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 Bihar Board 12th Maths Important Questions Long Answer Type Part 3

Bihar Board 12th Maths Important Questions Long Answer Type Part 3

Continuity and Differentiability

Question 1.
Find the relation b/w a & b so that the function (x) defined by
f(x)={ax+bbx+3 if x≤3 if x>3 is continuous at x = 3
Solution:

= 3a + b
f(3) = a.3 + 8 = 3a + 8

R.H.L. = Ltx→3f(x) Ltx→3(b x+3)=b.3 + 3
= 3b + 3
∵ f(x) is continuous at x = 3.
∴ LHX =RHL= f(3)
∵ 3a + b = 3b + 3′
⇒ 3a + b – 3b = 3
⇒ 3a – 2b = 3

Question 2.
Find k if f(x) = f(x)={kcosxπ−2x3 if x≠π2 if x=π2 is continuous at x = π2
Solution:
Given

Question 3.
Find dydx
(i) y = (cos x . cos 2x) . cos 3x
(ii) x = 2at2 y = at4
(iii) x = a cosθ, y = bsecθ
(iv) x = 4t, y = 4t
(v) x = cos θ – cos2θ, y = sinθ – sin2θ
(vi) x = a(cost + log tan t2), y = a sin t
Solution:
(i) ∵ y = (cos x . cos 2x) . cos 3x
Diff w.r. to x
dydx = (cosx . cos2x) . d.c. of cos 3x + cos 3x . d.c. of (cos x . cos 2x)
or, dydx = – cosx . cos 2x . sin 3x . 3 + cos 3x . (cos x d.c. of cos 2x + cos 2x d.c. of cos x)
or, dydx = – 3cos x . cos2x . sin3x + cos3x . (-cos x . sin2x – cos2x.sinx)
or, dydx = -3cosx . cos2x sin3x – cos3x . sin(2x + x)
∴ dydx = -3cosx . cos2x sin3x – sin3x cos3x

(ii) x = 2at2,
Diff w.r. to t
dxdt = 4at …………..(i)

y = at4
Diff w.r. to t
dydt = 3at3 ………… (ii)

Eqn. (ii) ÷ Eqn. (i)

(iii) x = a cos θ
Diff w.r. to θ
dxdt = -asin θ……………. (i)

y = b sec θ
Diff w.r. to θ
dydt = b secθ . tan θ……………. (ii)

Eqn. (ii) ÷ Eqn. (i)
dydx=bsecθ⋅tanθ−asinθ

(iv) x = 4t
Diff w.r. to t
dxdt = 4 …………… (i)

y = 4t
Diff w.r. to t
dydt=−4t2 ……………….. (ii)

Eqn. (ii) ÷ Eqn. (i)

(v) x = cos θ – cos 2 θ
Diff. w.r. to θ
∴ dxdθ = -sinθ + sin 2θ.2 ……….(i)

& y = sinθ – sin 2θ
Diff. w.r. to θ
∴ dydθ = cosθ – cos2θ.2 ……….(ii)
Eqn. (ii) ÷ Eqn. (i)

(vi) x = a(cos t + log tan t2), y = a sin t
Diff w.r.to t

Question 4.
Show that f(x) = {12x−13,x≤32x2+5,x>3x=3 is differentiable at x = 3. Also find f'(3).
Solution:

∵ L.H.S = R.H.S at x = 3
∴ f(x) is differentiable at x = 3 and f'(3) = 12.

Question 5.
For whal choice of a & b is the function :
f(x)={x2,x≤cax+b,x>c is differentiable at x = c.
Solution:
It is given that f(x) is differentiable at x = c
∴ f(x) will be continuous of x = c.

f(c) = c2
∵ L.H.S. = R.H.S.=/(c)
∴ c2 = ac + b = c2
Now f(x) is differentiable at x = c

= Ltx→c a(x−c)(x−c)= a
∴ L.H.S. = R.H.S.
∴ a = 2c …(ii)
from eqn. (i) & eqn. (ii)
c2 = 2c2 + b
=> b = c2 & a = 2c

Question 6.
IF f(x) = {x2+3x+a,x≤1bx+2,x>1 is every where differentiable. Find the values of a & b.
Solution:
For x ≤ 1, f(x) = x2 + 3x + a = polynomial
For x > 1,
f(x) = bx + 2 = polynomial,
∵ Polynomial function is every where differentiable.
∴ Therefore f(x) is differentiable for all x > 1 and also for all x < 1.
∵ f(x) is continuous at x = 1
∴ L.H.S. = R.H.S. = f(1)
or Ltx→1 (x2 + 3x + a) = Ltx→1(bx + 2) = 12 + 3.1 + a
or, 12 + 3.1 + a = b.1 + 2 = 12 + 3.1 + a
or, 4 + a = b + 2
⇒ a – b + 2 = 0 ….(i)
Again f(x) is differentiable at x = 1
∴ L.H.S. at x = 1 = R.H.S. at x = 1

or, 5 = b ⇒ B = 5
From eqn. (i),
a – 5 + 2 = 0
or, a – 3 = 0
∴ a = 3

Differentiation

Question 1.
Differentiate the following functions w.r. to x from 1st principles.
(i) sinx2+ax+1−−−−−−−−−√
(ii) tanx
Solution:

Question 2.
If y = [x+x2+a2−−−−−−√]n then prove that dydx=nyx2+a2√
Solution:
Let y = [x+x2+a2−−−−−−√]n
Diff w.r.to x.

Question 3.
If y = 1−x1+x−−−√ then prove that (1 – x2)dydx + y = 0
Solution:
Given y = 1−x1+x−−−√
Diff w.r.to x

Question 4.
Prove that d{x⋅a2−x2√+a22⋅sin−1xdx=a2−x2−−−−−−√
Solution:

Question 5.
If xy = ex-y prove that dydx=logx(1+logx2
Solution:
Given xy = ex-y
or eylogx = ex-y
ylogx = x – y
ylog x +y = x
or y(1 + log x) = x
or y = 1+logx
Diff. w.r.to x

Question 6.
If (cosx)y = (sin y)x, find dydx
Solution:
Given (cos x)2 = (sin y)x
or, log (cos x)y = log(sin y)x
or, y. log (cos x) = x log (siny)
Diff . w.r.to x
y.cosx. (-sinx) + log(cos x)dydx = x.siny . cos y. dydx + log(sin y)
or, -y tan x + log(cosx)dydx x cot y. dydx + log(sin y)
or dydx. [log(cos x) – xcot y] = x cot y] = log(sin y) + y tan x
∴ dydx=log(siny)+ytanxlog(cosx)−xcoty

Question 7.
Differentiate sin-1(2x1+x2) w. r. to tan-1 x, -1 < x < 1
Solution:
Let u = sin-1(2x1+x2)
or u = 2tan-1 x and v = tan-1 x&
Diff . w.r.to x
∴ dudx=2⋅11+x2 ……………… (i)
dvdx=11+x2 …………………… (ii)
Eqn(i) / Eqn(ii)
dudv = 2

Question 8.
If y = logtanx then dydx = ?
Solution:
Given y = logtanx then dydx

Question 9.
If y = sin{tan(log x)}, find dydx
Solution:
∵ y = sin{tan(log x)}
⇒ dydx=ddx [sin{tan(log x)}
= cos{tan(logx)) x ddx{tan(logx)}
= cos{tan(logx)) x sec2(logx) x ddx {log x}
= cos{tan(Iog x)) x sec2(log x) x 1x
= 1xsec2(logx)cos{tan(logx)).

Question 10.
Integrate ∫π/40cos2xdx
Solution:

Question 11.
If y = sinx2, find dydx
Solution:
∵ y = sinx2
Diff w.r. to x we get
∴ dydx=ddx (sin x2) = cos x2 ddx (x2) = 2cos x2.

Question 12.
If y = tan3 θ find dydθ
Solution:
y = tan3 θ
Diff w.r. to θ we get
dydθ=ddθ (tan3 θ) = 3tan3 θ ddθ (tan θ) = 3 tan2θ sec2θ

Question 13.
If y = cot-1 1−x1+x then prove that dydx=11+x2
Solution:
∵ y = cot-1 1−x1+x
put x = tan θ

 

Question 14.
If y = tan−11+sinx1−sinx−−−−−√ then find the value of dydx
Solution:
y = y=tan−11+sinx1−sinx−−−−−√