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Plus Two Maths Chapter 5 Continuity and Differentiability Chapter Wise Question and Answers
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Question 1.
Consider the function g(x) = \(\frac{x2}{x2}\)
 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 the denominator is zero.
i.e., x – 2 = 0 ⇒ x = 2.
∴ domain is R – {2}.
g(x) = \(\frac{x2}{x2}\) = \(\left\{\begin{array}{c}{\frac{x2}{x2}, \quad x2>0} \\{\frac{(x2)}{x2}, \quad x2<0}\end{array}=\left\{\begin{array}{ll}{1,} & {x>2} \\{1,} & {x<2}\end{array}\right.\right.\)
∴ 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) = \(\left\{\begin{array}{c}{x+x+1, x \geq 0} \\{xx+1, \quad x<0}
\end{array}=\left\{\begin{array}{c}{2 x+1, x \geq 0} \\{1, x<0}\end{array}\right.\right.\)
(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) = \(\left\{\begin{array}{ll}{\frac{x2}{x2},} & {x \neq 2} \\{0,} & {x=2}\end{array}\right.\)
(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}\(\left(\frac{3 xx^{3}}{13 x^{2}}\right)\)
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}\(\left(\frac{1x^{2}}{1+x^{2}}\right)\)
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}\(\sqrt{x}\)
\(\frac{d y}{d x}=\frac{1}{\sqrt{1x^{2}}}+\frac{1}{\sqrt{1x}} \times \frac{1}{2 \sqrt{x}}\).
(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 \(\lim _{x \rightarrow 0} \frac{k \cos x}{\pi2 x}\) (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) \(\lim _{x \rightarrow \pi / 2} \frac{k \cos x}{\pi2 x}=k \lim _{x \rightarrow \pi / 2} \frac{\cos x}{\pi2 x}\),
Put π – 2x = y when Put x → π/2, y → 0
(ii) Since f (x) is continuous
Question 13.
If f(x) \(f(x)=\left\{\begin{array}{cc}{x[x]} & {, x<.2} \\{0} & {; x=2} \\{3 x5} & {, x>2}
\end{array}\right.\)
 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} \(\left(\frac{1x^{2}}{1+x^{2}}\right)\), 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) cosx
(d) sinx
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) cosx
(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 \(f(x)=\left\{\begin{array}{ll}{x+5,} & {x \leq 1} \\{x5,} & {x>1}\end{array}\right.\) is continuous or not. (2)
(ii) If x = ^{sin1t}, y = ^{cos1t}, a > 0, show that \(\frac{d y}{d x}=\frac{y}{x}\)
Answer:
f(x) is not continuous.
Question 7.
(i) If \(f(x)=\left\{\begin{array}{ll}{1x,} & {0 \leq x \leq 1} \\{1+x,} & {1<x \leq 2}\end{array}\right.\) 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 \(\sqrt{1x^{2}}\)
(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
\(f(x)=\left\{\begin{array}{cc}{x+3,} & {x \leq3} \\{2 x,} & {3<x<3} \\{6 x+2,} & {x \geq 3}\end{array}\right.\)
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 = \(\frac{\sin ^{1} x}{\sqrt{1x^{2}}}\) 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 = \(\frac{2 \sin ^{1} x}{\sqrt{1x^{2}}}\)
(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 xaxis?
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 xaxis.
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 xaxis.
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 xcoordinate 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 \(\frac{x1}{x3}\) with respect to x.(2)
 Differentiate \(\sqrt{\frac{(x1)(x2)}{(x3)(x4)(x5)}}\) with respect to x. (4)
Answer:
1. Let y = \(\frac{x1}{x3}\) Differentiate w.r.t. x, we get;
2. Given, y = \(\sqrt{\frac{(x1)(x2)}{(x3)(x4)(x5)}}\)
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 \(\frac{d y}{d x}=\frac{x+y}{xy}\) (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 \(f(x)=\left\{\begin{array}{cc}{1x} & {, \quad x<0} \\{1} & {x=0} \\ {1+x} & {, \quad x>0}\end{array}\right.\)
(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 \(\sqrt{1x^{2}}+\sqrt{1y^{2}}=a(xy)\)
(i) Simplify the above equation to sin^{1}x – sin^{1}y = 2cot^{1} a by giving suitable substitution.
(ii) Prove that \(\frac{d y}{d x}=\sqrt{\frac{1y^{2}}{1x^{2}}}\)
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 cos1x}, then show that (1 – x^{2})y_{2} – xy_{1} – a^{2}y = 0 (2)
Answer:
(i)
(ii) Given, y = e^{a cos1x} ____(1)
Differentiating w.r.to x,
Again differentiating w. r.to x
⇒ (1 – x^{2})y_{2} – xy_{1} = a^{2}. e^{a cos1x}
⇒ (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,
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