# Free Objective Test 01 Practice Test - 11th and 12th

The distance moved by the particle in time t is given by x=t312t2+6t+8. At the instant when its acceleration is zero, then the velocity is

A.

42

B.

-42

C.

48

D.

-48

#### SOLUTION

Solution : B

We have,
x=t312t2+6t+8
dxdt=3t224t+6andd2xdt2=6t24
Now, Acceleration =0
d2xdt2=06t24=0t=4
Att=4, we have
Velocity =(dxdt)r4=3×4224×4+6=42.

Hence (b) is the correct answer.

For what values of x is the rate of increase of x35x2+5x+8 is twice rate of increase of x ?

A.

3,13

B.

3,13

C.

3,13

D. 3,13

#### SOLUTION

Solution : D

Let y=x35x2+5x+8. Then,
dydx=(3x210x+5)dxdtWhendydt=2dxdt,wehave(3x210x+5)dxdt=2dxdt3x210x+3=0(3x1)(x3)=0x=3,13.

Hence (d) is the correct answer.

The  two curves x33xy2+2=0 and 3x2yy32=0

A.

Cut at right angles

B.

Touch each other

C.

Cut at an angle π/3

D.

Cut at an angle π/4

#### SOLUTION

Solution : A

x33xy2+2=0...(1)3x2yy32=0...(2)
On differentiating equations (1) and (2) w.r.t x, we obtain
(dydx)c1=x2y22xy and (dydx)c2=2xyx2y2
Since m1.m2=1.Therefore the two curves cut at right angles.
Hence (a) is the correct answer.

Number of possible tangents to the curve y=cos(x+y),3πx3π  that are parallel to the line x+2y = 0, is

A.

1

B.

2

C.

3

D.

4

#### SOLUTION

Solution : C

We have, y = cos (x + y)
dydx=sin(x+y)(1+dydx)
Since, the tangents are parallel to the line x + 2y = 0
12=sin(x+y)(112)sin(x+y)=1x+y=π2,5π2,3π21y1.
Hence (c) is the correct answer.

If the tangent to the curve x+y=a at any point on it cuts the axes OX and OY at P and Q respectively, then OP +OQ is

A.

a2

B.

a

C.

2a

D.

4a

#### SOLUTION

Solution : B

x+y=a.....(i)
12x+12ydydx=0 dydx=yx
Equation of tangent at (x1y1)isyy1=y1x1(xx1)
xx1+yy1=a;op=ax1,OQ=ay1OP+OQ=a

If the function f(x)=2x39ax2+12a2 x+1, where a > 0, attains its maximum and minimum at p and q respectively such that p2=q, then a equals

A.

3

B.

1

C.

2

D. 12

#### SOLUTION

Solution : C

We have, f(x)=2x39ax2+12a2 x+1f(x)=6x218ax+12a2=06[x23ax+2a2]=0x23ax+2a2=0x22axax+2a2=0x(x2a)a(x2a)=0(xa)(x2a)=0x=a,x=2a
Now, f(x)=12x18a
f(a)=12a18a=6a<0f(x) will be maximum at x = a
i.e. p = a
Also, f(2a)=24a18a=6af(x)will be minimum at x = 2a
i.e.q = 2a
Given, p2=q
a2=2aa=2.
Hence (c) is the correct answer.

A function f such that f(a)=f′′(a)=......f2n(a)=0 and f has a local maximum value b at x = a, if f (x) is

A.

(xa)2n+2

B.

b1(x+1a)2n+1

C.

b(xa)2n+2

D.

(xa)2n+2b.

#### SOLUTION

Solution : C

For local maximum or local minimum odd derivative must be equal to zero.
For local maxima, even derivative must be negative.
Since maximum value at x = a is b.

f(x)=b(xa)2n+2(f2n+2(a)=ve)
Hence (c) is the correct answer.

The number of values of x where the  function f(x) = 2 (cos 3x + cos 3x attains its maximum, is

A.

1

B.

2

C.

0

D.

Infinite

#### SOLUTION

Solution : A

We have,
f(x)=2(cos 3x+cos3x)=4 cos(3+32)x cos(332)x4
and it is equal to 4 when both cos (3+32) x and cos(332)

Are equal to 1 for a value of x. This is possible only when x = 0.
Hence (a) is the correct answer.

The point in the interval [0,2π] where f(x)=ex sin x has maximum slope, is

A. π4
B. π2
C.

π

D. 3π2

#### SOLUTION

Solution : B

We have, f(x)=ex+cos x+sin x exAnd f(x)=sin x ex+cos xex+cos x ex+sin x cos xex.Now,f(x)=2 cos x cos x ex=0cos x=0x=π2.Also,f(x)=2 sin xex+2 cos xex=ve
Slope is maximum at x=π2.

Hence (b) is the correct answer.

The approximate value of square root of 25.2 is

A.

5.01

B.

5.02

C.

5.03

D.

5.04

#### SOLUTION

Solution : B

Let f (x) = x
Now, f(x+δ x)f(x)=f(x).δ x=δx2x
We may write, 25.2 = 25 + 0.2
Taking x = 25 and δx=0.2 We have
f(25.2)f(25)=0.2225=0.02f(25.2)=f(25)+0.02=25+0.02=5.02(25.2)=5.02

If f (x) is differentiable in the interval [2, 5], where f (2)=15 and f (5)=12, then there exists a number c, 2 < c < 5 for which f ' (c) is equal to

A.

12

B.

15

C.

110

D.

7

#### SOLUTION

Solution : C

As f (x) is differentiable in [2 , 5], therefore, it is also continuos in [2, 5]. Hence, by mean value theorem, there exists a real number c in (2, 5) such that
f(c)=f(5)f(2)52f(c)=12153=110.

Hence (c) is the correct answer.

The equation x log x = 3 - x has, in the interval (1, 3),

A.

Exactly one root

B.

Atmost one root

C.

Atleast one root

D.

No root

#### SOLUTION

Solution : C

Let f (x) = (x - 3) log x
Then, f (1) = - 2 log 1 = 0 and f (3) = (3-3) log 3 = 0. As, (x-3) and log x are continuos and differentiable in [1, 3], therefore (x-3) log x = f (x) is also continuos and differentiable in [1, 3]. Hence, by Rolle's theorem, there exists a value of x in (1, 3) such that
f ' (x) = 0  log x+(x-3) 1x = 0
x log x = 3 - x.
Hence (c) is the correct answer.

Between any two real roots of the equation ex sin x = 1, the equation ex cos x = - 1 has

A.

Atleast one root

B.

Exactly one root

C.

Atmost one root

D.

No root

#### SOLUTION

Solution : A

Let ,β(<β) be any two real roots of
f(x) = e - x - sin x
Then, f()=0=f(β)
Moreover, f(x) is continuos and differentiable for x ε[,β].
Hence, from Rolle's thereom, thereom, there exists atleast one x in ,β such t
f(x)=0excos x=0ex(1+ex cos x)=0ex cos x=1.
Hence (a) is the correct answer.

Let f (x) = sinx + ax + b. Then f(x) = 0 has

A. only  one  real root which is  positive if a > 1, b < 0
B. only  one  real root which is negative if a > 1, b < 0
C. only one real root which is negative if a < 1, b > 0
D. CAN'T SAY ANYTHING

#### SOLUTION

Solution : A

f'(x) = - cosx + a, if a > 1,then f(x) entirely increasing. So f(x) =0 has only one real root, which is positive if f(0) < 0 and negative if f(0) > 0.

Similarly when a < -1. Then f(x) entirely decreasing. So f(x) has only one real root which is negative if f(0) < 0 and positive if f(0) > 0

Let f(x) = {1 + sin x, x < 0x2  x + 1, x  0. Then

A. f has a local maximum at x = 0
B. f has a local minimum at x = 0
C. f is increasing every where
D. f is decreasing everywhere

#### SOLUTION

Solution : A

f is continuous at ‘0’ and f' (0-) > 0 and f' ( 0 +) < 0 . Thus f has a local maximum at ‘0’.