12th Grade > Mathematics
APPLICATION OF DERIVATIVES MCQs
Total Questions : 58
| Page 4 of 6 pages
Answer: Option A. -> 1
:
A
We have,
f(x)=2(cos3x+cos√3x)=4cos(3+√32)xcos(3−√32)x⩽4
and it is equal to 4 when both cos (3+√32)xandcos(3−√32)
Are equal to 1 for a value of x. This is possible only when x = 0.
Hence (a) is the correct answer.
:
A
We have,
f(x)=2(cos3x+cos√3x)=4cos(3+√32)xcos(3−√32)x⩽4
and it is equal to 4 when both cos (3+√32)xandcos(3−√32)
Are equal to 1 for a value of x. This is possible only when x = 0.
Hence (a) is the correct answer.
Answer: Option B. -> π2
:
B
We have, f′(x)=ex+cosx+sinxexAndf′(x)=−sinxex+cosxex+cosxex+sinxcosxex.Now,f′(x)=2cosxcosxex=0⇒cosx=0⇒x=π2.Also,f′(x)=−2sinxex+2cosxex=−ve
∴ Slope is maximum at x=π2.
Hence (b) is the correct answer.
:
B
We have, f′(x)=ex+cosx+sinxexAndf′(x)=−sinxex+cosxex+cosxex+sinxcosxex.Now,f′(x)=2cosxcosxex=0⇒cosx=0⇒x=π2.Also,f′(x)=−2sinxex+2cosxex=−ve
∴ Slope is maximum at x=π2.
Hence (b) is the correct answer.
Answer: Option C. -> b−(x−a)2n+2
:
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.
:
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.
Answer: Option C. -> Atleast one root
:
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.
:
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.
Answer: Option A. -> f has a local maximum at x = 0
:
A
f is continuous at ‘0’ and f'(0-) > 0 and f'( 0 +) < 0 . Thus f has a local maximum at ‘0’.
:
A
f is continuous at ‘0’ and f'(0-) > 0 and f'( 0 +) < 0 . Thus f has a local maximum at ‘0’.
Answer: Option A. -> Atleast one root
:
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 xε[∝,β].
Hence, from Rolle's thereom, thereom, there exists atleast one x in ∝,β such t
f′(x)=0⇒−e−x−cosx=0⇒−e−x(1+excosx)=0⇒excosx=−1.
Hence (a) is the correct answer.
:
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 xε[∝,β].
Hence, from Rolle's thereom, thereom, there exists atleast one x in ∝,β such t
f′(x)=0⇒−e−x−cosx=0⇒−e−x(1+excosx)=0⇒excosx=−1.
Hence (a) is the correct answer.
Answer: Option C. -> 110
:
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.
:
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.
Answer: Option B. -> 5.02
:
B
Let f (x) = √x
Now, f(x+δx)−f(x)=f′(x).δx=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
:
B
Let f (x) = √x
Now, f(x+δx)−f(x)=f′(x).δx=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
Answer: Option A. -> Atleast one root
:
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 xε[∝,β].
Hence, from Rolle's thereom, thereom, there exists atleast one x in ∝,β such t
f′(x)=0⇒−e−x−cosx=0⇒−e−x(1+excosx)=0⇒excosx=−1.
Hence (a) is the correct answer.
:
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 xε[∝,β].
Hence, from Rolle's thereom, thereom, there exists atleast one x in ∝,β such t
f′(x)=0⇒−e−x−cosx=0⇒−e−x(1+excosx)=0⇒excosx=−1.
Hence (a) is the correct answer.
Answer: Option C. -> 2
:
C
We have, f(x)=2x3−9ax2+12a2x+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.
:
C
We have, f(x)=2x3−9ax2+12a2x+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.
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