Find the principal values of `cos^(-1)(sqrt(3))/2` and `cos^(-1)(-1/2)`
Correct Answer - C `int_(-1)^(1//2)(e^(x)(2-x^(2))dx)/((1-x)sqrt(1-x^(2)))=int_(-1)^(1//2)(e^(x)(1-x^(2)+1))/((1-x)sqrt(1-x^(2)))` `=int_(-1)^(1//2) e^(x)[(sqrt(1+x))/(1-x)+1/((1-x)sqrt(1-x^(2)))]dx` `=e^(x)sqrt(1+x)/(1-x)|_(-1)^(1//2)=sqrt(3e)`
2 Answers 1 viewsCorrect Answer - D `(a)/(sqrt(bc))-2 = sqrt((b)/(c)) + sqrt((c)/(b))` `"or " a=b+c+2sqrt(bc)` `"or " a=(sqrt(b)+sqrt(c))^(2)` `"or " (sqrt(a)-sqrt(b)-sqrt(c))(sqrt(a)+sqrt(b)+sqrt(c))=0` `"or " sqrt(a)-sqrt(b)-sqrt(c)=0` `"since " sqrt(a)+sqrt(b)+sqrt(c)ne0 " " ("as "a,b,c gt 0).` `"Comparing...
2 Answers 1 viewsCorrect Answer - A::C The equations of lines passing through (1,0) are given by y=m(x-1). Its distance from the origin is `sqrt(3)//2`. Hence, `|(-m)/(sqrt(1+m^(2)))| = (sqrt(3))/(2) " or "m=+-sqrt(3)` Hence, the...
2 Answers 2 views`sqrt(|x|-2)` we know that square roots are defined for non- negative values only . It implies that we must have `|x|-2 le 0 ` Thus `sqrt(|x|-2) ge 0 ` (ii)...
2 Answers 1 views(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`...
2 Answers 1 views`E = sqrt(sin^(-1)x_(1)) sqrt(cos^(-1) x_(2)) + sqrt(sin^(-1) x_(2)) sqrt(cos^(-1) x_(3)) + sqrt(sin^(-1) x_(3)) sqrt(cos^(-1) x_(4)) +...+ sqrt(sin^(-1) x_(28)) sqrt(cos^(-1) x_(1))` `x_(i) in [0,1] AA i = 1, 2, 3, ..,...
2 Answers 1 viewsCorrect 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...
2 Answers 1 viewsArea of qudrilateral AbCD is maximum when area of ACD is maximum. Area of triangle ACD. `Delta_1=(1)/(2)||{:(3/sqrt2,,sqrt2,,),(3/sqrt2,,-sqrt2,,),(3sintheta,,2costheta,,),(3/sqrt2,,sqrt2,,):}||` `=|-3sqrt(2)costheta+3sqrt(2)sintheta|` `therefore Delta_1("max")=6`, when`theta=(7pi)/(4)` (as `theta epsilon(3pi//2,2pi`)) Maximum area is 12 sq.units (as...
2 Answers 1 viewsCorrect Answer - i) `-pi/4`, ii) `(5pi)/6`, iii) `-pi/3`, iv) `(2pi)/3`, v) `-pi/4`, vi) `(2pi)/3`
2 Answers 1 views`(dy)/(dx)=(e^(sqrt(x)))/(2sqrt(x))-(e^(-sqrt(x)))/(2sqrt(x))=(e^(sqrt(x)-)e^(-sqrt(x)))/(2sqrt(x))` `=(sqrt((e^(sqrt(x))+e^(-sqrt(x)))^(2))-4)/(2sqrt(x))` `=(sqrt(y^(2)-4))/(2sqrt(x))`
2 Answers 1 views