Application Of Derivatives(11th And 12th > Mathematics ) Questions and Answers
Explanation:-
Answer: Option B. -> -42:
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.
Explanation:-
Answer: Option A. -> Cut at right angles:
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.
Explanation:-
Answer: Option C. -> 3:
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.
Explanation:-
Answer: Option C. -> 2:
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.
Explanation:-
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.
Explanation:-
Answer: Option B. -> π2:
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.
Explanation:-
Answer: Option A. -> 1:
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.