The set of all real numbers is
A) Infinite set
B) Φ
C) Finite set
D) Equal set
Correct Answer - C `a^(2)+b^(2)=81` `x^(2)+y^(2)=121` `ax+by=99` `(a^(2)+b^(2))(x^(2)+y^(2))=(81)(121)`…(i) and `(ax+by)^(2)=(99)^(2)` ….(2) `(1)-(2)` `(ay-bx)^(2)=0` `ay - bx = 0`
2 Answers 1 viewsCorrect Answer - D `x^(2)+2ax+b^(2)=0" "x^(2)+2bx+c^(2)=0` `D_(1)gt0" "D_(2)gt0` `4a^(2)+b^(2)gt0" "4b^(2)-4c^(2)gt0` `a^(2)gtb^(2)…..(1)" "b^(2)gtc^(2)…….(2)` From (1) and (2) `a^(2)gtb^(2)gtc^(2)rArra^(2)gtc^(2)rArrc^(2)-a^(2)lt0` `x^(2)+2cx+a^(2)=0` `D=4c^(2)-4a^(2)lt0" No real roots"`
2 Answers 1 viewsCorrect Answer - D (A) `[x+y]le[x]+[y]` let x=0.1 y=0.9 `[0.1+0.9]le[0.1]+[0.9]` `1 le 9+0` wrong (B) `[xy]le[x]+[y]` `x=2,y=(1)/(2)` `[2*(1)/(2)]le[2][(1)/(2)]` `rArr1le0` wrong (C) `[2^(x)]le2^([x])` `x=0.99[2^(0.99)]le2^([0.99])` `[2^(0.99)]le2^(@)=1` wrong (D) `[(x)/(y)]le([x])/([y])` given `x,yge1` if `xlty[(x]/(y)]=0"...
2 Answers 1 viewsCorrect Answer - B By talking different values of a & k. options (B) is possible.
2 Answers 1 viewsCorrect Answer - B `(sin(lambdaalpha))/(sinalpha)-(cos(lambdaalpha))/(cosalpha)=lambda-1` By observation `sin(lambdaalpha)cosalpha-cos(lambdaalpha)sinalpha=(lambda-1)sinalphacosalpha` `sin(alpha-1)alpha=(lambda-1)sinalphacosalpha` clearly `lambda=1, lambda=3` is solution
2 Answers 1 viewsSymmetric but neither reflexive nor transitive
2 Answers 1 viewsCorrect Answer - C `f(x)+(x+1/2)f(1-x)=1` `f(l-x)+(1-x+1/2)f(1-(1-x))=1` `f(1-x)+(3/2-x)f(x)=1` `(1-f(x))/(x+1/2)+(3/2-x)f(x)=1` `1-f(x)+(3/2x-x^(2)+3/4-x/2)f(x)=x+1/2` `f(x)(x-x^(2)-1/4)=x-1/2` `f(x)(4x-4x^(2)-1)=4x-2` `f(x)=(-2+4x)/(4x-4x^(2)-1)` `2f(0)+3f(1)=2((-2+0)/(0-0-1))+3((-2+4)/(4-4-1))` `=+4+(3(+2))/-1=+4-6=-2` Alternate Put x = 0 `f(0)+1/2f(1)=1rArr2f(0)+f(1)=2` put x = 1 `f(1)+3/2f(0)=1rArr2f(1)+3f(0)=2` solving above `f(0)=2" and "f(1)=-2` `:....
2 Answers 1 viewsCorrect option is B) 90°, 45°, 45°
2 Answers 1 viewsCorrect option is B) The set of all multiples of 5
2 Answers 1 viewsCorrect option is C) Set of even prime numbers
2 Answers 1 views