## Maths Algebra Handwritten Notes Free PDF

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Q.1: यदि a + b = p, ab = q है , तो (a4 + b4 ) का मान ज्ञात कीजिए I
(A) p4 – 4p2q + q2
(B) p4 – 4p2q2 + 2q2
(C) p4 – 2p2q2 + q2
(D) p4 – 4p2q + 2q2
(SSC CHSL Aug 2021)

Ans : (D) p4 – 4p2q + 2q2
a + b = p , ab = q
(a + b)2 = p2
a2 + 2ab + b2 = p2
a2 + b2 = p2 – 2q
(a2 + b2)2 = (p2 – 2q)2
a4 + b4 + 2a2b2 = p4 – 4p2q + 4q2 [ab = q, a2 b2 = q2]
a4 + b4 = p4 – 4p2q + 2q2

Q.2: यदि $(x +\frac 1x)^3 = 27$ है , तो $(x^2 +\frac {1}{x^2})$ का मान क्या होगा ? दिया गया है कि जहां x वास्तविक है I
(A) 9
(B) 25
(C) 7
(D) 11
(SSC CHSL Aug 2021)

Ans : (C) 7
$(x +\frac 1x)^3 = 27$
$x +\frac 1x = (27)^{1/3} = 3$
$x^2 +\frac {1}{x^2} + 2 = 9$
$x^2 +\frac {1}{x^2} = 9 - 2 = 7$

Q.3: यदि $x -\frac 2x = 4$ है , तो $x^2 + \frac {4}{x^2}$ का मान ज्ञात करें I
(A) 18
(B) 8
(C) 12
(D) 20
(SSC CHSL Aug 2021)

Ans : (D) 20
$({x -\frac 2x)^2} = 4$
$x^2 +\frac {4}{x^2} - 2\times2 = 16$
$x^2 +\frac {4}{x^2} = 16 + 4 = 20$

Q.4: यदि $\sqrt{x} + \frac {1} {\sqrt x} = \sqrt 6$ है , तो $x^6 + \frac {1}{x^6}$ का मान ज्ञात करें I
(A) 2712
(B) 2502
(C) 2270
(D) 2702
(SSC CHSL Aug 2021)

Ans : (D) 2702
$\sqrt{x} + \frac {1} {\sqrt x} = \sqrt6$
$(\sqrt{x} + \frac {1} {\sqrt x})^2 = (\sqrt6)^2$
$x +\frac 1x +2 = 6$
$x +\frac 1x = 4$
$(x +\frac 1 x)^2 = (4)^2$
$x^2 +\frac {1}{x^2} = 16 - 2 = 14$
$(x^2 +\frac {1}{x^2})^3 = (14)^3$
$x^6 +\frac {1}{x^6} + 3x^2\times\frac {1}{x^2}\times (x^2 + (\frac {1}{x^2}) = 2744$
$x^6 +\frac {1}{x^6} = 2744 - 42 = 2702$

Q.5: यदि x2 + 1 – 2x = 0, x > 0 है, तो x2(x– 2) का मान ज्ञात करें I
(A) 0
(B) -1
(C) 1
(D) $\sqrt2$
(SSC CHSL Aug 2021)

Ans : (B) -1
x2 + 1 – 2x = 0
x > 0
x = 1, so 1 + 1 – 2 = 0
Put x = 1 in x2(x– 2)
1 (1 – 2)
= -1

Q.6: यदि $x^2 -3\sqrt2x + 1 =0$ है , तो $x^3 + (\frac {1}{x^3})$ का मान ज्ञात करें I
(A) $15\sqrt6$
(B) $30\sqrt6$
(C) $45\sqrt2$
(D) $30\sqrt2$
(SSC CHSL Aug 2021)

Ans : (C) $45\sqrt2$
$x^2 - 3\sqrt2x + 1 = 0$
$x - 3\sqrt2 +\frac1x = 0$
$x +\frac 1x =3\sqrt2$
$(x +\frac 1x)^3 = (3\sqrt2)^3$
$x^3 +\frac {1}{x^3} + 3x\times\frac1x\times (x +\frac1x) = 54\sqrt2$
$x^3 +\frac{1}{x^3} = 54\sqrt2 - 9\sqrt2$
$= 45\sqrt2$

Q.7: यदि x – y = 4 और xy = 3 है , तो x3 – y3 का मान ज्ञात करें I
(A) 88
(B) 100
(C) 64
(D) 28
(SSC CHSL Aug 2021)

Ans : (B) 100
x – y = 4
(x – y)3 = 43
x3 – y3 – 3xy (x – y) = 64 [Put xy = 3, x – y =4]
x3 – y3 – 3 x 3 x 4 = 64
x3 – y3 = 64 + 36 = 100

Q.8: यदि $x -\frac 1x = 2\sqrt2$ है , तो $x^3 -\frac {1}{x^3}$ का मान ज्ञात करें I
(A) $12\sqrt2$
(B) $10\sqrt2$
(C) $20\sqrt2$
(D) $22\sqrt2$
(SSC CHSL Aug 2021)

Ans : (D) $22\sqrt2$
$x -\frac 1x =2\sqrt2$
$(x -\frac 1x)^3 = (2\sqrt2)^3$
$x^3 -\frac {1}{x^3} - 3\times x\times\frac 1x (x -\frac1x) = 16\sqrt2$
$x^3 -\frac {1}{x^3} - 3\times 2\sqrt2 = 16\sqrt2$
$x^3 -\frac {1}{x^3} = 16\sqrt2 + 6\sqrt2$
$= 22\sqrt2$

Q.9: यदि x + 2y = 19 और x3 + 8y3 = 361 है , तो xy का मान क्या होगा ?
(A) 58
(B) 56
(C) 55
(D) 57
(SSC CHSL Aug 2021)

Ans : (D) 57
x + 2y = 19
(x + 2y)3 = (19)3
x3 + 8y3 + 6xy (x + 2y) = 6859
x3 + 8y3 + 6xy x 19 = 6859
6xy x 19 = 6859 – 361 = 6498
xy =$\frac {6498}{19\times6}$
xy = 57

Q.10: यदि $(x^2 + \frac{1}{49x^2}) = 15\frac 57$ है , तो $(x +\frac{1}{7x})$ का मान क्या होगा ?
(A) 4
(B) $\pm 7$
(C) $\pm 4$
(D) 7
(SSC CHSL Aug 2021)

Ans : (C) $\pm 4$
$(x^2 + \frac{1}{49x^2}) = 15\frac57$
$(x^2 + \frac{1}{7x})^2 + 2\times x\times \frac {1}{7x} =\frac{110}{7} + \frac27$
$(x + \frac{1}{7x})^2 = \frac{112}{7} = 16$
$x + \frac{1}{7x} =\pm 4$

Q.11: यदि x + y = 27 और x2 + y2 = 425 है , तो (x – y)2 का मान ज्ञात करें
(A) 121
(B) 225
(C) 169
(D) 144
(SSC CHSL Aug 2021)

Ans : (A) 121
x + y = 27
x2 + y2 = 425
(x + y)2 = (27)2
x2 + y2 + 2xy = 729
2xy = 729 – 425
2xy = 304
(x – y)2 = x2 + y2 – 2xy
= 425 – 304
(x – y)2 = 121

Q.12: यदि 3x + y = 12 और xy = 9 है, तो (3x – y) का मान क्या होगा ?
(A) 6
(B) 5
(C) 3
(D) 4
(SSC CHSL Aug 2021)

Ans : (A) 6
3x + y = 12
xy = 9
put x = 3, y = 3 Both Equation
So its satify
3x – y = 3 x 3 – 3
= 9 – 3 = 6

Q.13: यदि a2 + b2 + c2 = 576 और (ab + bc + ca) = 50 है, तो (a + b + c) का मान क्या होगा, यदि a+b+c < 0 है ?
(A) -24
(B) $\pm 24$
(C) $\pm 26$
(D) -26
(SSC CHSL Aug 2021)

Ans : (D) -26
(a + b + c)2 = a2 + b2 +c2 + 2 (ab + bc + ca)
576 + 2 x 50
(a + b + c)2 = 676
a + b + c = $\pm 26$
a + b + c < 0
so a+b+c = -26

Q.14: यदि $x +\frac{1}{3x} = 5$ है , तो $27x^3 +\frac{1}{x^3}$ का मान ज्ञात करें I
(A) 3024
(B) 3420
(C) 3042
(D) 3240
(SSC CHSL Aug 2021)

Ans : (D) 3240
$x +\frac{1}{3x} = 5$
$3x +\frac{1}{x} = 15$ [ Multiply by 3]
$(3x +\frac{1}{x})^3 = 15^3$
$27x^3 +\frac{1}{x^3} + 135 = 3375$
$27x^3 +\frac{1}{x^3} = 3375 - 135$
=3240

Q.15: यदि 3x + 5y = 14 और xy = 6 है, तो 9x2 + 25y2का मान कितना होगा ?
(A) 182
(B) 16
(C) 14
(D) 20
(SSC CHSL Aug 2021)

Ans : (B) 16
(3x + 5y)2 = (14)2
9x2 + 25y2 + 30xy = 196
9x2 + 25y2 + 30 x 6 = 196
9x2 + 25y2 = 196 – 180
9x2 + 25y2 = 16

Q.16: यदि a – b = 7 और a2 + b2 = 169 है, जहाँ a,b >0 है , तो 3a + b का मान ज्ञात करें
(A) 41
(B) 46
(C) 38
(D) 44
(SSC CHSL Aug 2021)

Ans : (A) 41
a – b = 7
(a – b)2 = 72
a2 + b2 – 2ab = 49
169 – 2ab = 49
2ab = 169 – 49
2ab = 120
ab = 60 [12 x 5]
a = 12, b = 5
3a + b
= 3 x 12 + 5
= 41

Q.17: यदि a + 5b = 25 और ab = 20 है, तो (a – 5b) का एक मान _________ होगा
(A) 16
(B) 15
(C) 13
(D) 14
(SSC CHSL Aug 2021)

Ans : (B) 15
a + 5b = 25, ab = 20
Put a = 20
b = 1
so a – 5b
= 20 – 5 x 1 = 15

Q.18: यदि $\sqrt {x} + \frac{1}{x} =2\sqrt{3}$ है, तो $x^4 +\frac{1}{x^4}$ का मान ज्ञात करें
(A) 10406
(B) 10402
(C) 9602
(D) 9606
(SSC CHSL Aug 2021)

Ans : (C) 9602
$\sqrt {x} +\frac{1}{x} = 2\sqrt3$
$(\sqrt {x} +\frac{1}{x})^2 = (2\sqrt3)^2$
$x +\frac{1}{x} + 2 = 12$
$x +\frac{1}{x} = 10$
$x^2 + \frac{1}{x^2} = 100 - 2 = 98$
$x^4 + \frac{1}{x^4} = 98^2 - 2$
= 9604 – 2
$x^4 +\frac{1}{x^4} = 9602$

Q.19: यदि (7x – 10y) = 8 और xy = 5 है, तो 49x2 + 100y2 का मान क्या होगा ?
(A) 632
(B) 623
(C) 746
(D) 764
(SSC CHSL Aug 2021)

Ans : (D) 764
7x – 10y = 8
(7x – 10y)2 = (8)2
49x2 + 100y2 – 140xy = 64
49x2 + 100y2 = 64 + 140xy
= 64 + 140xy
= 64 + 140 x 5
= 764

Q.20: यदि $x^2 +(4 - \sqrt {3})x - 1 = 0$ है, तो $x^2 +\frac{1}{x^2}$ का मान ज्ञात करें
(A) $21- 8\sqrt3$
(B) $17- 8\sqrt3$
(C) $9- 8\sqrt3$
(D) $21- 12\sqrt3$
(SSC CHSL Aug 2021)

Ans : (A) $21- 8\sqrt3$
$x^2 +(4 -\sqrt3)x - 1 = 0$
$\div x$ Both side
$x +(4 -\sqrt{3}) - \frac1x = 0$
$x -\frac1x = \sqrt3 - 4$
square Both side
$(x -\frac1x)^2 = (\sqrt 3 - 4)^2$
$x^2 +\frac{1}{x^2} - 2 = 3 + 16 - 8\sqrt3$
$x^2 +\frac{1}{x^2} = 21 - 8\sqrt3$

NUMBER OF PAGES – 64

General Science Notes

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