Free Objective Test 02 Practice Test - 11th and 12th
Question 1
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
1m/s
2m/s
3m/s
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=LMPQ⇒x+yx=52⇒y=32x∴dydt=32dxdt⇒3=32dxdt⇒dxdt=2m/s
Question 2
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
13
23
43
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=0∴dydt=−xydxdt=−x√400−x2.dxdt=−16√400−(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.
Question 3
The angle of intersection of the curves y=2sin2 x and y= cos 2x at x =π6 is
π4
π3
π2
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]x−π6=4(12)√32=√3=m1(say)
On differentiating equation (2) w.r.t x, we get
dydx=−2 sin 2x[dydx]x−π6=−2 sin π3=−√3=m2(say)
Hence, angle between the two curves is
θ=±tan−1(m1−m21+m1 m2)=±tan−1√3=π3or2π3
Hence (b) is the correct answer.
Question 4
The slope of the tangent to the curve x=t2+3t−8,y=2t2−2t−5 at the point t = 2 is
56
67
1
SOLUTION
Solution : C
We have,dxdt=2t+3 and dydt=4t−2dydx=dy/dtdx/dt=4t−22t+3
Thus, slope of the tangent to the curve at the point t = 2 is
[dydx]t−2=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
Question 5
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
12
1
32
2
SOLUTION
Solution : B
∴xmy=bm
Taking logarithm
m loge x+loge y=mloge b∴mx+1ydydx=0∴dydx=−myx
Equation of tangent at (x,y) is
Y−y=−myx(X−x)xY−xy=−myX+mxy⇒myX=xY=xy(1+m)⇒Xx(1+m)m+xx(1+m)=1
∴ Area of triangle OAB
=12.OA.OB
=12∣∣x(1+m)m∣∣|y(1+m)|=|xy|(1+m)22|m|For m=1,=|xy|(4)2=2|xy|(∵xy=b)=2|b|= constant
Question 6
The minimum value of the function defined by f(x) = minimum {x, x+1,2 - x } is
0
1
SOLUTION
Solution : D
∵ f(x) = maximum {x, x + 1, 2-x}
∴ Minimum value of function =32
Question 7
The value of (127)1/3 to four decimal places is
5.0267
5.4267
5.5267
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 x⇒Δ 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.
Question 8
If a + b +c = 0, then the equation 3ax2+2bx+c=0 has, in the interval (0, 1)
Atleast one root
Atmost one root
No root
exactly one root
SOLUTION
Solution : A
Let f(x)=an Xn+an−1 Xn−1+...+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)=nanxn−1+(n−1)an−1xn−2+...+a1=0
has atleast one root between ∞ and β.
Hence (c) is the correct answer.
Question 9
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)
Atleast one root
Atmost one root
No root
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.
Question 10
f(x) = cos (πx), x ≠ 0 increases in the interval
SOLUTION
Solution : A
f(x) = cos (πx)
Thus
f(x) = cos (πx)
increases in
13→12
generalizing it
(12k+1,12k)
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Question 11
f(x) = xloge x, x ≠ 1, is decreasing in interval
SOLUTION
Solution : A
f'(x) = logex.1−x.1x(logex)2
=(logex−1)(logex)2<0
It is decreasing at (0, e) – {1}
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Question 12
If f"(x) > 0 ∀ x ϵ R then for any two real numbers x1 and x2 , (x1 ≠ x2)
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)
Question 13
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)
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.
Question 14
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
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.
Question 15
The tangent to the curve x = a√cos2θcosθ,y=a√cos2θsinθ at the point corresponding to θ=π6 is
SOLUTION
Solution : A
dxdθ = −a √cos 2θsin θ + −a cos θ sin θ√cos 2θ= −a(cos 2θ sin θ + cos θ sin 2 θ)√cos 2θ = −a sin 3θ√cos 2θ= dydθ = a √cos 2θ cos θ − a sin θ sin 2θ√cos 2θ = a cos 3θ√cos 2θ
Hence dydx=−cot3θ⇒dydx|θ=π6 = 0
So the tangent to the curve at θ=π6 is parallel to the x-axis.