Category: Class 11

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 9 Differentiation Ex 9.2 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2

(I) Differentiate the following w.r.t. x

Question 1.
y = \(x^{\frac{4}{3}}+e^{x}-\sin x\)
Solution:

Question 2.
y = √x + tan x – x³
Solution:

Question 3.
y = log x – cosec x + \(5^{x}-\frac{3}{x^{\frac{3}{2}}}\)
Solution:

Question 4.
y = \(x^{\frac{7}{3}}+5 x^{\frac{4}{5}}-\frac{5}{x^{\frac{2}{5}}}\)
Solution:

Question 5.
y = 7^x + x⁷ – \(\frac{2}{3}\) x√x – log x + 7⁷
Solution:

Question 6.
y = 3 cot x – 5e^x + 3 log x – \(\frac{4}{x^{\frac{3}{4}}}\)
Solution:

(II) Diffrentiate the following w.r.t. x

Question 1.
y = x⁵ tan x
Solution:

Question 2.
y = x³ log x
Solution:

Question 3.
y = (x² + 2)² sin x
Solution:

Question 4.
y = e^x log x
Solution:

Question 5.
y = \(x^{\frac{3}{2}} e^{x} \log x\)
Solution:

Question 6.
y = \(\log e^{x^{3}} \log x^{3}\)
Solution:

(III) Diffrentiate the following w.r.t. x

Question 1.
y = x²√x + x⁴ log x
Solution:

Question 2.
y = e^x sec x – \(x^{\frac{5}{3}}\) log x
Solution:

Question 3.
y = x⁴ + x√x cos x – x² e^x
Solution:

Question 4.
y = (x³ – 2) tan x – x cos x + 7^x . x⁷
Solution:

Question 5.
y = sin x log x + e^x cos x – e^x √x
Solution:

Question 6.
y = e^x tan x + cos x log x – √x 5^x
Solution:

(IV) Diffrentiate the following w.r.t.x.

Question 1.
y = \(\frac{x^{2}+3}{x^{2}-5}\)
Solution:

Question 2.
y = \(\frac{\sqrt{x}+5}{\sqrt{x}-5}\)
Solution:

Question 3.
y = \(\frac{x e^{x}}{x+e^{x}}\)
Solution:

Question 4.
y = \(\frac{x \log x}{x+\log x}\)
Solution:
y = \(\frac{x \log x}{x+\log x}\)

Question 5.
y = \(\frac{x^{2} \sin x}{x+\cos x}\)
Solution:

Question 6.
y = \(\frac{5 e^{x}-4}{3 e^{x}-2}\)
Solution:

(V).

Question 1.
If f(x) is a quadratic polynomial such that f(0) = 3, f'(2) = 2 and f'(3) = 12, then find f(x).
Solution:
Let f(x) = ax² + bx + c …..(i)
∴ f(0) = a(0)² + b(0) + c
∴ f(0) = c
But, f(0) = 3 …..(given)
∴ c = 3 …..(ii)
Differentiating (i) w.r.t. x, we get
f'(x) = 2ax + b
∴ f'(2) = 2a(2) + b
∴ f'(2) = 4a + b
But, f'(2) = 2 …..(given)
∴ 4a + b = 2 …..(iii)
Also, f'(3) = 2a(3) + b
∴ f'(3) = 6a + b
But, f'(3) = 12 …..(given)
∴ 6a + b = 12 …..(iv)
equation (iv) – equation (iii), we get
2a = 10
∴ a = 5
Substituting a = 5 in (iii), we get
4(5) + b = 2
∴ b = -18
∴ a = 5, b = -18, c = 3
∴ f(x) = 5x² – 18x + 3

Check:
If f(0) = 3, f'(2) = 2 and f'(3) = 12, then our answer is correct.
f(x) = 5x² – 18x + 3 and f'(x) = 10x – 18
f(0) = 5(0)² – 18(0) + 3 = 3
f'(2) = 10(2) – 18 = 2
f'(3) = 10(3) – 18 = 12
Thus, our answer is correct.

Question 2.
If f(x) = a sin x – b cos x, f'(\(\frac{\pi}{4}\)) = √2 and f'(\(\frac{\pi}{6}\)) = 2, then find f(x).
Solution:
f(x) = a sin x – b cos x
Differentiating w.r.t. x, we get
f'(x) = a cos x – b (- sin x)
∴ f'(x) = a cos x + b sin x


Now, f(x) = a sin x – b cos x
∴ f(x) = (√3 + 1) sin x + (√3 – 1) cos x

VI. Fill in the blanks. (Activity Problems)

Question 1.
y = e^x . tan x
Diff. w.r.t. x

Solution:

Question 2.
y = \(\frac{\sin x}{x^{2}+2}\)
diff. w.r.t. x

Solution:

Question 3.
y = (3x² + 5) cos x
Diff. w.r.t. x

Solution:

Question 4.
Differentiate tan x and sec x w.r.t. x using the formulae for differentiation of \(\frac{u}{v}\) and \(\frac{1}{v}\) respectively.
Solution:

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 9 Differentiation Ex 9.1 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1

Question 1.
Find the derivatives of the following w.r.t. x by using the method of the first principle.
(a) x² + 3x – 1
Solution:
Let f(x) = x² + 3x – 1
∴ f(x + h) = (x + h)² + 3(x + h) – 1
= x² + 2xh + h² + 3x + 3h – 1
By first principle, we get

(b) sin(3x)
Solution:
Let f(x) = sin 3x
f(x + h) = sin3(x + h) = sin(3x + 3h)
By first principle, we get

(c) e^2x+1
Solution:

(d) 3^x
Solution:

(e) log(2x + 5)
Solution:
Let f(x) = log(2x + 5)
∴ f(x + h) = log[2(x + h) + 5] = log (2x + 2h + 5)
By first principle, we get

(f) tan(2x + 3)
Solution:
Let f(x) = tan(2x + 3)
∴ f(x + h) = tan[2(x + h) + 3] = tan(2x + 2h + 3)
By first principle, we get

(g) sec(5x – 2)
Solution:
Let f(x) = sec(5x – 2)
f(x + h) = sec[5(x + h) – 2] = sec(5x + 5h – 2)
By first principle, we get

(h) x√x
Solution:

Question 2.
Find the derivatives of the following w.r.t. x. at the points indicated against them by using the method of the first principle.
(i) \(\sqrt{2 x+5}\) at x = 2
Solution:

(ii) tan x at x = \(\frac{\pi}{4}\)
Solution:

(iii) 2^3x+1 at x = 2
Solution:

(iv) log(2x + 1) at x = 2
Solution:
Let f(x) = log(2x + 1)
∴ f(2) = log [2(2) + 1] = log 5 and
f(2 + h) = log [2(2 + h) + 1] = log(2h + 5)
By first principle, we get

(v) e^3x-4 at x = 2
Solution:

(vi) cos x at x = \(\frac{5 \pi}{4}\)
Solution:

Question 3.
Show that the function f is not differentiable at x = -3,
where f(x) = x² + 2 for x < -3
= 2 – 3x for x ≥ -3
Solution:


∴ L f'(-3) ≠ R f'(-3)
∴ f is not differentiable at x = -3.

Question 4.
Show that f(x) = x² is continuous and differentiable at x = 0.
Solution:

Question 5.
Discuss the continuity and differentiability of
(i) f(x) = x |x| at x = 0
Solution:

(ii) f(x) = (2x + 3) |2x + 3| at x = \(-\frac{3}{2}\)
Solution:

Question 6.
Discuss the continuity and differentiability of f(x) at x = 2.
f(x) = [x] if x ∈ [0, 4). [where [ ] is a greatest integer (floor) function]
Solution:
Explanation:
x ∈ [0, 4)
∴ 0 ≤ x < 4
We will plot graph for 0 ≤ x < 4
not for x < 0 and upto x = 4 making on X-axis.

f(x) = [x]
∴ Greatest integer function is discontinuous at all integer values of x and hence not differentiable at all integers.
∴ f is not continuous at x = 2.
∵ f(x) = 1, x < 2
= 2, x ≥ 2
x ∈ neighbourhood of x = 2.
∴ L.H.L. = 1, R.H.L. = 2
∴ f is not continuous at x = 2.
∴ f is not differentiable at x = 2.

Question 7.
Test the continuity and differentiability of
f(x) = 3x + 2 if x > 2
= 12 – x² if x ≤ 2 at x = 2.
Solution:

Question 8.
If f(x) = sin x – cos x if x ≤ \(\frac{\pi}{2}\)
= 2x – π + 1 if x > \(\frac{\pi}{2}\)
Test the continuity and differentiability of f at x = \(\frac{\pi}{2}\).
Solution:

Question 9.
Examine the function
f(x) = x² cos(\(\frac{1}{x}\)), for x ≠ 0
= 0, for x = 0
for continuity and differentiability at x = 0.
Solution:

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 8 Continuity Miscellaneous Exercise 8 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8

(I) Select the correct answer from the given alternatives.

Question 1.
f(x) = \(\frac{2^{\cot x}-1}{\pi-2 x}\), for x ≠ \(\frac{\pi}{2}\)
= log √2, for x = \(\frac{\pi}{2}\)
(A) f is continuous at x = \(\frac{\pi}{2}\)
(B) f has a jump discontinuity at x = \(\frac{\pi}{2}\)
(C) f has a removable discontinuity
(D) \(\lim _{x \rightarrow \frac{\pi}{2}} f(x)=2 \log 3\)
Answer:
(A) f is continuous at x = \(\frac{\pi}{2}\)
Hint:

Question 2.
If f(x) = \(\frac{1-\sqrt{2} \sin x}{\pi-4 x}\), for x ≠ \(\frac{\pi}{4}\) is continuous at x = \(\frac{\pi}{4}\), then f(\(\frac{\pi}{4}\)) =
(A) \(\frac{1}{\sqrt{2}}\)
(B) \(-\frac{1}{\sqrt{2}}\)
(C) \(-\frac{1}{4}\)
(D) \(\frac{1}{4}\)
Answer:
(D) \(\frac{1}{4}\)
Hint:

Question 3.
If f(x) = \(\frac{(\sin 2 x) \tan 5 x}{\left(e^{2 x}-1\right)^{2}}\), for x ≠ 0 is continuous at x = 0, then f(0) is
(A) \(\frac{10}{e^{2}}\)
(B) \(\frac{10}{e^{4}}\)
(C) \(\frac{5}{4}\)
(D) \(\frac{5}{2}\)
Answer:
(D) \(\frac{5}{2}\)
Hint:

Question 4.
f(x) = \(\frac{x^{2}-7 x+10}{x^{2}+2 x-8}\), for x ∈ [-6, -3]
(A) f is discontinuous at x = 2
(B) f is discontinuous at x = -4
(C) f is discontinuous at x = 0
(D) f is discontinuous at x = 2 and x = -4
Answer:
(B) f is discontinuous at x = -4
Hint:
f(x) = \(\frac{x^{2}-7 x+10}{x^{2}+2 x-8}\), for x ∈ [-6, -3]
= \(\frac{x^{2}-7 x+10}{(x+4)(x-2)}\)
Here f(x) is a rational function and is continuous everywhere except at the points Where denominator becomes zero.
Here, denominator becomes zero when x = -4 or x = 2
But x = 2 does not lie in the given interval.
∴ x = -4 is the point of discontinuity.

Question 5.
If f(x) = ax² + bx + 1, for |x – 1| ≥ 3 and
= 4x + 5, for -2 < x < 4
is continuous everywhere then,
(A) a = \(\frac{1}{2}\), b = 3
(B) a = \(-\frac{1}{2}\), b = -3
(C) a = \(-\frac{1}{2}\), b = 3
(D) a = \(\frac{1}{2}\), b = -3
Answer:
(A) a = \(\frac{1}{2}\), b = 3
Hint:
f(x) = ax² + bx + 1, |x – 1| ≥ 3
= 4x + 5; -2 < x < 4
The first interval is
|x – 1| ≥ 3
∴ x – 1 ≥ 3 or x – 1 ≤ -3
∴ x ≥ 4 or x ≤ -2
∴ f(x) is same for x ≤ -2 as well as x ≥ 4.
∴ f(x) is defined as:
f(x) = ax² + bx + 1; x ≤ -2
= 4x + 5; -2 < x < 4
= ax² + bx + 1; x ≥ 4
f(x) is continuous everywhere.
∴ f(x) is continuous at x = -2 and x = 4.
As f(x) is continuous at x = -2,
\(\lim _{x \rightarrow-2^{-}} f(x)=\lim _{x \rightarrow-2^{+}} f(x)\)
∴ \(\lim _{x \rightarrow-2}\left(a x^{2}+b x+1\right)=\lim _{x \rightarrow-2}(4 x+5)\)
∴ a(-2)² + b(-2) + 1 = 4(-2) + 5
∴ 4a – 2b + 1 = -3
∴ 4a – 2b = -4
∴ 2a – b = -2 …..(i)
∵ f(x) is continuous at x = 4,
\(\lim _{x \rightarrow 4^{-}} \mathrm{f}(x)=\lim _{x \rightarrow 4^{+}} \mathrm{f}(x)\)
∴ \(\lim _{x \rightarrow 4}(4 x+5)=\lim _{x \rightarrow 4}\left(a x^{2}+b x+1\right)\)
4(4) + 5 = a(4)² + b(4) + 1
16a + 4b + 1 = 21
16a + 4b = 20
4a + b = 5 …..(ii)
Adding (i) and (ii), we get
6a = 3
∴ a = \(\frac{1}{2}\)
Substituting a = \(\frac{1}{2}\) in (ii), we get
4(\(\frac{1}{2}\)) + b = 5
∴ 2 + b = 5
∴ b = 3
∴ a = \(\frac{1}{2}\), b = 3

Question 6.
f(x) = \(\frac{\left(16^{x}-1\right)\left(9^{x}-1\right)}{\left(27^{x}-1\right)\left(32^{x}-1\right)}\), for x ≠ 0
= k, for x = 0
is continuous at x = 0, then ‘k’ =
(A) \(\frac{8}{3}\)
(B) \(\frac{8}{15}\)
(C) \(-\frac{8}{15}\)
(D) \(\frac{20}{3}\)
Answer:
(B) \(\frac{8}{15}\)
Hint:

Question 7.
f(x) = \(\frac{32^{x}-8^{x}-4^{x}+1}{4^{x}-2^{x+1}+1}\), for x ≠ 0
= k, for x = 0,
is continuous at x = 0, then value of ‘k’ is
(A) 6
(B) 4
(C) (log 2) (log 4)
(D) 3 log 4
Answer:
(A) 6
Hint:

Question 8.
If f(x) = \(\frac{12^{x}-4^{x}-3^{x}+1}{1-\cos 2 x}\), for x ≠ 0 is continuous at x = 0 then the value of f(0) is
(A) \(\frac{\log 12}{2}\)
(B) log 2 . log 3
(C) \(\frac{\log 2 \cdot \log 3}{2}\)
(D) None of these
Answer:
(B) log 2 . log 3
Hint:

Question 9.
If f(x) = \(\left(\frac{4+5 x}{4-7 x}\right)^{\frac{4}{x}}\), for x ≠ 0 and f(0) = k, is continuous at x = 0, then k is
(A) e⁷
(B) e³
(C) e¹²
(D) \(e^{\frac{3}{4}}\)
Answer:
(C) e¹²
Hint:

Question 10.
If f(x) = \(\lfloor x\rfloor\) for x ∈ (-1, 2), then f is discontinuous at
(A) x = -1, 0, 1, 2
(B) x = -1, 0, 1
(C) x = 0, 1
(D) x = 2
Answer:
(C) x = 0, 1
Hint:
f(x) = \(\lfloor x\rfloor\), x ∈ (-1, 2)
This function is discontinuous at all integer values of x between -1 and 2.
∴ f(x) is discontinuous at x = 0 and x = 1.

II. Discuss the continuity of the following functions at the point(s) or on the interval indicated against them.

Question 1.
f(x) = \(\frac{x^{2}-3 x-10}{x-5}\), for 3 ≤ x ≤ 6, x ≠ 5
= 10, for x = 5
= \(\frac{x^{2}-3 x-10}{x-5}\), for 6 < x ≤ 9
Solution:

Question 2.
f(x) = 2x² – 2x + 5, for 0 ≤ x ≤ 2
= \(\frac{1-3 x-x^{2}}{1-x}\), for 2 < x < 4
= \(\frac{x^{2}-25}{x-5}\), for 4 ≤ x ≤ 7 and x ≠ 5
= 7, for x = 5
Solution:
The domain of f(x) is [0, 7].
(i) For 0 ≤ x ≤ 2
f(x) = 2x² – 2x + 5
It is a polynomial function and is Continuous at all point in [0, 2).

(ii) For 2 < x < 4
f(x) = \(\frac{1-3 x-x^{2}}{1-x}\)
It is a rational function and is continuous everwhere except at points where its denominator becomes zero.
Denominator becomes zero at x = 1
But x = 1 does not lie in the interval.
f(x) is continuous at all points in (2, 4).

(iii) For 4 ≤ x ≤ 7, x ≤ 5
f(x) = \(\frac{x^{2}-25}{x-5}\)
It is a rational function and is continuous everywhere except at points where its denominator becomes zero.
Denominator becomes zero at x = 5
But x = 5 does not lie in the interval.
∴ f(x) is continuous at all points in (4, 7] – {5}.

(iv) For continuity at x = 2:

∴ f(x) is continuous at x = 2.

(v) For continuity at x = 4:

∴ f(x) is continuous at x = 4.

(vi) For continuity at x = 5.
f(5) = 7

∴ f(x) is discontinuous at x = 5.
Thus, f(x) is continuous at all points on its domain except at x = 5.

Question 3.
f(x) = \(\frac{\cos 4 x-\cos 9 x}{1-\cos x}\), for x ≠ 0
f(0) = \(\frac{68}{15}\), at x = 0 on \(-\frac{\pi}{2}\) ≤ x ≤ \(\frac{\pi}{2}\)
Solution:
The domain of f(x) is [\(-\frac{\pi}{2}\), \(\frac{\pi}{2}\)]
(i) For [\(-\frac{\pi}{2}\), \(\frac{\pi}{2}\)] – {0}:
f(x) = \(\frac{\cos 4 x-\cos 9 x}{1-\cos x}\)
It is a rational function and is continuous everywhere except at points where its denominator becomes zero.
Denominator becomes zero when cos x = 1,
i.e., x = 0
But x = 0 does not lie in the interval.
∴ f(x) is continuous at all points in [\(-\frac{\pi}{2}\), \(\frac{\pi}{2}\)] – {0}

(ii) For continuity at x = 0:


∴ \(\lim _{x \rightarrow 0} f(x) \neq f(0)\)
∴ f(x) is discontinuous at x = 0.

Question 4.
f(x) = \(\frac{\sin ^{2} \pi x}{3(1-x)^{2}}\), for x ≠ 1
= \(\frac{\pi^{2} \sin ^{2}\left(\frac{\pi x}{2}\right)}{3+4 \cos ^{2}\left(\frac{\pi x}{2}\right)}\), for x = 1, at x = 1
Solution:

Question 5.
f(x) = \(\frac{|x+1|}{2 x^{2}+x-1}\), for x ≠ -1
= 0, for x = -1, at x = -1
Solution:

Question 6.
f(x) = [x + 1] for x ∈ [-2, 2)
Where [*] is greatest integer function.
Solution:
f(x) = [x + 1], x ∈ [-2, 2)
∴ f(x) = -1, x ∈ [-2, -1)
= 0, x ∈ [-1, 0)
= 1, x ∈ [0, 1)
= 2, x ∈ [1, 2)

∴ \(\lim _{x \rightarrow-1^{-}} \mathrm{f}(x)=\lim _{x \rightarrow-1^{+}} \mathrm{f}(x)\)
∴ f(x) is discontinuous at x = -1.
Similarly, f(x) is discontinuous at the points x = 0 and x = 1.

Question 7.
f(x) = 2x² + x + 1, for |x – 3| ≥ 2
= x² + 3, for 1 < x < 5
Solution:
|x – 3| ≥ 2
∴ x – 3 ≥ 2 or x – 3 ≤ -2
∴ x ≥ 5 or x ≤ 1
∴ f(x) = 2x² + x + 1, x ≤ 1
= x² + 3, 1 < x < 5
= 2x² + x + 1, x ≥ 5
Consider the intervals
x < 1 , i.e., (-∞, 1)
1 < x < 5, i.e., (1, 5) x > 5, i.e., (5, ∞)
In all these intervals, f(x) is a polynomial function and hence is continuous at all points.
For continuity at x = 1:

∴ f(x) is discontinuous at x = 5.
∴ f(x) is continuous for all x ∈ R, except at x = 5.

III. Identify discontinuities if any for the following functions as either a jump or a removable discontinuity on their respective domains.

Question 1.
f(x) = x² + x – 3, for x ∈ [-5, -2)
= x² – 5, for x ∈ (-2, 5]
Solution:
f(-2) has not been defined.
\(\lim _{x \rightarrow-2^{-}} f(x)=\lim _{x \rightarrow-2^{-}}\left(x^{2}+x-3\right)\)
= (-2)² + (-2) – 3
= 4 – 2 – 3
= -1
\(\lim _{x \rightarrow-2^{+}} f(x)=\lim _{x \rightarrow-2^{+}}\left(x^{2}-5\right)\)
= (-2)² – 5
= 4 – 5
= -1
∴ \(\lim _{x \rightarrow-2^{-}} f(x)=\lim _{x \rightarrow-2^{+}} f(x)\)
∴ \(\lim _{x \rightarrow-2} f(x) \text { exists. }\)
But f(-2) has not been defined.
∴ f(x) has a removable discontinuity at x = -2.

Question 2.
f(x) = x² + 5x + 1, for 0 ≤ x ≤ 3
= x³ + x + 5, for 3 < x ≤ 6
Solution:
\(\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{-}}\left(x^{2}+5 x+1\right)\)
= (3)² + 5(3) + 1
= 9 + 15 + 1
= 25
\(\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3^{+}}\left(x^{3}+x+5\right)\)
= (3)³ + 3 + 5
= 27 + 3 + 5
= 35
∴ \(\lim _{x \rightarrow 3^{-}} f(x) \neq \lim _{x \rightarrow 3^{+}} f(x)\)
∴ \(\lim _{x \rightarrow 3} f(x)\) does not exist.
∴ f(x) is discontinuous at x = 3.
∴ f(x) has a jump discontinuity at x = 3.

Question 3.
f(x) = \(\frac{x^{2}+x+1}{x+1}\), for x ∈ [0, 3)
= \(\frac{3 x+4}{x^{2}-5}\), for x ∈ [3, 6]
Solution:


∴ f(x) is continuous at x = 3.

IV. Discuss the continuity of the following functions at the point or on the interval indicated against them. If the function is discontinuous, identify the type of discontinuity and state whether the discontinuity is removable. If it has a removable discontinuity, redefine the function so that it becomes continuous.

Question 1.
f(x) = \(\frac{(x+3)\left(x^{2}-6 x+8\right)}{x^{2}-x-12}\)
Solution:
f(x) = \(\frac{(x+3)\left(x^{2}-6 x+8\right)}{x^{2}-x-12}\)
= \(\frac{(x+3)(x-2)(x-4)}{(x-4)(x+3)}\)
∴ f(x) is not defined at x = 4 and x = -3.
∴ The domain of function f = R – {-3, 4}.
For x ≠ -3 and 4,

f(x) is discontinuous at x = 4 and x = -3.
This discontinuity is removable.
∴ f(x) can be redefined as
f(x) = \(\frac{(x+3)\left(x^{2}-6 x+8\right)}{x^{2}-x-12}\), for x ≠ 4, x ≠ -3
= -5, for x ∈ R – {-3, 4}, x = -3
= 2, for x ∈ R – {-3, 4}, x = 4

Question 2.
f(x) = x² + 2x + 5, for x ≤ 3
= x³ – 2x² – 5, for x > 3
Solution:

∴ f(x) is discontinuous at x = 3.
This discontinuity is irremovable.

V. Find k if the following functions are continuous at the points indicated against them.

Question 1.
f(x) = \(\left(\frac{5 x-8}{8-3 x}\right)^{\frac{3}{2 x-4}}\), for x ≠ 2
= k, for x = 2 at x = 2.
Solution:
f(x) is continuous at x = 2. …..(given)

Question 2.
f(x) = \(\frac{45^{x}-9^{x}-5^{x}+1}{\left(k^{x}-1\right)\left(3^{x}-1\right)}\), for x ≠ 0
= \(\frac{2}{3}\), for x = 0, at x = 0
Solution:
f(x) is continuous at x = 0 …..(given)

VI. Find a and b if the following functions are continuous at the points or on the interval indicated against them.

Question 1.
f(x) = \(\frac{4 \tan x+5 \sin x}{a^{x}-1}\), for x < 0
= \(\frac{9}{\log 2}\), for x = 0
= \(\frac{11 x+7 x \cdot \cos x}{b^{x}-1}\), for x < 0
Solution:

Question 2.
f(x) = ax² + bx + 1, for |2x – 3| ≥ 2
= 3x + 2, for \(\frac{1}{2}\) < x < \(\frac{5}{2}\)
Solution:

VII. Find f(a), if f is continuous at x = a where,

Question 1.
f(x) = \(\frac{1+\cos (\pi x)}{\pi(1-x)^{2}}\), for x ≠ 1 and at a = 1.
Solution:

Question 2.
f(x) = \(\frac{1-\cos [7(x-\pi)]}{5(x-\pi)^{2}}\), for x ≠ π and at a = π.
Solution:

VIII. Solve using intermediate value theorem.

Question 1.
Show that 5^x – 6x = 0 has a root in [1, 2].
Solution:
Let f(x) = 5^x – 6x
5^x and 6x are continuous functions for all x ∈ R.
∴ 5^x – 6x is also continuous for all x ∈ R.
i.e., f(x) is continuous for all x ∈ R.
A root of f(x) exists, if f(x) = 0 for at least one value of x.
f(1) = 5¹ – 6(1) = -1 < 0
f(2) = (5)² – 6(2) = 13 > 0
∴ f(1) < 0 and f(2) > 0
∴ By intermediate value theorem, there has to be a point ‘c’ between 1 and 2 such that f(c) = 0.
∴ There is a root of the given equation in [1, 2].

Question 2.
Show that x³ – 5x² + 3x + 6 = 0 has at least two real roots between x = 1 and x = 5.
Solution:
Let f(x) = x³ – 5x² + 3x + 6
f(x) is a polynomial function and hence it is continuous for all x ∈ R.
A root of f(x) exists, if f(x) = 0 for at least one value of x.
Here, we have been asked to show that f(x) has at least two roots between x = 1 and x = 5.
f(1) = (1)³ – 5(1)² + 3(1) + 6
= 5 > 0
f(2) = (2)³ – 5(2)² + 3(2) + 6
= 8 – 20 + 6 + 6
= 0
∴ x = 2 is a root of f(x).
Also, f(3) = (3)³ – 5(3)² + 3(3) + 6
= 27 – 45 + 9 + 6
= -3 < 0
f(4) = (4)³ – 5(4)² + 3(4) + 6
= 64 – 80 + 12 + 6
= 2 > 0
∴ f(3) < 0 and f(4) > 0
∴ By intermediate value theorem, there has to be a point ‘c’ between 3 and 4 such that f(c) = 0.
∴ There are two roots, x = 2 and a root between x = 3 and x = 4.
Thus, there are at least two roots of the given equation between x = 1 and x = 5.

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 8 Continuity Ex 8.1 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1

Question 1.
Examine the continuity of
(i) f(x) = x³ + 2x² – x – 2 at x = -2
Solution:
Given, f(x) = x³ + 2x² – x – 2
f(x) is a polynomial function and hence it is continuous for all x ∈ R.
∴ f(x) is continuous at x = -2.

(ii) f(x) = sin x, for x ≤ \(\frac{\pi}{4}\)
= cos x, for x > \(\frac{\pi}{4}\), at x = \(\frac{\pi}{4}\)
Solution:

(iii) f(x) = \(\frac{x^{2}-9}{x-3}\), for x ≠ 3
= 8 for x = 3, at x = 3.
Solution:
f(3) = 8 ….(given)

∴ f(x) is discontinuous at x = 3.

Question 2.
Examine whether the function is continuous at the points indicated against them.
(i) f(x) = x³ – 2x + 1, if x ≤ 2
= 3x – 2, if x > 2, at x = 2.
Solution:

(ii) f(x) = \(\frac{x^{2}+18 x-19}{x-1}\), for x ≠ 1
= 20, for x = 1, at x = 1.
Solution:

(iii) f(x) = \(\frac{x}{\tan 3 x}+2\), for x < 0
= \(\frac{7}{3}\), for x ≥ 0, at x = 0.
Solution:

Question 3.
Find all the points of discontinuities of f(x) = [x] on the interval (-3, 2).
Solution:
f(x) = [x], x ∈ (-3, 2)
i.e., f(x) = -3, x ∈ (-3, -2)
= -2, x ∈ [-2, -1)
= -1, x ∈ [- 1, 0)
= 0, x ∈ [0, 1)
= 1, x ∈ [1, 2)

Similarly, f(x) is discontinuous at the points x = -1, x = 0, x = 1.
Thus all the integer values of x in the interval (-3, 2),
i.e., the points x = -2, x = -1, x = 0 and x = 1 are the required points of discontinuities.

Question 4.
Discuss the continuity of the function f(x) = |2x + 3|, at x = \(\frac{-3}{2}\).
Solution:

Question 5.
Test the continuity of the following functions at the points or intervals indicated against them.


Solution:

Question 6.
Identify discontinuities for the following functions as either a jump or a removable discontinuity.
(i) f(x) = \(\frac{x^{2}-10 x+21}{x-7}\)
Solution:
Given, f(x) = \(\frac{x^{2}-10 x+21}{x-7}\)
It is a rational function and is discontinuous if
x – 7 = 0, i.e., x = 7
∴ f(x) is continuous for all x ∈ R, except at x = 7.
∴ f(7) is not defined.

Thus, \(\lim _{x \rightarrow 7} \mathrm{f}(x)\) exist but f(7) is not defined.
∴ f(x) has a removable discontinuity.

(ii) f(x) = x² + 3x – 2, for x ≤ 4
= 5x + 3, for x > 4.
Solution:
f(x) = x² + 3x – 2, x ≤ 4
= 5x + 3, x > 4
f(x) is a polynomial function for both the intervals.
∴ f(x) is continuous for both the given intervals.
Let us test the continuity at x = 4.

∴ f(x) is discontinuous at x = 4.
∴ f(x) has a jump discontinuity at x = 4.

(iii) f(x) = x² – 3x – 2, for x < -3 = 3 + 8x, for x > -3.
Solution:
f(x) = x² – 3x – 2, x < -3 = 3 + 8x, x > -3
f(x) is a polynomial function for both the intervals.
∴ f(x) is continuous for both the given intervals.
Let us test the continuity at x = -3.

∴ f(x) is discontinuous at x = -3.
∴ f(x) has a jump discontinuity at x = -3

(iv) f(x) = 4 + sin x, for x < π = 3 – cos x for x > π.
Solution:
f(x) = 4 + sin x, x < π = 3 – cos x, x > π
sin x and cos x are continuous for all x ∈ R.
4 and 3 are constant functions.
∴ 4 + sin x and 3 – cos x are continuous for all x ∈ R.
∴ f(x) is continuous for both the given intervals.
Let us test the continuity at x = π.

But f(π) is not defined.
∴ f(x) has a removable discontinuity at x = π.

Question 7.
Show that the following functions have a continuous extension to the point where f(x) is not defined. Also, find the extension.
(i) f(x) = \(\frac{1-\cos 2 x}{\sin x}\), for x ≠ 0.
Solution:
f(x) = \(\frac{1-\cos 2 x}{\sin x}\), for x ≠ 0
Here, f(0) is not defined.
Consider,

But f(0) is not defined.
∴ f(x) has a removable discontinuity at x = 0.
∴ The extension of the original function is
f(x) = \(\frac{1-\cos 2 x}{\sin x}\) for x ≠ 0
= 0 for x = 0
∴ f(x) is continuous at x = 0.

(ii) f(x) = \(\frac{3 \sin ^{2} x+2 \cos x(1-\cos 2 x)}{2\left(1-\cos ^{2} x\right)}\), for x ≠ 0.
Solution:
f(x) = \(\frac{3 \sin ^{2} x+2 \cos x(1-\cos 2 x)}{2\left(1-\cos ^{2} x\right)}\), for x ≠ 0
Here, f(0) is not defined.
Consider,

But f(0) is not defined.
∴ f(x) has a removable discontinuity at x = 0.
∴ The extension of the original function is
f(x) = \(\frac{3 \sin ^{2} x+2 \cos x(1-\cos 2 x)}{2\left(1-\cos ^{2} x\right)}\), x ≠ 0
= \(\frac{7}{2}\), x = 0
∴ f(x) is continuous at x = 0.

(iii) f(x) = \(\frac{x^{2}-1}{x^{3}+1}\), for x ≠ -1
Solution:
f(x) = \(\frac{x^{2}-1}{x^{3}+1}\), for x ≠ -1
Here, f(-1) is not defined.
Consider,

But f(-1) is not defined.
∴ f(x) has a removable discontinuity at x = -1.
∴ The extension of the original function is
f(x) = \(\frac{x^{2}-1}{x^{3}+1}\), x ≠ -1
= \(-\frac{2}{3}\), x = -1
∴ f(x) is continuous at x = \(-\frac{2}{3}\)

Question 8.
Discuss the continuity of the following functions at the points indicated against them.

Solution:

Question 9.
Which of the following functions has a removable discontinuity? If it has a removable discontinuity, redefine the function so that it becomes continuous.


Solution:

Question 10.
(i) If f(x) = \(\frac{\sqrt{2+\sin x}-\sqrt{3}}{\cos ^{2} x}\), for x ≠ \(\frac{\pi}{2}\), is continuous at x = \(\frac{\pi}{2}\) then find f(\(\frac{\pi}{2}\)).
Solution:
f(x) is continuous at x = \(\frac{\pi}{2}\), …..(given)

(ii) If f(x) = \(\frac{\cos ^{2} x-\sin ^{2} x-1}{\sqrt{3 x^{2}+1}-1}\) for x ≠ 0, is continuous at x = 0 then find f(0).
Solution:
f(x) is continuous at x = 0, …..(given)

(iii) If f(x) = \(\frac{4^{x-\pi}+4^{\pi-x}-2}{(x-\pi)^{2}}\) for x ≠ π, is continuous at x = π, then find f(π).
Solution:
f(x) is continuous at x = π, …..(given)

Question 11.
(i) If f(x) = \(\frac{24^{x}-8^{x}-3^{x}+1}{12^{x}-4^{x}-3^{x}+1}\), for x ≠ 0
= k, for x = 0
is continuous at x = 0, then find k.
Solution:
f(x) is continuous at x = 0 …..(given)

(ii) If f(x) = \(\frac{5^{x}+5^{-x}-2}{x^{2}}\), for x ≠ 0
= k, for x = 0
is continuous at x = 0, then find k.
Solution:
f(x) is continuous at x = 0 …..(given)

(iii) If f(x) = \(\frac{\sin 2 x}{5 x}\) – a, for x > 0
= 4 for x = 0
= x² + b – 3, for x < 0
is continuous at x = 0, find a and b.
Solution:
f(x) is continuous at x = 0 ……(given)

(iv) For what values of a and b is the function
f(x) = ax + 2b + 18, for x ≤ 0
= x² + 3a – b, for 0 < x ≤ 2 = 8x – 2, for x > 2,
continuous for every x?
Solution:
f(x) is continuous for every x …..(given)
∴ f(x) is continuous at x = 0 and x = 2.
As f(x) is continuous at x = 0,
\(\lim _{x \rightarrow 0^{-}} \mathrm{f}(x)=\lim _{x \rightarrow 0^{+}} \mathrm{f}(x)\)

∴ (2)² + 3a – b = 8(2) – 2
∴ 4 + 3a – b = 14
∴ 3a – b = 10 …….(ii)
Subtracting (i) from (ii), we get
2a = 4
∴ a = 2
Substituting a = 2 in (i), we get
2 – b = 6
∴ b = -4
∴ a = 2 and b = -4

(v) For what values of a and b is the function
f(x) = \(\frac{x^{2}-4}{x-2}\), for x < 2
= ax² – bx + 3, for 2 ≤ x < 3
= 2x – a + b, for x ≥ 3
continuous for every x on R?
Solution:
f(x) is continuous for every x on R …..(given)
∴ f(x) is continuous at x = 2 and x = 3.
As f(x) is continuous at x = 2,

∴ a(3)² – b(3) + 3 = 2(3) – a + b
∴ 9a – 3b + 3 = 6 – a + b
∴ 10a – 4b = 3 …..(ii)
Multiplying (i) by 2, we get
8a – 4b = 2 ….(iii)
Subtracting (ii) from (iii), we get
-2a = -1
∴ a = \(\frac{1}{2}\)
Substituting a = \(\frac{1}{2}\) in (i), we get
4(\(\frac{1}{2}\)) – 2b = 1
∴ 2 – 2b = 1
∴ 1 = 2b
∴ b = \(\frac{1}{2}\)
∴ a = \(\frac{1}{2}\) and b = \(\frac{1}{2}\)

Question 12.
Discuss the continuity of f on its domain, where
f(x) = |x + 1|, for -3 ≤ x ≤ 2
= |x – 5|, for 2 < x ≤ 7
Solution:

Question 13.
Discuss the continuity of f(x) at x = \(\frac{\pi}{4}\) where,
f(x) = \(\frac{(\sin x+\cos x)^{3}-2 \sqrt{2}}{\sin 2 x-1}\), for x ≠ \(\frac{\pi}{4}\)
= \(\frac{3}{\sqrt{2}}\), for x = \(\frac{\pi}{4}\)
Solution:

Question 14.
Determine the values of p and q such that the following function is continuous on the entire real number line.
f(x) = x + 1, for 1 < x < 3
= x² + px + q, for |x – 2| ≥ 1.
Solution:
|x – 2| ≥ 1
∴ x – 2 ≥ 1 or x – 2 ≤ -1
∴ x ≥ 3 or x ≤ 1
∴ f(x) = x² + px + q for x ≥ 3 as well as x ≤ 1
Thus, f(x) = x² + px + q; x ≤ 1
= x + 1; 1 < x < 3 = x² + px + q; x > 3
f(x) is continuous for all x ∈ R.
∴ f(x) is continuous at x = 1 and x = 3.
As f(x) is continuous at x = 1,

Subtracting (i) from (ii), we get
2p = -6
∴ p = -3
Substituting p = -3 in (i), we get
-3 + q = 1
∴ q = 4
∴ p = -3 and q = 4

Question 15.
Show that there is a root for the equation 2x³ – x – 16 = 0 between 2 and 3.
Solution:
Let f(x) = 2x³ – x – 16
f(x) is a polynomial function and hence it is continuous for all x ∈ R.
A root of f(x) exists, if f(x) = 0 for at least one value of x.
f(2) = 2(2)³ – 2 – 16 = -2 < 0
f(3) = 2(3)³ – 3 – 16 = 35 > 0
∴ f(2) < 0 and f(3) > 0
∴ By intermediate value theorem,
there has to be point ‘c’ between 2 and 3 such that f(c) = 0.
∴ There is a root of the given equation between 2 and 3.

Question 16.
Show that there is a root for the equation x³ – 3x = 0 between 1 and 2.
Solution:
Let f(x) = x³ – 3x
f(x) is a polynomial function and hence it is continuous for all x ∈ R.
A root of f(x) exists, if f(x) = 0 for at least one value of x.
f(1) = (1)³ – 3(1) = -2 < 0
f(2) = (2)³ – 3(2) = 2 > 0
∴ f(1) < 0 and f(2) > 0
∴ By intermediate value theorem,
there has to be point ‘c’ between 1 and 2 such that f(c) = 0.
There is a root of the given equation between 1 and 2.

Question 17.
Let f(x) = ax + b (where a and b are unknown)
= x² + 5 for x ∈ R
Find the values of a and b, so that f(x) is continuous at x = 1.

Solution:
f(x) = x² + 5, x ∈ R
∴ f(1) = 1 + 5 = 6
If f(x) = ax + b is continuous at x = 1, then
f(1) = \(\lim _{x \rightarrow 1}(a x+b)\) = a + b
∴ 6 = a + b where, a, b ∈ R
∴ There are infinitely many values of a and b.

Question 18.
Activity: Suppose f(x) = px + 3 for a ≤ x ≤ b
= 5x² – q for b < x ≤ c
Find the condition on p, q, so that f(x) is continuous on [a, c], by filling in the boxes.

Solution:

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 4 Methods of Induction and Binomial Theorem Ex 4.5 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Methods of Induction and Binomial Theorem Ex 4.5

Question 1.
Show that C₀ + C₁ + C₂ + ….. + C₈ = 256
Solution:
Since C₀ + C₁ + C₂ + C₃ + ….. + C_n = 2ⁿ
Putting n = 8, we get
C₀ + C₁ + C₂ + ….. + C₈ = 2⁸
∴ C₀ + C₁ + C₂ + ….. + C₈ = 256

Question 2.
Show that C₀ + C₁ + C₂ + …… + C₉ = 512
Solution:
Since C₀ + C₁ + C₂ + C₃ + ….. + C_n = 2ⁿ
Putting n = 9, we get
C₀ + C₁ + C₂ + ….. + C₉ = 2⁹
∴ C₀ + C₁ + C₂ + …… + C₉ = 512

Question 3.
Show that C₁ + C₂ + C₃ + ….. + C₇ = 127
Solution:
Since C₀ + C₁ + C₂ + C₃ + ….. + C_n = 2ⁿ
Putting n = 7, we get
C₀ + C₁ + C₂ + ….. + C₇ = 2⁷
∴ C₀ + C₁ + C₂ +….. + C₇ = 128
But, C₀ = 1
∴ 1 + C₁ + C₂ + ….. + C₇ = 128
∴ C₁ + C₂ + ….. + C₇ = 128 – 1 = 127

Question 4.
Show that C₁ + C₂ + C₃ + ….. + C₆ = 63
Solution:
Since C₀ + C₁ + C₂ + C₃ + ….. + C_n = 2ⁿ
Putting n = 6, we get
C₀ + C₁ + C₂ + ….. + C₆ = 2⁶
∴ C₀ + C₁ + C₂ + …… + C₆ = 64
But, C₀ = 1
∴ 1 + C₁ + C₂ + ….. + C₆ = 64
∴ C₁ + C₂ + ….. + C₆ = 64 – 1 = 63

Question 5.
Show that C₀ + C₂ + C₄ + C₆ + C₈ = C₁ + C₃ + C₅ + C₇ = 128
Solution:
Since C₀ + C₁ + C₂ + C₃ + …… + C_n = 2ⁿ
Putting n = 8, we get
C₀ + C₁ + C₂ + C₃ + …… + C₈ = 2⁸
But, sum of even coefficients = sum of odd coefficients
∴ C₀ + C₂ + C₄ + C₆ + C₈ = C₁ + C₃ + C₅ + C₇
Let C₀ + C₂ + C₄ + C₆ + C₈ = C₁ + C₃ + C₅ + C₇ = k
Now, C₀ + C₁ + C₂ + C₃ + C₄ + C₅ + C₆ + C₇ + C₈ = 256
∴ (C₀ + C₂ + C₄ + C₆ + C₈) + (C₁ + C₃ + C₅ + C₇) = 256
∴ k + k = 256
∴ 2k = 256
∴ k = 128
∴ C₀ + C₂ + C₄ + C₆ + C₈ = C₁ + C₃ + C₅ + C₇ = 128

Question 6.
Show that C₁ + C₂ + C₃ + ….. + C_n = 2ⁿ – 1
Solution:
Since C₀ + C₁ + C₂ + C₃ + ….. + C_n = 2ⁿ
But, C₀ = 1
∴ 1 + C₁ + C₂ + C₃ + …… + C_n = 2ⁿ
∴ C₁ + C₂ + C₃ + ….. + C_n = 2ⁿ – 1

Question 7.
Show that C₀ + 2C₁ + 3C₂ + 4C₃ + ….. + (n + 1)C_n = (n + 2) 2^n-1
Solution:

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 4 Methods of Induction and Binomial Theorem Ex 4.4 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Methods of Induction and Binomial Theorem Ex 4.4

Question 1.
State, by writing the first four terms, the expansion of the following, where |x| < 1.
(i) (1 + x)^-4
Solution:

(ii) (1 – x)^1/3
Solution:

(iii) (1 – x²)^-3
Solution:

(iv) (1 + x)^-1/5
Solution:

(v) (1 + x²)^-1
Solution:

Question 2.
State by writing first four terms, the expansion of the following, where |b| < |a|.
(i) (a – b)^-3
Solution:
(a – b)^-3 = \(\left[a\left(1-\frac{b}{a}\right)\right]^{-3}\)

(ii) (a + b)^-4
Solution:

(iii) (a + b)^1/4
Solution:

(iv) (a – b)^-1/4
Solution:
(a – b)^-1/4 = \(\left[a\left(1-\frac{b}{a}\right)\right]^{\frac{-1}{4}}\)

(v) (a + b)^-1/3
Solution:

Question 3.
Simplify the first three terms in the expansion of the following:
(i) (1 + 2x)^-4
Solution:

(ii) (1 + 3x)^-1/2
Solution:

(iii) (2 – 3x)^1/3
Solution:

(iv) (5 + 4x)^-1/2
Solution:

(v) (5 – 3x)^-1/3
Solution:

Question 4.
Use the binomial theorem to evaluate the following upto four places of decimals.
(i) √99
Solution:

= 10 [1 – 0.005 – 0.0000125 – ……]
= 10(0.9949875)
= 9.94987 5
= 9.9499

(ii) \(\sqrt[3]{126}\)
Solution:

(iii) \(\sqrt[4]{16.08}\)
Solution:

(iv) (1.02)^-5
Solution:

(v) (0.98)^-3
Solution:

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 4 Methods of Induction and Binomial Theorem Ex 4.3 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Methods of Induction and Binomial Theorem Ex 4.3

Question 1.
In the following expansions, find the indicated term.
(i) \(\left(2 x^{2}+\frac{3}{2 x}\right)^{8}\), 3rd term
Solution:

(ii) \(\left(x^{2}-\frac{4}{x^{3}}\right)^{11}\), 5th term
Solution:

(iii) \(\left(\frac{4 x}{5}-\frac{5}{2 x}\right)^{9}\), 7th term
Solution:

(iv) In \(\left(\frac{1}{3}+a^{2}\right)^{12}\), 9th term
Solution:

(v) In \(\left(3 a+\frac{4}{a}\right)^{13}\), 10th term
Solution:

Question 2.
In the following expansions, find the indicated coefficients.
(i) x³ in \(\left(x^{2}+\frac{3 \sqrt{2}}{x}\right)^{9}\)
Solution:

(ii) x⁸ in \(\left(2 x^{5}-\frac{5}{x^{3}}\right)^{8}\)
Solution:

(iii) x⁹ in \(\left(\frac{1}{x}+x^{2}\right)^{18}\)
Solution:

(iv) x^-3 in \(\left(x-\frac{1}{2 x}\right)^{5}\)
Solution:

(v) x^-20 in \(\left(x^{3}-\frac{1}{2 x^{2}}\right)^{15}\)
Solution:

Question 3.
Find the constant term (term independent of x) in the expansion of
(i) \(\left(2 x+\frac{1}{3 x^{2}}\right)^{9}\)
Solution:

(ii) \(\left(x-\frac{2}{x^{2}}\right)^{15}\)
Solution:

(iii) \(\left(\sqrt{x}-\frac{3}{x^{2}}\right)^{10}\)
Solution:

(iv) \(\left(x^{2}-\frac{1}{x}\right)^{9}\)
Solution:

(v) \(\left(2 x^{2}-\frac{5}{x}\right)^{9}\)
Solution:

Question 4.
Find the middle terms in the expansion of
(i) \(\left(\frac{x}{y}+\frac{y}{x}\right)^{12}\)
Solution:

(ii) \(\left(x^{2}+\frac{1}{x}\right)^{7}\)
Solution:

(iii) \(\left(x^{2}-\frac{2}{x}\right)^{8}\)
Solution:

(iv) \(\left(\frac{x}{a}-\frac{a}{x}\right)^{10}\)
Solution:

(v) \(\left(x^{4}-\frac{1}{x^{3}}\right)^{11}\)
Solution:

Question 5.
In the expansion of (k + x)⁸, the coefficient of x⁵ is 10 times the coefficient of x⁶. Find the value of k.
Solution:

Question 6.
Find the term containing x⁶ in the expansion of (2 – x) (3x + 1)⁹.
Solution:

Question 7.
The coefficient of x² in the expansion of (1 + 2x)^m is 112. Find m.
Solution:
The coefficient of x² in (1 + 2x)^m = ^mC₂ (2²)
Given that the coefficient of x² = 112
∴ ^mC₂ (4) = 112
∴ ^mC₂ = 28
∴ \(\frac{\mathrm{m} !}{2 !(\mathrm{m}-2) !}=28\)
∴ \(\frac{m(m-1)(m-2) !}{2 \times(m-2) !}=28\)
∴ m(m – 1) = 56
∴ m² – m – 56 = 0
∴ (m – 8) (m + 7) = 0
As m cannot be negative.
∴ m = 8

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 3 Complex Numbers Ex 3.2 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.2

Question 1.
Find the square root of the following complex numbers:
(i) -8 – 6i
Solution:
Let \(\sqrt{-8-6 i}\) = a + bi, where a, b ∈ R
Squaring on both sides, we get
-8 – 6i = (a + bi)²
-8 – 6i = a² + b²i² + 2abi
-8 – 6i = (a² – b²) + 2abi …..[∵ i² = -1]
Equating real and imaginary parts, we get

(ii) 7 + 24i
Solution:
Let \(\sqrt{7+24 i}\) = a + bi, where a, b ∈ R
Squaring on both sides, we get
7 + 24i = (a + bi)²
7 + 24i = a² + b²i² + 2abi
7 + 24i = (a² – b²) + 2abi …..[∵ i² = -1]
Equating real and imaginary parts, we get

(iii) 1 + 4√3i
Solution:
Let \(\sqrt{1+4 \sqrt{3} i}\) = a + bi, where a, b ∈ R
Squaring on both sides, we get
1 + 4√3i = (a + bi)²
1 + 4√3i = a² + b²i² + 2abi
1 +4√3i = (a² – b²) + 2abi ……[∵ i² = -1]
Equating real and imaginary parts, we get

(iv) 3 + 2√10i
Solution:
Let \(\sqrt{3+2 \sqrt{10}} i\) = a + bi, where a, b ∈ R
Squaring on both sides, we get
3 + 2√10i = (a + bi)²
3 + 2√10i = a² + b²i² + 2abi
3 + 2√10i = (a² – b²) + 2abi …..[∵ i² = -1]
Equating real and imaginary parts, we get
a² – b² = 3 and 2ab = 2√10
a² – b² = 3 and b = \(\frac{\sqrt{10}}{a}\)

(v) 2(1 – √3i)
Solution:
Let \(\sqrt{2(1-\sqrt{3} i)}\) = a + bi, where a, b ∈ R
Squaring on both sides, we get
2(1 – √3i) = (a + bi)²
2(1 – √3i) = a² + b²i² + 2abi
2 – 2√3i = (a² – b²) + 2abi …..[∵ i² = -1]
Equating real and imaginary parts, we get

Question 2.
Solve the following quadratic equations.
(i) 8x² + 2x + 1 = 0
Solution:
Given equation is 8x² + 2x + 1 = 0
Comparing with ax² + bx + c = 0, we get
a = 8, b = 2, c = 1
Discriminant = b² – 4ac
= (2)² – 4 × 8 × 1
= 4 – 32
= -28 < 0
So, the given equation has complex roots.
These roots are given by

∴ the roots of the given equation are \(\frac{-1+\sqrt{7} \mathrm{i}}{8}\) and \(\frac{-1-\sqrt{7} \mathrm{i}}{8}\)

(ii) 2x² – √3x + 1 = 0
Solution:
Given equation is 2x² – √3x + 1 = 0
Comparing with ax² + bx + c = 0, we get
a = 2, b = -√3, c = 1
Discriminant = b² – 4ac
= (-√3)² – 4 × 2 × 1
= 3 – 8
= -5 < 0
So, the given equation has complex roots.
These roots are given by

∴ the roots of the given equation are \(\frac{\sqrt{3}+\sqrt{5} i}{4}\) and \(\frac{\sqrt{3}-\sqrt{5} i}{4}\)

(iii) 3x² – 7x + 5 = 0
Solution:
Given equation is 3x² – 7x + 5 = 0
Comparing with ax² + bx + c = 0, we get
a = 3, b = -7, c = 5
Discriminant = b² – 4ac
= (-7)² – 4 × 3 × 5
= 49 – 60
= -11 < 0
So, the given equation has complex roots.
These roots are given by

∴ the roots of the given equation are \(\frac{7+\sqrt{11} i}{6}\) and \(\frac{7-\sqrt{11} i}{6}\)

(iv) x² – 4x + 13 = 0
Solution:
Given equation is x² – 4x + 13 = 0
Comparing with ax² + bx + c = 0, we get
a = 1, b = -4, c = 13
Discriminant = b² – 4ac
= (-4)² – 4 × 1 × 13
= 16 – 52
= -36 < 0
So, the given equation has complex roots.
These roots are given by

∴ the roots of the given equation are 2 + 3i and 2 – 3i.

Question 3.
Solve the following quadratic equations.
(i) x² + 3ix + 10 = 0
Solution:
Given equation is x² + 3ix + 10 = 0
Comparing with ax² + bx + c = 0, we get
a = 1, b = 3i, c = 10
Discriminant = b² – 4ac
= (3i)² – 4 × 1 × 10
= 9i² – 40
= -9 – 40 …..[∵ i² = -1]
= -49
So, the given equation has complex roots.
These roots are given by

∴ x = 2i or x = -5i
∴ the roots of the given equation are 2i and -5i.
Check:
If x = 2i and x = -5i satisfy the given equation, then our answer is correct.
L.H.S. = x² + 3ix + 10
= (2i)² + 3i(2i) + 10i
= 4i² + 6i² + 10
= 10i² + 10
= -10 + 10 ……[∵ i² = -1]
= 0
= R.H.S.
L.H.S. = x² + 3ix + 10
= (-5i)² + 3i(-5i) + 10
= 25i² – 15i² + 10
= 10i² + 10
= -10 + 10 …..[∵ i² = -1]
= 0
= R.H.S.
Thus, our answer is correct.

(ii) 2x² + 3ix + 2 = 0
Solution:
Given equation is 2x² + 3ix + 2 = 0
Comparing with ax² + bx + c = 0, we get
a = 2, b = 3i, c = 2
Discriminant = b² – 4ac
= (3i)² – 4 × 2 × 2
= 9i² – 16
= -9 – 16
= -25 < 0
So, the given equation has complex roots.
These roots are given by

∴ the roots of the given equation are \(\frac{1}{2}\)i and -2i.

(iii) x² + 4ix – 4 = 0
Solution:
Given equation is x² + 4ix – 4 = 0
Comparing with ax² + bx + c = 0, we get
a = 1, b = 4i, c = -4
Discriminant = b² – 4ac
= (4i)² – 4 × 1 × -4
= 16i² + 16
= -16 + 16 …..[∵ i² = -1]
= 0
So, the given equation has equal roots.
These roots are given by

∴ the roots of the given equation are -2i and -2i.

(iv) ix² – 4x – 4i = 0
Solution:
ix² – 4x – 4i = 0
Multiplying throughout by i, we get
i²x² – 4ix – 4i² = 0
∴ -x² – 4ix + 4 = 0 ……[∵ i² = -1]
∴ x² + 4ix – 4 = 0
Comparing with ax² + bx + c = 0, we get
a = 1, b = 4i, c = -4
Discriminant = b² – 4ac
= (4i)² – 4 × 1 × -4
= 16i² + 16
= -16 + 16 …..[∵ i² = -1]
= 0
So, the given equation has equal roots.
These roots are given by

∴ the roots of the given equation are -2i and -2i.

Question 4.
Solve the following quadratic equations.
(i) x² – (2 + i) x – (1 – 7i) = 0
Solution:
Given equation is x² – (2 + i)x – (1 – 7i) = 0
Comparing with ax² + bx + c = 0, we get
a = 1, b = -(2 + i), c = -(1 – 7i)
Discriminant = b² – 4ac
= [-(2 + i)]² – 4 × 1 × -(1 – 7i)
= 4 + 4i + i² + 4 – 28i
= 4 + 4i – 1 + 4 – 28i …….[∵ i² = -1]
= 7 – 24i
So, the given equation has complex roots.
These roots are given by

(ii) x² – (3√2 + 2i) x + 6√2i = 0
Solution:
Given equation is x² – (3√2 + 2i) x + 6√2i = 0
Comparing with ax² + bx + c = 0, we get
a = 1, b = -(3√2 + 2i), c = 6√2i
Discriminant = b² – 4ac
= [-(3√2 + 2i)]² – 4 × 1 × 6√2i
= 18 + 12√2i + 4i² – 24√2i
= 18 – 12√2i – 4 …..[∵ i² = -1]
= 14 – 12√2i
So, the given equation has complex roots.
These roots are given by

(iii) x² – (5 – i) x + (18 + i) = 0
Solution:
Given equation is x² – (5 – i)x + (18 + i) = 0
Comparing with ax² + bx + c = 0, we get
a = 1, b = -(5 – i), c = 18 + i
Discriminant = b² – 4ac
= [-(5 – i)]² – 4 × 1 × (18 + i)
= 25 – 10i + i² – 72 – 4i
= 25 – 10i – 1 – 72 – 4i …..[∵ i² = -1]
= -48 – 14i
So, the given equation has complex roots.
These roots are given by

(iv) (2 + i) x² – (5 – i) x + 2(1 – i) = 0
Solution:
Given equation is
(2 + i) x² – (5 – i) x + 2(1 – i) = 0
Comparing with ax² + bx + c = 0, we get
a = 2 + i, b = -(5 – i), c = 2(1 – i)
Discriminant = b² – 4ac
= [-(5 – i)]² – 4 × (2 + i) × 2(1 – i)
= 25 – 10i + i² – 8(2 + i)(1 – i)
= 25 – 10i + i² – 8(2 – 2i + i – i²)
= 25 – 10i – 1 – 8(2 – i + 1) …..[∵ i² = -1]
= 25 – 10i – 1 – 16 + 8i – 8
= -2i
So, the given equation has complex roots.
These roots are given by

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 3 Complex Numbers Ex 3.1 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.1

Question 1.
Write the conjugates of the following complex numbers:
(i) 3 + i
(ii) 3 – i
(iii) -√5 – √7i
(iv) -√-5
(v) 5i
(vi) √5 – i
(vii) √2 + √3i
Solution:
(i) Conjugate of (3 + i) is (3 – i)
(ii) Conjugate of (3 – i) is (3 + i)
(iii) Conjugate of (-√5 – √7i) is (-√5 + √7i)
(iv) -√-5 = -√5 × √-1 = -√5i
Conjugate of -√-5 is √5i
(v) Conjugate of 5i is -5i
(vi) Conjugate of √5 – i is √5 + i
(vii) Conjugate of √2 + √3i is √2 – √3i

Question 2.
Express the following in the form of a + ib, a, b ∈ R, i = √-1. State the values of a and b:
(i) (1 + 2i)(-2 + i)
(ii) \(\frac{\mathrm{i}(4+3 \mathrm{i})}{(1-\mathrm{i})}\)
(iii) \(\frac{(2+i)}{(3-i)(1+2 i)}\)
(iv) \(\frac{3+2 i}{2-5 i}+\frac{3-2 i}{2+5 i}\)
(v) \(\frac{2+\sqrt{-3}}{4+\sqrt{-3}}\)
(vi) (2 + 3i)(2 – 3i)
(vii) \(\frac{4 i^{8}-3 i^{9}+3}{3 i^{11}-4 i^{10}-2}\)
Solution:

Question 3.
Show that (-1 + √3i)³ is a real number.
Solution:
(-1 + √3i)³
= (-1)³ + 3(-1)² (√3i) + 3(-1)(√3i)² +(√3i)³ [∵ (a + b)³ = a³ + 3a²b + 3ab² + b³]
= -1 + 3√3i – 3(3i²) + 3√3 i³
= -1 + 3√3i – 3(-3) – 3√3i [∵ i² = -1, i³ = -1]
= -1 + 9
= 8, which is a real number.

Question 4.
Evaluate the following:
(i) i³⁵
(ii) i⁸⁸⁸
(iii) i⁹³
(iv) i¹¹⁶
(v) i⁴⁰³
(vi) \(\frac{1}{i^{58}}\)
(vii) i³⁰ + i⁴⁰ + i⁵⁰ + i⁶⁰
Solution:
We know that, i² = -1, i³ = -i, i⁴ = 1
(i) i³⁵ = (i⁴)⁸ (i²) i = (1)⁸ (-1) i = -i
(ii) i⁸⁸⁸ = (i⁴)²²² = (1)²²² = 1
(iii) i⁹³ = (i⁴)²³ . i = (1)²³ . i = i
(iv) i¹¹⁶ = (i⁴)²⁹ = (1)²⁹ = 1
(v) i⁴⁰³ = (i⁴)¹⁰⁰ (i²) i = (1)¹⁰⁰ (-1) i = -i
(vi) \(\frac{1}{i^{88}}=\frac{1}{\left(i^{4}\right)^{14} \cdot i^{2}}=\frac{1}{(1)^{14}(-1)}=-1\)
(vii) i³⁰ + i⁴⁰ + i⁵⁰ + i⁶⁰
= (i⁴)⁷ i² + (i⁴)¹⁰ + (i⁴)¹² i² + (i⁴)¹⁵
= (1)⁷ (-1) + (1)¹⁰ + (1)¹² (-1) + (1)¹⁵
= -1 + 1 – 1 + 1
= 0

Question 5.
Show that 1 + i¹⁰ + i²⁰ + i³⁰ is a real number.
Solution:
1 + i¹⁰ + i²⁰ + i³⁰
= 1 + (i⁴)² . i² + (i⁴)⁵ + (i⁴)⁷ . i²
= 1 + (1)² (-1) + (1)⁵ + (1)⁷ (-1) [∵ i⁴ = 1, i² = -1]
= 1 – 1 + 1 – 1
= 0, which is a real number.

Question 6.
Find the value of
(i) i⁴⁹ + i⁶⁸ + i⁸⁹ + i¹¹⁰
(ii) i + i² + i³ + i⁴
Solution:
(i) i⁴⁹ + i⁶⁸ + i⁸⁹ + i¹¹⁰
= (i⁴)¹² . i + (i⁴)¹⁷ + (i⁴)²² . i + (i⁴)²⁷ . i²
= (1)¹² . i + (1)¹⁷ + (1)²² . i + (1)²⁷(-1) ……[∵ i⁴ = 1, i² = -1]
= i + 1 + i – 1
= 2i

(ii) i + i² + i³ + i⁴
= i + i² + i² . i + i⁴
= i – 1 – i + 1 [∵ i² = -1, i⁴ = 1]
= 0

Question 7.
Find the value of 1 + i² + i⁴ + i⁶ + i⁸ + …… + i²⁰.
Solution:
1 + i² + i⁴ + i⁶ + i⁸ + ….. + i²⁰
= 1 + (i² + i⁴) + (i⁶ + i⁸) + (i¹⁰ + i¹²) + (i¹⁴ + i¹⁶) + (i¹⁸ + i²⁰)
= 1 + [i² + (i²)²] + [(i²)³ + (i²)⁴] + [(i²)⁵ + (i²)⁶] + [(i²)⁷ + (i²)⁸] + [(i²)⁹ + (i²)¹⁰]
= 1 + [-1 + (- 1)²] + [(-1)³ + (-1)⁴] + [(-1)⁵ + (-1)⁶] + [(-1)⁷ + (-1)⁸] + [(-1)⁹ + (-1)¹⁰] [∵ i² = -1]
= 1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + (-1 + 1) + (-1 + 1)
= 1 + 0 + 0 + 0 + 0 + 0
= 1

Question 8.
Find the values of x and y which satisfy the following equations (x, y ∈ R):
(i) (x + 2y) + (2x – 3y)i + 4i = 5
(ii) \(\frac{x+1}{1+\mathrm{i}}+\frac{y-1}{1-\mathrm{i}}=\mathrm{i}\)
Solution:
(i) (x + 2y) + (2x – 3y)i + 4i = 5
∴ (x + 2y) + (2x – 3y)i = 5 – 4i
Equating real and imaginary parts, we get
x + 2y = 5 ……..(i)
and 2x – 3y = -4 ………(ii)
Equation (i) × 2 – equation (ii) gives
7y = 14
∴ y = 2
Putting y- 2 in (i), we get
x + 2(2) = 5
∴ x + 4 = 5
∴ x = 1
∴ x = 1 and y = 2
Check:
If x = 1 and y = 2 satisfy the given condition, then our answer is correct.
L.H.S. = (x + 2y) + (2x – 3y)i + 4i
= (1 + 4) + (2 – 6)i + 4i
= 5 – 4i + 4i
= 5
= R.H.S.
Thus, our answer is correct.

(ii) \(\frac{x+1}{1+\mathrm{i}}+\frac{y-1}{1-\mathrm{i}}=\mathrm{i}\)

(x + y) + (y – x – 2)i = 2i
(x + y) + (y – x – 2)i = 0 + 2i
Equating real and imaginary parts, we get
x + y = 0 and y – x – 2 = 2
∴ x + y = 0 ……(i)
and -x + y = 4 ……..(ii)
Adding (i) and (ii), we get
2y = 4
∴ y = 2
Putting y = 2 in (i), we get
x + 2 = 0
∴ x = -2
∴ x = -2 and y = 2

Question 9.
Find the value of:
(i) x³ – x² + x + 46, if x = 2 + 3i
(ii) 2x³ – 11x² + 44x + 27, if x = \(\frac{25}{3-4 i}\)
Solution:
(i) x = 2 + 3i
∴ x – 2 = 3i
∴ (x – 2)² = 9i²
∴ x² – 4x + 4 = 9(-1) …..[∵ i² = -1]
∴ x² – 4x + 13 = 0 ……(i)

∴ x³ – x² + x + 46 = (x² – 4x + 13)(x + 3) + 7
= 0(x + 3) + 7 ……[From (i)]
= 7

(ii) x = \(\frac{25}{3-4 i}\)

∴ x = 3 + 4i
∴ x – 3 = 4i
∴ (x – 3)² = 16i²
∴ x² – 6x + 9 = 16(-1) …….[∵ i² = -1]
∴ x² – 6x + 25 = 0 …….(i)

∴ 2x³ – 11x² + 44x + 27
= (x² – 6x + 25) (2x + 1) + 2
= 0 . (2x + 1) + 2 ……[From (i)]
= 0 + 2
= 2

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 7 Limits Ex 7.5 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 7 Limits Ex 7.5

I. Evaluate the following:

Question 1.
\(\lim _{x \rightarrow \frac{\pi}{2}}\left[\frac{cosec x-1}{\left(\frac{\pi}{2}-x\right)^{2}}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow a} \frac{\sin x-\sin a}{\sqrt[5]{x}-\sqrt[5]{a}}\)
Solution:

Question 3.
\(\lim _{x \rightarrow \pi}\left[\frac{\sqrt{5+\cos x}-2}{(\pi-x)^{2}}\right]\)
Solution:

Question 4.
\(\lim _{x \rightarrow \frac{\pi}{6}}\left[\frac{\cos x-\sqrt{3} \sin x}{\pi-6 x}\right]\)
Solution:

Question 5.
\(\lim _{x \rightarrow 1}\left[\frac{1-x^{2}}{\sin \pi x}\right]\)
Solution:

II. Evaluate the following:

Question 1.
\(\lim _{x \rightarrow \frac{\pi}{6}}\left[\frac{2 \sin x-1}{\pi-6 x}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow \frac{\pi}{4}}\left[\frac{\sqrt{2}-\cos x-\sin x}{(4 x-\pi)^{2}}\right]\)
Solution:

Question 3.
\(\lim _{x \rightarrow \frac{\pi}{6}}\left[\frac{2-\sqrt{3} \cos x-\sin x}{(6 x-\pi)^{2}}\right]\)
Solution:

Question 4.
\(\lim _{x \rightarrow a}\left[\frac{\sin (\sqrt{x})-\sin (\sqrt{a})}{x-a}\right]\)
Solution:

Question 5.
\(\lim _{x \rightarrow \frac{\pi}{2}}\left[\frac{\cos 3 x+3 \cos x}{(2 x-\pi)^{3}}\right]\)
Solution: