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 =
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 =
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
If LHL = RHL = f
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
Answer:
Let y =
Question 2.
If y = cos (√3x), then find
Answer:
Given, y = cos (√3x)
Differentiating w.r.t x, we get
Question 3.
If f(x) = x + 1, find
Answer:
Given, f(x) = x + 1
⇒ f(f(x)) = f(x) + 1
⇒ fof(x) = x + 1 + 1
⇒ fof(x) = x + 2
Now,
Question 4.
If f(x) = x + 7 and g(x) = x – 7, x ∈ R, then find the values of
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
Question 5.
If y = x|x|, find
Answer:
We have, y = x|x|
When, x < 0, then |x| = – x
∴ y = x(- x) = – x2
⇒
Question 6.
Differentiate tan-1
Answer:
Question 7.
Differentiate tan-1
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
Answer:
we have sin2 y + cos xy = k
On differentiating both sides w.r.t x, we get
Question 10.
If y = sin-1 (6x
Answer:
Given, y = sin-1(6x
y = sin-1(2.3x
put 3x = sin θ, then
y = sin-1 (2 sin θ
⇒ y = sin-1 (2 sin θ. cos θ)
⇒ y = sin-1 (sin 2θ)
⇒ y = 2θ
⇒ y = 2 sin-1(3x) [∵ θ = sin-1(3x)]
⇒
⇒
Question 11.
If (cos x)y = (cos y)x, then find
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
Answer:
First, solve the given equation and convert it into y = f(x) form. Then, differentiate to get the required result.
To prove
Given equation is
where x ≠ y, we first convert the given equation into y = f(x) form.
Clearly, 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 =
On differentiating both sides w.r.t. x, we get
Question 13.
If y = (sin-1 x)2 prove that
(1 – x2)
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
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
Answer:
log (x2 + y2) = 2 tan-1
on differentiating both sides w.r.t. x, we get
Question 18.
If xy – yx = ab, find
Answer:
Question 19.
If x = cos t + log tan
Answer:
Question 20.
If y = sin (sin x), prove that
Answer:
Given y = sin (sin x) ….. (i)
On differentiating both sides w.r.t. x we get
Again. on differentiating both sides w.r.t. z.
we get
Question 21.
If (x2 + y2)2 = xy, find
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
Answer:
Question 23.
If sin y = x cos(a + y), then show that
Also, show that
Answer:
Question 24.
If x = a sec3 θ and y = a tan3 θ, find
Answer:
Question 25.
If y = etan-1x, prove that (1 + x2)
Answer:
we have, y = etan-1x
on differentiating both sides w.r.t. x, we get
Question 26.
If xy + yx = ab, then find
Answer:
Question 27.
If ey (x + 1) = 1, then show that
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
Again, differentiating both sides w.r.t. ‘x’, we get
Question 28.
If y = xx, then prove that (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
Answer:
Question 30.
If x = a sin 2t(1 + cos 2t)and
y = b cos 2t (1 – cos 2t), then find the values of
Or
If x = a sin2t(1 + cos 2t) and y = b cos 2t (1 – cos 2t), then show that at t =
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
[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
[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
Or
If cos y = x cos(α + y), where cos α ≠ ±1, prove that
Answer:
Question 32.
Find
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)
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
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 y2