Plus Two Maths Chapter 5 Continuity and Differentiability Chapter Wise Question and Answers PDF Download

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Plus Two Maths Chapter 5 Continuity and Differentiability Chapter Wise Question and Answers

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Board	Kerala Board
Study Materials	Chapter wise Question and Answers
For Year	2021
Class	12
Subject	Mathematics
Chapters	Maths Chapter 5 Continuity and Differentiability
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Plus Two Maths Chapter 5 Continuity and Differentiability Question and Answers PDF Download

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Question 1.
Consider the function g(x) = \(\frac{|x-2|}{x-2}\)

1. Find the domain and range of the function g(x). (2)
2. Check whether the g(x) is continuous at x = 2. (1)

Answer:
1. The function is not defined at points where the denominator is zero.
i.e., x – 2 = 0 ⇒ x = 2.
∴ domain is R – {2}.
g(x) = \(\frac{|x-2|}{x-2}\) = \(\left\{\begin{array}{c}{\frac{x-2}{x-2}, \quad x-2>0} \\{\frac{-(x-2)}{x-2}, \quad x-2<0}\end{array}=\left\{\begin{array}{ll}{1,} & {x>2} \\{-1,} & {x<2}\end{array}\right.\right.\)
∴ range is {-1, 1}

2. The function g(x) is not defined at x = 2. Therefore discontinuous.

Question 2.
(i) If f(x) = x+|x| + 1, then which of the follow represents f (x) (1)


(ii) Test whether f (x) is continuous at x=0. Explain. (2)
Answer:
(i) (b) Since, f(x) = \(\left\{\begin{array}{c}{x+x+1, x \geq 0} \\{x-x+1, \quad x<0}
\end{array}=\left\{\begin{array}{c}{2 x+1, x \geq 0} \\{1, x<0}\end{array}\right.\right.\)

(ii) We have ,f (0) = 1,

Continuous at x = 0.

Question 3.
Consider the function f(x) = \(\left\{\begin{array}{ll}{\frac{\sin x}{x}} & {, x<0} \\{x+1} & {, x \geq 0}\end{array}\right.\)

1. Find \(\lim _{x \rightarrow 0} f(x)\) (2)
2. Is f (x) continuous at x= 0? (1)

Answer:
1. To find \(\lim _{x \rightarrow 0} f(x)\) we have to find f(0^– )and f(0⁺)
f(0^–) = \(\lim _{x \rightarrow 0} \frac{\sin x}{x}=1\),
f(0⁺) = \(\lim _{x \rightarrow 0}\) + 1 = 0 + 1 = 1
f(0^–) = f(0⁺) = 1 .Therefore \(\lim _{x \rightarrow 0} f(x)\) = 1

2. Here, f (0) = 0 + 1 = 1 = f(0^–) = f(0⁺) = 1
Therefore continuous at x = 0.

Question 4.
Consider the figure and answer the following questions.

(i) Identify the graphed function. (1)

(ii) Discuss the continuity of the above function at x = 2. (2)
Answer:
(i) (b) f(x) = \(\left\{\begin{array}{ll}{\frac{|x-2|}{x-2},} & {x \neq 2} \\{0,} & {x=2}\end{array}\right.\)

(ii) From the figure we can see that
f(2^–) = 1, f(2⁺) = -1 and f(2) = 0
Therefore, f(2^–) = 1 ≠ f(2⁺) = -1 ≠ f(2) = 0. Discontinuous.

Question 5.
Consider f(x) = \(\left\{\begin{array}{ll}{2 x} & {\text { if } x<2} \\{2} & {\text { if } x=2} \\ {x^{2}} & {\text { if } x>2}\end{array}\right.\)
(a) Find \(\lim _{x \rightarrow 2^{-}} f(x)\) and \(\lim _{x \rightarrow 2^{+}} f(x)\) (2)
(b) f(x) is continuous. If not so, how can you make it continuous. (1)
Answer:

Therefore f(x) is not continuous at x = 2.
If f(2) = 4, then f(x) becomes continuous.

Question 6.
If y = log₁₀x + log_x10 + log_xx + log₁₀10. Find \(.
Answer:
y = log_e x log₁₀e + log_e10 log_xe + 1 + 1

Question 7.
Examine the continuity of the function

Answer:
In both the intervals x = 1 and x < 1 the function f(x) is a polynomial so continuous. So we have to check the continuity at x = 1.

f(1) = 2
Since

f(x) is continuous at x = 1.

Question 8.
Find [latex]\frac{d y}{d x}\) of the following (3 score each)

1. 2x + 3y = sinx
2. xy + y² = tanx + y
3. x³ + x²y + xy² + y³ = 81
4. sin²x + cos²y = 1
5. \(\sqrt{x}\) + \(\sqrt{y}\) = 1
6. x² + xy + y² = 7
7. x²(x – y) = y²(x + y)
8. xy² + x²y = 2
9. sin y = xcos (a + y)

Answer:
1. Given; 2x + 3y = sinx
Differentiating with respect to x;

2. Given; xy + y² = tanx + y
Differentiating with respect to x;

3. Given; x³ + x²y + xy² + y³ = 81
Differentiating with respect to x;

4. Given; sin²x + cos²y = 1
Differentiating with respect to x;

5. \(\sqrt{x}\) + \(\sqrt{y}\) = 1
Differentiating with respect to x;

6. x² + xy + y² = 7
Differentiating with respect to x;

7. x²(x – y) = y²(x + y)
Differentiating with respect to x;

8. xy² + x²y = 2
Differentiating with respect to x;

9. sin y = xcos (a + y)
⇒ x = \(\frac{\sin y}{\cos (a+y)}\)
Diff. with respect to y.

Question 9.
Find \(\frac{d y}{d x}\) of the following (3 score each)

(viii) x = a(t – sint), y = a(1 – cost)
(ix) y = e^t cost, x = e^t sint.
Answer:
(i) We know; y = sin^-1\(\left(\frac{2 x}{1+x^{2}}\right)\)
⇒ y = = 2 tan^-1(x)
Differentiating with respect to x;

(ii) We know; y = tan^-1\(\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right)\)
Put x = tanθ ⇒ θ = tan^-1x

⇒ y = tan^-1(tan3θ)
⇒ y = 3 tan^-1(x)
Differentiating with respect to x;

(iii) We know; y = sin^-1\(\left(\frac{1-x^{2}}{1+x^{2}}\right)\)
Put x = tanθ ⇒ θ = tan^-1x

(iv) We know; y = sec^-1 = \(\left(\frac{1}{2 x^{2}-1}\right)\)
Put x = cosθ ⇒ θ = cos^-1x

⇒ y = sec^-1 sec2θ = 2θ
⇒ y = 2cos^-1(x)
Differentiating with respect to x;

(v) ∴ y = tan^-1\(\sqrt{\tan ^{2} x / 2}\) = tan^-1 tanx/2 = x/2
\(\frac{d y}{d x}=\frac{1}{2}\).

= π – 2tan^-1x
\(\frac{d y}{d x}=\frac{-2}{1+x^{2}}\).

(vii) Let x = sinθ and \(\sqrt{x}\) = sinφ

= sin^-1 (sinθcosφ + cosφsinφ)
= sin^-1 (sin(θ + φ)) = θ + φ
= sin^-1x + sin^-1\(\sqrt{x}\)
\(\frac{d y}{d x}=\frac{1}{\sqrt{1-x^{2}}}+\frac{1}{\sqrt{1-x}} \times \frac{1}{2 \sqrt{x}}\).

(ix) y = e^tcost ⇒ \(\frac{d y}{d t}\) = – e^t sin t + e^t
x = e^t sint ⇒ \(\frac{d x}{d t}\) = e^t cos t + e^t sin t

Question 10.
If y = log\(\left(\frac{1}{x}\right)\), Show that \(\frac{d y}{d x}\) + e^y = 0.
Answer:
Given, y = log\(\left(\frac{1}{x}\right)\) ? \(\left(\frac{1}{x}\right)\) = e^y …….(1)

Question 11.
If e^y (x+1) = 1. Show that

1. \(\frac{d y}{d x}\) = -e^y (2)
2. \(\frac{d^{2} y}{d x^{2}}=\left(\frac{d y}{d x}\right)^{2}\) (1)

Answer:
1. e^y (x+1) = 1
Differentiating e^y +e^y(x +1) \(\frac{d y}{d x}\) = 0

2.

Question 12.
(i) Evaluate \(\lim _{x \rightarrow 0} \frac{k \cos x}{\pi-2 x}\) (2)
(Hint: Put π – 2x = y , where Iris a constant)
(ii) Find the value of k if f (x) is a continuous function given by (1)

Answer:
(i) \(\lim _{x \rightarrow \pi / 2} \frac{k \cos x}{\pi-2 x}=k \lim _{x \rightarrow \pi / 2} \frac{\cos x}{\pi-2 x}\),
Put π – 2x = y when Put x → π/2, y → 0

(ii) Since f (x) is continuous

Question 13.
If f(x) \(f(x)=\left\{\begin{array}{cc}{x-[x]} & {, x<.2} \\{0} & {; x=2} \\{3 x-5} & {, x>2}
\end{array}\right.\)

1. Find \(\lim _{x \rightarrow 2} f(x)\) (2)
2. Is f(x) continuous at x = 2? (1)

Answer:
1. To find \(\lim _{x \rightarrow 2} f(x)\)
we have to find f(2^–) and f(2⁺)
f(2^–) = \(\lim _{x \rightarrow 2} x-[x]\) = 2 -1 = 1,
f(2⁺) = \(\lim _{x \rightarrow 2}\) 3x – 5 = 6 -5 = 1
f(2^–) = f(2⁺) = 1.
Therefore \(\lim _{x \rightarrow 2}\) f(x) = 1

2. Here, f(2) = 0 ≠ f(2^–) = f(2⁺) = 1.
Therefore discontinuity at x = 2.

Plus Two Maths Continuity and Differentiability Four Mark Questions and Answers

Question 1.
If x = 2cosθ; y = 3sinθ

1. Find \(\frac{d y}{d x}\).
2. Find \(\frac{d^{2} y}{d x^{2}}\)

Answer:
1. x = 2cosθ ⇒ \(\frac{d x}{d θ}\) = -2sinθ
y = 3sinθ ⇒ \(\frac{d x}{d θ}\) = 3cosθ

2.

Question 2.
If y = (tan^-1 x)², show that (x² +1)² y₂ + 2x(x² +1) y₁ = 2.
Answer:
y = (tan^-1 x)²
⇒ y₁ = 2(tan^-1 x) \(\frac{1}{1+x^{2}}\)
⇒ (1 + x²)y₁ = 2(tan^-1 x)
⇒ (1 + x²)y₂ + y₁2x = 2 \(\frac{1}{1+x^{2}}\)
⇒ (1 + x²)² y₂ + x(1 + x²)y₁ = 2.

Question 3.
Find \(\frac{d y}{d x}\) if y = sin^-1 \(\left(\frac{1-x^{2}}{1+x^{2}}\right)\), 0 < x < 1.
Answer:
Put x = tanθ

Question 4.
Let f(x) = \(\left\{\begin{array}{ll}{\cos x,} & {0 \leq x \leq c} \\{\sin x,} & {c<x \leq \pi}\end{array}\right.\)

1. Find the value of c if / is continuous on [0, π].
2. Show that is f not differentiable at the point c.

Answer:
1. Since f is continuous on [0, π], we have;
\(\lim _{x \rightarrow c^{+}}\) f(x) = \(\lim _{x \rightarrow c^{-}}\) f(x) = f(c)
⇒ \(\lim _{x \rightarrow c^{+}}\) sinx = \(\lim _{x \rightarrow c^{-}}\) cosx = cosc
⇒ sinc = cosc ⇒ c = \(\frac{π}{4}\).

2. \(f^{\prime}(x)=\left\{\begin{array}{ll}{-\sin x,} & {0 \leq x \leq c} \\{\cos x,} & {c<x \leq \pi}\end{array}\right. \)
Left derivative at \(\frac{\pi}{4}\) = – sin \(\frac{\pi}{4}\) = –\(\frac{1}{\sqrt{2}}\)
Right derivative at \(\frac{\pi}{4}\) = cos \(\frac{\pi}{4}\) = \(\frac{1}{\sqrt{2}}\)
Left derivative at \(\frac{\pi}{4}\) ≠ Right derivative at \(\frac{\pi}{4}\)
Therefore is not differentiable at the point c.

Question 5.

1. Find \(\frac{d y}{d x}\) if x = 2sinθ; y = 3cosθ
2. Which among the following functions is differentiable on R?

(a) |sinx|
(b) |cosx|
(c) cos|x|
(d) sin|x|
Answer:
1. \(\frac{d x}{d θ}\) = 2cosθ; \(\frac{d y}{d θ}\) = -3 sinθ ⇒ \(\frac{d y}{d x}\) = \(-\frac{3}{2}\)tanθ

2. (c) cos|x|
(Since cos x is an even function, it treats x and -x in the same way).

Question 6.
(i) Examine whether the function defined by \(f(x)=\left\{\begin{array}{ll}{x+5,} & {x \leq 1} \\{x-5,} & {x>1}\end{array}\right.\) is continuous or not. (2)
(ii) If x = ^sin-1t, y = ^cos-1t, a > 0, show that \(\frac{d y}{d x}=-\frac{y}{x}\)
Answer:

f(x) is not continuous.

Question 7.
(i) If \(f(x)=\left\{\begin{array}{ll}{1-x,} & {0 \leq x \leq 1} \\{1+x,} & {1<x \leq 2}\end{array}\right.\) then which of the following is not true (1)
(a) f is continuous in ( 0, 1 )
(b) f is continuous in (1, 2 )
(c) f is continuous in [ 0, 2 ]
(d) f is continuous in [ 0,1 ]
if \(\left\{\begin{array}{cc}{1,} & {x \leq 3} \\{a x+b} & {, \quad 3<x<5} \\{7,} & {5 \leq x}\end{array}\right.\)
(ii) Find f(3⁺) and f(5^–) (1)
(iii) Hence find the value of ‘a’ and ‘b’ so that f(x) is continuous. (2)
Answer:
(i) (c) Since f is not continuous at x = 1.

(iii) Since f (x) is continuous, it is continuous at x = 3 and x = 5
∴ f(3⁺) = f(3) ⇒ 3a + b = 1 ____(1)
and f(5^–) = f(5) ⇒ 5a + b = 7 ____(2)
(2) – (1) ⇒ 2a = 6, a = 3
(1) ⇒ b = 1 – 3 a ⇒ b = -8
∴ a = 3, b = – 8.

Question 8.
Consider f(x) = \(\left\{\begin{array}{ll}{2 x+3,} & {x \leq 2} \\{x+2 k,} & {x>2}\end{array}\right.\)

1. Find f(2) (1)
2. Evaluate \(\lim _{x \rightarrow 2^{+}}\)f(x) (1)
3. Find the value of k, if is continuous at x = 2. (2)

Answer:
1. f(2) = 2(2) + 3 = 7

2. Here, f(x) = x + 2k for x > 2.
\(\lim _{x \rightarrow 2^{+}}\)f(x) = \(\lim _{x \rightarrow 2}\)(x + 2k) = 2 + 2k.

3. Since f (x) is continuous at x = 2
We have, f(2) = \(\lim _{x \rightarrow 2^{+}}\)f(x)
⇒ 7 = 2 + 2k ⇒ k = \(\frac{5}{2}\)

Question 9.
Find \(\frac{d y}{d x}\) of the following (4 score each)

1. y = (logx)^cosx
2. x = 2at², y = at⁴
3. x = a(cosθ + θsinθ), y = a(sinθ – θcosθ)
4. y=x^x
5. y =(x log x)^log(logx)
6. y = \(\sqrt{\sin x \sqrt{\sin x+\sqrt{\sin x+\ldots .}}}\)
7. y^x = x^siny
8. y =(log x)^x + x^logx
9. y = (sinx)^x + sin^-1 \(\sqrt{x}\)

Answer:
1. Given; y = (logx)^cosx, taking log on both sides;
log y = cosxlog(logx),
Differentiating with respect to x;

2. Given; x = 2at² ⇒ \(\frac{d x}{d t}\) = 4at

3. Given; x = a(cosθ + θsinθ)
\(\frac{d x}{d \theta}\) = a(-sinθ + θcosθ + sinθ) = aθcosθ
y = a(sinθ – θcosθ)
\(\frac{d x}{d \theta}\) = a(cosθ – θ(-sinθ) – cosθ) = aθ sinθ
\(\frac{d y}{d x}=\frac{a \theta \sin \theta}{a \theta \cos \theta}=\tan \theta\)

4. y= x^x; Taking log on both sides;
log y = x log x

5. y = (x log x)^{log logx}
Taking log on both sides;
log y = (log log x) [log (xlogx)]

6. y = \(\sqrt{\sin x+y}\) ⇒ y² = sinx + y
⇒ y² – y = sinx

7. y^x = x^{sin y};
Taking log on both sides;
xlogy = siny log x

8. y = (log x)^x + x^logx = u + v

9. y = (sinx)^x + sin^-1\(\sqrt{x}\)
Let u = (sinx)^x ⇒ log u = x log sinx

Question 10.
Find \(\frac{d^{2} y}{d x^{2}}\) of the following

1. y = x² + 3x + 2 (2)
2. y = tan^-1x (2)

Answer:
1. Given; y = x² + 3x + 2
Differentiating with respect to x;
\(\frac{d y}{d x}\) = 2x + 3
Differentiating again with respect to x;
\(\frac{d^{2} y}{d x^{2}}\) = 2.

2. Given; y = tan^-1x
Differentiating with respect to x; \(\frac{d y}{d x}\) = \(\frac{1}{1+x^{2}}\)
Differentiating again with respect to x;
\(\frac{d^{2} y}{d x^{2}}\) = \(-\frac{1}{\left(1+x^{2}\right)^{2}} \cdot 2 x\).

Question 11.
Match the following. (4)

Answer:

Question 12.
If x – sint and y = sinmt show that
(i) y = sin(m sin^-1x) (1)
(ii) \(\frac{d y}{d x}\) (1)
(iii) (1 – x²) y₂ – xy₁ + m²y = 0 (2)
Answer:
(i) x = sint, y = sinmt
t = sin^-1x ⇒ y = sin(msin^-1x).

multiplying with \(\sqrt{1-x^{2}}\)
(1 – x²) y₂ – xy₁ = -m²y
(1 – x²) y₂ – xy₁ + m²y = 0.

Question 13.
Consider the function f(x) = x(x – 2), x ∈ [1, 3]. Verify mean value theorem for the function in[1, 3].
Answer:
f(x) = x(x – 2) = x² – 2x ⇒ f'(x) = 2x – 2.
As f is a polynomial, it is continuous in the interval [1, 3] and differentiable in the interval (1, 3).
Therefore two conditions of MVT are satisfied and so there exists c ∈ (1,3)such that.

Hence MVT is verified.

Question 14.
Verify Lagranges’ Mean value theorem for the function f(x) = 2x² – 10x + 29 in [2, 9]
Answer:
f(x) = 2x² – 10x + 29; f'(x) = 4x – 10.
As f is a polynomial, it is continuous in the interval [2, 9] and differentiable in the interval (2, 9).
Therefore two conditions of MVT are satisfied and so there exists c ∈ (2, 9) such that.

Hence MVT is verified.

Question 15.
Let f(x) = x(x – 1)(x – 2), x ∈ [0, 2]

1. Find f(0) and f(2) (1)
2. Find f'(x) (1)
3. Find the values of x when f'(x) = 0 verify Rolle’s theorem. (2)

Answer:
1. f(0) = 0, f(2) = 2(2 – 1)(2 – 2) = 0

2. We have, f(x) = x³ – 3x² + 2x
⇒ f'(x) = 3x² – 6x + 2.

3. f'(x) = 3x² – 6x + 2 = 0

Clearly all the three conditions of Rolle’s theorem are satisfied and 1 ± \(\frac{1}{\sqrt{3}}\) ∈ (0, 2).

Question 16.
Verify Rolle’s Theorem for the function
f(x) = x² + 2x – 8, x ∈ [-4, 2]
Answer:
f(x) = x² + 2x – 8, f'(x) = 2x + 2.
As f is a polynomial, it is continuous in the interval [-4, 2] and differentiable in the interval (-4, 2).
f(-4) = 16 – 8 – 8 = 0
f(2) = 4 + 4 – 8 = 0
f'(c) = 0 ⇒ 2c + 2 = 0 ⇒ c = -1 ∈ (-4, 2) Hence Rolle’s Theorem is verified.

Question 17.
Examine that Rolle’s Theorem is applicable to the following function in the given intervals, justify your answer.

1. f(x) = [x], x ∈ [5, 9]
2. f(x) = x² – 1, x ∈ [1, 2]

Answer:
1. The function f(x) = [x] is not differentiable and continuous at integral values. So in the given interval [5, 9] the function is neither differentiable nor continuous at x = 6, 7 ,8. Therefore Rolle’s Theorem is not applicable.

2. The function f(x) = x² – 1 is a polynomial function so differentiable and continuous.
f(1) = 1 – 1 = 0, f(2) = 4 – 1 = 3
f(1) ≠ f(2) . Therefore Rolle’s Theorem is not applicable.

Question 18.
Examine the continuity of the function
\(f(x)=\left\{\begin{array}{cc}{|x|+3,} & {x \leq-3} \\{-2 x,} & {-3<x<3} \\{6 x+2,} & {x \geq 3}\end{array}\right.\)
Answer:
In the intervals x ≤ -3, f(x) is the sum of a constant function and modulus function so continuous. In the intervals -3 < x < 3 and x ≥ 3the function f(x) is a polynomial so continuous. Hence we have to check the continuity at x = -3, x = 3.
At x = -3
f(-3) = 6

f(x) is continuous at x = -3.
At x = 3
f(3) = 6(3) + 2 = 20

Since \(\lim _{x \rightarrow 3^{-}}\)f(x) = f(3), f(x) is not continuous at x = 3.

Question 19.
Test continuity for the following functions.

Answer:

= 0 × a finite quantity between -1 and 1 = 0 Also f(0) = 0
Therefore f(x) is continuous at x = 0.

Therefore f(x) is discontinuous at x = 0

But f(1) = 2
∴ f(x) is discontinuous at x = 1.

Question 20.
If y = \(\frac{\sin ^{-1} x}{\sqrt{1-x^{2}}}\) prove that
(i) (1 – x²) y² = (sin^-1x)² (1)
(ii) (1 – x²)y₁ – xy = 1 (1)
(iii) (1 – x²) y₂ – 3xy₁ – y = 0 (2)
Answer:

(ii) Differentiating
(1 – x²) 2y.y₁ + y²x – 2x = \(\frac{2 \sin ^{-1} x}{\sqrt{1-x^{2}}}\)
(1 – x²)2yy₁ – 2xy² = 2y
(1 – x²) y₁ – xy = 1.

(iii) Again differentiating
(1 – x²) y₂ + y₁ x – 2x – xy₁ – y = 0
(1 – x²) y₂ – 3xy₁ – y = 0.

Question 21.
At what point on the curve y = x², x ∈ [-2, 2] at which the tangent is parallel to x-axis?
Answer:
Y = x², a continuous function on [-2, 2] and differentiable on [-2, 2] f(2) = 4 = f(-2). All conditions of Rolles theorem is satisfied. Given the tangent is parallel to x-axis.
f¹ (x) = 2x
f¹(c) = 2c
f¹(c) = 0 ⇒ 2c = 0 ⇒ c = 0 ∈ [-2, 2]
where c = 0, y = 0
Therefore (0, 0) is the required point.

Question 22.

is continuous in the interval [-1 1].
(a) Find \(\lim _{x \rightarrow 0}\)f(x) (2)
(b) Find f(0). (1)
(c) Find P. (1)
Answer:

(c) Since f is continuous in [-1 1] it is continuous at 0.
Therefore P = \(-\frac{1}{2}\).

Question 23.
If ax² + 2hxy + by² = 1

1. Find \(\frac{d y}{d x}\) (1)
2. Find \(\frac{d^{2} y}{d x^{2}}\) (3)

Answer:
1. We have, ax² + 2hxy + by² = 1 ___(1)
Differentiating w.r.t.x, we get,

2. Differentiating (2) w.r.tx, we get,

Question 24.
Consider the function y = x^x \(\sqrt{x}\)

1. Express the above function as logy = \(\left(x+\frac{1}{2}\right)\) logx (2)
2. Find \(\frac{d y}{d x}\) (2)

Answer:
1. Given, y = x^x \(\sqrt{x}\). Take log on both sides,

2. We have, logy = \(\left(x+\frac{1}{2}\right)\) logx
Differentiating w.r.t x, we get,

Plus Two Maths Continuity and Differentiability Six Mark Questions and Answers

Question 1.

1. Verify mean value theorem for the function f(x) = (x – 2)² in [1, 4].
2. Find a point on the curve y = (x – 2)² at which the tangent is parallel to the chord joining the points (1, 1) and (4, 2)
3. Find a point on the above curve at which the tangent is parallel to the x-axis.

Answer:
1. f(x) = (x – 1)², x ∈ [1, 4]
f(x) is continuous in [1, 4]
f'(x) = 2(x – 2) is differentiable in [1, 4]
Then there exists c ∈ [1, 4] so that

Hence Mean Value Theorem is verified.

2. c = \(\frac{5}{2}\) will be the x-coordinate to the point of contact of tangent and the curve, then y = (x – 2)² ⇒ y = (\(\frac{5}{2}\) – 2)² = \(\frac{1}{4}\)
Therefore the point is (\(\frac{5}{2}\), \(\frac{1}{4}\)).

3. The tangent parallel to x- axis will have
f'(c) = 0 ⇒ 2(c – 2) = 0 ⇒ c = 2
Then; x = 2 ⇒ y = (2 – 2)² = 0
Therefore the point is (2, 0).

Question 2.

1. Differentiate x^sinx w.r.t.x (2)
2. If x = at², y = 2at, then find \(\frac{d y}{d x}\) (2)
3. If y = sin^-1(cosx) + cos^-1(sinx), then find \(\frac{d y}{d x}\). (2)

Answer:
1. Let y = x^sinx, take log on both sides,
log y = sinx logx differentiate w. r.t.x, we get

2.

3. Given, y = sin^-1(cosx) + cos^-1(sinx)

Differentiate w.r.t. x, we get \(\frac{d y}{d x}\) = -2.

Question 3.

1. Differentiate \(\frac{x-1}{x-3}\) with respect to x.(2)
2. Differentiate \(\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\) with respect to x. (4)

Answer:
1. Let y = \(\frac{x-1}{x-3}\) Differentiate w.r.t. x, we get;

2. Given, y = \(\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\)
Take log on both sides;

Differentiate w.r.t. x, we get;

Question 4.
(i) Define |x|
(a) |x| = \(\sqrt{x^{2}}\)
(b) |x| = x
(c) |x| = -x
(d) |x| = x²
(ii) At which point \(\frac{d}{d x}\)|x| does not exist?
Find \(\frac{d}{d x}\) |x|. (2)
(iii) Find \(\frac{d}{d x}\)|x³ – 7x| . Also, find the point at which the derivative exists. (3)
Answer:
(i) (a) |x| = \(\sqrt{x^{2}}\).

(ii) At x = 0, \(\frac{d}{d x}\) |x| does not exist.

Does not exist at
x³ – 7x = 0 ⇒ x(x² – 7) = 0
⇒ x = 0, x² – 7 = 0 ⇒ x = ±\(\sqrt{7}\).

Question 5.
(i) Match the following (4)

(ii) If log (x² + y²) = 2 tan^-1\(\left(\frac{y}{x}\right)\), then, show that \(\frac{d y}{d x}=\frac{x+y}{x-y}\) (2)
Answer:
(i)

(ii) Given, log (x² + y²) = 2 tan^-1\(\left(\frac{y}{x}\right)\).
Differentiate w.r.to x, we get;

Question 6.
If x = a sec³θ and y = a tan³θ

Answer:
(i) Given, x = a sec³θ
Differentiate w.r.to θ, we get;
\(\frac{d x}{d \theta}\) = 3a sec²θ. secθ. tanθ = 3a sec³ θ. tan θ
Given, y = a tan³θ .
Differentiating w.r.to θ, we get
\(\frac{d x}{d \theta}\) = 3a tan² θ. sec²θ.

(iii) We have, \(\frac{d y}{d x}\) = sinθ
Differentiating w.r.to x, we get

(iv) We have,

Question 7.
Consider the function \(f(x)=\left\{\begin{array}{cc}{1-x} & {, \quad x<0} \\{1} & {x=0} \\ {1+x} & {, \quad x>0}\end{array}\right.\)
(i) Compete the following table. (2)

(ii) Draw a rough sketch of f (x). (2)
(iii) What is your inference from the graph about Its continuity. Verify your answer using limits. (2)
Answer:

Since, f (- 2) = 1 – (- 2) = 3, f (-1) = 1 – (-1) = 2,
f(1) = 1 + (1) = 2, f (2) = 1 + (2) = 3.

(ii)

(iii) From the graph we can see that there is no break or jump at x = 0. Therefore continuous.
From the figure we can see that
f(0^–) = 1 f(0⁺) = 1 and f(0) = 1
Hence, f(0^–) = f(0⁺) = f(0) = 1.
Therefore continuous.

Question 8.
Consider the equation \(\sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a(x-y)\)
(i) Simplify the above equation to sin^-1x – sin^-1y = 2cot^-1 a by giving suitable substitution.
(ii) Prove that \(\frac{d y}{d x}=\sqrt{\frac{1-y^{2}}{1-x^{2}}}\)
Answer:

(ii) We have; sin^-1 x – cos^-1y = 2cot^-1a.
Differentiating w.r.t x, we get,

Question 9.
(i) Match the following. (4)

(ii) If y = e^{a cos-1x}, then show that (1 – x²)y₂ – xy₁ – a²y = 0 (2)
Answer:
(i)

(ii) Given, y = e^{a cos-1x} ____(1)
Differentiating w.r.to x,

Again differentiating w. r.to x

⇒ (1 – x²)y₂ – xy₁ = a². e^{a cos-1x}
⇒ (1 – x²)y₂ – xy₁ = a²y
⇒ (1 – x²)y₂ – xy₁ – a²y = 0.

Question 10.
(i) Match the following (3)

Differentiate the following
(ii) y =\(\frac{1}{5 x^{2}+3 x+7}\) (1)
(iii) y = 3cosec⁴(7x) (1)
(iv) y = e^{2log tan 5x} (1)
Answer:
(i)

(iii) Given,

(iv) Given,

Plus Two Mathematics All Chapters Question and Answers

• Plus Two Mathematics Chapter Wise Question and Answers PDF
• Plus Two Maths Chapter 1 Relations and Functions Chapter Wise Question and Answers PDF
• Plus Two Maths Chapter 2 Inverse Trigonometric Functions Chapter Wise Question and Answers PDF
• Plus Two Maths Chapter 3 Matrices Chapter Wise Question and Answers PDF
• Plus Two Maths Chapter 4 Determinants Chapter Wise Question and Answers PDF
• Plus Two Maths Chapter 5 Continuity and Differentiability Chapter Wise Question and Answers PDF
• Plus Two Maths Chapter 6 Application of Derivatives Chapter Wise Question and Answers PDF
• Plus Two Maths Chapter 7 Integrals Chapter Wise Question and Answers PDF
• Plus Two Maths Chapter 8 Application of Integrals Chapter Wise Question and Answers PDF
• Plus Two Maths Chapter 9 Differential Equations Chapter Wise Question and Answers PDF
• Plus Two Maths Chapter 10 Vector Algebra Chapter Wise Question and Answers PDF
• Plus Two Maths Chapter 11 Three Dimensional Geometry Chapter Wise Question and Answers PDF
• Plus Two Maths Chapter 12 Linear Programming Chapter Wise Question and Answers PDF
• Plus Two Maths Chapter 13 Probability Chapter Wise Question and Answers PDF

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