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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 2.5

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). \(\frac{m}{\sqrt{1-x^{2}}}\)

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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:

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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 9 Limits and Continuity Ex 9.2

Evaluate the following limits:
Question 1.

Solution:

Question 2.
m and n are integers.
Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.

Solution:

= \(\frac{1}{4}-\frac{1}{2}\)
= \(\frac{-1}{4}\)

Question 11.

Solution:

Question 12.

Solution:

Question 13.

Solution:

∴ Limit does not exist

Question 14.

Solution:

Question 15.

Solution:

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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 9 Limits and Continuity Ex 9.1

In problems 1-6, complete the table using calculate and use the result to estimate the limit.
Question 1.

Solution:

Question 2.

Solution:

∴ Limit is 0.25

Question 3.

Solution:

∴ Limit is 0.288

Question 4.

Solution:

∴ Limit is -0.25

Question 5.

Solution:

∴ Limit is 1

Question 6.

Solution:

∴ Limit is 0

In exercise problems 7-15, use the graph to find the limits (if it exists). If the limit does not exist, explain why?
Question 7.
\(\lim _{x \rightarrow 3}\)(4 – x)
Solution:

Limit exist and is equal to 1

Question 8.
\(\lim _{x \rightarrow 1}\)(x² + 2)
Solution:

Limit exist and is equal to = 3

Question 9.

Solution:

Question 10.

Solution:

Question 11.

Solution:

Limit does not exist

Question 12.

Solution:
When x → 5, (x – 5) = -(x – 5)
∴ \(\lim _{x \rightarrow 5^{-}} \frac{-(x-5)}{(x-5)}\) = -1
When x → 5⁺, (x – 5) = (x – 5)

Question 13.
\(\lim _{x \rightarrow 1}\) sin(πx)
Solution:

Question 14.
\(\lim _{x \rightarrow 0}\) (sec x)
Solution:

Question 15.
\(\lim _{x \rightarrow \frac{\pi}{2}}\) tan x
Solution:

Limit does not exist

Sketch the graph of f, then identify the values of x₀ for which \(\lim _{x \rightarrow x_{0}}\) f(x) exists.
Question 16.

Solution:

Question 17.

Solution:

Limit exists except at x₀ = π

Question 18.
Sketch the graph of a function f that satisfies the given values:

Solution:

Question 19.
Write a brief description of the meaning of the notation \(\lim _{x \rightarrow 8}\) f(x) = 25
Solution:

Question 20.
If f(2) = 4, can you conclude anything about the limit of f(x) as x approaches 2?
Solution:
No, f(x) = 4, It is the value of the function at x = 2
This limit doesn’t exists at x = 2
Since f(2) = 4
It need not imply that \(\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{+}} f(x)\)
∴ we cannot conclude at x = 2

Question 21.
If the limit of f(x) as z approaches 2 is 4, can you conclude anything about f(2)?
Explain reasoning.
Solution:
\(\lim _{x \rightarrow 2}\) f(x) 4 , \(\lim _{x \rightarrow 2^{-}}\) f(x) = \(\lim _{x \rightarrow 2^{+}}\) f(x) = 4
When x approaches 2 from the left or from the right f(x) approaches 4.
Given that \(\lim _{x \rightarrow 2^{-}}\) f(x) = \(\lim _{x \rightarrow 2^{+}}\) f(x) = 4
The existence or non existence at x =2 has no leaving on the existence of the limit of f(x) as x approaches to 2.
∴ We cannot conclude the value of f(2)

Question 22.
Evaluate: \(\lim _{x \rightarrow 3} \frac{x^{2}-9}{x-3}\) if it exists by finding f(3^–) and f(3⁺).
Solution:

Question 23.
Verify the existence of
Solution:

Limit does not exist

Samacheer Kalvi 11th Maths Solutions Chapter 9 Limits and Continuity Ex 9.1 Additional Questions

Question 1.
Suppose . What are possible values of a and b?
Solution:

Question 2.

Solution:
We have,

Question 3.

Solution:
We have,

Question 4.

Solution:

Question 5.
Let a₁, a₂ …………… a_n be fixed real numbers such that f(x) = (x – a₁) , (x – a₂), ………. (x – a_n) what \(\lim _{x \rightarrow a}\) f(x) For a ≠ a₁, a₂, ………… a_n compute \(\lim _{x \rightarrow a}\) f(x)
Solution:
We have,

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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.2

Question 1.
Solve for x.

(i) |3 – x| < 7
Solution:
⇒ -7 < 3 – x < 7 3 – x > -7
-x > -7 -3 (= -10)
-x > -10 ⇒ x < 10
3 – x < 7
– x < 7 – 3 (= 4)
– x < 4x > -4 … .(2)
From (1) and (2) ⇒ x > -4 and x < 10
⇒ -4 < x < 10

(ii) |4x – 5| ≥ -2
Solution:

(iii)

Solution:

(iv) |x| – 10 < -3
Solution:
|x| < -3 + 10 (= 7)
|x| < 7 ⇒ -7 < x < 7

Question 2.
Solve \(\frac{1}{|2 x-1|}<6\) and express the solution using the interval notation.
Solution:

Question 3.
Solve -3|x| + 5 ≤ – 2 and graph the solution set in a number line.
Solution:
-3|x| + 5 ≤ – 2
⇒ -3 |x| ≤ – 2 – 5 (= -7)
-3|x| ≤ – 7 ⇒ 3 |x| ≥ 7

Question 4.
Solve 2|x + 1| – 6 ≤ 7 and graph the solution set in a number line.
Solution:

Question 5.

Solution:

Question 6.
Solve |5x – 12| < -2
Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.2 Additional Questions

Question 1.

Solution:

Question 2.

Solution:

⇒ x – 2< – 1 (or) x – 2 > 1 and – 2 < x – 2 < 2
⇒ x < 1 (or) x > 3 and -2 + 2 < x < 2 + 2
⇒ x < 1 (or) x > 3 and 0 < x < 4 Hence, the required solution is (0, 1) ∪ (3, 4)

Question 3.

Solution:

Question 4.
Solve: |x – 1| ≤ 5, |x| ≥ 2
Solution:
|x – 1| ≤ 5 and |x| ≥ 2
⇒ -5 ≤ x – 1 ≤ 5 and x ≤ -2 (or) x > 2
⇒ – 5 + 1 ≤ x ≤ 5 + 1
⇒ -4 ≤ x ≤ 6 and x ≤ -2 (or) x ≥ 2
Hence x < [-4, -2] ∪ [2, 6]

Question 5.

Solution:

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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 9 Limits and Continuity Ex 9.5

Question 1.
Prove that f(x) = 2x² + 3x – 5 is continuous at all points in R.
Solution:
Polynomial functions are continuous at every points of R.

Question 2.
Examine the continuity of the following:
(i) x + sin x
Solution:
f(x) = x + sin x
The Domain of the function (-∞, ∞)
∴ f(x) is continuous in (-∞, ∞)(i.e.,) for all x ∈ R

(ii) x² cos x
Solution:
f(x) = x² cos x
The Domain of the function (-∞, ∞)

f(x) is continuous in R

(iii) e^x tan x
Solution:
The Domain of the function in R – {(2n + 1) π/2}
∴ The functions is continuous for all x ∈ R – (2n + 1) \(\frac{\pi}{2}\), n ∈ Z

(iv) e^2x + x²
f(x) = e^2x + x² = 1 + 2x + \(\frac{(2 x)^{2}}{2 !}\) + …………. + x²
Solution:
∴ The functions is continuous for all x ∈ R

(v) x.ln x
Solution:
Thus f(x) is continuous for (0, ∞)

(vi) \(\frac{\sin x}{x^{2}}\)
Solution:
Thus f(x) is continuous for all x ∈ R – {0}

(vii) \(\frac{x^{2}-16}{x+4}\)
Solution:
f(x) = \(\frac{x^{2}-16}{x+4}=\frac{(x-4)(x+4)}{x+4}\)
The function f(x) is continuous for all x ∈ R – {-4}

(viii) |x + 2| + |x – 1|
Solution:
f(x) is continuous for x ∈ R

(ix) \(\frac{|x-2|}{|x+1|}\)
Solution:
The function is continuous for all x ∈ R – {-1}

(x) cot x + tan x
Solution:
The function is continuous for all x ∈ R – \(\frac{n \pi}{2}\), n ∈ z.

Question 3.
Find the points of discontinuity of the function f, where,
(i)

Solution:
f(3) = 12 + 5 = 17

∴ f(x) is discontinuous at x = 3

(ii)

Solution:
f(x) = 4

∴ f(x) is continuous for all x ∈ R

(iii)

Solution:
f(x) = 8 – 3 = 5

∴ f(x) is continuous for all x ∈ R

(iv)

Solution:

∴ f(x) is continuous for all x ∈ [0, π/2]

Question 4.
At the given points x₀ discover whether the given function is continuous or discontinuous citing the reasons for your answer.
(i)

Solution:
Given f(x₀) = 1

∴ f(x) is continuous at x₀ = 1

(ii)

Solution:

∴ f(x) is not continuous at x₀ = 3

Question 5.
Show that the function is continuous on (-∞, ∞)
Solution:

Given that f(1) = 3
∴ f(x) is continuous for all x ∈ R

Question 6.
For what value of α is this function f(x) = continuous at x = 1?
Solution:

∵ f(x) is continuous at x = 1, α = 4

Question 7.
Let Graph the function. Show that f(x) continuous on (-∞, ∞)
Solution:


∴ f(x) is continuous in (-∞, ∞)

Question 8.
If f and g are continuous function with f(3) = 5 and find g(3).
Solution:
Since f and g are continuous

2f(3) – g(3) = 4
2(5) – g(3) = 4
10 – 4 = g(3)
g(3) = 6

Question 9.
Find the points at which f is discontinuous. At which of these points f is continuous from the right, from the left, or neither? Sketch the graph of f.
(i)

∴ f(x) is not continuous at x = 1
Solution:

f(x) is not continuous at x = 1

(ii)

Solution:

∴ f(x) is not continuous at x = 0

Question 10.
A function f is defined as follows:

Is the function continuous?
Solution:

From (i), (ii) and (iii)
f(x) is continuous at x = 0, 1, 3

Question 11.
Which of the following functions f has removable discontinuity at x = x₀? If the discontinuity is removable, find a function g that agrees with f for x ≠ x₀ and is continuous on R.
(i) f(x) = \(\frac{x^{2}-2 x-8}{x+2}\), x₀ = -2
Solution:

(ii) f(x) = \(\frac{x^{3}+64}{x+4}\), x₀ = -4
Solution:

(iii) f(x) = \(\frac{3-\sqrt{x}}{9-x}\), x₀ = 9
Solution:

Question 12.
Find the constant b that makes g continuous on (-∞, ∞)

Solution:
Since g(x) is continuous,

Question 13.
Consider the function f(x) = x sin \(\frac{\pi}{x}\). What value must we give f(0) in order to make the function continuous everywhere?
Solution:

so to make the function f(x) is continuous at f(0) = 0

Question 14.
The function f(x) = \(\frac{x^{2}-1}{x^{3}-1}\) is not defined at x = 1. What value must we give f(1) in order to make f(x) continuous at x = 1?
Solution:

Question 15.
State how continuity is destroyed at x = x₀ for each of the following graphs.

Solution: