# Free Objective Test 03 Practice Test - 11th and 12th

The value of limx0(1x+2x+3x+...+nxn)a/x is

A. (2n!)an
B. (n!)2an
C. (n!)an
D. 0

#### SOLUTION

Solution : C

limx0(1x+2x+3x+...+nxn)a/x(1form)
=elimx0(1x+2x+3x+...+nxn1).an
=elimx0(1x+2x+3x+...+nxnn).an
=elimx0(1x+2x+3x+...+nxnx).an
=elimx0{1x1x+2x1x+...+nx1x}an
We know limx0{ax1x}=a
So, =e(log1+log2+log3+...logn)an
ea/n(log(1.2.3...n)) =  e(logen!)a/n=(n!)a/n

If α is a repeated root of ax2+bx+c=0 then limxαsin(ax2+bx+c)(xα)2 is

A. 0
B. a
C. b
D. c

#### SOLUTION

Solution : B

limxαsin(ax2+bx+c)(xα)2=limxαsina(xα)(xα)(xα)2
=limxαsina(xα)2a(xα)2×a=a

limx0(cosx)1/x4

A. 0
B. 1
C. e
D. limit does not exist

#### SOLUTION

Solution : A

limx0(cosx)1/x4=elimx0cosx1x4
elimx02sinx2x2.sinx2x2×14x2=e=0

Let f  be a function such that f(x+y)=f(x)+f(y)  for all x  and y  and f(x)=(2x2+3x)g(x) for all x  where g(x)  is continuous and g(0)=9  then f'(0)  is equals to

A. 9
B. 3
C. 27
D. 6

#### SOLUTION

Solution : C

f(x)=limh0f(x+h)f(x)h
=limh0f(x)+f(h)f(x)h
=limh0(2h2+3h)g(h)h
=limh0(2h+3)g(h)
=3g(0)
=27

ABC is an isosceles triangle inscribed in a circle of radius r. If AB = AC and h is the altitude from A to BC.The triangle ABC has perimeter P=2[(2hrh2)+2hr] and A be the area of the triangle .Find limh0AP3

A. 1r
B. 164r
C. 1128r
D. 12r

#### SOLUTION

Solution : C

In ABC,AB=AC
ADBC(D is mid point of BC)
Let r = radius of circumcircle
OA = OB = OC = r
Now BD = BO2OD2
=r2(hr)2=2rhh2
BC=22rhh2
=h2rhh2
Also limh0Ap3=h2rhh28(2rhh2+2hr)3
=limh0h3/22rh8h3/2(2rh+2r)3
=limh02rh8[2rh+2r]3
=2r8(2r+2r)3=2r8.8.2r2r=1128r

limn 20x=1 cos 2n(x10) is equal to

A.

0

B.

1

C.

19

D.

20

#### SOLUTION

Solution : B

limn cos 2nx = {1,x=rπ,rI0,xrπ,rI

Here, for x = 10, limncos2n(x10) = 1

and in all other cases it is zero

limn20x=1cos2n(x10) = 1

The value of limxx+cos xx+sin xis

A.

-1

B.

0

C.

1

D.

2

#### SOLUTION

Solution : C

limxx+cos xx+sin x[Puttingx=1h;as x,h 0]=limh 01h+cos(1h)1h+sin(1h)=limh 01+h cos 1h1+h cos 1h=1+01+0⎢ ⎢ ⎢1 sin 1h1 and 1 cos1h1, where h0,hcos 1h0 and hsin1h0⎥ ⎥ ⎥=1

limx0(1+x)13(1x)13x=

A.

2/3

B.

1/3

C.

1

D.

0

#### SOLUTION

Solution : A

limx0(1+x)13(1x)13x=limx0(1+x)13(1x)13(1+x)(1x).2=2.13

If limx0kx cosec x = limx0x cosec kx , then k =

A.

1

B.

-1

C.

D.

#### SOLUTION

Solution : C

limx0kx cosec x = limx0 x cosec kx

= k limx0xsinx =1klimx0kxsinkx =k=1k=k=± 1

limx0tan1xsin1xx3is equal to

A.

1/2

B.

-1/2

C.

1

D.

-1

#### SOLUTION

Solution : B

limx0tan1xsin1xx3   (00form)limx011+x211x23x2limx01x2(1+x2)3x2(1+x2)1x2limx0(1x2)(1+x2)23x2(1+x2)1x2[1x2+(1+x2)]=36=12

The value oflimx031+sinx31sinxxis

A.

2/3

B.

-2/3

C.

3/2

D.

-3/2

#### SOLUTION

Solution : A

limx031+sinx31sinxx=limx0(1+sinx)(1sin s)(1+sin x)2/3+(1+sin x)2/3+(1sin x)2/3(1sinx)2/3×1x=limx02(1+sin x)2/3+(1+sin x)1/3+(1sin x)1/3+(1sinx)2/3×sin(x)x
= 23.1=23

Iflimx(x2x+1axb)=0  then the values of a and b are given by -

A.

a=1,b=12

B.

a=1,b=12

C.

a=1,b=12

D.

a=-1, b=-1/2

#### SOLUTION

Solution : A

We have,limx(x2x+1axb)=0     [putting x=-y; x,y]limy(y2+y+1+ayb)=0limy[y(1+1y+1y2)1/2+ayb]=0limy[y{1+12(1y+1y2+.........)}+ayb]=0limy[y(1+a)+(12b)+12y+........]=01+a=0 and (1/2)b=0         a=1 and b=1/2

The values of constants a and b so thatlimx(x2+1x+1axb)=12,are

A.

a=1,b=32

B.

a=1,b=32

C.

a=0, b=0

D.

a =2, b= -1

#### SOLUTION

Solution : A

limx(x2+1x+1axb)=12limx(x2+1)(ax+b)(x+1)x+1=12limxx2(1a)(a+b)xb+1x+1=121a=0 and a+b=12a=1 b=32

The function f(x)=log(1+ax)log(1bx)x is not defined at x-0 . The value which should be assigned to f at x = 0 so that it is continuos at x=0 , is

A. a-b
B. a+b
C. log a+log b
D. log a-log b

#### SOLUTION

Solution : B

Since limit of a function is a + b as x  0, therefore to be continuous at a function, its value must be
a + b at x = 0  f(0) = a + b.

If the derivative of the function f(x)=bx2+ax+4;x1ax2+b;x<1, is everywhere continuous, then

A. a = 2, b = 3
B. a = 3, b = 2
C. a = - 2, b = - 3
D. a = - 3, b = - 2

#### SOLUTION

Solution : A

We have, f(x)={ax2+b,x<1bx2+ax+4,x1f(x)={2ax,<12bx+a,x1Since, f(x) is differentiable at x=1, therefore it is continuous at x=1 and hence,limx1f(x)=limx1+f(x)a+b=ba+4a=2and also, limx1f(x)=limx1+f(x)2a=2b+a3a=2bb=3     (a=2)Hence, a=2,b=3