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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 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

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 2.5

Question 1.
Solve 2x² + x – 15 ≤ 0.
Solution:
To find the solution of the inequality
ax² + bx + c ≥ 0 or ax² + bx +c ≤ 0 (for a > 0)
First we have to solve the quadratic equation ax² + bx + c = 0
Let the roots be a and P (where a < P)
So for the inequality ax² + bx + c ≥ 0 the roots lie outside α and β
(i.e.,) x ≤ α and x ≥ β
So for the inequality ax² + bx + c ≤ 0. The roots lie between α and β
(i.e.,) x > α and x < β (i.e.) a ≤ x ≤ β

Question 2.
Solve -x² + 3x – 2 ≥ 0
Solution:
-x² + 3x – 2 ≥ 0 ⇒ x² – 3x + 2 ≤ 0
(x – 1) (x – 2) ≤ 0
[(x – 1) (x – 2) = 0 ⇒ x = 1 or 2. Here α = 1 and β = 2. Note that α < β]
So for the inequality (x – 1) (x – 2) ≤ 2
x lies between 1 and 2
(i.e.) x ≥ 1 and x ≤ 2 or x ∈ [1, 2] or 1 ≤ x ≤ 2

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.5 Additional Questions

Question 1.
Solve for x.

Solution:

Select the intervals in which (3x +1) (3x – 2) is positive
(3x + 1) > 0 and (3x – 2) > 0 or
3x +1 < 0 and 3x – 2 < 0

Question 2.

Solution:

Question 3.

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 10.2

Find the derivatives of the following functions with respect to corresponding independent variables.
Question 1.
f(x) = x – 3 sinx
Solution:
f(x) = x – 3 sinx
= f'(x) = 1 – 3 (cos x)
= 1 – 3 cos x

Question 2.
y = sin x + cos x
Solution:
\(\frac{d y}{d x}\) = cosx + (-sinx) = cos x – sin x

Question 3.
f(x) = x sin x
Solution:
f(x) = uv
⇒ f'(x) = uv’ + vu’ = u\(\frac{d u}{d x}\) + v\(\frac{d v}{d x}\)
Now u = x ⇒ u’ = 1
v = sin x ⇒ v’ cos x
f'(x) = x (cos x) + sin x(1)
= x cos x + sin x

Question 4.
y = cos x – 2 tan x
Solution:
\(\frac{d y}{d x}\) = -sin x = 2 (sec²x)
= – sin x – 2 sec²x

Question 5.
g(t) = t³ cos t
Solution:
g(t) = t³ cost (i.e.) u = t³ and v = cos t
let u’ = \(\frac{d u}{d x}\) and v’= \(\frac{d v}{d x}\) = (-sint)
g'(t) = uv’ + vu’
g'(t) = t³ (-sin t) + cos t (3t²)
= -t³ sin t + 3t² cos t

Question 6.
g(t) = 4 sec t + tan t
Solution:
g{t) = 4 sect + tan t
g'(t) = 4(sec t tan t) + sec²t
= 4sec t tan t + sec²t

Question 7.
y = e^x sin x
Solution:
y = e^x sin x
⇒ y = uv’ + vu’
Now u = e^x ⇒ u’ = \(\frac{d u}{d x}\) e^x
v = sin x ⇒ v’ = \(\frac{d v}{d x}\) cos x
i.e. y’ = e^x (cos x) + sin x (e^x)
= e^x [sin x + cos x]

Question 8.
y = \(\frac{\tan x}{x}\)
Solution:

Question 9.
y = \(\frac{\sin x}{1+\cos x}\)
Solution:
y = \(\frac{\sin x}{1+\cos x}=\frac{u}{v}\) (say)
u = sin x v = 1 + cosx
u’ = cos x v’ = -sin x
y = \(\frac{u}{v} \Rightarrow y^{\prime}=\frac{v u^{\prime}-u v^{\prime}}{v^{2}}\)

Question 10.
y = \(\frac{x}{\sin x+\cos x}\)
Solution:

Question 11.
y = \(\frac{\tan x-1}{\sec x}\)
Solution:

Question 12.
y = \(\frac{\sin x}{x^{2}}\)
Solution:

Question 13.
y = tan θ (sin θ + cos θ)
Solution:

Question 14.
y = cosex x. cot x
Solution:
y = u v ⇒ y’ = uv’ + vu’
u = cosec x ⇒ u’ = -cosec x cot x
v = cot x ⇒ v’ = – cosec² x
(cosec x)(-cosec²x) + cot x(-coseç x cot x)
= cosec³x – cosec x cot²x
= – cosec x (cosec²x + cot²x)
= \(-\frac{1}{\sin x}\left(\frac{1+\cos ^{2} x}{\sin ^{2} x}\right)=-\frac{\left(1+\cos ^{2} x\right)}{\sin ^{3} x}\)

Question 15.
y = x sin x cos x
Solution:

Question 16.
y = e^-x. log x
Solution:
y = e^-x logx = uv (say)
Here u = e^-x and v = log x
⇒ u’ = -e^-x and v’ = \(\frac{1}{x}\)
Now y = uv ⇒ y’ = uv’ + vu’
(i.e.) \(\frac{d y}{d x}\) = e^-x \(\left(\frac{1}{x}\right)\) + log x(-e^-x)
= e^-x(\(\frac{1}{x}\) – log x)

Question 17.
y = (x² + 5) log (1 + x)e^-3x
Solution:

Question 18.
y = sin x⁰
Solution:

Question 19.
y = log₁₀x
Solution:

Question 20.
Draw the function f'(x) if f(x) = 2x² – 5x + 3
Solution:
f(x) = 2x² – 5x + 3
f'(x) = 4x – 5 which is a linear function
(i.e.) y = 4x – 5

Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.2 Additional Questions

Question 1.
Find the derivation of following functions
Question 1.
3 sin x + 4 cos x – e^x
Solution:
y = 3 sin x + 4 cos x – e^x
\(\frac{d y}{d x}\) = 3 (cos x) + 4 (- sin x) – (e^x)
= 3 cos x – 4 sin x – e^x

Question 2.
sin 5 + log₁₀x + 2 secx
Solution:
y = sin 5 + log₁₀x + 2 secx
\(\frac{d y}{d x}\) = 0 + \(\left(\frac{1}{x}\right)\) log₁₀ e + 2[sec x + tan x] = \(\frac{\log _{10} e}{x}\) + 2 sec x tan x

Question 3.
6 sin x log₁₀x + e
Solution:

Question 4.
(x⁴ – 6x³ + 7x² + 4x + 2) (x³ – 1)
Solution:
Let u = x⁴ – 6x³ + 7x² + 4x + 2 and v = x³ – 1
u’ = 4x³ – 6 (3x²) + 7 (2x) + 4 (1) + 0
= 4x³ – 18x² + 14x + 4
v’= 3x³
y = uv’ + vu’
i.e. \(\frac{d y}{d x}\) = (x⁴ – 6x³ + 7x² + 4x + 2) (3x²) + (x³ – 1) (4x³ – 18x² + 14x + 4)
= 3x⁶ – 18x⁵ + 21x⁴ + 12x³ + 6x² + 4x⁶ – 18x⁵ + 14x⁴ + 4x³ – 4x³ + 18x² – 14x – 4
= 7x⁶ – 36x⁵ + 35x⁴ + 12x³ + 24x² – 14x – 4

Question 5.
(3x² + 1)²
Solution:
y = (3x² + 1)² = (3x² + 1) (3x² + 1)
Let u = 3x² + 1 and v = 3x² + 1
∴ u’ = 3(2x) = 6x and v’ = 6x
y’ = uv’ + vu’
(i.e.,) \(\frac{d y}{d x}\) = (3x² + 1) (6x) + (3x² + 1) 6x = 12x (3x² + 1)

Question 6.
(3 sec x – 4 cosec x) (2 sin x + 5 cos x)
Solution:
y = (3 sec x – 4 cosec x) (2 sin x + 5cos x)
Let u = 3 secx-4 cosecx and v = 2 sinx + 5 cosx
u’ = 3 (sec x tan x) – 4 (-cosec x cot x) ; v’ = 2 (cos x) + 5 (- sin x)
u’ = 3 sec x tan x + 4 cosec x cot x); v’ = 2 cos x – 5 sin x .
∴ y’ = uv’ + vu’
So \(\frac{d y}{d x}\) = (3 sec x – 4 cosec x) (2 cos x – 5 sinx) + (2 sin x + 5 cos x) (3 sec x tan x + 4 cosec x cot x) = 6 sec x cos x – 15 sec x sin x – 8 cosec x cos x + 20 cosec x sin x + 6 sinx secx tanx + 8 sinx cosecx cotx+ 15 cosx secx tanx + 20 cos x cosec x cot x
= 6 \(\frac{1}{\cos x}\) cosx – 15 \(\frac{1}{\cos x}\)sin x – 8 \(\frac{1}{\sin x}\) cos x + 20 \(\frac{1}{\sin x}\) sin x + 6 sin x \(\frac{1}{\cos x}\) tan x + 8 sin x \(\frac{1}{\sin x}\) cot x + 15 cos x \(\frac{1}{\cos x}\) tan x + 20 cos x \(\frac{1}{\sin x}\) cot x
= 6 – 15tan x – 8cot x + 20 + 6 tan²x + 8 cot x + 15 tan x + 20cot²x
= 26 + 6 tan²x + 20 cot²x

Question 7.
x² e^x sinx
Solution:
y = x² e^x sin x
Let u = x², v = e^x and w = sinx
u’ = 2x, v’ = e^x and w’ = cos x
y’ = uvw’ + vwu’ + uwv’
= (x² e^x) cos x + (e^x sin x)(2x) + (x² sin x)e^x
= x² e^x cos x + 2xe^x sin x + x² e^x sin x
= xe^x {x cos x + 2 sin x + x sin x}

Question 8.
\(\frac{\cos x+\log x}{x^{2}+e^{x}}\)
Solution:
y = \(\frac{\cos x+\log x}{x^{2}+e^{x}}\)
Let u = cos x + log x and v = x² + e^x
∴ u’ = – sin x + \(\frac{1}{x}\), v’ = 2x + e^x

Question 9.
\(\frac{\tan x+1}{\tan x-1}\)
Solution:

Question 10.
\(\frac{\sin x+x \cos x}{x \sin x-\cos x}\)
Solution:
y = \(\frac{\sin x+x \cos x}{x \sin x-\cos x}\)
Let u = sinx + x cosx and v = x sin x – cos x
u’ = cos x + x( – sin x) + cos x (1)
= cos x – x sin x + cos x = 2 cos x – x sin x
v’=x (cos x) + sin x(1) – (- sin x)
= x cos x + sin x + sin x = 2 sin x + x cos x
y = \(\frac{u}{v}\) ∴ y’ = \(\frac{v u^{\prime}-u v^{\prime}}{v^{2}}\)

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 10 Differentiability and Methods of Differentiation Ex 10.1

Question 1.
Find the derivatives of the following functions using first principle.
(i) f(x) = 6
Solution:
Given f(x) = 6
f(x + h) = 6

[h → 0 means h is very nears to zero from left to right but not zero]

(ii) f(x) = -4x + 7
Solution:
Given f(x) = -4x + 7
f(x + h) = -4(x + h) + 7
= -4x – 4h + 7

(iii) f(x) = -x² + 2
Given f(x) = -x² + 2
f(x + h) = -(x + h)² + 2
= -x² – h² – 2xh + 2

Question 2.
Find the derivatives from the left and from the right at x = 1 (if they exist) of the following functions. Are the functions differentiable at x = 1?
(i) f(x) = |x – 1|
Solution:


f'(1) does not exist
∴ ‘f’ is not differentiable at x = 1.

(ii) f(x) = \(\sqrt{1-x^{2}}\)
Solution:

∴ ‘f’ is not differentiable at x = 1.

(iii)

Solution:

‘f’ is not differentiable at x = 1

Question 3.
Determine whether the following functions is differentiable at the indicated values.
(i) f(x) = x |x| at x = 0
Solution:


Limits exists
Hence ‘f’ is differentiable at x = 0.

(ii) f(x) = |x² – 1| at x = 1
Solution:

f(x) is not differentiable at x = 1.

(iii) f(x) = |x| + |x – 1| at x = 0, 1
Solution:

∴ f(x) is not differentiable at x = 0.

∴ f(x) is not differentiable at x = 1.

(iv) f(x) = sin |x| at x = 0
Solution:

∴ f(x) is not differentiable at x = 0.

Question 4.
Show that the following functions are not differentiable at the indicated value of x.
(i)

Solution:


f(x) is not differentiable at x = 2.

(ii)

Solution:

f(x) is not differentiable at x = 0.

Question 5.
The graph off is shown below. State with reasons that x values (the numbers), at which f is not differentiable.
Solution:

(i) at x = – 1 and x = 8. The graph ‘ is not differentiable since ‘ has vertical tangent at x = -1 and x = 8(also At x = -1. The graph has shape edge v] and at x = 8;The graph has shape peak ^]
(ii) At x = 4: The graph f is not differentiable, since at x =4. The graph f’ is not continuous.
(iii) At x = 11; The graph f’ is not differentiable, since at x = 11. The tangent line of the graph is perpendicular.

Question 6.
If f(x) = |x + 100| + x², test whether f’ (-100) exists.
Solution:
f(x) = |x + 100| + x²

Question 7.
Examine the differentiability of functions in R by drawing the diagrams.
(i) |sin x|
Solution:

Limit exist and continuous for all x ∈ R clearly, differentiable at R — {nπ n ∈ z) Not differentiable at x = nπ , n ∈ z.

(ii) |cos x|
Solution:

Limit exist and continuous for all x ∈ R clearly, differentiable at R {(2n + 1)π/2/n ∈ z} Not differentiable at x = (2n + 1) \(\frac{\pi}{2}\), n ∈ Z.

Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.1 Additional Questions

Question 1.
Is the function f(x) = |x| differentiable at the origin. Justify your answer.
Solution:

Question 2.
Discuss the differentiability of the functions:

Solution:

∴ f(2) is not differentiable at x = 2. Similarly, it can be proved for x = 4.

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 9 Limits and Continuity Ex 9.6

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

(a) 1
(b) 0
(c) ∞
(d) -∞
Solution:
(b) 0

Question 2.

(a) 2
(b) 1
(c) -2
(d) 0
Solution:
(c) -2

Question 3.

(a) 0
(b) 1
(c) 2
(d) does not exist
Solution:
(d) does not exist

Question 4.

(a) 1
(b) -1
(c) 0
(d) 2
Solution:
(a) 1

Question 5.

(a) e⁴
(b) e²
(c) e³
(d) 1
Solution:
(a) e⁴

Question 6.

(a) 1
(b) 0
(c) -1
(d) \(\frac{1}{2}\)
Solution:
(d) \(\frac{1}{2}\)

Question 7.

(a) log ab
(b) log \(\left(\frac{a}{b}\right)\)
(c) log \(\left(\frac{b}{a}\right)\)
(d) \(\frac{a}{b}\)
Solution:
(b) log \(\left(\frac{a}{b}\right)\)

Question 8.

(a) 2 log 2
(b) 2 (log 2)²
(c) log 2
(d) 3 log 2
Solution:
(b) 2 (log 2))²

Question 9.
If f(x) = \(x(-1)^{ \left\lfloor \frac { 1 }{ x } \right\rfloor }\), x ≤ θ, then the value of is equal to …………….
(a) -1
(b) 0
(c) 2
(d) 4
Solution:
(b) 0

Question 10.

(a) 2
(b) 3
(c) does not exist
(d) 0
Solution:
(c) does not exist

Limit does not exist

Question 11.
Let the function f be defined f(x) = then ……………

Solution:

Limit does not exist

Question 12.
If f: R → R is defined by f(x) = \(\lfloor x-3\rfloor+|x-4|\) for x ∈ R, then
is equal to …………..
(a) -2
(b) -1
(c) 0
(d) 1
Solution:
(c) 0

Question 13.

(a) 1
(b) 2
(c) 3
(d) 0
Solution:
(d) 0

Question 14.
If then the value of p is ………….
(a) 6
(b) 9
(c) 12
(d) 4
Solution:
(c) 12

Question 15.

(a) \(\sqrt{2}\)
(b) \(\frac{1}{\sqrt{2}}\)
(c) 1
(d) 2
Solution:
(a) \(\sqrt{2}\)

Question 16.

(a) \(\frac{1}{2}\)
(b) 0
(c) 1
(d) ∞
Solution:
(a) \(\frac{1}{2}\)

Question 17.

(a) 1
(b) e
(c) \(\frac{1}{e}\)
(d) 0
Solution:
(a) 1

Question 18.

(a) 1
(b) e
(c) \(\frac{1}{2}\)
(d) 0
Solution:
(a) 1

Question 19.
The value of is ……………
(a) 1
(b) -1
(c) 0
(d) ∞
Solution:
(d) ∞
Hint:

So limit does not exist

Question 20.
The value of where k is an integer is …………..
(a) -1
(b) 1
(c) 0
(d) 2
Solution:
(b) 1

Question 21.
At x = \(\frac{3}{2}\) the function f(x) = \(\frac{|2 x-3|}{2 x-3}\) is ………….
(a) Continuous
(b) discontinuous
(c) Differentiate
(d) non-zero
Solution:
(b) discontinuous

Question 22.
Let f: R → R be defined by f(x) = then f is ……………
(a) Discontinuous at x = \(\frac{1}{2}\)
(b) Continuous at x = \(\frac{1}{2}\)
(c) Continuous everywhere
(d) Discontinuous everywhere
Solution:
(b) Continuous at x = \(\frac{1}{2}\)

Question 23.
The function f(x) = is not defined for x = -1. The value of f(-1) so that the function extended by this value is continuous is …………..
(a) \(\frac{2}{3}\)
(b) \(-\frac{2}{3}\)
(c) 1
(d) 0
Solution:
(b) \(-\frac{2}{3}\)
Hint: For the function to be continuous at x = 1

Question 24.
Let f be a continuous function on [2, 5]. If f takes only rational values for all x and f(3) = 12, then f(4.5) is equal to ……………
(a) \(\frac{f(3)+f(4.5)}{7.5}\)
(b) 12
(c) 17.5
(d) \(\frac{f(4.5)-f(3)}{1.5}\)
Solution:
(b) 12

Question 25.
Let a function f be defined by f(x) = \(\frac{x-|x|}{x}\) for x ≠ 0 and f(0) = 2. Then f is …………..
(a) Continuous nowhere
(b) Continuous everywhere
(c) Continuous for all x except x = 1
(d) Continuous for all x except x = 0
Solution:
(d) Continuous for all x except x = 0
Hint:

∴ f(x) is not continuous at x = 0
⇒ f(x) is continuous for all except x = 0

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 10.4

Find the derivatives of the following functions
Question 1.
y = x^{cos x}
Solution:
y = x^{cos x}
Taking log on both sides
log y = log x^{cos x} = cos x log x
differentiating w.r.to x we get

Question 2.
y = x^logx + (logx)^x
Solution:
y = x^logx + (logx)^x
Let y = u + v
Then \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\)
u = x^logx
Taking log on both sides
log u = log x log x = log (x)²
differentiating w.r.to x

Taking log on both sides
log u = log (logx)^x = x log (log x)
differentiating w.r.to x

Question 3.
\(\sqrt{x y}\) = e^{(x – y)}
Solution:

Question 4.
x^y = y^x
Solution:
x^y = y^x
Taking log on both sides
logx^y = logy^x
(i.e.) y log x = x log y
differentiating w.r.to x

Question 5.
(cos x)^{log x}
Solution:
y = (cos x)^{log x}
Taking log on both sides
log y = log (cos x)^{log x} = log x (log cos x)

Question 6.
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1
Solution:
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1
Differentiating w.r.to x

Question 7.
\(\sqrt{x^{2}+y^{2}}=\tan ^{-1}\left(\frac{y}{x}\right)\)
Solution:
\(\sqrt{x^{2}+y^{2}}=\tan ^{-1}\left(\frac{y}{x}\right)\)
Differentiating w.r.to x

Question 8.
tan (x + y) + tan (x – y) = x
Solution:
tan (x + y) + tan (x – y) = x
Differentiating w.r.to x we get

Question 9.
If cos (xy) = x, show that \(\frac{d y}{d x}=\frac{-(1+y \sin (x y))}{x \sin x y}\)
Solution:
cos (xy) = x
Differentiating w.r.to x

Question 10.
\(\tan ^{-1} \sqrt{\frac{1-\cos x}{1+\cos x}}\)
Solution:

Question 11.
\(\tan ^{-1}\left(\frac{6 x}{1-9 x^{2}}\right)\)
Solution:

Question 12.
cos [2 \(\tan ^{-1} \sqrt{\frac{1-x}{1+x}}\)]
Solution:

Question 13.
x = a cos³t; y = a sin²t
Solution:

Question 14.
x = a (cos t + t sin t); y = a [sin t – t cos t]
Solution:

Question 15.
x = \(\frac{1-t^{2}}{1+t^{2}}\); y = \(\frac{2 t}{1+t^{2}}\)
Solution:

Question 16.
\(\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\)
Solution:

Question 17.
sin^-1 (3x – 4x³)
Solution:

Question 18.
\(\tan ^{-1}\left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)\)
Solution:

Question 19.
Find the derivative of sin x² with respect to x²
Solution:

Question 20.
Find the derivative of \(\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\) with respect to tan^-1 x.
Solution:
Let u = \(\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\) and v = tan^-1 x
Now we have to find \(\frac{d u}{d v}\)

Question 21.

Solution:

Question 22.
Find the derivative with with respect to
Solution:

Question 23.
If y = sin^-1 then find y”.
Solution:

Question 24.
If y = e^tan-1x, show that (1 + x²) y” + (2x – 1) y’ = 0
Solution:
y = e^tan-1x
y = e^tan-1x \(\left(\frac{1}{1+x^{2}}\right)\)
⇒ y’ = \(\frac{y}{1+x^{2}}\) ⇒ y'(1 + x²) = y
differentiating w.r.to x
y’ (2x) + (1 + x²) (y”) = y’
(i.e.) (1 + x²) y” + y’ (2x) – y’ = 0
(i.e.) (1 + x²) y” + (2x – 1) y’ = 0

Question 25.
If y = \(\frac{\sin ^{-1} x}{\sqrt{1-x^{2}}}\) show that (1 – x²) y₂ – 3xy₁ – y = 0
Solution:

-xy + (1 – x²) y₁ = 1
differentiating both sides again w.r.to x
-[x y₁ + y (1)] + (1 – x²) (y₂) + y₁ (-2x) = 0
(i.e.) -xy₁ – y + (1 – x²) y₂ – 2xy₁ = 0
(1 – x²) y₂ – 3xy₁ – y = 0

Question 26.
If x = a (θ + sin θ), y = a (1 – cos θ) then prove that at θ = \(\frac{\pi}{2}\), y” = \(\frac{1}{a}\)
Solution:

Question 27.
If sin y = x sin (a + y) Then prove that \(\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin a}\), a ≠ nπ
Solution:

Question 28.
If y = (cos^-1 x)², prove that (1 – x²) \(\frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}\) – 2 = 0. Hence find y₂ when x = 0.
Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.4 Additional Problems

Question 1.
If y = A cos4x + B sin 4x, A and B are constants then Show that y₂ + 16y = 0
Solution:

Question 2.
If y = cos (m sin^-1 x), prove that (1 – x²) y₃ – 3xy₂ + (m² – 1) y₁ = 0
Solution:
We have y = cos (m sin^-1 x)
y₁ = sin (m sin^-1x). \(\frac{m}{\sqrt{1-x^{2}}}\)

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 10 Differentiability and Methods of Differentiation Ex 10.3

Differentiate the following
Question 1.
y = (x² + 4x + 6)⁵
Solution:
Let = u = x² + 4x + 6
⇒ \(\frac{d u}{d x}\) = 2x + 4
Now y = u⁵ ⇒ \(\frac{d y}{d x}\) = 5u⁴
∴ \(\frac{d y}{d x}=\frac{d y}{d u} \times \frac{d u}{d x}\) = 5u⁴ (2x + 4)
= 5(x² + 4x + 6)⁴ (2x + 4)
= 5 (2x + 4) (x² + 4x + 6)⁴

Question 2.
y = tan 3x
Solution:
y = tan 3x
put u = 3x
\(\frac{d u}{d x}\) = 3
Now y = tan u
⇒ \(\frac{d u}{d x}\) = sec² u
So \(\frac{d y}{d x}=\frac{d y}{d u} \times \frac{d u}{d x}\) = (sec² u) (3)
= 3 sec² 3x

Question 3.
y = cos (tan x)
Solution:
Put u = tan x
\(\frac{d u}{d x}\) = sec²x
Now y = cos u ⇒ \(\frac{d u}{d x}\) = – sin u
Now \(\frac{d y}{d x}=\frac{d y}{d u} \times \frac{d u}{d x}\)
= (-sin u) (sec²x)
= – sec² (sin (tan x))

Question 4.
y = \(\sqrt[3]{1+x^{3}}\)
Solution:

Question 5.
y = \(e^{\sqrt{x}}\)
Solution:

Question 6.
y = sin (e^x)
Solution:
y = sin (e^x)
Let u = e^x

Question 7.
F(x) = (x³ + 4x)⁷
Solution:
F(x) = (x³ + 4x)⁷
Put u = x³ + 4x

Question 8.
h(t) = \(\left(t-\frac{1}{t}\right)^{\frac{3}{2}}\)
Solution:

Question 9.
f(t) = \(\sqrt[3]{1+\tan t}\)
Solution:

Question 10.
y = cos (a³ + x³)
Solution:

Question 11.
y = e^-mx
Solution:

Question 12.
y = 4 sec 5x
Solution:

Question 13.
y = (2x – 5)⁴ (8x² – 5)^-3
Solution:

= \(\frac{8(2 x-5)^{3}}{\left(8 x^{2}-5\right)^{4}}\) (-4x² + 30x – 5)

Question 14.
y = (x² + 1) \(\sqrt[3]{x^{2}+2}\)
Solution:

Question 15.
y = xe^-x2
Solution:
y = xe^-x2
y = uv where u = x and v = e^-x2
Now u’ = 1 and v’ = e^-x2 (-2x)
v’ = – 2xe^-x2
Now y = uv ⇒ y’ = uv’ + vu’
(i.e.) \(\frac{d y}{d x}\) = x[-2xe^-x2] + e^-x2 (1)
= e^-x2 (1 – 2x²)

Question 16.
s(t) = \(\sqrt[4]{\frac{t^{3}+1}{t^{3}-1}}\)
Solution:

Question 17.
f(x) = \(\frac{x}{\sqrt{7-3 x}}\)
Solution:

Question 18.
y = tan (cos x)
Solution:
y = tan (cos x)

Question 19.
y = \(\frac{\sin ^{2} x}{\cos x}\)
Solution:

Question 20.
y = \(5^{-\frac{1}{x}}\)
Solution:

Question 21.
y = \(\sqrt{1+2 \tan x}\)
Solution:

Question 22.
y = sin³x + cos³x
Solution:
y = sin³x + cos³x
Here u = sin³ x = (sin x)³
⇒ \(\frac{d u}{d x}\) = 3 (sin x)² (cos x)
= 3sin²x cos x
v = cos³x = (cos x)³
⇒ \(\frac{d v}{d x}\) = 3 (cos x)² (-sin x) = -3 sin x cos²x
Now y = u + v ⇒ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\)
= 3 sin²x cos x – 3sin x cos²x
= 3 sin x cos x (sin x – cos x)

Question 23.
y = sin² (cos kx)
Solution:

Question 24.
y = (1 + cos²x)⁶
Solution:

Question 25.
y =\(\frac{e^{3 x}}{1+e^{x}}\)
Solution:

Question 26.
y = \(\sqrt{x+\sqrt{x}}\)
Solution:

Question 27.
y = e^{x cos x}
Solution:

Question 28.
y = \(\sqrt{x+\sqrt{x+\sqrt{x}}}\)
Solution:

Question 29.
y = \(\sin (\tan (\sqrt{\sin x}))\)
Solution:

Question 30.
y = sin^-1\(\left(\frac{1-x^{2}}{1+x^{2}}\right)\)
Solution: