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Sunday, September 12, 2021

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

Plus Two Maths Chapter 5 Continuity and Differentiability Chapter Wise Question and Answers PDF Download: Students of Standard 12 can now download Plus Two Maths Chapter 5 Continuity and Differentiability chapter wise question and answers pdf from the links provided below in this article. Plus Two Maths Chapter 5 Continuity and Differentiability Question and Answer pdf will help the students prepare thoroughly for the upcoming Plus Two Maths Chapter 5 Continuity and Differentiability exams.


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

Plus Two Maths Chapter 5 Continuity and Differentiability question and answers consists of questions asked in the previous exams along with the solutions for each question. To help them get a grasp of chapters, frequent practice is vital. Practising these questions and answers regularly will help the reading and writing skills of students. Moreover, they will get an idea on how to answer the questions during examinations. So, let them solve Plus Two Maths Chapter 5 Continuity and Differentiability chapter wise questions and answers to help them secure good marks in class tests and exams.


Board

Kerala Board

Study Materials

Chapter wise Question and Answers

For Year

2021

Class

12

Subject

Mathematics

Chapters

Maths Chapter 5 Continuity and Differentiability

Format

PDF

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Spandanam Blog


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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)
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 3M Q2
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 3M Q2.1
(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,
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 3M Q2.2
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.
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 3M Q4
(i) Identify the graphed function. (1)
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 3M Q4.1

Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 3M Q4.2
(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:
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 3M Q5

Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 3M Q5.1
Therefore f(x) is not continuous at x = 2.
If f(2) = 4, then f(x) becomes continuous.

Question 6.
If y = log10x + logx10 + logxx + log1010. Find \(.
Answer:
y = logx log10e + loge10 logxe + 1 + 1
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 3M Q6

Question 7.
Examine the continuity of the function
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 3M Q7
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.
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 3M Q7.1
f(1) = 2
Since
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 3M Q7.2
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;
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 3M Q8

2. Given; xy + y2 = tanx + y
Differentiating with respect to x;
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 3M Q8.1

3. Given; x3 + x2y + xy2 + y3 = 81
Differentiating with respect to x;
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 3M Q8.2

4. Given; sin2x + cos2y = 1
Differentiating with respect to x;
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 3M Q8.3

5. \(\sqrt{x}\) + \(\sqrt{y}\) = 1
Differentiating with respect to x;
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 3M Q8.4

6. x2 + xy + y2 = 7
Differentiating with respect to x;
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 3M Q8.5

7. x2(x – y) = y2(x + y)
Differentiating with respect to x;
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 3M Q8.6

8. xy2 + x2y = 2
Differentiating with respect to x;
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 3M Q8.7

9. sin y = xcos (a + y)
⇒ x = \(\frac{\sin y}{\cos (a+y)}\)
Diff. with respect to y.
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 3M Q8.8

Question 9.
Find \(\frac{d y}{d x}\) of the following (3 score each)
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 3M Q9
(viii) x = a(t – sint), y = a(1 – cost)
(ix) y = et cost, x = et 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;
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 3M Q9.1

(ii) We know; y = tan-1\(\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right)\)
Put x = tanθ ⇒ θ = tan-1x
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 3M Q9.2
⇒ y = tan-1(tan3θ)
⇒ y = 3 tan-1(x)
Differentiating with respect to x;
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 3M Q9.3

(iii) We know; y = sin-1\(\left(\frac{1-x^{2}}{1+x^{2}}\right)\)
Put x = tanθ ⇒ θ = tan-1x
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 3M Q9.4

(iv) We know; y = sec-1 = \(\left(\frac{1}{2 x^{2}-1}\right)\)
Put x = cosθ ⇒ θ = cos-1x
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 3M Q9.5
⇒ y = sec-1 sec2θ = 2θ
⇒ y = 2cos-1(x)
Differentiating with respect to x;
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 3M Q9.6

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

Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 3M Q9.7
= π – 2tan-1x
\(\frac{d y}{d x}=\frac{-2}{1+x^{2}}\).

(vii) Let x = sinθ and \(\sqrt{x}\) = sinφ
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 3M Q9.8
= sin-1 (sinθcosφ + cosφsinφ)
= sin-1 (sin(θ + φ)) = θ + φ
= sin-1x + sin-1\(\sqrt{x}\)
\(\frac{d y}{d x}=\frac{1}{\sqrt{1-x^{2}}}+\frac{1}{\sqrt{1-x}} \times \frac{1}{2 \sqrt{x}}\).

Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 3M Q9.9

(ix) y = etcost ⇒ \(\frac{d y}{d t}\) = – et sin t + et
x = et sint ⇒ \(\frac{d x}{d t}\) = et cos t + et sin t
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 3M Q9.10

Question 10.
If y = log\(\left(\frac{1}{x}\right)\), Show that \(\frac{d y}{d x}\) + ey = 0.
Answer:
Given, y = log\(\left(\frac{1}{x}\right)\) ? \(\left(\frac{1}{x}\right)\) = ey …….(1)
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 3M Q10

Question 11.
If ey (x+1) = 1. Show that

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

Answer:
1. ey (x+1) = 1
Differentiating ey +ey(x +1) \(\frac{d y}{d x}\) = 0
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 3M Q11

2.
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 3M Q11.1

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)
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 3M Q12
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
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 3M Q12.1

(ii) Since f (x) is continuous
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 3M Q12.2

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
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 4M Q1

2.
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 4M Q1.1

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
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 4M Q3

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:
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 4M Q6
f(x) is not continuous.
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 4M Q6.1

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.

Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 4M Q7

(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=xx
  5. y =(x log x)log(logx)
  6. y = \(\sqrt{\sin x \sqrt{\sin x+\sqrt{\sin x+\ldots .}}}\)
  7. yx = xsiny
  8. y =(log x)x + xlogx
  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;
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 4M Q9

2. Given; x = 2at2 ⇒ \(\frac{d x}{d t}\) = 4at
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 4M Q9.1

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= xx; Taking log on both sides;
log y = x log x
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 4M Q9.2

5. y = (x log x)log logx
Taking log on both sides;
log y = (log log x) [log (xlogx)]
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 4M Q9.3

6. y = \(\sqrt{\sin x+y}\) ⇒ y2 = sinx + y
⇒ y2 – y = sinx
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 4M Q9.4

7. yx = xsin y;
Taking log on both sides;
xlogy = siny log x
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 4M Q9.5

8. y = (log x)x + xlogx = u + v
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 4M Q9.6

Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 4M Q9.7

9. y = (sinx)x + sin-1\(\sqrt{x}\)
Let u = (sinx)x ⇒ log u = x log sinx
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 4M Q9.8

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)
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 4M Q11
Answer:
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 4M Q11.1

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).
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 4M Q12

Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 4M Q12.1
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.
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 4M Q13
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.
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 4M Q14
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
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 4M Q15
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
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 4M Q18
f(x) is continuous at x = -3.
At x = 3
f(3) = 6(3) + 2 = 20
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 4M Q18.1
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.
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 4M Q19
Answer:
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 4M Q19.1
= 0 × a finite quantity between -1 and 1 = 0 Also f(0) = 0
Therefore f(x) is continuous at x = 0.
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 4M Q19.2
Therefore f(x) is discontinuous at x = 0
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 4M Q19.3
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:
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 4M Q20
(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.
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 4M Q22
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:
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 4M Q22.1

Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 4M Q22.2
(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,
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 4M Q23

2. Differentiating (2) w.r.tx, we get,
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 4M Q23.1

Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 4M Q23.2

Question 24.
Consider the function y = 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\(\sqrt{x}\). Take log on both sides,
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 4M Q24

2. We have, logy = \(\left(x+\frac{1}{2}\right)\) logx
Differentiating w.r.t x, we get,
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 4M Q24.1

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
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 6M Q1
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 xsinx 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 = xsinx, take log on both sides,
log y = sinx logx differentiate w. r.t.x, we get
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 6M Q2

2.
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 6M Q2.1

3. Given, y = sin-1(cosx) + cos-1(sinx)
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 6M Q2.2
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;
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 6M Q3

2. Given, y = \(\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\)
Take log on both sides;
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 6M Q3.1
Differentiate w.r.t. x, we get;
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 6M Q3.2

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.
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 6M Q4

Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 6M Q4.1
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)
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 6M Q5
(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)
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 6M Q5.1

(ii) Given, log (x2 + y2) = 2 tan-1\(\left(\frac{y}{x}\right)\).
Differentiate w.r.to x, we get;
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 6M Q5.2

Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 6M Q5.3

Question 6.
If x = a sec3θ and y = a tan3θ
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 6M Q6
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.
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 6M Q6.1

(iii) We have, \(\frac{d y}{d x}\) = sinθ
Differentiating w.r.to x, we get
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 6M Q6.2

(iv) We have,
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 6M Q6.3

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)
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 6M Q7
(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:
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 6M Q7.1
Since, f (- 2) = 1 – (- 2) = 3, f (-1) = 1 – (-1) = 2,
f(1) = 1 + (1) = 2, f (2) = 1 + (2) = 3.

(ii)
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 6M Q7.2

(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:
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 6M Q8

Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 6M Q8.1

(ii) We have; sin-1 x – cos-1y = 2cot-1a.
Differentiating w.r.t x, we get,
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 6M Q8.2

Question 9.
(i) Match the following. (4)
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 6M Q9
(ii) If y = ea cos-1x, then show that (1 – x2)y2 – xy1 – a2y = 0 (2)
Answer:
(i)
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 6M Q9.1

(ii) Given, y = ea cos-1x ____(1)
Differentiating w.r.to x,
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 6M Q9.2
Again differentiating w. r.to x
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 6M Q9.3
⇒ (1 – x2)y2 – xy1 = a2. ea cos-1x
⇒ (1 – x2)y2 – xy1 = a2y
⇒ (1 – x2)y2 – xy1 – a2y = 0.

Question 10.
(i) Match the following (3)
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 6M Q10
Differentiate the following
(ii) y =\(\frac{1}{5 x^{2}+3 x+7}\) (1)
(iii) y = 3cosec4(7x) (1)
(iv) y = e2log tan 5x (1)
Answer:
(i)
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 6M Q10.1

Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 6M Q10.2

(iii) Given,
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 6M Q10.3

(iv) Given,
Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability 6M Q10.4


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