# Free Objective Test 02 Practice Test - 11th and 12th

A man of height 2m walks directly away from a lamp of height 5m, on a level road at 3 m/s. The rate at which the length of his shadow is increasing is

A.

1m/s

B.

2m/s

C.

3m/s

D.

4m/s

#### SOLUTION

Solution : B

Let be the lamp and PQ be the man and OQ=x metre be his shadow and let MQ =y metre. dydt=speed of the man
=3 m/s (given)
ΔOPQ and ΔOLM are similar
OMOQ=LMPQx+yx=52y=32xdydt=32dxdt3=32dxdtdxdt=2m/s

A ladder20 ft long has one end on the ground and the other end in contact with a vertical wall. The lower end slips along the ground. If the lower end of the ladder is 16 ft away from the wall, upper end is moving λ  times as fast as the lower end, then λ is

A.

13

B.

23

C.

43

D.

53

#### SOLUTION

Solution : C

Let OC be the wall. Let AB be the position of the ladder at any time t such that OA =x and OB=y. Length of the ladder AB =20 ft.
In ΔAOB, x2+y2=(20)2
2xdxdt+2ydydt=0dydt=xydxdt=x400x2.dxdt=16400(16)2.dxdt=43dxdt
-ve sign indicates, that when X increases with time, y decreases. Hence, the upper end is moving 43 times as fast as the lower end.

The angle of intersection of the curves y=2sin2 x and y= cos 2x at x =π6 is

A.

π4

B.

π3

C.

π2

D.

2π3

#### SOLUTION

Solution : B and D

We have,
y=sin2x...(1)y=cos 2x...(2) And
On differentiating equation (1) w.r.t x, we get
dydx=4 sin x cos x[dydx]xπ6=4(12)32=3=m1(say)
On differentiating equation (2) w.r.t x, we get
dydx=2 sin 2x[dydx]xπ6=2 sin π3=3=m2(say)
Hence, angle between the two curves is
θ=±tan1(m1m21+m1 m2)=±tan13=π3or2π3
Hence (b) is the correct answer.

The slope of the tangent to the curve x=t2+3t8,y=2t22t5 at the point t = 2 is

A. 76
B.

56

C.

67

D.

1

#### SOLUTION

Solution : C

We have,

dxdt=2t+3 and dydt=4t2dydx=dy/dtdx/dt=4t22t+3
Thus, slope of the tangent to the curve at the point t = 2 is
[dydx]t2=4(2)22(2)+3=67
Thus, slope of the tangent to the curve at the point t = 2 is
Hence (c) is the correct answer

The value of m for which the area of the triangle included  between the axes  and any tangent to the curve xm y=bm is constant is

A.

12

B.

1

C.

32

D.

2

#### SOLUTION

Solution : B

xmy=bm
Taking logarithm
m loge x+loge y=mloge bmx+1ydydx=0dydx=myx
Equation of tangent at (x,y) is
Yy=myx(Xx)xYxy=myX+mxymyX=xY=xy(1+m)Xx(1+m)m+xx(1+m)=1
Area of triangle OAB
=12.OA.OB =12x(1+m)m|y(1+m)|=|xy|(1+m)22|m|For m=1,=|xy|(4)2=2|xy|(xy=b)=2|b|= constant

The minimum value of the function defined by f(x) = minimum {x, x+1,2 - x } is

A.

0

B. 12
C.

1

D. 32

#### SOLUTION

Solution : D

f(x) = maximum {x, x + 1, 2-x} Minimum value of function =32

The value of (127)1/3 to four decimal places is

A.

5.0267

B.

5.4267

C.

5.5267

D.

5.001

#### SOLUTION

Solution : A

Let y = x1/3,x=125 and x+Δx=127. Then,
dydx=13x2/3 and Δ x=0
When, x = 125, we have
y=5 and dydx=175
y=dydxΔ xΔ y175×2=275
(127)1/3=y+Δ y=5+275=5+83×1100
(127)1/3=5+(2.6667)100=5.02667=5.0267
Hence (a) is the correct answer.

If  a + b +c = 0, then the equation 3ax2+2bx+c=0  has, in the interval (0, 1)

A.

Atleast one root

B.

Atmost one root

C.

No root

D.

exactly one root

#### SOLUTION

Solution : A

Let f(x)=an Xn+an1 Xn1+...+a2 x2+a1 x+a0
Which is a polynomial function in x of degree n. Hence f(x)  is continuos and differentiable for all x.
Let <β  . We given, f ()=0=f(β).
By Rolle's theorem, f' (c) = 0 for some value c,
<c<β.
Hence the equation
F(x)=nanxn1+(n1)an1xn2+...+a1=0

has atleast one root between  and β.
Hence (c) is the correct answer.

If f(x)=∣ ∣sin xsin asin bcos xcos acos btan xtan atan b∣ ∣,

where 0<a<b<π2
then the equation
f(x)=0 has in the interval (a,b)

A.

Atleast one root

B.

Atmost one root

C.

No root

D.

exactly one root

#### SOLUTION

Solution : A

Here f(a)=∣ ∣sin asin asin bcos acos acos btan atan atan b∣ ∣=0.Also f(b)=0.
Moreover, as sin x, cos x and tan x are continuos and differentiable in (a, b) for 0 < a < b < π2, therefore f(x) is also continuos and differentiable in [a, b]. Hence, by Rolle's theorem, there exists atleast one real number c in (a, b) such that f ' (c) = 0.
Hence (a) is the correct answer.

f(x) = cos (πx), x  0 increases in the interval

A. (12k+1, 12k)
B. (12k+2, 12k+1)
C. (12k+1, 12k+3)
D. (12k+1, 12k+3)

#### SOLUTION

Solution : A

f(x) = cos (πx)
Thus
f(x) = cos (πx)
increases in
1312
generalizing it
(12k+1,12k) f(x) = xloge x, x  1, is decreasing in interval

A. (0, e)
B. (1, e)
C. (e, )
D. R

#### SOLUTION

Solution : A

f'(x) = logex.1x.1x(logex)2
=(logex1)(logex)2<0
It is decreasing at (0, e) – {1} If f"(x) > 0  x ϵ R then for any two real numbers x1 and x2 , (x1  x2)

A. f(x1 + x22) > f(x1) + f(x2)2
B. f(x1 + x22) < f(x1) + f(x2)2
C. f(x1 + x22) > f(x1) + f(x2)2
D. f(x1 + x22) < f(x1) + f(x2)2

#### SOLUTION

Solution : B

Let A = (x1,f(x1)) and B = (x2,f(x2)) be any two points on the graph of y = f(x).
Since f"(x) > 0, in the graph of the function tangent will always lie below the curve. Hence chord AB will lie completely above the graph of y = f(x).
Hence f(x1)+f(x2)2>f(x1+x22)

f(x) = x2  4|x| and g(x) = {min{f(t) : 6  t  x},   x ϵ [6, 0]max{f(t) : 0  t  x},   x ϵ [0, 6], than g(x)

A. Exactly one point of local minima
B. Exactly one point of local maxima
C. No point of local maxima but exactly one point of local minima
D. Neither a point of local maxima nor minima

#### SOLUTION

Solution : D

Bold line represents the graph of y = g(x) clearly g(x) has neither a point of local maxima nor a point of local minima. Let S be the set of real values of parameter λ  for which the equation f(x) = 2x3  3(2+λ)x2 + 12λ x has exactly one local maximum and exactly one local minimum. Then S is a subset of

A. (4, )
B. (3, 3)
C. (3, )
D. R

#### SOLUTION

Solution : C

f(x) = 2x3  3(2+λ)x2 + 12λ x f(x) = 6x2  6(2+λ)x + 12λf(x) = 0  x = 2, λ
If f(x) has exactly one local maximum and exactly one local minimum, then λ  2.

The tangent to the curve x =  acos2θcosθ,y=acos2θsinθ at the point corresponding to θ=π6 is

A. parallel to the x-axis
B. parallel to the y-axis
C. parallel to line y = x
D. 3X-4Y+2=0

#### SOLUTION

Solution : A

dxdθ = a cos 2θsin θ + a cos θ sin θcos 2θ= a(cos 2θ sin θ + cos θ sin 2 θ)cos 2θ = a sin 3θcos 2θ= dydθ = a cos 2θ cos θ  a sin θ sin 2θcos 2θ = a cos 3θcos 2θ
Hence dydx=cot3θdydx|θ=π6 = 0
So the tangent to the curve at θ=π6 is parallel to the x-axis.