## 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) = – x^{2}

⇒ 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) = x^{3} – 3x in [-√3, 0]. (All India 2017)

Answer:

Given, f(x) = x^{3} – 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) = 3x^{2} – 3 [from Eq. (i)]

⇒ f’(c) = 3c^{2} – 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 sin^{2} y + cos xy = K. (Delhi 2017)

Answer:

we have sin^{2} 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 x^{n} = 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

⇒ x^{2} (1 + y) = y^{2} (1 + x)

⇒ x^{2} + x^{2}y = y^{2} + y^{2}x

⇒ x^{2} – y^{2} = y^{2}x – x^{2}y

⇒ (x – y) (x + y) = – xy (x – y)

[∵ a^{2} – b^{2} = (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 – x^{2})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} = c^{2}, for some c > 0,

prove that

independent of a and b. (All India 2019)

Answer:

Given (x – a)^{2} + (y – b)^{2} = c^{2}

Differentiating on w.r.t x, we get

Question 15.

If x = ae^{t}(sin t + cos t) and y = ae^{t}(sin t – cos t), then prove that dydx=x+yx−y (All India 2019)

Answer:

Given x = x = ae^{t}(sin t + cos t)

and y = ae^{t}(sin t – cos t)

Question 16.

Differentiate x^{sin x} + (sin x)^{cos x} with respect to x. (All IndIa 2019)

Answer:

Question 17.

If log (x^{2} + y^{2}) = 2 tan^{-1}(yx) show that dydx=x+yx−y (Delhi 2019)

Answer:

log (x^{2} + y^{2}) = 2 tan^{-1}(yx)

on differentiating both sides w.r.t. x, we get

Question 18.

If x^{y} – y^{x} = 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 (x^{2} + y^{2})^{2} = xy, find dydx. (CBSE 2018)

Answer:

We have (x^{2} + y^{2})^{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 sec^{3} θ and y = a tan^{3} θ, find d2ydx2 at θ = π3. (CBSE 2018C)

Answer:

Question 25.

If y = e^{tan-1}x, prove that (1 + x^{2})d2ydx2 + (2x – 1)dydx = 0. (CBSE 2018 C)

Answer:

we have, y = e^{tan-1}x

on differentiating both sides w.r.t. x, we get

Question 26.

If x^{y} + y^{x} = a^{b}, then find dydx. (All India 2017)

Answer:

dydx = −xy−1⋅y−yxlogyxylogx+yx−1⋅x

Question 27.

If e^{y} (x + 1) = 1, then show that d2ydx2=(dydx)2. (All India 2017)

Answer:

Given, e^{y} (x + 1) = 1

On taking log both sides, we get

log [e^{y} (x + 1) = log]

y + log(x + 1) = log 1 [∵ log e^{y} = 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 = x^{x}, then prove that (Delhi 2016, 2014)

d2ydx2−1y(dydx)2−yx = 0 (Delhi 2016, 2014)

Answer:

Given y = x^{x}

On taking log both sides, we get

log y = log x^{x}

⇒ 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 sin^{2} 2t + 2 cos 2t + 2 cos^{2} 2t)

= a[2(cos^{2} 2t – sin^{2} 2t) + 2 cos 2t]

= a (2 cos 4t + 2 cos 2t) = 2a (cos 4t + cos 2t)

[∵ cos^{2} 2θ – sin^{2} 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 – x^{2})d2ydx2 – xdydx + p^{2}y = 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√), x^{2} ≤ 1, then find dy/ dx. (Delhi 2015)

Answer:

First, put x^{2} = 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 y^{2}d2ydx2 – 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 = e^{m sin-1 x}, then show that

(1 – x^{2})d2ydx2 – xdydx – m^{2} y = 0. (All India 2015).

Answer:

Given y = e^{m 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 + x^{2})d2ydx2 + xdydx = n^{2}y. (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−1x1−x2√ – 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 e^{x} + e^{y} = e^{x + y}, prove that dydx + e^{y – x} = 0. (Foreign 2014)

Answer:

Given, e^{x} + e^{y} = e^{x + y} ………… (i)

On dividing Eq.(i) by e^{x + 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 = ae^{θ} (sin θ – cos θ) and

y = ae^{θ} (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 x^{m} y^{m} = (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 x^{m} y^{n} = (x + y)^{m + n}

On taking log both sides, we get

log (x^{m} y^{m}) = log(x + y)^{m + n}

⇒ log(x^{m}) + log(y^{n}) = (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 = Pe^{ax} + Qe^{bx}, then show that – (a + b) + aby = 0. (All India 2014)

Answer:

Question 54.

If x = cos t(3 – 2 cos^{2} t)and y = sin t (3 – 2 sin^{2} t), then find the value of dydx at t = π4. (All India 2014)

Answer:

Given, x = cos t(3 – 2 cos^{2} t)

⇒ x = 3 cos t – 2 cos^{3} t

On differentiating both sides w.r.t. t, we get

dxdt = 3(- sin t) – 2(3) cos^{2}t (- sin t)

⇒ dxdt =- 3 sin t + 6 cos t sin t ….. (i)

Also, y = sin t (3 – 2sin^{2} t)

⇒ y = 3 sin t – 2 sin^{3} t

On differentiating both sides w.r.t. t, we get

dydt = 3 cos t – 2 × 3 × sin^{2} 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} + y^{cot x} = 1, then find dy/dx. (All India 2014C)

Answer:

Let u = (tan^{-1} x)^{y} and v = y^{cot 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 x^{3}d2ydx2 = (xdydx−y)2. (Delhi 2013C)

Answer:

Question 61.

If x = cos θ and y = sin^{3} θ, then prove that yd2ydx2+(dydx)2 = 3 sin^{2} θ(5 cos^{2} θ – 1). (All India 2013C)

Answer:

Given x = cos θ ……. (i)

and y = sin^{3} 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} + x^{log x} (Delhi 2013)

Answer:

Let y = (log x)^{x} + x^{log x}

Also, let u = (log x)^{x} and v = x^{log 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 (x^{2} + a^{2})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 = −cosec2tacost

Question 66.

Differentiate the following with respect to x.

sin^{-1}[2x+1⋅3x1+(36)x] (All India 2013)

Answer:

First, put 6^{x} 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 cos^{3} θ and y = a sin^{3} 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 x^{y} = e^{x – y}, then prove that dydx=logx(1+logx)2 (All India 2013, Delhi 2010)

Or

If x^{y} = e^{x – 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, x^{y} = e^{x – y}

On taking log both sides, we get

y log_{e}x = (x – y)log_{e}e

⇒ y log_{e} x = x – y [∵ log_{e}e = 1]

⇒ y(1 + log x) = x

⇒ y = x1+logx

On differentiating both sides w.r.t x, we get

Question 70.

If y^{x} = e^{x – 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, x^{y} = e^{x – y}

On taking log both sides, we get

y log_{e}x = (x – y)log_{e}e

⇒ y log_{e} x = x – y [∵ log_{e}e = 1]

⇒ y(1 + log x) = x

⇒ y = x1+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^{-1}x, show that

(1 – x^{2})d2ydx2 – xdydx = 0.

Answer:

Given y = (sin^{-1} x)^{2}

Differentiating on w.r.t x, we get

Question 73.

If x = asin−1t−−−−−√ and y = acos−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 (x^{2} + 1)^{2} d2ydx2 + 2x(x^{2} + 1)dydx = 2 (Delhi 2012)

Answer:

Question 76.

If y = x^{sin 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 = sintsec4ta

Also, d2ydt2 = ddt(dydt) = ddt(a cos t) = – a sin t

Question 79.

Find dydx, when y = x^{cot x} + 2x2−3x2+x+2 (All IndIa 2012C)

Answer:

Question 80.

If x = tan(1alogy), then show that (1 + x^{2})d2ydx2 + (2x – a)dydx = 0 (All India 2011)

Answer:

Question 81.

Differentiate x^{x cos x} + x2+1x2−1 w.r.t x. (Delhi 2011)

Answer:

x^{x 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 + x^{2})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 y^{2} + (dydx)2 = a^{2} + b^{2}. (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 y^{2} + (dydx)2 = a^{2} + b^{2}

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 = y^{2} + (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}

= a^{2} sin^{2} x + b^{2} cos^{2} x + 2ab sin x cos x + a^{2} cos^{2} x + b^{2} sin^{2} x – 2ab sin x cos x

= a^{2} sin^{2} x + b^{2} cos^{2} x + a^{2} cos^{2} x + b^{2} sin^{2} x

= a^{2} (sin^{2} x + cos^{2} x) + b^{2} (sin^{2} x + cos^{2} x)

= a^{2} + b^{2} [∵ sin^{2} x + cos^{2} 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:

sec3θaθ

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

(x^{2} + 1)^{2} d2ydx2 + 2x(x^{2} + 1)dydx = 2. (Delhi 2010C)

Answer:

Question 93.

If y = cosec^{-1}x, x > 1, then show that x(x^{2} – 1)d2ydx2 + (2x^{2} – 1)dydx = 0 (All India 2010)

Answer:

Given y = cosec^{-1}x

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 = e^{a cos-1 x}, – 1 ≤ x ≤ 1, then show that

(1 – x^{2})d2ydx2 – x dydx – a^{2}y = 0. (All IndIa 2010)

Answer:

Given y = e^{m 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 m^{n} = n log m]

On differentiating both sides w.r.t. x, we get

Question 98.

If y = e^{x} 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 = x^{x}

On taking log both sides, we get

log u = log x^{x}

⇒ log u = x log x [∵ log m^{n} = n log m]

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