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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

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:

(∵ 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 fol_A l_Bof = f
Solution:
i) To prove that fol_A = f
Since f : A → B and ‘A : A → A, we have fol_A:
A → B defined on the same domain A

ii) To prove that l_Bof = f
Since f : A → B and l_B : B → B we have
l_Bof : A → B defined In the same domain A

Question 5.
If f : A → B is a bijective function, then prove that (i) fof^-1 = I_B (ii) f^-1of = I_A
Solution:
To prove that fof^-1 = I_B
Given f: A → B is a bijection then we have
f^-1 : B → A is also a bijection

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)

Question 7.
If f : A → B and g : B→ A are two functions such that gof = I_A and fog = I_B then g = f^-1
Solution:
i) To prove that f is one one.

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

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.

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) = 5^x
Solution:
Let y = 5^x = 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:

Question 12.
If the function of defined by

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, ∞)

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

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

∴ 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 2^x +2^y = 2.
Solution :

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)

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 :

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

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

then show that f [0, 3] ⊆ [0, 3] and find fof
Solution:

Question 19.
If f, g : R→ R are defined by

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

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.

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

(i) \(f(x)=\frac{1}{6 x-x^2-5}\)
Solution:

(ii) \(f(x)=\frac{1}{\sqrt{x^2-a^2}},(a>0)\)
Solution:

(iii) \(f(x)=\sqrt{(x+2)(x-3)}\)
Solution:
\(f(x)=\sqrt{(x+2)(x-3)} \in R\)

(iv) \(\mathbf{f}(\mathbf{x})=\sqrt{(\mathbf{x}-\alpha)(\beta-\mathbf{x})}, \quad(0<\alpha<\beta)\)
Solution:

(v) \(f(x)=\sqrt{2-x}+\sqrt{1+x}\)
Solution:

(vi) \(f(x)=\sqrt{x^2-1}+\frac{1}{\sqrt{x^2-3 x+2}}\)
Solution:

(vii) \(f(\mathbf{x})=\frac{1}{\sqrt{|\mathbf{x}|-\mathbf{x}}}\)
Solution:

(viii) \(\mathbf{f}(\mathbf{x})=\sqrt{|\mathbf{x}|-\mathbf{x}}\)
Solution:

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)}

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)}

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))

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:

(ii) \(f(x)=\frac{x}{1+x^2}\)
Solution:

(iii) \(f(x)=\sqrt{9-x^2}\)
Solution:


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:

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

(i) f(x) = a^x 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) = a^x a ^-x – sinx
= (a^x a ^-x + sin x) = – f(x)
∴ f is an odd function.

(ii) \(f(x)=x\left(\frac{e^x-1}{e^x+1}\right)\)
Solution :
Given \(f(x)=x\left(\frac{e^x-1}{e^x+1}\right)\)

(iii) f(x) = log \(\left(x+\sqrt{x^2+1}\right)\)
Solution :

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:

(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

(iii) \(f(x)=\sqrt{\log _{10}\left(\frac{3-x}{x}\right)}\)
Solution:

(iv) \(f(x)=\sqrt{x+2}+\frac{1}{\log _{10}(1-x)}\)
Solution:

(v) \(f(x)=\frac{\sqrt{3+x}+\sqrt{3-x}}{x}\)
Solution: