{"id":2315,"date":"2025-12-31T12:45:47","date_gmt":"2025-12-31T07:15:47","guid":{"rendered":"https:\/\/www.manabadi.co.in\/boards\/?p=2315"},"modified":"2026-01-03T11:34:58","modified_gmt":"2026-01-03T06:04:58","slug":"ts-inter-1st-year-maths-1a-solutions-cp-1-functions-ex-1a","status":"publish","type":"post","link":"https:\/\/www.manabadi.co.in\/boards\/ts-inter-1st-year-maths-1a-solutions-cp-1-functions-ex-1a\/","title":{"rendered":"TS Inter 1st Year Maths 1A Solutions Chapter 1 Functions Ex 1(a)"},"content":{"rendered":"\n

TS Inter 1st Year Maths 1A Functions Solutions Exercise 1(a)
I.
Question 1.
If the function f is defined by

then find the values of
(i) f(3)
(ii) f(0)
(iii) f(-1.5)
(iv) f(2) + f(- 2)
(v) f(- 5)
Answer:
(i) f(3), For x > 1; f(x) = x + 2
f(3) = 3 + 2 = 5
(ii) f(0), For \u2013 1 \u2264 x \u2264 1; f(0) = 2
(iii) f(-1.5), For \u2013 3 < x < \u2013 1; f(x) = x \u2013 1
\u2234 f(-1.5) = -1.5- 1 = \u2013 2.5
(iv) f(2) + f(-2); For x > 1, f(x) = x + 2
\u2234 f(2) = 2 + 2 = 4
For \u2013 3 < x < \u2013 1;
f(x) = x \u2013 1
f(-2) = -2 \u2013 1 = -3
f(2) + f (-2) = 4 \u2013 3 = 1
(v) f(-5); is not defined such domain of \u2018f\u2019 is {x \/ x > \u2013 3].
Question 2.
If f : R {0} \u2192 R defined by f(x) = x3<\/sup> \u2013 1\/x3<\/sup>, then show that f(x) + f(1\/x) = 0.
Answer:
Given f(x) = x3<\/sup> \u2013 1\/x3<\/sup>

Question 3.
If f: R \u2192 R defined by f(x) = 1\u2212x2<\/sup>\/1+x2<\/sup>, then show that f(tan \u03b8) = cos 2\u03b8
Answer:
Given f(x) = 1\u2212x2<\/sup>\/1+x2<\/sup> \u2200 x \u2208 R

= cos 2\u03b8
Question 4.
If f: R \u2013 (\u00b11) \u2192 R is defined by f(x) = log\u22231+x\/1\u2212x\u2223, then show that f(2x\/1+x2<\/sup>) = 2f(x).
Answer:
Given f: R \u2013 (\u00b11) \u2192 R defined by f(x) = log\u22231+x\/1\u2212x\u2223

Question 5.
If A = (-2, -1, 0, 1, 2) and f : A \u2192 B is a surjection defined by f(x) = x2<\/sup> + x + 1, then find B. (May 2014)
Answer:
A = {-2,-1,0,1,2} and f: A \u2192 B is a surjection and f(x) = x2<\/sup> + x + 1;
\u2234 f(-2) = (-2)2<\/sup> + (-2) + 1=3,
f(-1) = (-1)2<\/sup> + (-1) + 1 = 1
f(0) = 02<\/sup> + 0 + 1 = 1
f(1) =12<\/sup> + 1 + 1 = 3
f(2) = 22<\/sup> + 2 + 1 = 7
\u2234 B = f(A) = (1, 3, 7)
Question 6.
If A = {1, 2, 3, 4} and f: A \u2192 R is a function defined by f(x) = x2<\/sup>\u2212x+1\/x+1, then find the range of f.
Answer:
Given A = {1, 2, 3, 4} and f(x) = x2<\/sup>\u2212x+1\/x+1

Question 7.
If f (x + y) = f (xy) \u2200 x, y \u2208 R, then prove that f is a constant function.
Answer:
Given f (x + y) = f(x y) \u2200 x, y \u2208 R Suppose x = y = 0 then
f(0 + 0) = f(0 x 0)
\u21d2 f(0) = f(0) \u2026\u2026\u2026\u2026\u2026\u2026..(1)
Suppose x = 1, y = 0 then then f (1 + 0) = f(1 x 0)
\u21d2 f(D = f (0) \u2026\u2026\u2026\u2026\u2026(2)
Suppose x = 1, y = 1 then f (1 + 1) = f(1 x 1)
\u21d2 f(2) = f(1) \u2026\u2026\u2026\u2026\u2026.. (3)
f(0) = f(1) = f(2)
= f(0) = f(2)
Similarly f(3) = f(0), f(4) = f(0) \u2026\u2026\u2026\u2026. f(n) = f(0)
\u2234 f is a constant function.<\/p>\n\n\n\n

II.
Question 1.
If A = {x \/ \u2013 1 \u2264 x \u2264 11, f(x) = x2, g(x) = x3 Which of the following are surjections
(i) f : A \u2192 A
(ii) g : A \u2192 A.
Answer:
i) Given A {x \/ \u2013 1 \u2264 x \u2264 1}, f(x) = x2
and f : A \u2192 A
Suppose y \u2208 A
then x2 = y \u21d2 x = \u00b1 \u221ay
If x = \u221ay and if y = \u2013 1 then x = \u221a-1 \u2208 A
f : A \u2192 A is not a surjection.
ii) Given A = {x\/-1 \u2264 x \u2264 1), g(x) = x3
and g : A \u2192 A
Suppose ye A then x2 = y \u21d2 x = y\u221a3 \u2208 A
If y = -1 then x = -1 \u2208 A
y = 0 then x = 0 \u2208 A
y = 1 then x = 1 \u2208 A
g : A \u2192 A is a surjection.
Question 2.
Which of the following are injections or surjections or Bisections ? Justify your answers.
i) f : R \u2192 R defined by f(x) = 2x+1\/3
Answer:
Given f(x) = 2x+1\/3
Let a1<\/sub>, a2<\/sub> \u2208 R
\u2234 f(a1) = f(a2<\/sup>)
\u21d2 2a1+1\/3=2a2<\/sup>+1\/3
\u21d2 2a1 + 1 = 2a2<\/sup> + 1
\u21d2 a1 = a2<\/sup>
f(a1) = f(a2<\/sub>) \u21d2 a1<\/sub> = a2<\/sub> \u2200 a1<\/sub>, a2<\/sub> \u2208 R
f(x) = 2x+1\/3 is an injection.<\/p>\n\n\n\n

Suppose y \u2208 R (codomain of f) then
y = 2x+1\/3 \u21d2 x = 3y\u22121\/2
Then f(x) = f(3y\u22121\/2)=2(3y\u22121)\/2+1\/3 = y
f is a surjection f: R \u2192 R defined by f(x) = 2x+1\/3 is a bijection.<\/p>\n\n\n\n

ii) f : R \u2192 (0, \u221e) defined by f(x) = 2x
Answer:
Let a1, a2<\/sub> \u2208 R then f(a1) = f(a2)
\u21d2 2a1<\/sub> = 2a2<\/sub>
\u21d2 a1 = a2 \u2200 a1<\/sub>, a2<\/sub> \u2208 R
f(x) = 2x, f: R \u2192 (0, \u221e) is injection.
Let y \u2208 (0, \u221e) and y = 2x \u21d2 x = log2 y
then f(x) = 2x = 2 log2y = y
\u2234 f is a surjection.
Since f is injection and surjection, f is a bijection.
iii) f : (0, \u221e) \u2192 R defined by f(x) = logex.
Answer:
Let x1<\/sub>, x2<\/sub> \u2208 (0, \u221e)and = logex. then f(x1<\/sub>) = f(x2<\/sub>)
\u21d2 logex1 = logex2 \u21d2 x1<\/sub> = x2<\/sub>
\u2234 f(x1<\/sub>) = f(x2<\/sub>)
x1<\/sub> = x2<\/sub> and f is injection.<\/p>\n\n\n\n

Let y \u2208 R then y = logex \u21d2 x = ey
f(x) = logex = logeey = y and f is a surjection.
Since f is both injective and surjective, f is a bijection.<\/p>\n\n\n\n

iv) f : [0, \u221e) \u2192 [0, \u221e) defined by f(x) = x2<\/sup>
Answer:
Let x1<\/sub>, x1<\/sub> \u2208 [0, \u221e) given f(x) = x2<\/sup>
f(x1) = f(x2)
\u21d2 x1<\/sub> = x2<\/sub>
x1<\/sub> = x2<\/sub> (\u2235 x1<\/sub>, x2<\/sub> > 0)
f(x) = x2<\/sup>,
\u2234 f: [0, \u221e) \u2192 [0, \u221e) is an injection.<\/p>\n\n\n\n

Let y \u2208 [0, \u221e)then y = x2<\/sup> \u21d2 x = \u221ay (\u2235 y > 0)
f(x) = x2<\/sup> = (\u221ay)2 = y
and f is a surjection
\u2234 f is a bijection.<\/p>\n\n\n\n

v) f : R \u2192 [0, \u221e) defined by f(x) = x2<\/sub>
Answer:
Let x1<\/sub> x2<\/sub> \u2208 R and f(x) = x2<\/sup>
\u2234 f(x1<\/sub>) = f(x2<\/sub>)
\u21d2 x1<\/sub>2<\/sup> = x22<\/sup>
\u21d2 x1<\/sub> = \u00b1x2 (\u2235 x1<\/sub>, x2<\/sub> \u2208 R)
f is not an injection
Let y \u2208 [0, \u221e] then y = x2<\/sup> \u21d2 x = \u00b1\u221ay
where y \u2208 [0, \u221e] then f(x) = x2<\/sup> = (\u221ay )2<\/sup> = y.
\u2234 f is a surjection.
Since f is not injective and only surjective, we say that f is not a bijection.<\/p>\n\n\n\n

vi) f : R \u2192 R defined by f(x) = x2<\/sup>
Answer:
Let x1<\/sub> x2<\/sub> \u2208 R then f(x1) = f(x2<\/sup>)
\u21d2 x1<\/sub>2<\/sup> = x2<\/sub>2<\/sup>
\u21d2 x1<\/sub> = \u00b1 x2<\/sub> (\u2235 x1<\/sub>, x2<\/sub> \u2208 R)
f(x) is not an injection.
Let y \u2208 R then y = x2<\/sup>
\u21d2 x = \u00b1\u221ay
For elements that belong to (-\u221e, 0).
codomain R of f has no pre-image in f.
\u2234 f is not a surjection.
Hence f is not a bijection.<\/p>\n\n\n\n

Question 3.
If g = 1(1,1), (2, 3), (3, 5), (4, 7)) is a function from A = {1, 2, 3, 4} to B = {1, 3, 5, 7}. If this is given by the formula g(x) = ax + b then find a and b.
Answer:
A = {1, 2, 3, 4}, B = {1, 3, 5, 7}
g = {(1, 1), (2, 3), (3, 5), (4, 7)}
\u2235 g(1) = 1, g(2) = 3, g(3) = 5, g(4) = 7
Hence for an element a \u2208 A f \u2203 b \u2208 B such that g : A \u2192 B is a function.
Given g(x) = ax + b \u2200 x \u2208 A
g(1) = a + b = 1
g(2) = 2a + b = 3
solving a = 2, b = -1
Question 4.
If the function f : R \u2192 R defined by f(x) = 3x+3\u2212x\/2, then show that f (x+y) + f (x-y) = 2 f(x) f(y).
Answer:

Question 5.
If the function f : R \u2192 R defined by f(x) = 4x\/4x+2, then show that f (1 \u2013 x) = 1 \u2013 f(x) and hence reduce the value of f(14) + 2f(12) + f(34).
Answer:

Question 6.
If the function f : {-1, 1} \u2192 {0, 2} defined by f(x) = ax + b is a subjection, then find a and b.
Answer:
Since f: {-1, 1} \u2192 {0, 2} and f(x) = ax + b is a surjection.
Given f (-1) = 0, f (1) = 2 (or) f (-1) = 2, f (1)=0
Case I : f (-1) = 0, f (1) = 2
\u2234 \u2013 a + b = 0, a + b = 2
Solving b =1 , a = 1<\/p>\n\n\n\n

Case II : f (-1) = 2, and f (1) = 0
then \u2013 a + b = 2 and a + b = 0
Solving b = 1, a = -1
Hence a = + 1 and b = 1
Question 7.
If f(x) = cos (log x), then show that f(1\/x) f(1\/y) \u2013 (1\/2)[f(x\/y) + f(x\/y)] = 0
Answer:
Given f(x) = cos(log x)
then f(1\/x) = cos(log(1\/x))
= cos(-log x) = cos(log x) (\u2235 log 1 = 0)
Similarly f(1\/x) = cos(log y)
f(x\/y) = cos(log(xy)) = cos(log x \u2013 log y)
f(x\/y) = cos (log xy) = cos [log x + log y]
f(x\/y) + f(x y) = cos(log x \u2013 log y) + cos (log x + log y)
= 2 cos (log x) cos (log y) (\u2235 cos (A \u2013 B) + cos (A + B))
= 2 cos A cos B
f(1\/x) f(1\/y) \u2013 (1\/2)[f(x\/y) + f(x\/y)] = cos (log x) cos (log y) \u2013 1\/2 [2cos (log x) cos (logy)]
= 0<\/p>\n\n\n\n

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TS Inter 1st Year Maths 1A Functions Solutions Exercise 1(a)I.Question 1.If the function f is defined bythen find the values of(i) f(3)(ii) f(0)(iii) f(-1.5)(iv) f(2) + f(- 2)(v) f(- 5)Answer:(i) f(3), For x > 1; f(x) = x + 2f(3) = 3 + 2 = 5(ii) f(0), For \u2013 1 \u2264 x \u2264 1; f(0) 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