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You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.6

Integrate the following with respect to x.
Question 1.
\(\frac{x}{\sqrt{1+x^{2}}}\)
Solution:

Question 2.
\(\frac{x^{2}}{1+x^{6}}\)
Solution:

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

Question 4.

Solution:

Question 5.
\(\frac{\sin \sqrt{x}}{\sqrt{x}}\)
Solution:

Question 6.
\(\frac{\cot x}{\log (\sin x)}\)
Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.
\(\frac{\sqrt{x}}{1+\sqrt{x}}\)
Solution:

Question 11.
\(\frac{1}{x \log x \log (\log x)}\)
Solution:

Question 12.
αβx^α-1e^-βxα
Solution:

Question 13.
\(\tan x \sqrt{\sec x}\)
Solution:

Question 14.
x(1 – x)¹⁷
Solution:

Question 15.
sin⁵ x cos³ x
Solution:

Question 16.
\(\frac{\cos x}{\cos (x-a)}\)
Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.6 Additional Problems

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

Question 2.
\(\frac{x \sin ^{-1}\left(x^{2}\right)}{\sqrt{1-x^{4}}}\)
Solution:

Question 3.
sec4 x tan x
Solution:

Question 4.
\(\frac{\sin x}{\sin (x+a)}\)
Solution:

Question 5.
\(\frac{\sqrt{\tan x}}{\sin x \cos x}\)
Solution:

Question 6.
x(l – x)¹⁶
Solution:

Question 7.
x²(2 – x)¹⁵
Solution:

Question 8.
(x + 1)\(\sqrt{2 x+3}\)
Solution:

Question 9.
(x²)\(\sqrt{x+1}\)
Solution:

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.10

Determine the region in the plane determined by the inequalities:

Question 1.
x ≤ 3y, x ≥ y
Solution:
Given in equation are x ≤ 3y,x ≥ y
Suppose x = 3y ⇒ \(\frac{x}{3}=\) = y

If x = y

Question 2.
y ≥ 2x, -2x + 3y ≤ 6
Solution:
Suppose y = 2x

-2x + 3y = 6
-2x = 6 – 3y

Question 3.
3x + 5y ≥ 45, x ≥ 0, y ≥ 0.
Solution:
If 3x + 5y = 45


x ≥ 0 is nothing but the positive portion of Y-axis and y ≥ 0 is the positive portion of X-axis.
Shaded region is the required portions.

Question 4.
2x + 3y ≤ 35, y ≥ 2, x ≥ 5
Solution:
If 2x + 3y = 35 then

y = 2 is a line parallel to X-axis at a distance 2 units
x = 5 is a line parallel to Y-axis at a distance of 5 units

The required region is below 2x + 3y = 35, above y = 2 and to the right of x = 5

Question 5.
2x + 3y ≤ 6, x + 4y ≤ 4, x ≥ 0, y ≥ 0.
Solution:
If 2x + 3y = 6

x + 4y = 4

x ≥ 0, y ≥ 0 represents the area in the 1 quadrant.

The required region is below 2x + 3y = 6 and below x + 4y = 4 bounded by x-axis and y-axis.

Question 6.
x – 2y ≥ 0, 2x – y ≤ -2, x ≥ 0, y ≥ 0
Solution:
If x – 2y = 0

2x – y = -2


x ≥ 0, y ≥ 0 represents the portion in the 1 quadrant only.

Question 7.
2x + y ≥ 8, x + 2y ≥ 8, x + y ≤ 6.
Solution:
2x + y = 8

x + 2y = 8

x + y = 6

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.10 Additional Questions

Question 1.
3x + 2y ≤ 12, x ≥ 1, y ≥ 2
Solution:
The given inequality is 3x + 2y ≤ 12.
Draw the graph of the line 3x + 2y ≤ 12
Table of values satisfying the equation
3x + 2y ≤ 12

Putting (0, 0) in the given inequation, we have 3 × 0 + 2 × 0 ≤ 12
∴ Half plane of 3x + 2y ≤ 12 is towards origin
Also the given inequality is x ≥ 1.
Draw the graph of the line x = 1.

Putting (0, 0) in the given inequation, we have 0 ≥ 1 which is false.
∴ Half plane of x ≥ 1 is away from origin.
The given inequality is y ≥ 2.
Putting (0, 0) in the given inequation, we have 0 ≥ 2 which is false.
∴ Half plane of y ≥ 2 is away from origin.

Question 2.
x + y ≥ 4, 2x – y > 0
Solution:
The given inequality is x + y ≥ 4.
Draw the graph of the line x + y = 4.

Table of values satisfying the equation
x + y = 4

Putting (0, 0) in the given inequation, we have 0 + 0 ≥ 4 ⇒ 0 ≥ 4, which is false.
∴ Half plane of x + y ≥ 4 is away from origin.
Also the given inequality is 2x – y > 0.
Draw the graph of the line 2x – y = 0.
Table of values satisfying the equation
2x – y = 0

Putting (3, 0) in the given inequation, we have 2 × 3 – 0 > 0 ⇒ 6 > 0, which is true.
∴ Half plane of 2x – y > 0 containing (3, 0)

Question 3.
x + y ≤ 9, y > x, x ≥ 0
Solution:
The given inequality is x + y ≤ 9.
Draw the graph of the line x + y = 9.

Table of values satisfying the equation
x + y = 9

Putting (0, 0) in the given inequation, we have 0 + 0 ≤ 9⇒ 0 ≤ 9, is towards is origin.
∴ Half plane of x + y ≤ 9 is away from origin.
Also the given inequality is x – y < 0.
Draw the graph of the line x -y = 0.
Table of values satisfying the equation
x – y = 0

Putting (0, 3) in the given inequation, we have 0 – 3 – 0 < 0 ⇒ -3 < 0, which is true.
∴ Half plane of x – y < 0 containing the points (0, 3).

Question 4.
5x + 4y ≤ 20, x ≥ 1, y ≥ 2
Solution:
The given inequality is 5x + 4y ≤ 20.
Draw the graph of the line 5x + 4y = 20.

Table of values satisfying the equation
5x + 4y = 20

Putting (0, 0) in the given inequation, we have 5 × 0 + 4 × 0 ≤ 20 ⇒ 0 ≤ 20, which is true.
∴ Half plane of 5x + 4y ≤ 20 is away from origin.
Also the given inequality is x ≥ 1.
Draw the graph of the line x = 1.
Putting (0, 0) in the given inequation, we have 0 ≥ 1, which is false.
∴ Half plane of x ≥ 1 is y ≥ 2.
Draw the graph of the line y = 2.
Putting (0, 0) in the given inequation, we have 0 ≥ 2, which is false.
∴ Half plane of y ≥ 2 is away from origin.

Question 5.
3x +4y ≤ 60, x + 3y ≤ 30, x ≥ 0, y ≥ 0
Solution:
The given inequality is 3x + 4y ≤ 60.
Draw the graph of the line 3x + 4y = 60.
Table of values satisfying the equation
3x + 4y = 60

Putting (0, 0) in the given inequation, we have 3 × 0 + 4 × 0 ≤ 60 ⇒ 0 ≤ 60, which is true.
∴ Half plane of 3x + 4y ≤ 60 is towards origin.
Also the given inequality is x + 3y ≤ 30.
Draw the graph of the line x + 3y = 30.
Table of values satisfying the equation .
x + 3y = 30


Putting (0, 0) in the given inequation, we have 0 + 3 × 0 ≤ 30 ⇒ 0 ≤ 30, which is true.
∴ Half plane of x + 3y ≤ 30 is towards origin.

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.4

Question 1.
If f'(x) = 4x – 5 and f(2) = 1; find f(x)
Solution:
\(\int f^{\prime}(x) d x=\int(4 x-5) d x\)
f(x) = \(\frac{4 x^{2}}{2}\) – 5x + c
f(x) = 2x² – 5x + c
But f(2) = 1
2(2)² – 5(2) + c = 1
8 – 10 + c = 1
c = 3
Thus, f(x) = 2x² – 5x + 3

Question 2.
If f'(x) = 9x² – 6x and f(0) = -3; find f(x)
Solution:
f'(x) = \(\int\left(9 x^{2}-6 x\right) d x\)
f(x) = \(\frac{9 x^{3}}{3}-\frac{6 x^{2}}{2}\) + c
f(x) = 3x³ – 3x² + c
But f(0) = -3
3(0)³ – 3(0)² + c = -3
c = -3
Thus, f(x) = 3x³ – 3x² – 3
f(x) = 3(x³ – x² – 1)

Question 3.
If f'(x) = 12x – 6 and f(1) = 30, f'(1) = 5 find f(x)
Solution:
\(\)
f'(x) = \(\) – 6x + c
f'(x) = 6x² – 6x + c
But f'(1) = 5
6(1)² – 6(1) + c = 5
c = 5
f” (x) = 6x² – 6x + 5
\(\int f^{\prime \prime}(x) d x=\int\left(6 x^{2}-6 x+5\right) d x\)
f(x) = \(\frac{6 x^{3}}{3}-\frac{6 x^{2}}{2}\) + 5x + c
f(x) = 2x³ – 3x² + 5x + c
But f(1) = 30
2(1)³ – 3(1)² + 5(1) + c = 30
2 – 3 + 5 + c = 30
c = 26
f(x) = 2x³ – 3x² + 5x + 26

Question 4.
A ball is thrown vertically upward from the ground with an initial velocity of 39.2 mísec. If the only force considered is that attributed to the acceleration due to gravity, find
(i) how long will it take for the ball to strike the ground?
(ii) the speed with which will it strike the ground? and
(iii) how high the ball will rise?
Solution:
We know that

To find c₁ by substituting the initial conditions
x = 0 at t = 0 in the above equations,
We get C₁ = 0
∴ x = \(\frac{a t^{2}}{2}\) + 39.2t ………….. (3)
(i) While the ball strike the ground x = 0

(ii) The speed with which will it strike the ground?
At t = 8; (1) ⇒ v = -9.8(8) + 39.2
= -78.4 + 39.2
= -39.2
∴ The speed with which the ball will strike the ground is = 39.2 m/s
(iii) At maximum height v = 0
⇒ (2) ⇒ -9.8 t + 39.2 = 0
t = 4 sec
⇒ (3) ⇒ x = \(-\frac{9.8 \times 16}{2}\) + 39.2 × 4
= -78.4 + 156.8 = 78.4 m/s

Question 5.
A wound is healing in such a way that t days since Sunday the area of the wound has been decreasing at a rate of \(\frac{-6}{(t+2)^{2}}\)cm² per day where 0 < r ≤ 8. If on Monday the area of the wound was 1 .4 cm²
(i) What was the area of the wound on Sunday?
(ii) What is the anticipated area of the wound on Thursday if it continues to heal at the
same rate?
Solution:
Let A be the area of wound at time ‘t’.

A = \(\frac{6}{t+2}\) + c
By the given, condition area of the wound on Monday is 1.4 cm²
⇒ A = 1.4; t = 1
⇒ 1.4 = \(\frac{6}{t+2}\) + c
1.4 = \(\frac{6}{3}\) + c
c = -0.6
∴ Area of wound at any day.
⇒ A = \(\frac{6}{t+2}\) – 0.6
(i) The area of a wound on Sunday
t = 0 ⇒ A = \(\frac{6}{2}\) – 0.6 = 3 – 0.6 = 2.4 cm²
(ii) The area of the wound on Thrusday
t = 4 ⇒ A = \(\frac{6}{6}\) – 0.6 = 1 – 0.6 = 0.4 cm²

Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.4 Additional Problems

Question 1.
If f'(x) = 2x – 7 and f(1) = 0 find f(x)
Solution:
Given f’ (x) = 2x – 7
⇒ f (x) = \(\int(2 x-7) d x\)
= x² – 7x + c
Given f(1) = 0 ⇒ 1 – 7 + c = 0
⇒ c = 6
So f(x) = x² – 7x + 6

Question 2.
Given f”(x) = 6x + 6, f'(0) = -5 and f(1) = 6 find f(x)
Solution:
f” (x) = 6x + 6
⇒ f ‘ (x) = (6x + 6) dx
= \(\frac{6 x^{2}}{2}\) + 6x + c
= 3x² + 6x + c
Given f ‘(0) = -5 ⇒ 0 + c = -5
⇒ c = -5
∴ f ‘(x) = 3x² + 6x – 5
So f(x) = \(\int\left(3 x^{2}+6 x-5\right) d x\)
= \(\frac{3 x^{3}}{3}+\frac{6 x^{2}}{2}\) – 5x + c
(i.e.,) f(x) = x³ + 3x² – 5x + c
Given f(1) = 6
⇒ 1 + 3 + 5 + c = 6
⇒ c – 1 = 6 ⇒ c = 7
So f(x) = x³ + 3x² – 5x + 7

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.3

Integrate the following w.r.to x.
Question 1.
(x + 4)⁵ + \(\frac{5}{(2-5 x)^{4}}\) – cosec²(3x – 1)
Solution:

Question 2.
4cos(5 – 2x) + 9e^3x-6 + \(\frac{24}{6-4 x}\)
Solution:

Question 3.
sec²\(\frac{x}{5}\) + 18 cos 2x + 10 sec (5x + 3) tan (5x + 3)
Solution:

Question 4.
\(\frac{8}{\sqrt{1-(4 x)^{2}}}+\frac{27}{\sqrt{1-9 x^{2}}}-\frac{15}{1+25 x^{2}}\)
Solution:

Question 5.

Solution:

Question 6.

Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.3 Additional Problems

Question 1.
5x⁴ + 3(2x + 3)⁴ – 6(4 – 3x)⁵
Solution:

Question 2.
4 – \(\frac{5}{x+2}\) + 3 cos 2x
Solution:

Question 3.
9 cosec²(px – q) – 6(1 – x)⁴ + 4e^{3 – 4x}
Solution:

Question 4.
\(\frac{4}{(3+4 x)}\) + (10x + 3)⁹ – 3cosec(2x + 3) cot (2x + 3)
Solution:

Question 5.
a sec²(bx + c) + \(\frac{q}{e^{l-m x}}\)
Solution:

Question 6.

Solution:

Question 7.

Solution:

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.2

Integrate the following functions w.r.to ‘x’.
Question 1.
(i) (x + 5)⁶
(ii) \(\frac{1}{(2-3 x)^{4}}\)
(iii) \(\sqrt{3 x+2}\)
Solution:

Question 2.
(i) sin 3x
(ii) cos (5 – 11x)
(iii) cosec² (5x – 7)
Solution:

Question 3.
(i) e^{3x – 6}
(ii) e^{8 – 7x}
(iii) \(\frac{1}{6-4 x}\)
Solution:

Question 4.
(i) sec²\(\frac{x}{5}\)
(ii) cosec (5x + 3) cot (5x + 3)
(iii) sec (2 – 15x) tan (2 – 15x)
Solution:

Question 5.
(i) \(\frac{1}{\sqrt{1-(4 x)^{2}}}\)
(ii) \(\frac{1}{\sqrt{1-81 x^{2}}}\)
(iii) \(\frac{1}{1+36 x^{2}}\)
Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.2 Additional Problems

Question 1.
(3x + 4)⁶
Solution:

Question 2.
\(\frac{1}{(x+5)^{4}}\)
Solution:

Question 3.
\(\frac{1}{p+q x}\)
Solution:

Question 4.
cos(4x + 5)
Solution:

Question 5.
cosec²(7 – 11x)
Solution:

Question 6.
sec(3 + x) tan(3 + x)
Solution:

Question 7.
cosec(3 – 2x) cot(3 – 2x)
Solution:

Question 8.
e^{3x + 2}
Solution:

Question 9.
\(\frac{1}{\sin ^{2}(l-m x)}\)
Solution:

Question 10.
(lx + m)^1/2
Solution:

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.1

Integrate the following with respect to ‘x’:
Question 1.
(i) x¹¹
(ii) \(\frac{1}{x^{7}}\)
(iii) \(\sqrt[3]{x^{4}}\)
(iv) \(\left(x^{5}\right)^{\frac{1}{8}}\)
Solution:

Question 2.
(i) \(\frac{1}{\sin ^{2} x}\)
(ii) \(\frac{\tan x}{\cos x}\)
(iii) \(\frac{\cos x}{\sin ^{2} x}\)
(iv) \(\frac{1}{\cos ^{2} x}\)
Solution:

Question 3.
(i) 12³
(ii) \(\frac{x^{24}}{x^{25}}\)
(iii) e^x
Solution:

Question 4.
(i) (1 + x²)^-1
(ii) \(\left(1-x^{2}\right)^{-\frac{1}{2}}\)
Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.1 Additional Problems

Integrate the following with respect to x.
Question 1.
(i) \(\sqrt{x^{7}}\)
(ii) (x¹⁰)^1/7
Solution:

Question 2.
(i) \(\frac{1}{x^{5}}\)
(ii) x^-1
(iii) \(\frac{1}{x^{5 / 2}}\)
Solution:

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.8

Question 1.
Find all values of x for which \(\frac{x^{3}(x-1)}{(x-2)}\) > 0
Solution:

Now we have to find the signs of
x³, x – 1 and x – 2 as follows
x³ = 0; x – 1 = 0 ⇒ x = 1; x – 2 = 0 ⇒ x = 2
Plotting the points in a number line and finding intervals

So the solution set = (0, 1) ∪ (2, ∞)

Question 2.

Solution:

Plotting the points 3/2, 2 and 4 on the number line and taking the intervals.

Question 3.

Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.8 Additional Questions

Question 1.

Solution:

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.5

Choose the correct or the most suitable answer from the given four alternative
Question 1.

Solution:
(b)

Question 2.
If y = f(x² + 2) and f'(3) = 5, then \(\frac{d y}{d x}\) at x = 1 is …………….
(a) 5
(b) 25
(c) 15
(d) 10
Solution:
(d)

Question 3.
If y = \(\frac{1}{4}\)u⁴, u = \(\frac{2}{3}\)x³ + 5, then \(\frac{d y}{d x}\) is ……………

Solution:
(c)

Question 4.
If f(x) = x² – 3x, then the points at which f(x) = f'(x) are …………………..
(a) both positive integers
(b) both negative integers
(c) both irrational
(d) one rational and another irrational
Solution:
(c)
f(x) = x² – 3x
f'(x) = 2x – 3
Given f(x) = f'(x)
⇒ x² – 3x = 2x – 3
⇒ x² – 5x + 3 = 0
x = \(\frac{5 \pm \sqrt{25-12}}{2}=\frac{5 \pm \sqrt{13}}{2}\)
⇒ The roots are irrational

Question 5.
If y = \(\frac{1}{a-z}\), then \(\frac{d z}{d y}\) is ……………….
(a) (a – z)²
(b) -(z – a)²
(c) (z + a)²
(d) -(z + a)²
Solution:
(a)

Question 6.
If y = cos (sin x²), then \(\frac{d y}{d x}\) at x = \(\sqrt{\frac{\pi}{2}}\) is …………..
(a) -2
(b) 2
(c) -2\(\sqrt{\frac{\pi}{2}}\)
(d) 0
Solution:
(d)
y = cos (sin x²)
\(\frac{d y}{d x}\) = – sin (sin x²) [cos (x²)] (2x)
∴ \(\frac{d y}{d x}\) at x = \(\sqrt{\frac{\pi}{2}}\) = -sin (1) [0] = 0

Question 7.
If y = mx + c and f(0) = f'(0) = 1, then f(2) is ………………
(a) 1
(b) 2
(c) 3
(d) -3
Solution:
(c)
y = mx+c
\(\frac{d y}{d x}\) = m
y = x + c (i.e.) f(x) = x + c
y(a tx = 0) = f(0) 0 + c = 1 ⇒ c = 1
y = x + 1 ⇒ f(x) = x + 1
f(2) = 2 + 1 = 3

Question 8.
If f(x) = x tan^-1x, then f'(1) is ……………
(a) 1 + \(\sqrt{\frac{\pi}{4}}\)
(b) \(\frac{1}{2}+\frac{\pi}{4}\)
(c) \(\frac{1}{2}-\frac{\pi}{4}\)
(d) 2
Solution:
(b)
f(x) = x tan^-1 x

Question 9.
\(\frac{d}{d x}\)(e^x+5logx) is ……………..
(a) e^x.x⁴ (x + 5)
(b) e^x.x (x + 5)
(c) e^x + \(\frac{5}{x}\)
(d) e^x – \(\frac{5}{x}\)
Solution:
(a)
y = e^x+5logx = e^x.e^5logx = e^x.e^logx5
= x⁵ e^x
∴ \(\frac{d y}{d x}\) = x⁵ (e^x) + e^x (5x⁴)
= e^x. x⁴ (x + 5)

Question 10.
If the derivative of (ax – 5) e^3x at x = 0 is -13, then the value of a is …………….
(a) 8
(b) -2
(c) 5
(d) 2
Solution:
(d)
y = (ax – 5)e^3x
\(\frac{d y}{d x}\) = y’ = (ax – 5) (3e^3x) + e^3x (a)
= e^3x[3ax – 15 + a]
Given \(\frac{d y}{d x}\) = -13 at x = 0
⇒ [-15 + a] = -13
⇒ a = -13 + 15
a = 2

Question 11.
x = \(\frac{1-t^{2}}{1+t^{2}}\), y = \(\frac{2 t}{1+t^{2}}\) then \(\frac{d y}{d x}\) is …………..

Solution:
(c)
Given x = \(\frac{1-t^{2}}{1+t^{2}}\) and y = \(\frac{2 t}{1+t^{2}}\)
when we put t = tan θ
Then x = cos 2θ and y = sin 2θ

Question 12.
If x = a sin θ and y = b cos θ, then \(\frac{d^{2} y}{d x^{2}}\) is …………..

Solution:
(c)

Question 13.
The differential coefficient of log₁₀x with respect to log_x 10 is …………….
(a) 1
(b) -(log₁₀x)²
(c) (log_x 10)²
(d) \(\frac{x^{2}}{100}\)
Solution:
(b)

Question 14.
If f(x) = x + 2, then f'(f(x)) at x = 4 is ……………..
(a) 8
(b) 1
(c) 4
(d) 5
Solution:
(b)
f(x) = x + 2
f'(x) = 1
f'(x) (at x = 4) = 1

Question 15.
If y = \(\frac{(1-x)^{2}}{x^{2}}\), then \(\frac{d y}{d x}\) is ………………

Solution:
(d)

Question 16.
If pv = 81, then \(\frac{d p}{d v}\) at v = 9 is ………….
(a) 1
(b) -1
(c) 2
(d) -2
Solution:
(b)

Question 17.
If f(x) = then the right hand derivative of f(x) at x = 2 is ……………….
(a) 0
(b) 2
(c) 3
(d) 4
Solution:
(c)

Question 18.
It is given that f'(a) exists, then is ……………..
(a) f(a) – af'(a)
(b) f ‘(a)
(c) -f ‘(a)
(d) f(a) + af ‘(a)
Solution:
(a)

Question 19.
If f(x) = then f ‘(2) is ………………
(a) 0
(b) 1
(c) 2
(d) does not exist
Solution:
(d)

∴ f ‘(2) does not exist

Question 20.
If g(x) = (x² + 2x + 3) f(x) and f(0) = 5 and then g ‘(θ) is ……………
(a) 20
(b) 22
(c) 18
(d) 12
Solution:
(b) 22

Question 21.
If f(x) = , then at x = 3, f ‘(x) is ………………
(a) 1
(b) -1
(c) 0
(d) does not exist
Solution:
(d)

as LHS ≠ RHS limit does not exist

Question 22.
The derivative of f(x) = x|x| at x = -3 is …………..
(a) 6
(b) -6
(c) does not exist
(d) 0
Solution:
(a)
f(x) = x|x|
f(x) = x(-x) ⇒ f(x) = – x²
f ‘(x) = -(2x)
f ‘(-3) = -(2) (-3) = 6

Question 23.
If f(x) = , then which one of the following is true?
(a) f(x) is not differentiable at x = a
(b) f(x) is discontinuous at x = a
(c) f(x) is continuous for all x in R
(d) f(x) is differentiable for all x ≥ a
Solution:
(a)

f(x) is not differentiable at x = a

Question 24.
If f(x) = is differentiable at x = 1, then ………………

Solution:
(c)

Question 25.
Then number of points in R in which the function f(x) = |x – 1| + |x – 3| + sin x is not differentiable, is ……………..
(a) 3
(b) 2
(c) 1
(d) 4
Solution:
(b) 2
f(x) = |x – 1| + |x – 3| + sin x is not differentiable at x = 1, and x = 3

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.7

Question 1.
Factorize: x⁴ + 1. (Hint: Try completing the square.)
Solution:

Question 2.
If x² + x + 1 is a factor of the polynomial 3x³ + 8x² + 8x + a, then find the value of a.
Solution:
Let 3x³ + 8x² + 8x + a = (x² + x + 1) (3x + a) .
Equating coefficient of x
8 = a + 3
8 – 3 = a
a = 5

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.7 Additional Questions Solved

Question 1.
Solve for x² – 7x³ + 8x² + 8x – 8 = 0. given 3 – \(\sqrt{5}\) is a root
Solution:
when 3 – \(\sqrt{5}\) is a root, 3 + \(\sqrt{5}\) is the other root.
S.o.r. = (3 – \(\sqrt{5}\)) + (3 + \(\sqrt{5}\)) = 6

The equation is x² – 6x + 4 = 0
Now x⁴ – 7x³ + 8x² + 8x – 8 = (x² – 6x + 4) (x² + px – 2)
Equating co-eff of x
12 + 4p = 8
4p = 8 – 12 = -4
So the other factor is x² – x – 2
Now solving x² – x – 2 = 0

Question 2.
Solve the equation x³ + 5x² – 16x – 14 = 0. given x + 7 is a root
Solution:
x³ + 5x² – 16x – 14 = (x + 7) (x² + px – 2)
Equating co-eff of x
7p – 2 = -16
7p = -16 + 2 = -14
⇒ p = -2
So the other factor is x² – 2x – 2

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.6

Question 1.
Find the zeros of the polynomial function f(x) = 4x² – 25
Solution:

Question 2.
If x = -2 is one root of x³ – x² – 17x = 22, then find the other roots of equation.
Solution:
x = – 2 is one root
So applying synthetic division

Question 3.
Find the real roots of x⁴ = 16
Solution:
x⁴ = 16
⇒ x⁴ – 16 = 0
(i.e.,) x⁴ – 4² = 0
⇒ (x² + 4)(x² – 4) = 0
x² + 4 = 0 will have no real roots
so solving x² – 4 = 0
x² = 4

Question 4.
Solve (2x + 1)² – (3x + 2)² = 0
Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.6 Additional Questions

Question 1.
Find the zeros of the polynomial function f(x) = 9x² – 36
Solution:

Question 2.
If x = 2 is one root of x³ + 2x² – 5x – 6 = 0 then find the other roots of the equation
Solution:
x = 2 is a root
so applying synthetic division

∴ The other factor is x² + 4x + 3
Now x³ + 2x² – 5x – 6 = (x – 2)(x² + 4x + 3)
∴ x³ + 2x² – 5x – 6 = 0 ⇒ (x – 2)(x² + 4x + 3)
x – 2 = 0 or x² + 4x + 3 = 0
x = 2 or (x + 1)(x + 3) = 0
⇒ x = -1 or -3
so the roots are x = -1, 2, -3