If `0
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Correct Answer - 9
`1 + sin (cos^(-1) x) + sin^(2) (cos^(-1) x) + ...oo = 2`
`rArr (1)/(1-sin (cos^(-1)x)) = 2`
or `(1)/(2) = 1 - sin (cos^(-1) x)`
or `sin (cos^(-1)x) = (1)/(2)`
or `sin (cos^(-1) x) = (1)/(2)`
or `cos^(-1) x = (pi)/(6)`
or `x = (sqrt3)/(2) " or " 12x^(2) = 9`
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Let `g(x)` be differentiable on `R` and `int_(sint)^1x^2g(x)dx=(1-sint),` where `t in (0,pi/2)dot` Then the value of `g(1/(sqrt(2)))` is____
Correct Answer - 2 We have `int_(sint)^(1) x^(2)g(x)dx=(1sint)`…………1 Differentiating both the sides of 1 with respect to `t` we get `0-(sin^(2)t)g(sint)(cost)=-cost` or `g(sint)=1/(sin^(2)t)`…………….2 Putting `t=(pi)/4` in (2) we get `g(1/(sqrt(2)))=2`.
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If the value of the definite integral `int_0^1^(2007)C_7x^(2000)dot(1-x)^7dx` is equal to `1/k ,w h e r ek in N ,` then the value of `k/(26)` is____
Correct Answer - 208 Let `I=int_(0)^(1).^(207)C_(7).ubrace(x^(200))_(II).ubrace((1-x)^(7))_(I)dx` `=.^(207)C_(7)[(1-x)^(7).(x^(201))/201|_(0)^(1)+7/201int_(0)^(1)(1-x)^(6).x^(201)dx]` `=.^(207)C_(7) . 1/207 int_(0)^(1)(1-x)^(6) . x^(201)dx` Integrating by parts againsix times more we get `I=.^(207)C_(7).(7!)/(201.202.203.204.205.206.207)int_(0)^(1)x^(207)dx` `=((207)!)/(7!(200)!) . (7!)/(201.202......207) . 1/208` `=((207)!)/((207)!7!) . (7!)/208=1/208=1/k` or...
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Prove the orthogonal matrices of order two are of the form `[(cos theta,-sin theta),(sin theta,cos theta)]` or `[(cos theta,sin theta),(sin theta,-cos
Let matrix `A=[(a,b),(c,d)]` is orthogonal matrix. `:. [(a,b),(c,d)][(a,c),(b,d)]=[(1,0),(0,1)]` `implies [(a^(2)+b^(2),ac+bd),(ac+bd,c^(2)+d^(2))]=[(1,0),(0,1)]` `implies a^(2)+b^(2)=1` (1) `c^(2)+d^(2)=1` (2) `ac+bd=0` (3) `implies a/d= (-b)/c= k` (let) `implies c^(2)+d^(2)=(a^(2)+b^(2))/k=1//k^(2)` or `k^(2)=1` or `k= pm 1`...
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If `theta-phi=pi/2,` prove that, `[(cos^2 theta,cos theta sin theta),(cos theta sin theta,sin^2 theta)] [(cos^2 phi,cos phi sin phi),(cos phi sin phi,
Correct Answer - C `AB=[(cos^(2) theta,),(cos theta sin theta,)][(cos^(2) phi,cos phi sin phi),(cos phi sin phi,sin^(2) phi)]` `=[(cos^(2) theta cos^(2) phi+cos theta cos phi sin theta sin phi ,cos^(2)theta cos phi...
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If `F(x)=[("cos"x,-sin x,0),(sin x,cos x,0),(0,0,1)]` and `G(y)=[(cos y,0,sin y),(0,1,0),(-sin y,0,cos y)]`, then `[F(x) G(y)]^(-1)` is equal to
Correct Answer - B We have. `[F(x) G(y)]^(-1)=[G(y)]^(-1) [F(x)]^(-1)` `=G(-y) F(-x)`
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Find the principal values of the following (i) `sin^(-1) (sin 1)` (ii) `sin^(-1) (sin 2)` (iii) `sin^(-1) (sin 10)` (iv) `sin^(-1) (sin 20)` (v) `sin^
(i) Since `1 in [-pi//2, pi//2], sin^(-1) (sin 1) = 1` (ii) Since `2 in [-pi//2, pi//2], sin^(-1) (sin 2) != 2` `:. Sin^(-1) (sin 2) = sin^(-1) (sin (pi)...
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Find the principal value of the following (i) `cos^(-1) (cos 3)` (ii) `cos^(-1) (cos 4)` (iii) `cos^(-1) (cos 15)` (iv) `cos^(-1) (cos 30)` (v) `cos^(
(i) Since `3 in [0, pi], cos^(-1) (cos 3) = 3` (ii) Since `4 !in [0, pi], cos^(-1) (cos 4) != 4` `:. Cos^(-1) (cos 4) = 2pi - 4`...
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The value of `sin^(-1)[xsqrt(1-x)-sqrt(x)sqrt(1-x^2)]` is equal to `sin^(-1)x+sin^(-1)sqrt(x)` `sin^(-1)x-sin^(-1)sqrt(x)` `sin^(-1)sqrt(x)-sin^(-1)x`
Correct Answer - B Let `x = sin theta and sqrtx = sin phi, " where " x in [0, 1]` `rArr theta, phi in [0, pi//2]` `rArr theta - phi...
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Let `a = cos^(-1) cos 20, b = cos^(-1) cos 30 and c = sin^(-1) sin (a + b)` then The largest integer x for which `sin^(-1) (sin x) ge |x -(a + b + c)|
Correct Answer - C `a = 20 - 6pi, b = 10 pi - 30, c = sin^(-1) sin(4pi - 10) = 10 - 3pi` So, `a + b + c...
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सरल कीजिए, ` cos theta [{:( cos theta,, sin theta ), (-sin theta,, cos theta ):}] + sin theta [ {:(sin theta ,, - cos theta ), (cos theta ,, sin theta
Correct Answer - `[{:(,1,0),(,1,1):}]`
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