Continuity and Differentiability Class 12 Maths Important Questions Chapter 5

 

Continuity and Differentiability Class 12 Important Questions with Solutions Previous Year Questions

Continuity

Question 1.
Determine the value of ‘k’ for which the following function is continuous at x = 3: (All India 2017)

Answer:

Question 2.
Determine the value of the constant ‘k’ so that the function

is continuous at x = 0. (Delhi 2017)
Answer:

Question 3
Find the values of p and q for which

is continuous at x = π2. (Delhi 2016)
Answer:

Question 4.
If

is continuous at x = 0, then find the values of a and b. (All India 2015)
Answer:

Question 5.
Find the value of k, so that the function

is continuous at x = 0. (All India 2014C).

Alternate Method:

Question 6.
If

and f is continuous at x = 0, then find the value of a. (Delhi 2013C)
Answer:

Question 7.
Find the value of k, for which

is continuous at x = 0. (All India 2013)
Answer:

Question 8.
Find the value of k, so that the following function is continuous at x = 2. (Delhi 2012C)

Answer:

Question 9.
Find the value of k, so that the function f defined by

is continuous at x = π2. (Delhi 2012C; Foregin 2011)
Answer:

Question 10.
Find the value of a for which the function f is defined as

is continuous at x = 0. (Delhi 2011)
Answer:

Question 11.
If the function f(x) given by

is continuous at x = 1, then find the values of a and b. (Delhi 2011; All India 2010)
Answer:

On substituting these values in Eq. (i), we get
5a – 2b = 3a + b = 11
⇒ 3a + b = 11 …… (ii)
and 5a – 2b = 11 ……. (iii)
On subtracting 3 × Eq. (iii) from 5 × Eq. (ii), we get
15a + 5b – 15a + 6b = 55 – 33
⇒ 11b = 22 ⇒ b = 2
On putting the value of b in Eq. (ii). we get
3a + 2 = 11 ⇒ 3a = 9 = a = 3
Hence, a = 3 and b = 2

Question 12.
Find the values of a and b such that the following function f(x) is a continuous function. (Delhi 2011)

Answer:

is a continuous function. So, it is continuous at x = 2 and at x = 10.
∴ By definition.
(LHL)x=2 = (RHL)x=2 = f(2) …… (i)
and (LHL)x=10 = (RHL)x=10 = f(10) …… (ii)
Now, let us calculate LHL and RHL at x = 2.

Now, from Eq. (ii), we have
LHL= RHL
⇒ 10a + b = 21 ….. (iv)
On subtracting Eq. (iv) from Eq. (iii), we get
– 8a = – 16
⇒ a = 2
On putting a = 2 in Eq. (iv), we get
2a + b = 21 ⇒ b = 1
Hence, a = 2 and b = 1

Question 13.
Find the relationship between a and b, so that the function f defined by

is continuous at x = 3. (All India 2011)
Answer:
let

is a continuous at x = 3.
Then, LHL = RHL = f(3) ……. (i)

⇒ RHL = 3b + 3
From Eq.(i), we have
LHL = RHL ⇒ 3a + 1 = 3b + 3
Then, 3a – 3b = 2, which is the required relation between a and b.

Question 14.
Find the value of k, so that the function f defined by

is continuous at x = π. (Foreign 2011)
Answer:

Question 15.
For what values of λ, is the function

is continuous at x = 0? (Foreign 2011)
Answer:

∵ LHL ≠ RHL, which is a contradiction to Eq. (i).
∴ There is no value of λ. for which f(x) is continuous at x = 0.

Question 16.
Discuss the continuity of the function f(x) at x = 1/2 , when f(x) is defined as follows. (Delhi 2011C)

Answer:
Here, we find LHL, RHL and f(12).
If LHL = RHL = f(12) then we say that f(x) is continuous at x = 12, otherwise f(x) discontinuous at x = 12.

Given function is

Question 17.
Find the value of α, if the function f(x) defined by

is continuous at x = 2. Also, discuss the continuity of f(x) at x = 3. (All India 2011C)
Answer:

Question 18.
Find the values of a and b such that the function defined as follows is continuous. (Delhi 2010, 2010C)

Answer:
a = 3 and b = – 2

Question 19.
For what value of k, is the function defined by

continuous at x = 0?
Also, find whether the function is continuous at x = 1. (Delhi 2010, 2010C)
Answer:

Question 20.
Find all points of discontinuity of f, where f is defined as follows.

Answer:
First, verify continuity of the given function at x = – 3 and x = 3. Then, point at which the given function is discontinuous will be the point of discontinuity.

⇒ RHL = 6
Also, f(- 3) = value of f(x) at x = – 3
= – (- 3) + 3
= 3 + 3 = 6
∵ LHL = RHL f(- 3)
∴ f(x) is continuous at x = – 3 So, x = – 3 is the point of continuity.

Continuity at x = 3

⇒ RHL = 20
∵ LHL ≠ RHL
∴ f is discontinuous at x = 3
Now, as f (x) is a polynomial function for x < – 3, – 3 < x < 3 and x > 3, so it is continuous in these intervals.
Hence, only x = 3is the point of discontinuity of f(x).

Differentiability

Question 1.
Differentiate e3x√, with respect to x. (All India 2019)
Answer:
Let y = e3x√

Question 2.
If y = cos (√3x), then find dydx. (All India 2019)
Answer:
Given, y = cos (√3x)
Differentiating w.r.t x, we get

Question 3.
If f(x) = x + 1, find ddx (fof) (x). (Delhi 2019)
Answer:
Given, f(x) = x + 1
⇒ f(f(x)) = f(x) + 1
⇒ fof(x) = x + 1 + 1
⇒ fof(x) = x + 2
Now, ddx (fof)(x) = ddx(x + 2) = 1

Question 4.
If f(x) = x + 7 and g(x) = x – 7, x ∈ R, then find the values of ddx (fog) x. (Delhi 2019)
Answer:
Given, f(x) = x + 7,
g(x) = x – 7, x ∈ R
Now, (fog) (x) = f[g(x)] = f(x – 7) = (x – 7) + 7
(fog) (x) = x
On differentiate w.r.t. x, we get
ddx (fog)(x) = ddx (x) ⇒ ddx (fog) (x) = 1

Question 5.
If y = x|x|, find dydx for x < 0. (All India 2019)
Answer:
We have, y = x|x|
When, x < 0, then |x| = – x
∴ y = x(- x) = – x2
⇒ dydx = – 2x

Question 6.
Differentiate tan-1 (1+cosxsinx) with respect to x. (CBSE 2018)
Answer:

Question 7.
Differentiate tan-1 (cosx−sinxcosx+sinx) with respect to x. (CBSE 2018 C)
Answer:

Question 8.
Find the value of c in Rolle’s theorem for the function f(x) = x3 – 3x in [-√3, 0]. (All India 2017)
Answer:
Given, f(x) = x3 – 3x in [-√3, 0]
We know that, according to Rolles theorem, if f(x) is continuous in [a, b] differentiable in (a, b) and f(a) = f(b), then there exist c ∈ (a, b) such that f’(c) = 0.
Here f(x), being a polynomial function, is continuous in [-√3, 0] and differentiable in (-√3, 0).
Also, f(-√3) = 0 = f(0)
∴ f'(c) = 0, for some c ∈ (- √3, 0) …… (ii)
Now, f’(x) = 3x2 – 3 [from Eq. (i)]
⇒ f’(c) = 3c2 – 3 = 0 [from Eq. (ii)]
⇒ c = ± 1
But C ∈ (-√3, 0) so neglecting positive value of c.
∴ c = – 1

Question 9.
Find dydx at x = 1, y = π4 if sin2 y + cos xy = K. (Delhi 2017)
Answer:
we have sin2 y + cos xy = k
On differentiating both sides w.r.t x, we get

Question 10.
If y = sin-1 (6x1−9x2−−−−−−√), < 132√ x < 132√ then find dydx. (Delhi 2017)
Answer:
Given, y = sin-1(6x 1−9x2))
y = sin-1(2.3x 1−(3x)2−−−−−−−−√)
put 3x = sin θ, then
y = sin-1 (2 sin θ1−sin2θ)
⇒ y = sin-1 (2 sin θ. cos θ)
⇒ y = sin-1 (sin 2θ)
⇒ y = 2θ
⇒ y = 2 sin-1(3x) [∵ θ = sin-1(3x)]
⇒ dydx=21−9x2√
⇒ dydx=61−9x2√

Question 11.
If (cos x)y = (cos y)x, then find dydx. (All India 2019; Delhi 2012)
Answer:
First, take log on both sides, then differentiate both sides by using product rule.
Given, (cos x)y = (cos y)x
On taking log both sides, we get
log (cos x)y = log (cos y)x
⇒ y log (cos x) = x log(cos y)
[∵ log xn = n log x]
On differentiating both sides w.r.t. x, we get

Question 12.
If 1+y−−−−√+y1+x−−−−√=0 = 0,(x ≠ y), then prove that dydx=−1(1+x)2. (All India 2019: Foreign 2012; Delhi 2011C)
Answer:
First, solve the given equation and convert it into y = f(x) form. Then, differentiate to get the required result.
To prove dydx=−1(1+x)2
Given equation is 1+y−−−−√+y1+x−−−−√ = 0,
where x ≠ y, we first convert the given equation into y = f(x) form.
Clearly, x 1+y−−−−√ = – y 1+x−−−−√
On squaring both sides, we get
⇒ x2 (1 + y) = y2 (1 + x)
⇒ x2 + x2y = y2 + y2x
⇒ x2 – y2 = y2x – x2y
⇒ (x – y) (x + y) = – xy (x – y)
[∵ a2 – b2 = (a – b) (a + b)]
⇒ (x – y) (x + y) + xy (x – y) = 0
⇒ (x – y) (x + y + xy) = 0
∴ Either x – y = 0 or x + y + xy = 0
Now, x – y = 0 ⇒ x = y
But it is given that x ≠ y.
So, it is a contradiction.
∴ x – y = 0 is rejected.
Now, consider y + xy + x = 0
⇒ y(1 + x) = – x ⇒ y = −x1+x
On differentiating both sides w.r.t. x, we get

Question 13.
If y = (sin-1 x)2 prove that
(1 – x2)d2ydx2 – x dydx – 2 = 0 (Delhi 2019)
Answer:
Given y = (sin-1 x)2
Differentiating on w.r.t x, we get

Question 14.
If (x – a)2 + (y – b)2 = c2, for some c > 0,
prove that

independent of a and b. (All India 2019)
Answer:
Given (x – a)2 + (y – b)2 = c2
Differentiating on w.r.t x, we get

Question 15.
If x = aet(sin t + cos t) and y = aet(sin t – cos t), then prove that dydx=x+yx−y (All India 2019)
Answer:
Given x = x = aet(sin t + cos t)
and y = aet(sin t – cos t)

Question 16.
Differentiate xsin x + (sin x)cos x with respect to x. (All IndIa 2019)
Answer:

Question 17.
If log (x2 + y2) = 2 tan-1(yx) show that dydx=x+yx−y (Delhi 2019)
Answer:
log (x2 + y2) = 2 tan-1(yx)
on differentiating both sides w.r.t. x, we get

Question 18.
If xy – yx = ab, find dydx. (Delhi 2019)
Answer:

Question 19.
If x = cos t + log tan(t2), y = sin t, then find the values of d2ydt2 and d2ydx2 at t = π4. (Delhi 2019; All IndIa 2012 C)
Answer:

Question 20.
If y = sin (sin x), prove that
d2ydx2 + tan x dydx + y cos x = 0. (CBSE 2018)
Answer:
Given y = sin (sin x) ….. (i)
On differentiating both sides w.r.t. x we get
dydx = cos (sin x) . cos x ….. (ii)
Again. on differentiating both sides w.r.t. z.
we get
d2ydx2 = cos (sin x) . (- sin x) + cos x (- sin (sin x)) . cos x

Question 21.
If (x2 + y2)2 = xy, find dydx. (CBSE 2018)
Answer:
We have (x2 + y2)2 = xy
on differentiating both sides w.r.t. x, we get

Question 22.
If x = a(2θ – sin 2θ) and y = a(1 – cos 2θ), find dydx when θ = π3. (CBSE 2018)
Answer:

Question 23.
If sin y = x cos(a + y), then show that
dydx=cos2(a+y)cosa.
Also, show that dydx = cos a, when x = 0. (CBSE 2018 C)
Answer:

Question 24.
If x = a sec3 θ and y = a tan3 θ, find d2ydx2 at θ = π3. (CBSE 2018C)
Answer:

Question 25.
If y = etan-1x, prove that (1 + x2)d2ydx2 + (2x – 1)dydx = 0. (CBSE 2018 C)
Answer:
we have, y = etan-1x
on differentiating both sides w.r.t. x, we get

Question 26.
If xy + yx = ab, then find dydx. (All India 2017)
Answer:
dydx = −xy−1⋅y−yxlogy

Question 27.
If ey (x + 1) = 1, then show that d2ydx2=(dydx)2. (All India 2017)
Answer:
Given, ey (x + 1) = 1
On taking log both sides, we get
log [ey (x + 1) = log]
y + log(x + 1) = log 1 [∵ log ey = y]
On differentiating both sides w.r.t x, we get
dydx+1x+1 = 0 …… (i)
Again, differentiating both sides w.r.t. ‘x’, we get

Question 28.
If y = xx, then prove that (Delhi 2016, 2014)
d2ydx2−1y(dydx)2−yx = 0 (Delhi 2016, 2014)
Answer:
Given y = xx
On taking log both sides, we get
log y = log xx
⇒ log y = x log x
On differentiating both sides w.r.t x, we get

Question 29.
Differentiate tan-1 (1+x2√−1x) w.r.t. sin-1 (2x1+x2), when x ≠ 0. (Delhi 2016, 2014)
Answer:

Question 30.
If x = a sin 2t(1 + cos 2t)and
y = b cos 2t (1 – cos 2t), then find the values of dydx at t = π4 and t = π3. (Delhi 2016; All India 2014)
Or
If x = a sin2t(1 + cos 2t) and y = b cos 2t (1 – cos 2t), then show that at t = π4,dydx=ba (All India 2014).
Answer:
Given, x = a sin 21(1 + cos 2t)
and y = b cos 2t(1 – cos 2t)
On differentiating x and y separately w.r.t. t,
we get
dxdt = a[sin 2t ddt(1 + cos 2t) + (1 + cos 2t) ddt (sin 2t)
[by using product rule of derivative]
= a [sin2t × (0 – 2 sin 2t) + (1 + cos 2t) (2 cos 2t)]
= a (- 2 sin2 2t + 2 cos 2t + 2 cos2 2t)
= a[2(cos2 2t – sin2 2t) + 2 cos 2t]
= a (2 cos 4t + 2 cos 2t) = 2a (cos 4t + cos 2t)
[∵ cos2 2θ – sin2 2θ = cos 4θ]

= 4a cos 3t cos t
and dydt = b[cos 2t ddt (1 – cos 2t) + (1 – cos2t) ddt (cos 2t)]
[by using product rule of derivative]
= b [cos 2t × (0 + 2 sin 2t) + (1 – cos 2t) (- 2 sin 2t)]
= b (2 sin 2t cos 2t – 2 sin 2t + 2 sin 2 t cos 2t)
= 2b (2 sin 2t cos 2t – sin 2t)
= 2b (sin 4 t – sin 2 t) [∵ 2 sin 2θ cos 2θ = sin 4θ]

Question 31.
If x cos(a + y) = cos y, then prove that dydx=cos2(a+y)sina. Hence, show that sin α d2ydx2 + sin 2 (α + y) dydx = 0. (All India 2015).
Or
If cos y = x cos(α + y), where cos α ≠ ±1, prove that dydx=cos2(a+y)sina. (Foregin 2014).
Answer:

Question 32.
Find dydx, if y = sin-1 [6x−41−4x2√5] (All IndIa 2016).
Answer:

Question 33.
Find the values of a and b, if the function f defined by

is differentiable at x = 1. (Foreign 2016)
Answer:

From Eq. (i), we have
Lf'(1) = Rf'(1)
⇒ 5 = b
⇒ b = 5
Now, on substituting b = 5 in Eq. (ii), we get
5 – a – 2 = 0
⇒ a = 3
Hence, a = 3 and b = 5.

Question 34.
If x = sin t and y = sin pt, then prove that
(1 – x2)d2ydx2 – xdydx + p2y = 0. (Foreign 2015)
Answer:
Given, x = sin t and y = sin pt
On differentiating x and y separately w.r.t t, we get

Question 35
If y = tan-1(1+x2√+1−x2√1+x2√−1−x2√), x2 ≤ 1, then find dy/ dx. (Delhi 2015)
Answer:
First, put x2 = sin θ, then reduce it in simplest form.
Further, differentiate it.

Question 36.
If x = a cos θ + b sin θ, y = a sin θ – b cos θ, then show that y2d2ydx2 – xdydx + y = 0. (Delhi 2015. ForeIgn 2014 )
Answer:
Given x = a cos θ + b sin θ, ……. (i)
and y = a sin θ – b cos θ …….. (ii)
On differentiating both sides of Eqs. (i) and (ii) w.r.t. θ, we get

Question 37.
Show that the function f(x) = |x + 1| + |x – il, for all x ∈ R, is not differentiable at the points x = – 1 and x = 1. (All India 2015)
Answer:

Question 38.
If y = em sin-1 x, then show that
(1 – x2)d2ydx2 – xdydx – m2 y = 0. (All India 2015).
Answer:
Given y = em sin-1 x
On differentiating both sides w.r.t x, we get,

Question 39.
If f(x) = x2+1−−−−−√; g(x) = x+1x2+1 and h(x) = 2x – 3 then find f’[h’{g’(x)}]. (All India 2015).
Answer:

Question 40.
If y = (x+1+x2−−−−−√)n, then show that (1 + x2)d2ydx2 + xdydx = n2y. (Foregin 2015).
Answer:

Question 41.
Find whether the following function is differentiable at x = 1 and x = 2 or not. (Foreign 2015).

Answer:

∵ LHD = RHD
So, f(x) is differentiable at x = 2
Hence, f(x) is not differentiable at x = 1, but it differentiable at x = 2

Question 42.
For what value of λ, the function defined by

is continuous at x = 0? Hence, check the differentiability of f(x) at x = 0. (All India 2015C)
Answer:

Question 43.
If y = (sin x)x + sin-1 √x,then find ddydx. (Delhi 2015C, 2013C)
Answer:
Given, y = (sin x)x + sin-1 √x …… (i)
Let u = (sin x)x ……. (ii)
Then, Eq. (i) becomes, y = u + sin-1 √x ….. (iii)
On taking log both sides of Eq. (ii), we get
log u = x log sin x
On differentiating both sides w.r.t. x, we get

Question 44.
If y = xcos−1x – l0g1−x2−−−−−√, then prove that dydx=cos−1x(1−x2)3/2 (Delhi 2015C)
Answer:

Question 45.
Write the derivative of sin x with respect to cos x. (Delhi 2014C)
Answer:
Let u = sin x
On differentiating both sides w.r.t. X, we get
dudx = cos x ……. (i)
Also, let v = cos x
On differentiating both sides w.r.t. x, we get
dvdx = – sin x ……… (ii)
Now, dudv=dudx×dxdv=−cosxsinx [from Eqs. (i) and (ii)]
∴ dudv = – cot x

Question 46.
If y = sin-1 {x,1−x−−−−√ – 1−x2−−−−−√} and 0 < x < 1, then find dydx. (All India 2014C; Delhi 2010)
Answer:
First, convert the given expression in sin-1[x 1−y2−−−−−√ – y1−x2−−−−−√] form and then put x = sin Φ and y = sin Φ. Now, simplify the resulting expression and differentiate it.

Question 47.
If ex + ey = ex + y, prove that dydx + ey – x = 0. (Foreign 2014)
Answer:
Given, ex + ey = ex + y ………… (i)
On dividing Eq.(i) by ex + y, we get
e-y + e-x = 1 ………. (ii)
On differentiating both sides of Eq. (ii) w.r.t. x,
We get

Question 48.
Find the value of dydx at θ = π4, if
x = aeθ (sin θ – cos θ) and
y = aeθ (sin θ + cos θ). (All IndIa 2014)
Answer:

Question 49.
If x = α(cos t + log tant2) y = a sin t, then evaluate d2ydx2 at t = π3. (Delhi 2014C)
Answer:
83√a

Question 50.
If xm ym = (x + y)m + n, prove that dydx=yx. (Foreign 2014)
Answer:
First, take log on both sides. Further, differentiate it to prove the required result.
Given xm yn = (x + y)m + n
On taking log both sides, we get
log (xm ym) = log(x + y)m + n
⇒ log(xm) + log(yn) = (m + n) log(x + y)
⇒ m log x + n log y = (m + n) log (x + y)
On differentiating both sides W:r.t. x, we get

Question 51.
Differentiate tan-1(1−x2√x) w.r.t. cos-1(2x1−x2−−−−−√), when x ≠ 0. (Delhi 2014)
Answer:

Question 52.
Differentiate tan-1(x1−x2√) w.r.t. sin-1 (2x 1−x2−−−−−√). (Delhi 2014)
Answer:

Question 53.
If y = Peax + Qebx, then show that  – (a + b)  + aby = 0. (All India 2014)
Answer:

Question 54.
If x = cos t(3 – 2 cos2 t)and y = sin t (3 – 2 sin2 t), then find the value of dydx at t = π4. (All India 2014)
Answer:
Given, x = cos t(3 – 2 cos2 t)
⇒ x = 3 cos t – 2 cos3 t
On differentiating both sides w.r.t. t, we get
dxdt = 3(- sin t) – 2(3) cos2t (- sin t)
⇒ dxdt =- 3 sin t + 6 cos t sin t ….. (i)
Also, y = sin t (3 – 2sin2 t)
⇒ y = 3 sin t – 2 sin3 t
On differentiating both sides w.r.t. t, we get
dydt = 3 cos t – 2 × 3 × sin2 t cos t

Question 55.
If (x – y) exx−y = a, prove that y dydx + x = 2y. (Delhi 2014C)
Answer:

Question 56.
If x = a(cos t + t sin t)and y = a (sin t – t cos t), then find the value of d2ydx2 at t = π4. (Delhi 2014C)
Answer:
82√aπ

Question 57.
If y = tan-1 (ax) + log x−ax+a−−−√, prove that dydx=2a3x4−a4 (All India 2014C)
Answer:

Question 58.
If (tan-1 x)y + ycot x = 1, then find dy/dx. (All India 2014C)
Answer:
Let u = (tan-1 x)y and v = ycot x
Then, given equation becomes u + y = 1
On differentiating both sides w.r.t. x, we get
dudx+dvdx = 0 ……. (i)
Now, u = (tan-1 x)
On taking log both sides, we get
log u = y 1og(tan-1 x)
On differentiating both sides w.r.t. x, we get

Question 59.
If x = 2 cos θ – cos 2θ and y = 2 sin θ – sin 2θ, then prove that dydx = tan (3θ2). (Delhi 2013C)
Answer:
Given x = 2 cos θ – cos 2θ
and y = 2 sin θ – sin 2θ
On differentiating both sides w.r.t θ, we get

Question 60.
If y = x log (xa+bx), then prove that x3d2ydx2 = (xdydx−y)2. (Delhi 2013C)
Answer:

Question 61.
If x = cos θ and y = sin3 θ, then prove that yd2ydx2+(dydx)2 = 3 sin2 θ(5 cos2 θ – 1). (All India 2013C)
Answer:
Given x = cos θ ……. (i)
and y = sin3 θ ……. (ii)
On differentiating both sides of Eqs. (i) and (ii) w.r.t θ, we get

Question 62.
Differentiate the following function with respect to x.
(log x)x + xlog x (Delhi 2013)
Answer:
Let y = (log x)x + xlog x
Also, let u = (log x)x and v = xlog x, then y = u + v
⇒ dydx=dudx+dvdx …… (i)
Now, consider u = (log x)x
On taking log both sides, we get
log u = log (log x)x = x log(log x)
On differentiating both sides w.r.t. x, we get

Question 63.
If y = log[x + x2+a2−−−−−−√], then show that (x2 + a2)d2ydx2 + xdydx = 0 (Delhi 2013)
Answer:

Question 64.
Show that the function f(x) = |x – 3|, x ∈ R, is continuous but not differentiable at x = 3. (Delhi 2013)
Answer:

Question 65.
If x = a sin t and y = a[cos t + log tan (t/2)], then find d2ydx2 (Delhi 2013)
Answer:
d2ydx2 = acost

Question 66.
Differentiate the following with respect to x.
sin-1[2x+1⋅3x1+(36)x] (All India 2013)
Answer:
First, put 6x equal to tan θ. so that it becomes some standard trigonometric function. Then, simplify the expression and then differentiate by using chain rule.

Question 67.
If x = a cos3 θ and y = a sin3 θ, then find the value of d2ydx2 at θ = π6. (All lndia 2013)
Answer:

Question 68.
If x sin(a + y) + sin a cos(a + y) = 0, then prove that = dydx=sin2(a+y)sina. (All IndIa 2013)
Answer:

Question 69.
If xy = ex – y, then prove that dydx=logx(1+logx)2 (All India 2013, Delhi 2010)
Or
If xy = ex – y, then prove that (All India 2011)
dydx=logx{log(xe)}2
Answer:
First, take log on both sides and convert it into y = f(x) form. Then, differentiate both sides to get required result.
Given, xy = ex – y
On taking log both sides, we get
y logex = (x – y)logee
⇒ y loge x = x – y [∵ logee = 1]
⇒ y(1 + log x) = x
⇒ y = 1+logx
On differentiating both sides w.r.t x, we get

Question 70.
If yx = ex – y, then prove that
dydx=(1+logy)2logy (All India 2013)
Answer:
First, take log on both sides and convert it into y = f(x) form. Then, differentiate both sides to get required result.
Given, xy = ex – y
On taking log both sides, we get
y logex = (x – y)logee
⇒ y loge x = x – y [∵ logee = 1]
⇒ y(1 + log x) = x
⇒ y = 1+logx
On differentiating both sides w.r.t x, we get

Question 71.
If sin y = x sin(a + y), then prove that
dydx=sin2(a+y)sina (Delhi 2012)
Answer:

Question 72.
If y = sin-1x, show that
(1 – x2)d2ydx2 – xdydx = 0.
Answer:
Given y = (sin-1 x)2
Differentiating on w.r.t x, we get

Question 73.
If x = sin−1t and y = cos−1t then show that dydx=−yx. (All India 2012)
Answer:

Question 74.
Differentiate tan-1[1+x2√−1x] w.r.t. x. (All India 2012)
Answer:
l2(1+x2)

Question 75.
If y = (tan-1 x)2, then show that (x2 + 1)2 d2ydx2 + 2x(x2 + 1)dydx = 2 (Delhi 2012)
Answer:

Question 76.
If y = xsin x – cos x + , then find . (Delhi 2012C)
Answer:

Question 77.
If x = a(cos t + t sin t) and y = a(sin t – t cos t), then find d2xdt2,d2ydt2 and d2ydx2. (All India 2012)
Answer:
Given x = a(cos t + t sin t)
On differentiating both sides w.r.t t, we get

Question 78.
If x = a(cos t + log tan t2) and y = a sin t, find d2ydt2 and d2ydx2. (All India 2012)
Answer:
d2ydx2 = sintsec4t
Also, d2ydt2 = ddt(dydt) = ddt(a cos t) = – a sin t

Question 79.
Find dydx, when y = xcot x + 2x2−3x2+x+2 (All IndIa 2012C)
Answer:

Question 80.
If x = tan(1alogy), then show that (1 + x2)d2ydx2 + (2x – a)dydx = 0 (All India 2011)
Answer:

Question 81.
Differentiate xx cos x + x2+1x2−1 w.r.t x. (Delhi 2011)
Answer:
xx cos x [cos x – x log x sin x + log x cos x + 4x(x2−1)2

Question 82.
If x = a (θ – sin θ), y = a (1 + cos θ), then find d2ydx2. (Delhi 2011)
Answer:

Question 83.
Prove that
ddx[x2a2−x2−−−−−−√+a22sin−1(xa)] = a2−x2−−−−−−√ (Foregin 2011)
Answer:

Question 84.
If y = log[x + x2+1−−−−−√], then prove that (x + 1)d2ydx2 + xdydx = 0. (Foreign 2011)
Answer:

Question 85.
If log(1+x2−−−−−√ – x) = y1+x2−−−−−√, then show that (1 + x2)dydx + xy + 1 = 0. (All India 2011C)
Answer:

Question 86.
If x = a(θ + sin θ) and y = a(1 – cos θ),then find  (All India 2011C)
Answer:
14asec4θ2

Question 87.
If y = a sin x + b cos x, then prove that y2 + (dydx)2 = a2 + b2. (All India 2011C)
Answer:
First, we differentiate the given expression with respect to x and get first derivative of y. Then, put the value of y and first derivative of y in LHS of given expression and then solve it to get the required RHS.
To prove y2 + (dydx)2 = a2 + b2
Given, y = a sin x + b cos x ….. (ii)
On differentiating both sides of Eq. (ii) w.r.t. x,
we get
dydx = a cos x – b sin x
Now, Let us take LHS of Eq. (i).
Here, LHS = y2 + (dydx)2
On putting the value of y and dy/dx , we get
LHS = (a sin x + b cos x)2 + (a cos x – b sin x)2
= a2 sin2 x + b2 cos2 x + 2ab sin x cos x + a2 cos2 x + b2 sin2 x – 2ab sin x cos x
= a2 sin2 x + b2 cos2 x + a2 cos2 x + b2 sin2 x
= a2 (sin2 x + cos2 x) + b2 (sin2 x + cos2 x)
= a2 + b2 [∵ sin2 x + cos2 x = 1]
= RHS
Hence proved.

Question 88.
If x = a(cos θ + θ sin θ) and y = a(sin θ – θ cos θ), then find d2ydx2 (All India 2011C)
Answer:
sec3θ

Question 89.
If x = a(θ – sin θ) and y = a(1 + cos θ), then find dydx at θ = π3. (Delhi 2011C)
Answer:
Given x = a(θ – sin θ)
and y = a(1 + cos θ)
On differentiating both sides w.r.t. θ, we get

Question 90.
If y (sin x – cos x)(sin x – cos x), π4 < x < 3π4, then find dydx. (All India 2010C)
Answer:
First, take log on both sides and then differentiate to get the required value of dydx.

Question 91.
If y = cos-1[2x−31−x2√13√], then find dydx. (All India 2010C)
Answer:
In the given expression, put x = sin θ and simplify the resulting expression, then differentiate it.

Question 92.
If y = (cot-1 x)2, then show that
(x2 + 1)2 d2ydx2 + 2x(x2 + 1)dydx = 2. (Delhi 2010C)
Answer:

Question 93.
If y = cosec-1x, x > 1, then show that x(x2 – 1)d2ydx2 + (2x2 – 1)dydx = 0 (All India 2010)
Answer:
Given y = cosec-1x
On differentiating both sides w.r.t. x, we get

Question 94.
If y = cos-1(3x+41−x2√5), then find dydx. (All India 2010)
Answer:
11−x2√

Question 95.
Show that the function defined as follows, is continuous at x = 1, x = 2 but not differentiable at x = 2. (Delhi 2010)

Answer:

Continuity at x = 2:

⇒ RHL = 6
Also, f(2) = 2(2)2 – 2 = 8 – 2 = 6
Since, LHL = RHL = f(2)
∴ f(x) is continuous at x = 2
Hence, f(x) is continuous at all indicated points.
Now, let us verify differentiability of the given function at x = 2

Differentiability at x = 2:

Since, LHD ≠ RHD
So, f(x) is not differentiable at x = 2
Hence. f(x) is continuous at x = 1 and x = 2 but not differentiable at x = 2
Hence Proved.

Question 96.
If y = ea cos-1 x, – 1 ≤ x ≤ 1, then show that
(1 – x2)d2ydx2 – x dydx – a2y = 0. (All IndIa 2010)
Answer:
Given y = em sin-1 x
On differentiating both sides w.r.t x, we get,

Question 97.
Find dydx, if y = (cos x)x + (sin x)1/x. (Delhi 2010)
Answer:
Given, y = (cos x)x + (sin x)1/x
Let u = (cos x)x and v = (sin x)1/x
Then, given equation becomes
y = u + v
on differentiating both sides w.r.t x, we get
⇒ dydx=dudx+dvdx
Consider, u = (cos x)x
On taking log both sides, we get
log u = log (cos x)x
⇒ log u = x log (cos X)
[∵ log mn = n log m]
On differentiating both sides w.r.t. x, we get

Question 98.
If y = ex sin x, then prove that
d2ydx2 – 2dydx + 2y = 0. (All India 2010C)
Answer:
First, find dydx and d2ydx2 and then put their values along with value of y in LHS of proven expression.

Question 99.
If y = (x)x + (sin x)x, then find dydx (All India 2010C).
Answer:
Given, y = (x)x + (sin x)x
Let u = (x)x
and v = (sin x)x
Then, given equation becomes, y = u + v
On differentiating both sides w.r.t. x, we get
dydx=dudx+dvdx …….. (i)
Consider, u = xx
On taking log both sides, we get
log u = log xx
⇒ log u = x log x [∵ log mn = n log m]