5864d848de8edbd3712bdf3e58b138c9
2020.40
goprep.co
5864d848de8edbd3712bdf3e58b138c9
let’s fill up the table below :
let’s fill up the table below :
goprep.co
2020.40
(b)first part,given, (i) x4 – 4x3 + 6x2(ii) x2(i) × (ii)= (x4 – 4x3 + 6x2) × x2= x4x2 – 4x3x2 + 6x2x2we know (xaxb= xa+b)= x4+2– 4x3+2 + 6x2+2= x6 – 4x5 + 6x4(i) ÷ (ii)we know = x4-2 – 4x3-2 + 6x2-2= x2 – 4x + 6second partgiven, (i) 3m2n3 + 40m3n4 – 5m4n5(ii) 10m2n2(i) × (ii)= (3m2n3 + 40m3n4 – 5m4n5) × 10m2n2we know (xaxb= xa+b)= 30m2m2n3n2 + 400m3m2n4n2 – 50m4m2n5n2= 30m2+2n3+2 + 400m3+2n4+2 – 50m4+2n5+2= 30m4n5 + 400m5n6 – 50m6n7(i) ÷ (ii)we know (c)given,(i) 49l2 – 100m2(ii) (7l + 10m)(i) × (ii)= (49l2 – 100m2)(7l + 10m)we know (xaxb= xa+b)= 343l2l+ 490l2m – 700m2l – 1000m2m= 343l2+1 + 490l2m – 700lm2 – 1000m2+1= 343l3 + 490l2m – 700lm2 – 1000m3(i) ÷ (ii)∵ a2 – b2 = (a – b)(a + b), taking a = 7l, b = 10m we havecancelling out the common terms from numerator and denominator, we get= 7l – 10m(d)given,(i) 625a4 – 81b4(ii) 5a + 3b(i) × (ii)= (625a4 – 81b4)(5a + 3b)= 625a4a + 1875a4b – 405ab4 – 243b4bwe know (xaxb= xa+b)= 625a4+1 + 1875a4b – 405ab4 – 243b4+1= 3125a5 + 1875a4b – 405ab4 – 243b5(i) ÷ (ii)∵ a2 – b2 = (a – b)(a + b), taking a = 25a2, b = 9b2, we have∵ a2 – b2 = (a – b)(a + b), taking a = 5a, b = 3b we have=cancelling out the common terms from numerator and denominator, we get= (5a – 3b)(25a2 + 9b2)= 125aa2 + 45ab2 – 75a2b – 27bb2we know (xaxb= xa+b)= 125a2+1 + 45ab2 – 75a2b – 27b2+1= 125a3 + 45ab2 – 75a2b – 27b3
(b)first part,given, (i) x4 – 4x3 + 6x2(ii) x2(i) × (ii)= (x4 – 4x3 + 6x2) × x2= x4x2 – 4x3x2 + 6x2x2we know (xaxb= xa+b)= x4+2– 4x3+2 + 6x2+2= x6 – 4x5 + 6x4(i) ÷ (ii)we know = x4-2 – 4x3-2 + 6x2-2= x2 – 4x + 6second partgiven, (i) 3m2n3 + 40m3n4 – 5m4n5(ii) 10m2n2(i) × (ii)= (3m2n3 + 40m3n4 – 5m4n5) × 10m2n2we know (xaxb= xa+b)= 30m2m2n3n2 + 400m3m2n4n2 – 50m4m2n5n2= 30m2+2n3+2 + 400m3+2n4+2 – 50m4+2n5+2= 30m4n5 + 400m5n6 – 50m6n7(i) ÷ (ii)we know (c)given,(i) 49l2 – 100m2(ii) (7l + 10m)(i) × (ii)= (49l2 – 100m2)(7l + 10m)we know (xaxb= xa+b)= 343l2l+ 490l2m – 700m2l – 1000m2m= 343l2+1 + 490l2m – 700lm2 – 1000m2+1= 343l3 + 490l2m – 700lm2 – 1000m3(i) ÷ (ii)∵ a2 – b2 = (a – b)(a + b), taking a = 7l, b = 10m we havecancelling out the common terms from numerator and denominator, we get= 7l – 10m(d)given,(i) 625a4 – 81b4(ii) 5a + 3b(i) × (ii)= (625a4 – 81b4)(5a + 3b)= 625a4a + 1875a4b – 405ab4 – 243b4bwe know (xaxb= xa+b)= 625a4+1 + 1875a4b – 405ab4 – 243b4+1= 3125a5 + 1875a4b – 405ab4 – 243b5(i) ÷ (ii)∵ a2 – b2 = (a – b)(a + b), taking a = 25a2, b = 9b2, we have∵ a2 – b2 = (a – b)(a + b), taking a = 5a, b = 3b we have=cancelling out the common terms from numerator and denominator, we get= (5a – 3b)(25a2 + 9b2)= 125aa2 + 45ab2 – 75a2b – 27bb2we know (xaxb= xa+b)= 125a2+1 + 45ab2 – 75a2b – 27b2+1= 125a3 + 45ab2 – 75a2b – 27b3
5e8442ddfa78055acbcc5aebec0703e0
2021.21
goprep.co
5e8442ddfa78055acbcc5aebec0703e0
if <span lang="en
if <span lang="en
goprep.co
2021.21
given: to find: cos2xwe know that,cos 2x = 1 – 2sin2xputting the value, we get
given: to find: cos2xwe know that,cos 2x = 1 – 2sin2xputting the value, we get
17dccb3cf7d881d1c98204844d25e02a
2021.21
goprep.co
17dccb3cf7d881d1c98204844d25e02a
if <span lang="en
if <span lang="en
goprep.co
2021.21
r.h.s:)⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ = l.h.shence, provedii)r.h.s. = ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ tanθ/2 = l.h.shence,proved.
r.h.s:)⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ = l.h.shence, provedii)r.h.s. = ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ tanθ/2 = l.h.shence,proved.
dc5997a2d56fcaefa90164b83efbdcba
2020.45
goprep.co
dc5997a2d56fcaefa90164b83efbdcba
add:3a (2b + 5c), 3c (2a + 2b)
add:3a (2b + 5c), 3c (2a + 2b)
goprep.co
2020.45
here, 3a(2b + 5c) = 6ab + 15ac3c(2a + 2b) = 6ac + 6bchence the two polynomials to be added are 6ab + 15ac and6ac + 6bcthe result is as follows 6ab + 21ac + 6bc
here, 3a(2b + 5c) = 6ab + 15ac3c(2a + 2b) = 6ac + 6bchence the two polynomials to be added are 6ab + 15ac and6ac + 6bcthe result is as follows 6ab + 21ac + 6bc
809c42b1d130275605d54e423ba5b1a8
2021.17
goprep.co
809c42b1d130275605d54e423ba5b1a8
if <span lang="en
if <span lang="en
goprep.co
2021.17
we have, now , 
we have, now , 
136902346ca445559184b57d92ad2f47
2021.10
goprep.co
136902346ca445559184b57d92ad2f47
if <span lang="en
if <span lang="en
goprep.co
2021.10
using the chain rule of differentiation, (ans)
using the chain rule of differentiation, (ans)
e927ca1cb91d76c6fb859dc904d75f25
2021.17
goprep.co
e927ca1cb91d76c6fb859dc904d75f25
if <span lang="en
if <span lang="en
goprep.co
2021.17
λ = 3μ = 1so, λ + μ = 3 + 1= 4
λ = 3μ = 1so, λ + μ = 3 + 1= 4
1f9316dc2776d2c4c351c0059f9bf2dd
2020.45
goprep.co
1f9316dc2776d2c4c351c0059f9bf2dd
3x2 + 7ix + 6 = 0
3x2 + 7ix + 6 = 0
goprep.co
2020.45
given:3x2 + 7ix + 6 = 0⇒3x2 + 9ix-2ix + 6 = 0⇒⇒ …(i2 = -1)⇒ = 0⇒⇒⇒x + 3i = 0 & 3x-2i = 0⇒ x = 3i & ans: x = 3i and 
given:3x2 + 7ix + 6 = 0⇒3x2 + 9ix-2ix + 6 = 0⇒⇒ …(i2 = -1)⇒ = 0⇒⇒⇒x + 3i = 0 & 3x-2i = 0⇒ x = 3i & ans: x = 3i and 
72ed8817f39308dc19fceca9ec65f7f5
2021.17
goprep.co
72ed8817f39308dc19fceca9ec65f7f5
express <span lan
express <span lan
goprep.co
2021.17
here,…… eq 1multiplying eq 1 by 100 …… eq 2subtracting eq 1 from eq2⇒ 100 x – x = 234⇒ 99 x = 234 ……eq 3now, let ……eq 4multiplying eq 3 by 100⇒ ⇒ 100 y – y = 23⇒ 99 y = 23 …… eq 5from eq 3 and eq 5
here,…… eq 1multiplying eq 1 by 100 …… eq 2subtracting eq 1 from eq2⇒ 100 x – x = 234⇒ 99 x = 234 ……eq 3now, let ……eq 4multiplying eq 3 by 100⇒ ⇒ 100 y – y = 23⇒ 99 y = 23 …… eq 5from eq 3 and eq 5
90e1d2b358ac8ccf016d4c1e87907fd3
2021.25
goprep.co
90e1d2b358ac8ccf016d4c1e87907fd3
if x2 
if x2 
goprep.co
2021.25
⇒ ⇒ cot45 = 1⇒ ⇒ x2 = sin230 + 4cot245-sec260⇒ ⇒ ⇒ ⇒ 
⇒ ⇒ cot45 = 1⇒ ⇒ x2 = sin230 + 4cot245-sec260⇒ ⇒ ⇒ ⇒ 
2b88fefe3119d8b80f0dc7c8239d9340
2021.17
goprep.co
2b88fefe3119d8b80f0dc7c8239d9340
if <span lang="en
if <span lang="en
goprep.co
2021.17
let (given)⇒ ⇒ ⇒ k=17
let (given)⇒ ⇒ ⇒ k=17
ade9b1c2c320cb50d813f69062a72191
2021.17
goprep.co
ade9b1c2c320cb50d813f69062a72191
if <span lang="en
if <span lang="en
goprep.co
2021.17
given data:nowalso,3(2-3λ) + 4(1-4λ) - 5(4+5λ) = 0-50λ = 10using the above value,
given data:nowalso,3(2-3λ) + 4(1-4λ) - 5(4+5λ) = 0-50λ = 10using the above value,
dec2f7f7db774ed8da03510e3fe056de
2021.43
goprep.co
dec2f7f7db774ed8da03510e3fe056de
if <span lang="en
if <span lang="en
goprep.co
2021.43
given, we should find the value for [by using the formulae, sin2θ + cos2θ = 1 and sin2θ = 2 sinθcosθ]as already mentioned in the question, , so x lies in the 1st quadrant and both sine and cosine functions are positive.therefore, sin x + cos x
given, we should find the value for [by using the formulae, sin2θ + cos2θ = 1 and sin2θ = 2 sinθcosθ]as already mentioned in the question, , so x lies in the 1st quadrant and both sine and cosine functions are positive.therefore, sin x + cos x
612d2df43f7d907b844f65939da287fa
2021.39
goprep.co
612d2df43f7d907b844f65939da287fa
if 2nc
if 2nc
goprep.co
2021.39
given:⇒ we know that and also n! = n(n – 1)(n – 2)…………2.1⇒ ⇒ ⇒ 2 × (2n – 1) × 2 = 44⇒ 2n – 1 = 11⇒ 2n = 12⇒ ⇒ n = 6∴ the value of n is 6.
given:⇒ we know that and also n! = n(n – 1)(n – 2)…………2.1⇒ ⇒ ⇒ 2 × (2n – 1) × 2 = 44⇒ 2n – 1 = 11⇒ 2n = 12⇒ ⇒ n = 6∴ the value of n is 6.
997a754ac3d27d82b983ece47475d1df
2021.21
goprep.co
997a754ac3d27d82b983ece47475d1df
if <span lang="en
if <span lang="en
goprep.co
2021.21
5a a a conclusion: a 
5a a a conclusion: a 
399d83b1577324a5f99d445d3610776f
2021.43
goprep.co
399d83b1577324a5f99d445d3610776f
if <span lang="en
if <span lang="en
goprep.co
2021.43
proof:as cosθ+cos2 θ=1cos θ =1-cos2 θcos θ =sin2 θ ….(i)using(a + b)3 =a3+b3+3(a + b)now,sin12 θ+3sin10 θ+3sin8 θ+sin6 θ+2sin4 θ+2sin2 θ-2= (sin4 θ)3+sin4 θ-sin2 θ[sin4 θ+sin2 θ](sin2 θ)3+2(sin2 θ)2+2sin2 θ-2=(sin4 θ+sin2 θ)3+2(cos θ)2+2cosθ-2.=((sin2 θ)2+sin2 θ)3 +2cos2 θ+2cosθ-2from (1),=(cos2 θ +sin2 θ)3+2cos2 θ+2cosθ-2as sin2θ + cos2θ = 1= 1 +2cos2 θ+2sin2 θ-2= 1 +2(cos2 θ+sin2 θ)-2=1+2(1)-2=1hence proved.
proof:as cosθ+cos2 θ=1cos θ =1-cos2 θcos θ =sin2 θ ….(i)using(a + b)3 =a3+b3+3(a + b)now,sin12 θ+3sin10 θ+3sin8 θ+sin6 θ+2sin4 θ+2sin2 θ-2= (sin4 θ)3+sin4 θ-sin2 θ[sin4 θ+sin2 θ](sin2 θ)3+2(sin2 θ)2+2sin2 θ-2=(sin4 θ+sin2 θ)3+2(cos θ)2+2cosθ-2.=((sin2 θ)2+sin2 θ)3 +2cos2 θ+2cosθ-2from (1),=(cos2 θ +sin2 θ)3+2cos2 θ+2cosθ-2as sin2θ + cos2θ = 1= 1 +2cos2 θ+2sin2 θ-2= 1 +2(cos2 θ+sin2 θ)-2=1+2(1)-2=1hence proved.
f7b7f9b9e2d5977b37944e07fb7eddec
2020.50
goprep.co
f7b7f9b9e2d5977b37944e07fb7eddec
if p(2n – 1, n) : p(2n + 1, n – 1) = 22 : 7 find n.
if p(2n – 1, n) : p(2n + 1, n – 1) = 22 : 7 find n.
goprep.co
2020.50
given: p(2n – 1, n) : p(2n + 1, n – 1) = 22 : 7to find: value of nwe know,so, according to question:p(2n – 1, n) : p(2n + 1, n – 1) = 22 : 7⇒ 7(n + 2)(n + 1) = 22×2 (2n + 1)⇒ 7(n2 + n + 2n + 2) = 88n + 44⇒ 7(n2 + 3n + 2) = 88n + 44⇒ 7n2 + 21n + 14 = 88n + 44⇒ 7n2 + 21n – 88n + 14 – 44 = 0⇒ 7n2 – 67n – 30 = 0⇒ 7n2 – 70n + 3n – 30 = 0⇒ 7n(n – 10) + 3(n – 10) = 0⇒ (n – 10)(7n + 3) = 0∴ the value of n = 10
given: p(2n – 1, n) : p(2n + 1, n – 1) = 22 : 7to find: value of nwe know,so, according to question:p(2n – 1, n) : p(2n + 1, n – 1) = 22 : 7⇒ 7(n + 2)(n + 1) = 22×2 (2n + 1)⇒ 7(n2 + n + 2n + 2) = 88n + 44⇒ 7(n2 + 3n + 2) = 88n + 44⇒ 7n2 + 21n + 14 = 88n + 44⇒ 7n2 + 21n – 88n + 14 – 44 = 0⇒ 7n2 – 67n – 30 = 0⇒ 7n2 – 70n + 3n – 30 = 0⇒ 7n(n – 10) + 3(n – 10) = 0⇒ (n – 10)(7n + 3) = 0∴ the value of n = 10
d0b3e72c5b2f5ce1e1f88cf8a601d7ed
2021.21
goprep.co
d0b3e72c5b2f5ce1e1f88cf8a601d7ed
if <span lang="en
if <span lang="en
goprep.co
2021.21
explanation:given 
explanation:given 
1400b0f1841580439e7f314c2c992a90
2021.21
goprep.co
1400b0f1841580439e7f314c2c992a90
the value of tan 
the value of tan 
goprep.co
2021.21
tan 3a = tan(2a+a)⇒ tan 3a(1-tan 2a.tan a) = tan 2a + tan a⇒ tan 3a-tan 3a.tan 2a.tan a = tan 2a + tan a⇒ tan 3a- tan 2a - tan a = tan 3a.tan 2a.tan a
tan 3a = tan(2a+a)⇒ tan 3a(1-tan 2a.tan a) = tan 2a + tan a⇒ tan 3a-tan 3a.tan 2a.tan a = tan 2a + tan a⇒ tan 3a- tan 2a - tan a = tan 3a.tan 2a.tan a
164d4338a6ec7c009c47b2d633c0dde0
2020.45
goprep.co
164d4338a6ec7c009c47b2d633c0dde0
add:xy2z2 + 3x2y2z–4x2yz2,-9x2y2z + 3xy2z2 + x2yz2
add:xy2z2 + 3x2y2z–4x2yz2,-9x2y2z + 3xy2z2 + x2yz2
goprep.co
2020.45
the addition is as follows:the result is as follows 4xy2z2-6x2y2z-3x2yz2
the addition is as follows:the result is as follows 4xy2z2-6x2y2z-3x2yz2