2ae42368d8e9a0cc27902a08e5dbee6f
2021.43
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2ae42368d8e9a0cc27902a08e5dbee6f
सिद्ध कीजिए कि <br> (iii) `int(1)/(sqrt3sinx+cosx)dx=(1)/(2)log tan((x)/(2)+(pi)/(12))+c`
सिद्ध कीजिए कि <br> (iii) `int(1)/(sqrt3sinx+cosx)dx=(1)/(2)log tan((x)/(2)+(pi)/(12))+c`
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2021.43
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c3904a843b688b1352ce96299a138452
2021.25
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c3904a843b688b1352ce96299a138452
`int ((1-cotx))/((1+cotx))dx.`
`int ((1-cotx))/((1+cotx))dx.`
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2021.25




answer: `-log |sin x+cos x|c`
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7b71438affa89051af2fbbf454d8b0cb
2021.10
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7b71438affa89051af2fbbf454d8b0cb
binomials
binomials
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2021.10
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4c73c29e821843d905ee3844f367952f
2021.43
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4c73c29e821843d905ee3844f367952f
`y=(tanx)^(cotx)+(cotx)^(tanx)`
`y=(tanx)^(cotx)+(cotx)^(tanx)`
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2021.43




answer: `(tanx)^(cotx).cosec^(2)x(1-logtanx)+(cotx)^(tanx).sec^(2)x[log(cotx)-1]`
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72068ebb4a685df92fa225bb5d6a9b2f
2021.43
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72068ebb4a685df92fa225bb5d6a9b2f
गुणनखण्ड कीजिए : `x^(3)+6x^(2)+11x+6` 
गुणनखण्ड कीजिए : `x^(3)+6x^(2)+11x+6` 
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2021.43




solution: यहाँ, अचर पद = 6 <br> इसके गुणनखण्ड `=pm1,pm2,pm3,pm6` <br> माना `p(x)=x^(3)+6x^(2)+11x+6` <br> `x=-1` रखने पर <br> `p(-1)=(-1)^(3)+6(-1)^(2)+11(-1)+6=-1+6-11+6=0` <br> `:." "x+1,p(x)` का एक गुणनखण्ड है । <br> अब, `p(x)=x^(3)+6x^(2)+11x+6` <br> `=x^(2)(x+1)+5x^(2)+11x+6` <br> `=x^(2)(x+1)+5x(x+1)+6x+6` <br> `=x^(2)(x+1)+5x(x+1)+6(x+1)` <br> `=(x+1)(x^(2)+5x+6)` <br> `=(x+1)[x^(2)+2x+3x+6]` <br> `=(x+1)[x(x+2)+3(x+2)]` <br> `=(x+1)(x+2)(x+3)` 
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1e122e56bf3c11157e47013218be05d1
2021.25
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1e122e56bf3c11157e47013218be05d1
 `int\ (sin^6x + cos^6x)/(sin^2 x*cos^2x)\ dx`
(i) `tanx+ cotx+3x+c`
(ii) `tanx+ cotx-3x+c`
(iii)`tanx-cotx-3x+c`
(iv)`tanx-cotx+3x+c`
 `int\ (sin^6x + cos^6x)/(sin^2 x*cos^2x)\ dx`
(i) `tanx+ cotx+3x+c`
(ii) `tanx+ cotx-3x+c`
(iii)`tanx-cotx-3x+c`
(iv)`tanx-cotx+3x+c`
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2021.25
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ef2fa56aa9d6b406b905e4f6d6af0440
2021.43
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ef2fa56aa9d6b406b905e4f6d6af0440
factorise `(2a + 3b)^(2) - (3a - 2b)^(2)` . 
factorise `(2a + 3b)^(2) - (3a - 2b)^(2)` . 
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2021.43
`(5a + b) (5a - b)`
`(a + 5b) (a - 5b)`
`(5a + b) (5b - a)`
`(5a + b) (5b + a)`
answer: c
solution: `(2a + 3b)^(2) - (3a - 2b)^(2)`   
 `= [2a + 3b + 3a - 2b] [2a + 3b - 3a + 2b]`   
 = `(5a + b) (5b - a)`   
 hence , the correct option is (c) . 
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4122945cb338c621f01a14bdeff2c3b7
2021.43
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4122945cb338c621f01a14bdeff2c3b7
using binomial theorem, evaluate : `(102)^5`
using binomial theorem, evaluate : `(102)^5`
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2021.43
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b60301ae5d129e43110a3b966f4f5807
2021.43
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b60301ae5d129e43110a3b966f4f5807
for any positive integers m, n (with `n ge m`) <br> if `({:(n),(m):}) = .^(n)c\_(m)` prove that <br> `({:(n),(m):}) + ({:(n - 1),(m):}) + ({:(n - 2),(m):}) + … + ({:(m),(m):}) = ({:(n + 1),(m + 1):})` <br> prove that <br> `({:(n),(m):}) + 2 ({:(n + 1),(m):}) + 3 ({:(n - 2),(m):}) + .... + (n - m + 1)` <br> `({:(m),(m):}) = ({:(n + 2),(m + 2):})`
for any positive integers m, n (with `n ge m`) <br> if `({:(n),(m):}) = .^(n)c\_(m)` prove that <br> `({:(n),(m):}) + ({:(n - 1),(m):}) + ({:(n - 2),(m):}) + … + ({:(m),(m):}) = ({:(n + 1),(m + 1):})` <br> prove that <br> `({:(n),(m):}) + 2 ({:(n + 1),(m):}) + 3 ({:(n - 2),(m):}) + .... + (n - m + 1)` <br> `({:(m),(m):}) = ({:(n + 2),(m + 2):})`
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2021.43




solution: let `s = ((n)/(m))+((n-1)/(m)) + ((n-2)/(m))+ ((n-2)/(m)) +.....+((m)/(m)) =((n+1)/(m+1)).....(i)` <br> it is obvious that, `n ge m` <br> note : this question is based upon additive loop. <br> now ,`s =((m)/(m)) + ((m+1)/(m)) + ((m+2)/(m)) +.......+((n)/(m))` <br> ` ={((m+1)/(m+1))+((m+1)/(m))}[because ((m)/(m)) = 1= ((m+1)/(m+1))]` <br> `= ((m+2)/(m+1)) + ((m+2))/(m)) + ......+ ((n)/(m))" "[because""^(n)c\_(r)+ ""^(n)c\_(r+1) = ""^(n+1)c\_(r+1)]` <br> `=((m+2)/(m+1)) +......+((n)/(m))` <br> `=.............` <br> ` =((n)/(m+1))+((n)/(m)) = ((n+1)/(m+1))` which is ture ....(ii) <br> again, we have to prove that <br> `((n)/(m))+2((n-1)/(m)) + 3((n-2)/(m)) +......+ (n-m+1)((m)/(m)) = ((m+2)/(m+2))` <br> let `s\_(1) = ((n)/(m))+2((n-1)/(m)) +3((n-2)/(m)) +......+(n-m+1)((mm)/(m))` <br> `{:(= ((n)/(m)) + ((n-1)/(m)) + ((n-2)/(m)) +...+ ((m)/(m))), ( " "+ ((n-1)/(m)) + ((n-2)/(m)) +...+ ((m)/(m))), (" "+ ((n-2)/(m)) +...+((m)/(m)) ),(" "...) , (" "+ ((m)/(m))):}}n-m + 1` rows <br> now, sum of the first row is `((n+1)/(m+1))` <br> sum of the second row is `((n)/(m+1))` <br> sum of the third row is `((n+1)/(m+1))`, <br> .................... <br> sum of the last row is `((m)/(m)) = ((m+1)/(m+1))` <br> thus `s = ((n+1)/(m+1))+((n)/(m+1)) + ((n +1)/(m+1))+.......+ ((m+1)/(m+1)) = ((n+1+1)/(m+2)) = ((n+2)/(m+2))` <br> [from eq. (i) replacing n by n +1 and m by m + 1]
4fcc5f2b3fd2ac8ba1987452faccb50d
2021.43
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4fcc5f2b3fd2ac8ba1987452faccb50d
सिध्द कीजिए कि `(4x + 2y)^(3)-(4x-2y)^(3)=16y^(3)+192x^(2)y` 
सिध्द कीजिए कि `(4x + 2y)^(3)-(4x-2y)^(3)=16y^(3)+192x^(2)y` 
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2021.43
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05979933874e5bf183d4a33dd6d61978
2021.31
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05979933874e5bf183d4a33dd6d61978
यदि `"cos"^(-1)x+"cos"^(-1)y+"cos"^(-1)z=pi` तो सिद्ध कीजिए कि `x^(2)+y^(2)+z^(2)+2xyz=1`
यदि `"cos"^(-1)x+"cos"^(-1)y+"cos"^(-1)z=pi` तो सिद्ध कीजिए कि `x^(2)+y^(2)+z^(2)+2xyz=1`
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2021.31
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d37f9fd987425bfb30d90a2f24d7f4d7
2021.31
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d37f9fd987425bfb30d90a2f24d7f4d7
prove that : `|{:(x+a,x+2a,x+3a),(x+2a,x+3a,x+4a),(x+4a,x+5a,x+6a):}|=0`
prove that : `|{:(x+a,x+2a,x+3a),(x+2a,x+3a,x+4a),(x+4a,x+5a,x+6a):}|=0`
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2021.31
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826a8e9fc30d7f0d556c04fdc83dc1ca
2020.50
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826a8e9fc30d7f0d556c04fdc83dc1ca
prove that : `|{:(x+a,x+2a,x+3a),(x+2a,x+3a,x+4a),(x+4a,x+5a,x+6a):}|=0`
prove that : `|{:(x+a,x+2a,x+3a),(x+2a,x+3a,x+4a),(x+4a,x+5a,x+6a):}|=0`
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2020.50




solution: 
9fa04aee85ad2b04165938ec7512ea8d
2021.31
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9fa04aee85ad2b04165938ec7512ea8d
binomial nomenclature
binomial nomenclature
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2021.31
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b96376b56714c178ac3c2071cfa7bcc8
2021.25
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b96376b56714c178ac3c2071cfa7bcc8
`"if "y=(tanx)^(cotx)+(cotx)^(tanx)",prove that "(dy)/(dx)=(tanx)^(cotx)."cosec"^(2)x(1-logtanx)+(cotx)^(tanx).sec^(2)x[log(cotx)-1].`
`"if "y=(tanx)^(cotx)+(cotx)^(tanx)",prove that "(dy)/(dx)=(tanx)^(cotx)."cosec"^(2)x(1-logtanx)+(cotx)^(tanx).sec^(2)x[log(cotx)-1].`
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2021.25
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4569fc74d0fdff2a4b56f9fdab9f29af
2021.25
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4569fc74d0fdff2a4b56f9fdab9f29af
`int(dx)/(sin^2xcos^2x)` equals(a) `tanx+cotx+c` (b) `tanx-cotx+c`(c) `tanxcotx+c` (d) `tanx-cot2x+c`
`int(dx)/(sin^2xcos^2x)` equals(a) `tanx+cotx+c` (b) `tanx-cotx+c`(c) `tanxcotx+c` (d) `tanx-cot2x+c`
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2021.25
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0eacd82efd0293ec03bedcc3ec8b8c2f
2021.39
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0eacd82efd0293ec03bedcc3ec8b8c2f
if `sectheta=m, tantheta = n`, then `1/m{(m+n)+1/(m+n)}`
if `sectheta=m, tantheta = n`, then `1/m{(m+n)+1/(m+n)}`
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2021.39
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fa1d3c29249205354d6bae94eae2ff9a
2021.43
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fa1d3c29249205354d6bae94eae2ff9a
using binomial theorem, evaluate : `(96)^3`
using binomial theorem, evaluate : `(96)^3`
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2021.43




solution: `(96)^3=(100-4)^3`  

`(x+a)^n=sum\_(r=0)^(n).^nc\_ra^rr^(n-r)`  

`(100-4)^3= .^3c\_0(100)^3+.^3c\_1(100)^2(-4)^(1).+^3c\_2(100)^1(-4)^(2)+.^3c\_3(100)^0(-4)^(3)`  

=>`1000000-120000+4800-64`  

=>`884736`.


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ad79f3376bc5175aca6b606745bbf036
2021.39
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ad79f3376bc5175aca6b606745bbf036
`(2a^2-1)/a ,4 a-3/a ,(6a^3-5)/a` s of the series til 
`(2a^2-1)/a ,4 a-3/a ,(6a^3-5)/a` s of the series til 
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2021.39
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6131bf19c7c6f08c1f1e42a46d798543
2021.39
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6131bf19c7c6f08c1f1e42a46d798543
a quadrilateral whose each angle is a right angle is a 
square                         (b) 
 rectangle          
 rhombus                     (d)  parallelogram
a quadrilateral whose each angle is a right angle is a 
square                         (b) 
 rectangle          
 rhombus                     (d)  parallelogram
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2021.39
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f02a1fe7c61a2e2d9479910c8cf1bc2a
2021.43
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f02a1fe7c61a2e2d9479910c8cf1bc2a
मान ज्ञात कीजिये `int (2x^(2)+1)/(x^(2)(x^(2)+4))dx`
मान ज्ञात कीजिये `int (2x^(2)+1)/(x^(2)(x^(2)+4))dx`
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2021.43




solution: `-(1)/(4x)+(7)/(8)tan^(-1).(x)/(2)+c`
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