Free Application of Derivatives 01 Practice Test - 12th Grade - Commerce
Question 1
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 2
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 3
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 4
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 5
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 6
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
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)5−2⇒f′(c)=12−153=110.
Hence (c) is the correct answer.
Question 7
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 8
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 9
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 10
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’.