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 Bihar Board 12th Maths Important Questions Long Answer Type Part 4

Bihar Board 12th Maths Important Questions Long Answer Type Part 4

Application of derivatives

Question 1.
The volume of cube is increasing at a rate of 9 cubic centime Ires per second. How fast is the surface area increasing when the length of an edge is 10 centimetres ?
Solution:
Let x be the length of a side
v be the volume and
S be the surface areas of the cube
then v = x3
and S = 6x2, where x is a function of time t.
Given, dvdt = 9cm3/s

Hence x = 10cm.
dsdt = 3.6cm3/s

Question 2.
Find the intervals in which the function f given by f(x) = sinx + cosx, 0 ≤ x ≤ 2π is strictly increasing or strictly decreasing.
Solution:
We have
f(x) = sin x + cos x
f(x) = cosx-sinx
Now, f'(x) = 0
gives sin x= cos x which gives that
x = π4,5π4
0 ≤ x ≤ 2π
The point x = π4 and x = 5π4 divide the interval [0, 2π] into three disjoint intervals namely,

Interval	Sign of f’(x)	Nature of function
(0, π4 )	>0	f is strictly increasing
(π4,5π4 )	<0	f is strictly decreasîng
( 5π4, 2π )	>0	F is strictly increasing

Question 3.
Find the equation of the tangent to-the curve y = x−7(x−2)(x−3) at the point where it cuts’the x- axis.
Solution:
Note that on x-axis, y = 0 .
So that equation of the curve, when y = 0, gives x=7. Thus, the curve cuts the x-axis at (0,7). Now differentiating the equation of the curve with respect to x, we obtain.
dydx=(1−y)(2x−5)(x−2)(x−3)
or, dydx](7,0)=1−0(5)(4)=120
Therefore, the slope of the tangent at (7,0) is 120
Hence the equation of the tangent at (7,0) is
y – 0 = 120 (x – 7)
or, 20y – x + 7 = 0

Question 4.
Find the equations of the tangent and normal to be the curve x2/3 + y2/3 = 2 at (1, 1)
Solution:
Differentiating x2/3 + y2/3 = 2 with respect to x, we get

Therefore, the slope of tangent at (1,1) is dydx] = -1
So the equation of the tangent at (1,1) is
y – 1 = -1(x – 1)
or, y + x – 2 = 0

Also, the slope of the normal at ( 1,1) is given by
−1 Slope of tangent at (1,1) = 1
Therefore, the equation of the normal at (1,1) is
y – 1 = 1 (x – 1)
or, y – x = 0

Question 5.
Use differential to approximate 36−6−−−−−√
Solution:
Take y = x−−√
Let x = 36 and let Δx = 0.6 then

Now dy is approximate equal to Δy and is given by

(as y = x−−√
Thus, the approximate value of 36.6−−−−√ is 6 + 0 05 = 6.05.

Question 6.
Use differential to approximate (25)1/3
Solution:
Let y = x1/3
Let x = 27 and let Δx = -2 then
Δy = (x + Δr)1/3 – x1/3 = (25)1/3 – (27)1/3
= (25)1/3 – 3
or, (25)1/3 = 3 + Δy
Now dy is approximately equal to Δy and is given by

Thus, the approximate value of (25)1/3 is given by 3 + (-0.074) = 2.926

Question 7.
Find the approximate value of 3412, where f(x) = 3x2 + 5x + 3.
Solution:
Let x = 3 and Δx = 0.02 Then
f(3.02) = f(u + Δx)
= 3(x + Δx)2 + 5( x + Δx) + 3
Note that Δy = f(x+Δx) – f(x) ‘
Therefore, f(x + Δr)f(x)+Δy
= f(x) + f'(x)Δx (as dx = Δx)
f(3.02) = (3x2 + 5x + 3) + (6x + 5)Δx
= (3(3)2 + 5(3) + 3} + {6(3) + 5}(0.02)Δx
_ (as x = 3, Δx = 0.02)
= (27 + 15 + 3) + (18 + 5)(0.02)
= 45 + 0.46 = 45.46
Hence, approximate value of f( 3.62) is 45.46

Question 8.
Find the approximate change in the volume v of a cube of side x meters caused by increasing the side by 2%.
Solution:
Note that
y = x3
or dv = (dvdx)Δx = (3x2)Δx
= (3x2) (002x) = 006 x3 m3 at 2% of x is 002x.
Thus, the approximate change in yolume is 0 06 x3 m3

Question 9.
A car starts from a point p at time t = 0 seconds and stops a point Q. The distance x in metres covered by it in t seconds is given b x = t2(2 – t3)

Find the time taken by it to reach Q and also find distance between i and Q.
Solution:
Let v be the velocity of the car at t seconds.
Now x = t2(2 – t3)
∴ v = dxdt = 4r – t2 = t(4 – t)
Thus v = 0 gives t = 0 and/or t = 4 ’
Now v = 0 at P as well as at Q and at P, t- 0.
so, at Q. t = 4.
Thus the car will reach the point Q after 4 seconds. Also the distance travelled in 4 seconds is given by
[x]t = 4 = 42(2 – 43
= 16( 23 ) = 323m

Question 10.
A mail of height 2 metres walks at a uniform speed of 5 km/h s* way from a lamp post which is 6 metres high. Find the rate at which the length of his shadow increases.
Solution:
Let AB be the lamp-post, the lamp being at the position Band let MN be the man at a particular time t and Let AM= l metres. Then MS is the shadow of the man. Let MS = S metres.

Note that ΔMSN ~ ASB
or, MSAS=MNAB
AS = 3S(MN = 2 and AB = 6)
AM = 35 – 5 = 25 But AM = l
l = 2S
∴ dldt=2dSdt
Since dldt = 5km/h
Hence, the length of the shadow increase at the rate 52 km/ h

Question 11.
Find ∫dxx(x2+1)2
Solution:

Integrals

Question 1.
find the following integrals :
(i) ∫2x1+x2dx
(ii) ∫(logx)2xdx
(iii) ∫etan−1x1+x2dx
Solution:
(i) Put 1 + x2 = t then 2xdx = dt
Therefore ∫2xdx1+x2=∫dtt
= ∫1t⋅dt
= logt + c
= log(1 + x2) + c

(ii) Put log x = t (log x)2 so that 1xdx = dt
Therefore ∫1x (logx)2dx = ∫t2 dt
= t33+C
= (logx)3 + c

(iii) Put etan-1 x = t so that
etan-1 x 11+x2.dx = dt
Therefore 11+x2
= ∫ dt = t + c = etan-1 x + c

Question 2.
Find the following integrals :
(i) ∫ sin3 x.cos2 xdx
(ii) ∫sinxsin(x+a)dx
Solution:
(i) = ∫ sin2 x . cos3 x . (sin x).dx
= ∫ (1 – cos2 x). cos2 x . (sinx).dx
= ∫ (cos2x – cos4x)sinx dx
Let cos x = t
Differentiate w.r. t. x
-sin x = dtdx
⇒ sin x . dx = -dt
= ∫(t2 – t4)(-dt)
= -∫(t2 – t4)(dt)
= -∫(t2dt – ∫t4)(dt)
= t33+t45 + c
= −cos3x3+cos4x5 + c

(ii) Put x + a = t
Then dx = dt

= cosa∫dt – sina∫cott dt
= (cos a) t -sin a [ log | sin t| + c1]
= (cos a) ( x +a) – sin a [ log | sin {x + a)| + Cj]
= x cos a + a cos a – sin a log | sin (x+a) |+ c

Hence, ∫sinxsin(x+a)dx = xcosa – sinalog|sin(x + a)| + c
where, c = c1 sin a + a cos a is another arbitrary constant.

Question 3.
Find the following integrals:
(i) ∫cos2x.dx
(ii) ∫sin 2x. cos3x. dx
(iii) ∫sin3xdx
Solution:
(i) We know that
cos2x = 2cos2 x – 1
cos2 = cos2x+1
Then ∫ cos2x. dx = ∫cos2x+12dx

(ii) = 12∫ 2sin2.x-cos3.x-cfct
= 12∫[sin(2x + 3x) + sin(2x – 3x)]dx
= 12∫(sin5x – sinx)dx
= 12∫sin5x dx – 12∫sin x. dx

(iii) We know that,
sin 3x = 3 sin x – 4 sin3 x
sin3x = 3sinx−sin3x

Question 4.
Find the following integrals : dx t dx
(i) ∫dxx2−16
(ii) ∫dx2x−x2√
Solution:

(ii) = ∫dx1−(x−1)2√
Put x – 1 = t then dx = dt
Therefore , = ∫dx2x−x2√=∫dt1−t2√
sin-1(t) + c = din-1(x – 1) + c

Question 5.
Find the following intergrals
(i) 
(ii) 
Solution:
(i) We have x2 – 6x + 13 = x2-6x + 32 -32 + 13 . ; = (x -3)2 + 4
So ∫dxx2−6x+13=∫dx(x−3)2+4
Let x – 3 = t then dx = dt

Question 6.
Find the integrals : = ∫x2+1x2−5x+6
Solution:
x2+1x2−5x+6 is not proper rational junction, so we divide x2 + 1 by x2 – 5x + 6 a and find that

So that 5x-5 = (x- 3) A + (x-2) B
Equating the coefficients of x and constant 7 terms on both sides, we get A + B = 5 and 34 + 21? = 5 r
A = -5 and B = 10

= x – 5 tog |x – 2| + 10 tog |x – 3| + c

Question 7.
Find the integrals ; ∫x2+2x+5−−−−−−−−−√dx
Solution:

Question 8.
Find the integral : 3−2x−x2−−−−−−−−−√dx
Solution:
∫4−(x−1)2−−−−−−−−−−√dx
Put x + 1 = r then dx = dt.

Question 9.
Find the integrals :∫sin2x⋅cos2x9−cos4(2x)√dx
Solution:
Put cos2 (2x) = t
So that 4sin2x . cos2x .dx = -dt

Question 10.
Evaluate ∫π/20cosx√cosx√+sinx√dx
Solution:

Question 11.
Prove that ∫π/20logsinxdx=∫π/20logcosxdx=−π2log2.
Solution:
Let I = ∫π/20logsin dx = ∫π/20 log sin(π2 – x)dx
[ ∫α0f(x)dx=∫α0f(a−x)dx ]
[∵ latex]\int_{0}^{\pi / 2}[/latex] log cos x dx
I = ∫π/20 log cos xdx
Adding (1) and (2), we have
2I = ∫π/20 (log sin x + log cos x)dx = ∫π/20 logsin2x dx
= ∫π/20 log sin 2x dx – log 2∫π/20 dx
= ∫π/20 logsin 2x dx – π2log 2 …………….. (3)
Let 2x = z = dx = dzdz ∴ limit are 0 to π.
2 I = 12∫π/20logsinzdz−π2log2
∫π/20 logsin z dz – π2 lofg 2
= I – π2 log 2
∴ I = π2 log 2

Question 12.
Evaluate ∫∞0sin(sin−1x)1+x2dx
Solution:
Put tan-1 x = z ⇒ dx1+x2 = dz\
Limits are o to π2
I = ∫π/20 sin z dz = −[cosz]π/20
= = (cosπ2 – cos 0) = 1

Question 13.
Evaluate ∫10tan−1x1+x2dx
Solution:
Let t = tan-1 x then dt = 11+x2dx
The new limits are when x = 0, t = 0 and when x = 1, t = π4 thus a x varies form 0 to 1 , t, varies from 0 to π4

Question 14.
Evaluate ∫π/4−π/4sin2xdx
Solution:
∵ sin2 is an even function

Question 15.
Evaluate sin4θ cos4θ dθ
Solution: