac { 3 }{ 2 }[/latex])
(v) (3 – 2x) (3 + 2x)
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-20
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-3

UP Board Solutions

प्रश्न 2.
सीधे गुणा किए बिना निम्नलिखित गुणनफलों के मान ज्ञात कीजिए :
(i) 103 x 107
(ii) 95 x 96
(iii) 104 x 96
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-21

प्रश्न 3.
उपयुक्त सर्वसमिकाएँ प्रयोग करके निम्नलिखित का गुणनखण्डन कीजिए :
(i) 9x2 + 6xy + y2
(ii) 4y2 – 4y + 1
(iii) x2 – [latex]\frac { { y }^{ 2 } }{ 100 }[/latex]
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-22

प्रश्न 4.
उपयुक्त सर्वसमिकाओं का प्रयोग करके निम्नलिखित में से प्रत्येक का प्रसार कीजिए :
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-4
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-23

प्रश्न 5.
गुणनखण्डन कीजिए :
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-24

प्रश्न 6.
निम्नलिखित घनों को प्रसारित रूप में लिखिए :
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-25
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-26

प्रश्न 7.
उपयुक्त सर्वसमिकाएँ प्रयोग करके निम्नलिखित के मान ज्ञात कीजिए :
(i) (99)3
(ii) (102)3
(iii) (998)3
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-27

UP Board Solutions

प्रश्न 8.
निम्नलिखित में से प्रत्येक का गुणनखण्डन कीजिए।
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-5
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-28

प्रश्न 9.
सत्यापित कीजिए :
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-29

प्रश्न 10.
निम्नलिखित में से प्रत्येक का गुणनखण्डन कीजिए
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-30
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-6

प्रश्न 11.
गुणनखण्ड कीजिए : 27x3 + y3 + z3 – 9xyz
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-31

प्रश्न 12.
सत्यापित कीजिए :
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-32

प्रश्न 13.
यदि x + y + z = 0 हो तो दिखाइए कि x3 + y3 + z3 = 3xyz
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-33

प्रश्न 14.
घनों का परिकलन किए बिना निम्नलिखित में से प्रत्येक का मान ज्ञात कीजिए :
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-34

UP Board Solutions

प्रश्न 15.
नीचे दिए गए आयतों, जिनमें उनके क्षेत्रफल दिए गए हैं, में से प्रत्येक की लम्बाई और चौड़ाई के लिए सम्भव व्यंजक दीजिए।
(i) क्षेत्रफल : 25a2 – 35a + 12
(ii) क्षेत्रफल : 35y2 + 13y – 12
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-35

प्रश्न 16.
घनाभों (Cuboids), जिनके आयतन नीचे दिए गए हैं, की विमाओं के लिए सम्भव व्यंजक क्या हैं :
(i) आयतन : 3x2 – 12x
(ii) आयतन : 12ky2 + 8ky – 20k
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-36

We hope the UP Board Solutions for Class 9 Maths Chapter 2 Polynomials (बहुपद) help you. If you have any query regarding UP Board Solutions for Class 9 Maths Chapter 2 Polynomials (बहुपद), drop a comment below and we will get back to you at the earliest.

Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 परिमेयीकरण

Ex 3.2 Rationalisation अतिलघु उत्तरीय प्रश्न (Very Short Answer Type Questions)

निम्न के मान ज्ञात कीजिए- (प्रश्न 1 – 6)

प्रश्न 1.
[latex]\frac{1}{\sqrt{4}-\sqrt{3}}[/latex]
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 2.
[latex]\frac{1}{3+2 \sqrt{2}}[/latex]
हलः
हर का परिमेयीकरण करने पर,
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 3.
[latex]\frac{1}{\sqrt{7}-\sqrt{6}}[/latex] [NCERT]
हलः
हर का परिमेयीकरण करने पर,
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 4.
[latex]\frac{1}{\sqrt{5}+\sqrt{2}}[/latex] [NCERT]
हलः
हर का परिमेयीकरण करने पर,
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

In Algebra, 1/32 to decimals are one of the types of numbers, which has a whole number and the fractional part separated by a decimal point.

प्रश्न 5.
[latex]\frac{1}{\sqrt{7}-2}[/latex] [NCERT]
हलः
हर का परिमेयीकरण करने पर,
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 6.
[latex]\frac{1}{\sqrt{7}}[/latex] [NCERT]
हलः
हर का परिमेयीकरण करने पर,
[latex]\frac{1}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}}=\frac{\sqrt{7}}{7}[/latex]

Ex 3.2 Rationalisation लघु उत्तरीय प्रश्न (Short Answer Type Questions)

निम्न संख्याओं में हर का परिमेयीकरण कीजिए- (प्रश्न 7 – 10)

Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 7.
[latex]\frac{1}{8+5 \sqrt{2}}[/latex]
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 8.
[latex]\frac{6}{\sqrt{5}+\sqrt{2}}[/latex]
हलः
हर के संयुग्मी से गुणा व भाग करने पर,
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

Click here to get an answer to your question ✍️ How do you write 9/20 as a decimal?

प्रश्न 9.
[latex]\frac{2}{\sqrt{3}-\sqrt{5}}[/latex]
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 10.
[latex]\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}[/latex] [NCERT Exemplar]
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 11.
यदि x = 2 + [latex]\sqrt{15}[/latex] तब x + [latex]\frac{1}{x}[/latex] का मान ज्ञात कीजिए।
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 12.
यदि x = 2 + [latex]\sqrt{3}[/latex], तब x2 + [latex]\frac{1}{x^{2}}[/latex] का मान ज्ञात कीजिए।
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 13.
यदि x = 7 + [latex]4 \sqrt{3}[/latex], तब x + [latex]\frac{1}{x}[/latex] का मान ज्ञात कीजिए। .
हल:
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 14.
निम्न में से प्रत्येक को परिमेय हर के रूप में व्यक्त कीजिए।
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAQ 15

Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 15.
हर का परिमेयीकरण कर, निम्न को सरल कीजिए।
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2
हल:
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 16.
निम्न को सरल कीजिए।
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAQ 19
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 17.
निम्न में से प्रत्येक से a तथा b के मान ज्ञात कीजिए।
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2
हल:
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAQ 24
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 18.
यदि x = [latex]\frac{\sqrt{5}-2}{\sqrt{5}+2}[/latex] तथा y = [latex]\frac{\sqrt{5}+2}{\sqrt{5}-2}[/latex], तब निम्न के मान ज्ञात कीजिए।
(i) x2
(ii) y2
(iii) xy
(iv) x2 + y2 + xy
हल:
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 19.
यदि x = [latex]5-2 \sqrt{6}[/latex], तब निम्न के मान ज्ञात कीजिए।
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2
हल:
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 20.
यदि a = [latex]\frac{1}{3-\sqrt{8}}[/latex] , b = [latex]\frac{1}{3+\sqrt{8}}[/latex] तब निम्न के मान ज्ञात कीजिए।
(i) a2
(ii) b2
(iii) ab
(iv) 52 – 6ab + 3b2
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 21.
निम्न में से प्रत्येक का मान दशमलव के तीन स्थानों तक ज्ञात कीजिए।
यदि दिया है: [latex]\sqrt{2}[/latex] = 1.4142, [latex]\sqrt{3}[/latex] = 1.732, [latex]\sqrt{5}[/latex] = 2.2360, [latex]\sqrt{6}[/latex] = 2.4445, [latex]\sqrt{10}[/latex] = 3.162
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAQ 30
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAQ 32

प्रश्न 22.
सरल कीजिए।
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAQ 34
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 23.
यदि x = [latex]\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}[/latex] तथा y = [latex]\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}[/latex] तब 3x2 + 4xy-3y2 का मान ज्ञात कीजिए।
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 24.
a तथा b का मान ज्ञात कीजिए।
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAQ 37
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAQ 39

प्रश्न 25.
यदि [latex]\frac{\sqrt{3}+1}{\sqrt{3}-1}=a+b \sqrt{3}[/latex], तब a व b के मान ज्ञात कीजिए।
हलः
हर का परिमेयीकरण करने पर,
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAQ 41
दोनों पक्षों की तुलना से a = 2 व b = 1

प्रश्न 26.
यदि [latex]\frac{5-\sqrt{6}}{5+\sqrt{6}}=a-b \sqrt{6}[/latex], तब a व b के मान ज्ञात कीजिए।
हलः
हर का परिमेयीकरण करने पर
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 27.
यदि x =[latex]\frac{1}{3-2 \sqrt{2}}[/latex], y = [latex]\frac{1}{3+2 \sqrt{2}}[/latex], तब xy2 + x2y का मान ज्ञात कीजिए।
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 28.
यदि [latex]\sqrt{3}[/latex] = 1.732 व [latex]\sqrt{5}[/latex] = 2.236, तब [latex]\frac{6}{\sqrt{5}-\sqrt{3}}[/latex] का मान ज्ञात कीजिए।
हलः
हर का परिमेयीकरण करने पर
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

Ex 3.2 Rationalisation बहुविकल्पीय (Multiple Choice Questions)

सही विकल्प का चयन कीजिए
प्रश्न 1.
यदि [latex]x+\sqrt{15}=4[/latex] तब [latex]x+\frac{1}{x}[/latex]
(a) 8
(b) 4
(c) 15
(d) इनमें से कोई नहीं
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2
अतः विकल्प (a) सही है।

Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 2.
[latex]3 \sqrt{5}-\sqrt{7}[/latex] का सरलतम परिमेय गुणनखण्ड है
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 MCQ 46
हल:
[latex]3 \sqrt{5}-\sqrt{7}[/latex] का सरलतम परिमेय गुणनखण्ड = [latex]3 \sqrt{5}-\sqrt{7}[/latex]
अतः विकल्प (c) सही है।

प्रश्न 3.
यदि x = 7 + [latex]4 \sqrt{3}[/latex] व xy = 1, तब [latex]\frac{1}{x^{2}}+\frac{1}{y^{2}}=[/latex] =
(a) 194
(b) 28
(c) 1915
(d) इनमें से कोई नहीं
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2
अतः विकल्प (a) सही है।

प्रश्न 4.
यदि x = [latex]\sqrt[3]{2+\sqrt{3}}[/latex], तब [latex]x^{3}+\frac{1}{x^{3}}[/latex]
(a) 3
(b) 5
(c) 4
(d) 6
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2
अतः विकल्प (c) सही है।

प्रश्न 5.
[latex]\sqrt{3-2 \sqrt{2}}[/latex] का मान
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 MCQ 49
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2
अतः विकल्प (c) सही है।

प्रश्न 6.
यदि x = 4 – [latex]\sqrt{15}[/latex], तब x + [latex]\frac{\mathbf{1}}{\mathbf{x}}[/latex] =
(a) 8
(b) 4
(c) 15
(d) 12
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2
अतः विकल्प (a) सही है।

प्रश्न 7.
7 + [latex]\sqrt{48}[/latex], का धनात्मक वर्गमूल है।
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 MCQ 52
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2
अतः विकल्प (a) सही है।

Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 8.
यदि [latex]\sqrt{2}[/latex] = 1.414, तब [latex]\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}=[/latex] (NCERT Exemplar)
(a) 1.414
(b) 2.07
(c) 0.414
(d) इनमें से कोई नहीं-
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2
अतः विकल्प (c) सही है।

Ex 3.2 Rationalisation स्वमूल्यांकन परीक्षण (Self Assessment Test)

प्रश्न 1.
[latex]\frac{5}{\sqrt{3}-\sqrt{5}}[/latex] = के हर का परिमेयीकरण कीजिए।
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAT 55

प्रश्न 2.
[latex]\frac{1}{7+3 \sqrt{2}}[/latex] के हर का परिमेयीकरण कीजिए।
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 3.
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAT 57
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 4.
यदि a = 8 + [latex]3 \sqrt{7}[/latex] व b = [latex]\frac{1}{8+3 \sqrt{7}}[/latex], तब सिद्ध कीजिए कि a2 + b2 = 254
हलः
हर का परिमेयीकरण करने पर
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 5.
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAT 60
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 6.
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAT 62
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 7.
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAT 64
हल:
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 8.
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAT 66
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 9.
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAT 68
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 10
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAT 70
हल:
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 11.
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAT 72
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 12.
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAT 74
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 13.
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAT 76
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 14.
सिद्ध कीजिए कि
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAT 78
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 15.
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAT 80
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAT 81
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 16.
सिद्ध कीजिए कि
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAT 83
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 17.
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAT 85
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 18.
सिद्ध कीजिए कि
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAT 87
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

Balaji Publications Mathematics Class 9 Solutions