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## Kerala Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability

### Plus Two Maths Continuity and Differentiability Three Mark Questions and Answers

Question 1.

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

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

Answer:

1. The function is not defined at points where 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) = \(\left\{\begin{array}{ll}{\frac{\sin x}{x}} & {, x<0} \\{x+1} & {, x \geq 0}\end{array}\right.\)

- Find \(\lim _{x \rightarrow 0} f(x)\) (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 = log_{10}x + log_{x}10 + log_{x}x + log_{10}10. Find \(.

Answer:

y = log_{e }x log_{10}e + log_{e}10 log_{x}e + 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)

- 2x + 3y = sinx
- xy + y
^{2}= tanx + y - x
^{3}+ x^{2}y + xy^{2}+ y^{3}= 81 - sin
^{2}x + cos^{2}y = 1 - \(\sqrt{x}\) + \(\sqrt{y}\) = 1
- x
^{2}+ xy + y^{2}= 7 - x
^{2}(x – y) = y^{2}(x + y) - xy
^{2}+ x^{2}y = 2 - sin y = xcos (a + y)

Answer:

1. Given; 2x + 3y = sinx

Differentiating with respect to x;

2. Given; xy + y^{2} = tanx + y

Differentiating with respect to x;

3. Given; x^{3} + x^{2}y + xy^{2} + y^{3} = 81

Differentiating with respect to x;

4. Given; sin^{2}x + cos^{2}y = 1

Differentiating with respect to x;

5. \(\sqrt{x}\) + \(\sqrt{y}\) = 1

Differentiating with respect to x;

6. x^{2} + xy + y^{2} = 7

Differentiating with respect to x;

7. x^{2}(x – y) = y^{2}(x + y)

Differentiating with respect to x;

8. xy^{2} + x^{2}y = 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^{-1}x

⇒ 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^{-1}x

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

Put x = cosθ ⇒ θ = cos^{-1}x

⇒ 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^{-1}x

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

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

= sin^{-1} (sinθcosφ + cosφsinφ)

= sin^{-1} (sin(θ + φ)) = θ + φ

= sin^{-1}x + 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^{t}cost ⇒ \(\frac{d y}{d t}\) = – 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

- \(\frac{d y}{d x}\) = -e
^{y}(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.\)

- Find \(\lim _{x \rightarrow 2} f(x)\) (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

- Find \(\frac{d y}{d x}\).
- 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 (x^{2} +1)^{2} y_{2} + 2x(x^{2} +1) y_{1} = 2.

Answer:

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

⇒ y_{1} = 2(tan^{-1} x) \(\frac{1}{1+x^{2}}\)

⇒ (1 + x^{2})y_{1} = 2(tan^{-1} x)

⇒ (1 + x^{2})y_{2} + y_{1}2x = 2 \(\frac{1}{1+x^{2}}\)

⇒ (1 + x^{2})^{2} y_{2} + x(1 + x^{2})y_{1} = 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) = \(\left\{\begin{array}{ll}{\cos x,} & {0 \leq x \leq c} \\{\sin x,} & {c<x \leq \pi}\end{array}\right.\)

- Find the value of c if / is continuous on [0, π].
- 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.

- Find \(\frac{d y}{d x}\) if x = 2sinθ; y = 3cosθ
- 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) = \(\left\{\begin{array}{ll}{2 x+3,} & {x \leq 2} \\{x+2 k,} & {x>2}\end{array}\right.\)

- Find f(2) (1)
- Evaluate \(\lim _{x \rightarrow 2^{+}}\)f(x) (1)
- 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)

- y = (logx)
^{cosx} - x = 2at
^{2}, y = at^{4} - x = a(cosθ + θsinθ), y = a(sinθ – θcosθ)
- y=x
^{x} - y =(x log x)
^{log(logx)} - y = \(\sqrt{\sin x \sqrt{\sin x+\sqrt{\sin x+\ldots .}}}\)
- y
^{x}= x^{siny} - y =(log x)
^{x}+ x^{logx} - 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 = 2at^{2} ⇒ \(\frac{d x}{d t}\) = 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 = \(\sqrt{\sin x+y}\) ⇒ y^{2} = sinx + y

⇒ y^{2} – 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

- y = x
^{2}+ 3x + 2 (2) - y = tan
^{-1}x (2)

Answer:

1. Given; y = x^{2} + 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^{-1}x

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^{-1}x) (1)

(ii) \(\frac{d y}{d x}\) (1)

(iii) (1 – x^{2}) y_{2} – xy_{1} + m^{2}y = 0 (2)

Answer:

(i) x = sint, y = sinmt

t = sin^{-1}x ⇒ y = sin(msin^{-1}x).

multiplying with \(\sqrt{1-x^{2}}\)

(1 – x^{2}) y_{2} – xy_{1} = -m^{2}y

(1 – x^{2}) y_{2} – xy_{1} + m^{2}y = 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) = x^{2} – 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) = 2x^{2} – 10x + 29 in [2, 9]

Answer:

f(x) = 2x^{2} – 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]

- Find f(0) and f(2) (1)
- Find f'(x) (1)
- 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) = x^{3} – 3x^{2} + 2x

⇒ f'(x) = 3x^{2} – 6x + 2.

3. f'(x) = 3x^{2} – 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) = x^{2} + 2x – 8, x ∈ [-4, 2]

Answer:

f(x) = x^{2} + 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.

- f(x) = [x], x ∈ [5, 9]
- f(x) = x
^{2}– 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) = x^{2} – 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 – x^{2}) y^{2} = (sin^{-1}x)^{2} (1)

(ii) (1 – x^{2})y_{1} – xy = 1 (1)

(iii) (1 – x^{2}) y_{2} – 3xy_{1} – y = 0 (2)

Answer:

(ii) Differentiating

(1 – x^{2}) 2y.y_{1} + y^{2}x – 2x = \(\frac{2 \sin ^{-1} x}{\sqrt{1-x^{2}}}\)

(1 – x^{2})2yy_{1} – 2xy^{2} = 2y

(1 – x^{2}) y_{1} – xy = 1.

(iii) Again differentiating

(1 – x^{2}) y_{2} + y_{1} x – 2x – xy_{1} – y = 0

(1 – x^{2}) y_{2} – 3xy_{1} – y = 0.

Question 21.

At what point on the curve y = x^{2}, x ∈ [-2, 2] at which the tangent is parallel to x-axis?

Answer:

Y = x^{2}, 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.

f^{1} (x) = 2x

f^{1}(c) = 2c

f^{1}(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 ax^{2} + 2hxy + by^{2} = 1

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

Answer:

1. We have, ax^{2} + 2hxy + by^{2} = 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 }\(\sqrt{x}\)

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

- Verify mean value theorem for the function f(x) = (x – 2)
^{2}in [1, 4]. - 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) - 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.

- Differentiate x
^{sinx}w.r.t.x (2) - If x = at
^{2}, y = 2at, then find \(\frac{d y}{d x}\) (2) - 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.

- Differentiate \(\frac{x-1}{x-3}\) with respect to x.(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| = x^{2}

(ii) At which point \(\frac{d}{d x}\)|x| does not exist?

Find \(\frac{d}{d x}\) |x|. (2)

(iii) Find \(\frac{d}{d x}\)|x^{3} – 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

x^{3} – 7x = 0 ⇒ x(x^{2} – 7) = 0

⇒ x = 0, x^{2} – 7 = 0 ⇒ x = ±\(\sqrt{7}\).

Question 5.

(i) Match the following (4)

(ii) If log (x^{2} + y^{2}) = 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 (x^{2} + y^{2}) = 2 tan^{-1}\(\left(\frac{y}{x}\right)\).

Differentiate w.r.to x, we get;

Question 6.

If x = a sec^{3}θ and y = a tan^{3}θ

Answer:

(i) Given, x = a sec^{3}θ

Differentiate w.r.to θ, we get;

\(\frac{d x}{d \theta}\) = 3a sec^{2}θ. secθ. tanθ = 3a sec^{3} θ. tan θ

Given, y = a tan^{3}θ .

Differentiating w.r.to θ, we get

\(\frac{d x}{d \theta}\) = 3a tan^{2} θ. sec^{2}θ.

(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^{-1}x – sin^{-1}y = 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^{-1}y = 2cot^{-1}a.

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 – x^{2})y_{2} – xy_{1} – a^{2}y = 0 (2)

Answer:

(i)

(ii) Given, y = e^{a cos-1x} ____(1)

Differentiating w.r.to x,

Again differentiating w. r.to x

⇒ (1 – x^{2})y_{2} – xy_{1} = a^{2}. e^{a cos-1x}

⇒ (1 – x^{2})y_{2} – xy_{1} = a^{2}y

⇒ (1 – x^{2})y_{2} – xy_{1} – a^{2}y = 0.

Question 10.

(i) Match the following (3)

Differentiate the following

(ii) y =\(\frac{1}{5 x^{2}+3 x+7}\) (1)

(iii) y = 3cosec^{4}(7x) (1)

(iv) y = e^{2log tan 5x} (1)

Answer:

(i)

(iii) Given,

(iv) Given,