TS Inter 1st Year Maths 1A Functions Important Questions

Students must practice these TS Inter 1st Year Maths 1A Important Questions Chapter 1 Functions to help strengthen their preparations for exams.

TS Inter 1st Year Maths 1A Functions Important Questions

Very Short Answer Questions

Question 1.
If \(A=\left\{0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}\right\}\) surjection defined by f(x) = cos x, then find B
Solution:
f: A → B is a surjection
⇒ Codomain of B = Range f(A)
Given f(x) = cos x
TS Inter 1st Year Maths 1A Functions Important Questions 2

TS Inter 1st Year Maths 1A Functions Important Questions

Question 2.
find the domain of the real-valued function f(x)=\(\frac{1}{\log (2-x)}\)
Solution:
f(x)=\(\frac{1}{\log (2-x)}\) is defined for 2 – x > 0 and 2 – x ≠ 1
⇒ x-2<0 and 2-x
⇒ x<2 and x ≠ 1 ⇒ x ∈(-∞, 2)- {1}
∴ Domain of f = {x / x ∈ (-∞, 2)-{1}}

Question 3.
If f : A → B, g: B → C are two bijective functions, then prove that gof = A → C is also a bijective function.
Solution:
i) Given f, g are bijections, f, g are both one one and onto.
To prove that gof : A → C is one one:
TS Inter 1st Year Maths 1A Functions Important Questions 3
(∵ f : A → B is one one, and
g : B → C are one one)
∴ gof : A → C is one one

ii) To Prove that gof = A → C is onto:
Let C∈C; since g : B → C is on to ∀ c ∈ C ∃
b E B such that g(b) = c …………………….. (1)
Also f: A → B is on to for b∈B ∃ a∈A such that f (a) = b …………………….. (2)
∴ bc = g(b) = g[f(a)]
= (gof) (a)
Hence for c ∈ C, ∃ a ∈ A such that
gof : A → C is onto
Hence from the above two results
gof : A → C is a Bijection.

Question 4.
If f : A → B is a function and lA,IB are identity functions on A, B respectively then prove that folA lBof = f
Solution:
i) To prove that folA = f
Since f : A → B and ‘A : A → A, we have folA:
A → B defined on the same domain A
TS Inter 1st Year Maths 1A Functions Important Questions 4

ii) To prove that lBof = f
Since f : A → B and lB : B → B we have
lBof : A → B defined In the same domain A
TS Inter 1st Year Maths 1A Functions Important Questions 5

Question 5.
If f : A → B is a bijective function, then prove that (i) fof-1 = IB (ii) f-1of = IA
Solution:
To prove that fof-1 = IB
Given f: A → B is a bijection then we have
f-1 : B → A is also a bijection
TS Inter 1st Year Maths 1A Functions Important Questions 6

TS Inter 1st Year Maths 1A Functions Important Questions

Question 6.
lf f : A → B, g : B→C are two bijective functions then prove that (gof)-1 = f-1 og-1
Solution:
Given that f : A→ B and g: B → C are bijections
we have gof = A → C is a bijection.
∴ (gof)-1 : C → A is also a bijection.
Also since f : A→B and g : B → C are bijections then f-1: B → A and g-1: c → B are bijections and hence f-1og-1; c → A is also a bijection.
(gof)-1 and f-1og-1 are two functions defined or the same domain C.
let c E C and g : B → C is a bijection ∃ unique b E B. Such that g(b)=c ⇒ b=g-1 (c)
Also b ∈ B and f: A → B is a bijection, ∃ a unique a ∈ A such that f (a) b ⇒ a = f-1(b)
TS Inter 1st Year Maths 1A Functions Important Questions 7

Question 7.
If f : A → B and g : B→ A are two functions such that gof = IA and fog = IB then g = f-1
Solution:
i) To prove that f is one one.
TS Inter 1st Year Maths 1A Functions Important Questions 8
So these exists a reimage g(b) A for ‘b’ Under ‘f’
∴ f is onto.
Hence ‘f’ is one one, onto and hence a bijection.
∴ f : B → A exists and is also one one onto

TS Inter 1st Year Maths 1A Functions Important Questions

iii) To prove g = f-1
Now g : B → A and f-1: B → A
We have g and f-1 are del med in the same domain B.
Let a ∈ A and b be the f image of ‘a’ where b∈B.
TS Inter 1st Year Maths 1A Functions Important Questions 9

Question 8.
If f : A→ B, g : B→C and h : C→ Dare functions then ho (gof) = (hog) of
Solution :
Given f : A → B and g : B → C we have
gof : A→ C
Now gof: A → C and h: C → D we have
ho(gof) : A→ D, Also hog : B→D and f : A → B
We have (hog) of : A → D
Hence (hog) of and ho(gof) are defined in the same domain A.
Let a ∈ A then (ho(gof)] (a) = h [(gof) (a)]
= h [g [f(a)]
= (hog) [f(a)] = [(hog) of] (a)
∴ ho (gof) = (hog) of.

Question 9.
On what domain the functions f(x) = x2– 2x and g(x) = – x+6 are equal?
Solution:
f(x) = g(x)
x2– 2x = – x+6
= x2-x-6-0
= (x-3) (x+2) = 0 = x = -2,3
∴ f(x) and g(x) are equal on the domain { -2,3}

Question 10.
Find the inverse of the function f(x) = 5x
Solution:
Let y = 5x = f(x) then x = f-1(y)
Also x = log5y
∴ f1(y) = log5(y)
∴ f1(y) log5y ⇒ f-1(x) = log5x

Question 11.
If f : R – {o} → R is deflued by f(x) = x+\(\frac{1}{x}\)!,then prove that [f(x)]2 = f(x2) + f(1)
Solution:
TS Inter 1st Year Maths 1A Functions Important Questions 11

Question 12.
If the function of defined by
TS Inter 1st Year Maths 1A Functions Important Questions 12
then find the values If exist of f(4); f(2.5), f(-2), f(-4), f(0), f(-7)
Solution :
i) Since f(x)=3x-2 for x>3
f(4) = 3(4) – 2 = 10
Domain of f is
(- ∞, – 3)∪[-2,2] u(3, ∞)

TS Inter 1st Year Maths 1A Functions Important Questions

ii) f(2.5) does not exist since 2.5 does not belong to the domain of f.

iii) f(x) = x2– 2 for x [-2,2]
We have f(-2) = (2) -2 = 2

iv) f(x) = 2x + I for x < – 3
f( – 4)=2(- 4)+ 1= – 7

v)f(x)=x2 – 2 for x∈[-2,2] and f(0) = – 2.

vi) f(x) = 2x + 1, for x < – 3
f(-7) = 2(-7)+ 1 = – 14+ 1 = – 13

Question 13.
Determine whether the function f: R → R defined by
TS Inter 1st Year Maths 1A Functions Important Questions 17
is an injection or a surjection or a bijection?
Solution :
By definition of the function f(3) = 3
and f(1 )= 5(1) – 2 =3
∴ 1 and 3 have same f image
Hence f is not an injection.
Let y ∈ R then y>2 or y≤2
if y>2 take x = y ∈ R so that 1(x) = x = y
TS Inter 1st Year Maths 1A Functions Important Questions 13
∴ f is a surjection.
∴ Since f is not an injection it is not a bijection.

Question 14.
Find the domain of definition of the function y(x), given by the equation 2x +2y = 2.
Solution :
TS Inter 1st Year Maths 1A Functions Important Questions 14

Question 15.
If f : R→ R defined as f(x+y)=f(x)+f(y)∀x, y ∈ R and f(1) = 7, then find \(\sum_{r=1}^n f(r)\)
Solution :
Consider
f(2)=f(1+1) = f(1)+f(1)=2f(1)
f(3) = 1(2 + 1) f(2) + f(1) = 2f(1) + f(1) = 3f(1)
Sìmilarly f(r) = r f(1)
TS Inter 1st Year Maths 1A Functions Important Questions 15

Question 16.
If \(f(x)=\frac{\cos ^2 x+\sin ^4 x}{\sin ^2 x+\cos ^4 x}, \forall x \in R\) then show that f(2012) = 1.
Solution :
TS Inter 1st Year Maths 1A Functions Important Questions 16

TS Inter 1st Year Maths 1A Functions Important Questions

Question 17.
If f : R→ R, g: R → R defined by f(x) = 4x -1 and g(x)= x2+2 then find
(i) (gof) (x)
(ii) (gof) \(\left(\frac{a+1}{4}\right)\)
(iii) (fof) (x)
(iv) go (fof) (0)
Solution :
Given f(x) = 4x – 1 and g(x) = x2+2
Where f: R → R and g: R → R then

¡) (gof)(x) = g[f(x)] = g[4x – 1]
=(4x – 1 )2+2= 16 x 2- 8x+3

TS Inter 1st Year Maths 1A Functions Important Questions 18
iii) (fof) (x) = f [f(x)] = f[4x -1]
= 4 (4x – 1) -1 = 16x – 5

iv) [go (fof)] (0) = go [f(f(0)]
= go [f (-1)] = g[f(-1)]
= g[-5] = 25 + 2 = 27

Question 18.
If f : [0, 3] – [0, 3] is defined by
TS Inter 1st Year Maths 1A Functions Important Questions 19
then show that f [0, 3] ⊆ [0, 3] and find fof
Solution:
TS Inter 1st Year Maths 1A Functions Important Questions 20

Question 19.
If f, g : R→ R are defined by
TS Inter 1st Year Maths 1A Functions Important Questions 21
then find (fog)π + (gof) (e)
Solution:
We have g(π) = 0, and f(e) = 1
∴ (fog) π = f [g(π)] = f(0) = 0
(gof)(e)=g[f(e)]2g(1) = – 1
∴ (fog)(π)+(gof)(e) = 0 – 1 = – 1

TS Inter 1st Year Maths 1A Functions Important Questions

Question 20.
Let A = {1, 2,3} ,B = {a,b,c}, C = {p,q,r}.
If f : A+B, g : B → C are defined by
f= ((1, a), (2, c), (3, b))
g = ((a, q), (b, r), (c, p)) then show that f-1og-1 = (gof)-1
Solution:
Given f : A → Band g: B → C we have
f-1 {(a, 1), (c, 2), (b, 3)}
and g-1 = {(q, a), (r, b), (p, c)}
f-1og-1 {(q, 1), (r, 3), (p, 2)}
gof = {(1, q), (2, p), (3, r)}
(gof)-1 = ((q, 1), (p, 2), (r, 3))
∴ (gof)-1= f-1og-1

Question 21.
If f : Q → Q defined by f(x) = 5x + 4 ∀ x∈Q show that f is a bijection and find f-1.
Solution:
Let x1, x2 ∈ Q then f(x1) = f(x2)
⇒ 5x1 + 4 = 5x2 + 4
⇒ 5x1 = 5x2 ⇒ x1 = x2
∴ f is an injection.
TS Inter 1st Year Maths 1A Functions Important Questions 22

Question 22.
Find the domains of the following real-valued functions.

(i) \(f(x)=\frac{1}{6 x-x^2-5}\)
Solution:
TS Inter 1st Year Maths 1A Functions Important Questions 23

TS Inter 1st Year Maths 1A Functions Important Questions

(ii) \(f(x)=\frac{1}{\sqrt{x^2-a^2}},(a>0)\)
Solution:
TS Inter 1st Year Maths 1A Functions Important Questions 24

(iii) \(f(x)=\sqrt{(x+2)(x-3)}\)
Solution:
\(f(x)=\sqrt{(x+2)(x-3)} \in R\)
TS Inter 1st Year Maths 1A Functions Important Questions 25

(iv) \(\mathbf{f}(\mathbf{x})=\sqrt{(\mathbf{x}-\alpha)(\beta-\mathbf{x})}, \quad(0<\alpha<\beta)\)
Solution:
TS Inter 1st Year Maths 1A Functions Important Questions 26

(v) \(f(x)=\sqrt{2-x}+\sqrt{1+x}\)
Solution:
TS Inter 1st Year Maths 1A Functions Important Questions 27

(vi) \(f(x)=\sqrt{x^2-1}+\frac{1}{\sqrt{x^2-3 x+2}}\)
Solution:
TS Inter 1st Year Maths 1A Functions Important Questions 28

TS Inter 1st Year Maths 1A Functions Important Questions

(vii) \(f(\mathbf{x})=\frac{1}{\sqrt{|\mathbf{x}|-\mathbf{x}}}\)
Solution:
TS Inter 1st Year Maths 1A Functions Important Questions 29

(viii) \(\mathbf{f}(\mathbf{x})=\sqrt{|\mathbf{x}|-\mathbf{x}}\)
Solution:
TS Inter 1st Year Maths 1A Functions Important Questions 30

Question 23.
If f = {(4, 5), (5, 6), (6, – 4) and g = ((4, -4), (6, 5), (8, 5)} then find
i) f+g
ii) f – g
iii) 2f + 4g
iv) f+4
v) fg
vi) \(\frac{f}{g}\)
vii) \(|\mathbf{f}|\)
viii) \(\sqrt{f}\)
ix) f2
x) f3
Solution:
Given f {(4, 5), (5, 6), (6, – 4)] and g = ((4,-4), (6, 5), (8, 5)) then domain of f = {4, 5, 6) and Range of f = {4, 6, 8)
Domain of f ± g = A B = (4, 6)
= (domain of f) ∩ (domain of g)

i) f+g={(4,5,-4)(6,-4+5))
= {(4, 1), (6, 1)}

ii) f-g= {(4,5+4),(6,-4-5)}
= {(4,9), (6,-9)}

iii) Domain of 2f = {4, 5, 6}
Domain of 4g (4, 6, 8)
Domain of 2f + 4g = (4, 6)
∴2f = {(4, 10), (5, 12), (6, -8)}
4g = {(4, – 16), (6, 20), (8, 20)}
∴2f + 4g = {(4,-6),(6,12)}

TS Inter 1st Year Maths 1A Functions Important Questions

iv) Domain of f + 4 = {4,5, 6}
f + 4 = {(4, 9), (5, 10), (6, 0)}

v) Domain of fg = (domain of f) n (domain of g)
A∩B = {4, 6}
= {(4, (5) (-4), (6, (- 4), (5)}
= {(4, – 20), (6, – 20)}

TS Inter 1st Year Maths 1A Functions Important Questions 31

ix) Domain of f2 = (Domain of f(x)] (4, 5, 6)
∴ f2 = ((4, 25), (5, 36), (6, 16))

x) Domain of f3 = (4, 5, 6)
∴ f3 = ((4, 125), (5, 216), (6, -64))

TS Inter 1st Year Maths 1A Functions Important Questions

Question 24.
Find the domains and ranges of the following real valued functions.
(i) \(f(x)=\frac{2+x}{2-x}\)
(ii) \(f(x)=\frac{x}{1+x^2}\)
(iii)\(f(x)=\sqrt{9-x^2}\)
Solution:
TS Inter 1st Year Maths 1A Functions Important Questions 35
TS Inter 1st Year Maths 1A Functions Important Questions 33

(ii) \(f(x)=\frac{x}{1+x^2}\)
Solution:
TS Inter 1st Year Maths 1A Functions Important Questions 34

TS Inter 1st Year Maths 1A Functions Important Questions

(iii) \(f(x)=\sqrt{9-x^2}\)
Solution:
TS Inter 1st Year Maths 1A Functions Important Questions 35
TS Inter 1st Year Maths 1A Functions Important Questions 36
But f(x) posses only non negative values Range of f = [0, 3]

Question 25.
If f(x) = x2 and g(x) = I x find the following functions.
i) f+g
ii) f- g
iii) fg
iv) 2f
v) f2
vi) f+3
Solution:
TS Inter 1st Year Maths 1A Functions Important Questions 37

Question 26.
Determine whether the following functions are even or odd

(i) f(x) = ax a -x + sin x
Solution :
A function f is said to be even if [(-x) = f(x) and odd If f(-x) = – f(x)
f(x) = ax a -x – sinx
= (ax a -x + sin x) = – f(x)
∴ f is an odd function.

TS Inter 1st Year Maths 1A Functions Important Questions

(ii) \(f(x)=x\left(\frac{e^x-1}{e^x+1}\right)\)
Solution :
Given \(f(x)=x\left(\frac{e^x-1}{e^x+1}\right)\)
TS Inter 1st Year Maths 1A Functions Important Questions 38

(iii) f(x) = log \(\left(x+\sqrt{x^2+1}\right)\)
Solution :
TS Inter 1st Year Maths 1A Functions Important Questions 39

Question 27.
Find the domains of the following real-valued functions.

(i) \(f(\mathbf{x})=\frac{1}{\sqrt{[\mathbf{x}]^2-[\mathbf{x}]-2}}\)
Solution:
TS Inter 1st Year Maths 1A Functions Important Questions 40
TS Inter 1st Year Maths 1A Functions Important Questions 41

(ii) f(x) = log (x – [x])
Solution:
f(x) ∈ R
⇔ x – [x] >0 ⇔ x>[x]
⇔ x is not an integer.
∴ Domain of f is R – Z

TS Inter 1st Year Maths 1A Functions Important Questions

(iii) \(f(x)=\sqrt{\log _{10}\left(\frac{3-x}{x}\right)}\)
Solution:
TS Inter 1st Year Maths 1A Functions Important Questions 42

(iv) \(f(x)=\sqrt{x+2}+\frac{1}{\log _{10}(1-x)}\)
Solution:
TS Inter 1st Year Maths 1A Functions Important Questions 43

(v) \(f(x)=\frac{\sqrt{3+x}+\sqrt{3-x}}{x}\)
Solution:
TS Inter 1st Year Maths 1A Functions Important Questions 44

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