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

### Question 1

The distance moved by the particle in time t is given by x=t3−12t2+6t+8. At the instant when its acceleration is zero, then the velocity is

42

-42

48

-48

#### SOLUTION

Solution :B

We have,

x=t3−12t2+6t+8

⇒dxdt=3t2−24t+6andd2xdt2=6t−24

Now, Acceleration =0

⇒d2xdt2=0⇒6t−24=0⇒t=4

Att=4, we have

Velocity =(dxdt)r−4=3×42−24×4+6=−42.

Hence (b) is the correct answer.

### Question 2

For what values of x is the rate of increase of x3−5x2+5x+8 is twice rate of increase of x ?

−3,−13

−3,13

3,−13

#### SOLUTION

Solution :D

Let y=x3−5x2+5x+8. Then,

dydx=(3x2−10x+5)dxdtWhendydt=2dxdt,wehave(3x2−10x+5)dxdt=2dxdt⇒3x2−10x+3=0⇒(3x−1)(x−3)=0⇒x=3,13.

Hence (d) is the correct answer.

### Question 3

The two curves x3−3xy2+2=0 and 3x2y−y3−2=0

Cut at right angles

Touch each other

Cut at an angle π/3

Cut at an angle π/4

#### SOLUTION

Solution :A

x3−3xy2+2=0...(1)3x2y−y3−2=0...(2)

On differentiating equations (1) and (2) w.r.t x, we obtain

(dydx)c1=x2−y22xy and (dydx)c2=−2xyx2−y2

Since m1.m2=−1.Therefore the two curves cut at right angles.

Hence (a) is the correct answer.

### Question 4

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

1

2

3

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)(1−12)⇒sin(x+y)=1⇒x+y=π2,5π2,3π2−1≤y≤1.

Hence (c) is the correct answer.

### Question 5

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

a2

a

2a

4a

#### SOLUTION

Solution :B

√x+√y=√a.....(i)

⇒12√x+12√ydydx=0

∴dydx=−√y√x

Equation of tangent at (x1y1)isy−y1=−√y1√x1(x−x1)

⇒x√x1+y√y1=√a;⇒op=√a√x1,OQ=√a√y1∴OP+OQ=√a

### Question 6

If the function f(x)=2x3−9ax2+12a2 x+1, where a > 0, attains its maximum and minimum at p and q respectively such that p2=q, then a equals

3

1

2

#### SOLUTION

Solution :C

We have, f(x)=2x3−9ax2+12a2 x+1∴f(x)=6x2−18ax+12a2=0⇒6[x2−3ax+2a2]=0⇒x2−3ax+2a2=0⇒x2−2ax−ax+2a2=0⇒x(x−2a)−a(x−2a)=0⇒(x−a)(x−2a)=0⇒x=a,x=2a

Now, f′(x)=12x−18a

∴f′(a)=12a−18a=−6a<0∴f(x) will be maximum at x = a

i.e. p = a

Also, f′(2a)=24a−18a=6a∴f(x)will be minimum at x = 2a

i.e.q = 2a

Given, p2=q

⇒a2=2a⇒a=2.

Hence (c) is the correct answer.

### Question 7

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

(x−a)2n+2

b−1−(x+1−a)2n+1

b−(x−a)2n+2

(x−a)2n+2−b.

#### 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−(x−a)2n+2(∵f2n+2(a)=−ve)

Hence (c) is the correct answer.

### Question 8

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

1

2

0

Infinite

#### SOLUTION

Solution :A

We have,

f(x)=2(cos 3x+cos√3x)=4 cos(3+√32)x cos(3−√32)x⩽4

and it is equal to 4 when both cos (3+√32) x and cos(3−√32)

Are equal to 1 for a value of x. This is possible only when x = 0.

Hence (a) is the correct answer.

### Question 9

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

π

#### 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=0⇒cos x=0⇒x=π2.Also,f′(x)=−2 sin xex+2 cos xex=−ve

∴ Slope is maximum at x=π2.

Hence (b) is the correct answer.

### Question 10

The approximate value of square root of 25.2 is

5.01

5.02

5.03

5.04

#### SOLUTION

Solution :B

Let f (x) = √x

Now, f(x+δ x)−f(x)=f′(x).δ x=δx2√x

We may write, 25.2 = 25 + 0.2

Taking x = 25 and δx=0.2 We have

f(25.2)−f(25)=0.22√25=0.02∴f(25.2)=f(25)+0.02=√25+0.02=5.02⇒√(25.2)=5.02

### Question 11

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

12

15

110

7

#### SOLUTION

Solution :C

Asf(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)5−2⇒f′(c)=12−153=110.

Hence (c) is the correct answer.

### Question 12

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

Exactly one root

Atmost one root

Atleast one root

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.

### Question 13

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

Atleast one root

Exactly one root

Atmost one root

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)=0⇒−e−x−cos x=0⇒−e−x(1+ex cos x)=0⇒ex cos x=−1.

Hence (a) is the correct answer.

### Question 14

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

#### 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

### Question 15

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

#### SOLUTION

Solution :A

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