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Q.

Which of the following statement(s) is/are CORRECT?

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a

Let f(x) is a continuous and differentiable function on [0,1] and f(0)=0,f(1)=1 then the minimum value of 01(f'(x))2dx is 1

b

The value of x + y which satisfies the equation 3x24xy+3y2+10x10y+10=0 is 0

c

Given that x1+x2+x3=0,y1+y2+y3=0 and  x1y1+x2y2+x3y3=0  (x1,x2,x3,y1,y2,y3R)  then the value of x12x12+x22+x32+y12y12+y22+y32 is 1

d

The variance of  the observations 112, 116, 120, 125, 132 is 48.8

answer is A, B, D.

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Detailed Solution

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A)  f(0)=0,f(1)=1  (f'(x)1)20     (f'(x))22f'(x)+10      01(f'(x))2dx201f'(x)dx+011dx0        01(f'(x))2dx1

B)  3(1+x)2+3(1y)2+4(1+x)(1y)=0       (1+x)2+(1y)2+2((1+x)+(1y))2=0        x=1,y=1

C)   n¯1=(x1,x2,x3),n¯2=(y1,y2,y3),n¯3=(1,1,1)     n¯1.n¯3=0,n¯1.n¯2=0,n¯2.n¯3=0        Consider   p¯=(1,0,0)

p¯.n¯1|n¯1|=x1x12+x22+x32,p¯.n¯2|n¯2|=y1y12+y22+y32,p¯.n¯3|n¯3|=13       (p¯.n¯1|n¯1|)2+(p¯.n¯2|n¯2|)2+(p¯.n¯3|n¯3|)=|p¯|2      x12x12+x22+x32+y12y12+y22+y32+13=1

D) 112, 116, 120, 125, 132
Variance is independent of change of scale shift the origin to 120
The observations are –8, –4, 0, 5, 12
Variance  =σ2=i=1n(xix¯)2n=i=15(xi1)25
=2445=48.80

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