{"id":21925,"date":"2024-01-08T12:44:10","date_gmt":"2024-01-08T07:14:10","guid":{"rendered":"https:\/\/apboardsolutions.in\/?p=21925"},"modified":"2024-01-09T12:01:42","modified_gmt":"2024-01-09T06:31:42","slug":"inter-1st-year-maths-1a-functions-important-questions","status":"publish","type":"post","link":"https:\/\/apboardsolutions.in\/inter-1st-year-maths-1a-functions-important-questions\/","title":{"rendered":"Inter 1st Year Maths 1A Functions Important Questions"},"content":{"rendered":"

Students get through Maths 1A Important Questions<\/a> Inter 1st Year Maths 1A Functions Important Questions which are most likely to be asked in the exam.<\/p>\n

Intermediate 1st Year Maths 1A Functions Important Questions<\/h2>\n

I.
\nQuestion 1.
\nIf f: R – {0} \u2192 R is defined by f(x) = x + \\(\\frac{1}{x}\\), then prove that (f(x))2<\/sup> = f(x2<\/sup>) + f(1).
\nSolution:
\nf : R – {0} \u2192 R and
\nf(x) = x + \\(\\frac{1}{x}\\)
\nNow f(x2<\/sup>) + f(1) = (x2<\/sup> + \\(\\frac{1}{x^{2}}\\)) + (1 + \\(\\frac{1}{1}\\))
\n= x2<\/sup> + 2 + \\(\\frac{1}{x^{2}}\\) = (x + \\(\\frac{1}{x}\\))2<\/sup> = (f(x))2<\/sup>
\n\u2234 (f(x))2<\/sup> = f(x2<\/sup>) + f(1)<\/p>\n

\"Inter<\/p>\n

Question 2.
\nIf the function f is defined by
\n\"Inter
\nthen find the values, if exist, of f(4), f(2.5), f(-2), f(-4), f(0), f(-7).\u00a0 (Mar. ’14)
\nSolution:
\nDomain of f is(-\u221e, -3) \u222a [-2, 2] \u222a (3, \u221e)
\ni) Since f(x) = 3x – 2 for x > 3
\nf(4) = 3(4) – 2 = 10<\/p>\n

ii) 2.5 does not belong to domain off, hence
\nf(2.5) is not desired.<\/p>\n

iii) \u2235 f(x) = x2<\/sup> – 2 for -2 \u2264 x \u2264 2
\nf(-2) = (-2)2<\/sup> – 2 = 4 – 2 = 2<\/p>\n

iv) \u2235 f(x) = 2x + 1 for x < -3
\nf(-4) = 2(-4) + 1 = – 8 + 1 = -7<\/p>\n

v) \u2235 f(x) = x2<\/sup> – 2, for -2 \u2264 x \u2264 2
\nf(0)= (0)2<\/sup> – 2 = 0 – 2 = -2<\/p>\n

vi) \u2235 f(x) = 2x + 1 for x < -3
\nf(-7) = 2(-7) + 1 = -14 + 1 = -13<\/p>\n

Question 3.
\nIf A = {0, \\(\\frac{\\pi}{6}\\), \\(\\frac{\\pi}{4}\\), \\(\\frac{\\pi}{3}\\), \\(\\frac{\\pi}{2}\\)} and f : A \u2192 B is a surjection defined by f(x) = cos x, then find B.\u00a0 (A.P. Mar. ’16, ’11; May ’11)
\nSolution:
\n\u2235 f : A \u2192 B is a sujection defined by
\nf(x) = cos x
\nB = Range of f = f(A)
\n= {f(0), f(\\(\\frac{\\pi}{6}\\)), f(\\(\\frac{\\pi}{4}\\)), f(\\(\\frac{\\pi}{3}\\)), f(\\(\\frac{\\pi}{2}\\))}
\n= {cos 0, cos \\(\\frac{\\pi}{6}\\), cos \\(\\frac{\\pi}{4}\\), cos \\(\\frac{\\pi}{3}\\), cos \\(\\frac{\\pi}{2}\\)}
\n= {1, \\(\\frac{\\sqrt{3}}{2}\\), \\(\\frac{1}{\\sqrt{2}}\\), \\(\\frac{1}{2}\\), 0}<\/p>\n

Question 4.
\nDetermine whether the function f: R \u2192 R defined by f(x) = \\(\\frac{e^{|x|}-e^{-x}}{e^{x}+e^{-x}}\\) is an injection or a surjection or a bijection.
\nSolution:
\nf(x) = \\(\\frac{e^{|x|}-e^{-x}}{e^{x}+e^{-x}}\\)
\nf(0) = \\(\\frac{e^{0}-e^{-0}}{e^{0}+e^{-0}}\\) = \\(\\frac{1-1}{1+1}\\) = 0
\nf(-1) = \\(\\frac{e^{1}-e^{1}}{e^{-1}+e^{1}}\\) = 0
\n\u2235 f(0) = f(-1)
\n\u21d2 f is not an injection.
\nLet y = f(x) = \\(\\frac{e^{|x|}-e^{-x}}{e^{x}+e^{-x}}\\)
\nWhen y = 1, there is no x \u2208 R such that f(x) = 1
\n\u21d2 f is not a surjection
\nif there is such x \u2208 R then
\n\\(\\frac{e^{|x|}-e^{-x}}{e^{x}+e^{-x}}\\) = 1
\n\u21d2 e|x|<\/sup> – e-x<\/sup> = ex<\/sup> + e-x<\/sup>, clearly x \u2260 0
\nfor x > 0, this equation gives
\nex<\/sup> – e-x<\/sup> = ex<\/sup> + e-x<\/sup> \u21d2 -e-x<\/sup> = e-x<\/sup> which is not possible
\nfor x < 0, this equation gives
\ne-x<\/sup> – e-x<\/sup> = ex<\/sup> + e-x<\/sup>
\n\u21d2 e-x<\/sup> = ex<\/sup> which is also not possible.<\/p>\n

\"Inter<\/p>\n

Question 5.
\nDetermine whether the function f: R \u2192 R defined by
\n\"Inter
\nis an injection or a surjection or a bijection
\nSolution:
\n\u2235 f(x’) = x for x > 2
\n\u21d2 f(3) = 3
\n\u2235 f(x) = 5x – 2, for x \u2264 2
\n\u21d2 f(1) = 5(1) – 2 = 3
\n\u2235 1 and 3 have same f-image.
\nHence f is not an injection.
\nLet y \u2208 R then y > 2 (or) y \u2264 2.
\nIf y > 2 take x = y \u2208 R so that
\nf(x) = x = y
\nIf y \u2264 2 take x = \\(\\frac{y+2}{5}\\) \u2208 R and
\nx = \\(\\frac{y+2}{5}\\) < 1
\n\u2234 f(x) = 5x – 2 = 5\\(\\left[\\frac{y+2}{5}\\right]\\) – 2 = y
\n\u2234 f is a surjection
\n\u2235 f is not an injection \u21d2 It is not a bijection.<\/p>\n

Question 6.
\nFind the domain of definition of the function y(x), given by the equation 2x<\/sup> + 2y<\/sup> = 2.
\nSolution:
\n2x<\/sup> = 2 – 2y<\/sup> < 2 (\u2235 2y<\/sup> > 0)
\n\u21d2 log2<\/sub> 2x<\/sup> < log2<\/sub>2
\n\u21d2 x < 1
\n\u2234 Domain = (-\u221e, 1)<\/p>\n

Question 7.
\nIt f: R \u2192 R is defined as f(x + y) = f(x)+ f(y) \u2200 x, y \u2208R and f(1) = 7, then find \\(\\sum_{r=1}^{n} f(r)\\).
\nSolution:
\nConsider f(2) = f(1 + 1) = f (1) + f (1) = 2f(1)
\nf(3) =f(2 + 1) = f(2) + f(1) = 3f(1)
\nSimilarly f(r) = rf(1)
\n\u2234 \\(\\sum_{r=1}^{n} f(r)\\) = f(1) + f(2) + ….. + f(n)
\n= f(1) + 2f(1)+ ….. + nf(1)
\n= f(1) (1 + 2 + ….. + n)
\n= \\(\\frac{7 n(n+1)}{2}\\)<\/p>\n

Question 8.
\nIf f(x) = \\(\\frac{\\cos ^{2} x+\\sin ^{4} x}{\\sin ^{2} x+\\cos ^{4} x}\\) \u2200 x \u2208 R then show that f(2012) = 1.)
\nSolution:
\nf(x) = \\(\\frac{\\cos ^{2} x+\\sin ^{4} x}{\\sin ^{2} x+\\cos ^{4} x}\\)
\n= \\(\\frac{1-\\sin ^{2} x+\\sin ^{4} x}{1-\\cos ^{2} x+\\sin ^{4} x}\\)
\n\"Inter<\/p>\n

Question 9.
\nIf f : R \u2192 R, g : R \u2192 R are defined by f(x) = 4x – 1 and g(x) = x2<\/sup> + 2 then find
\ni) (gof)(x)
\nii) (gof)\\(\\left(\\frac{a+1}{4}\\right)\\)
\niii) (fof)(x)
\niv) go(fof)(0)\u00a0 (Mar. ’05)
\nSolution:
\nGiven f : R \u2192 R and g : R \u2192 R and
\nf(x) = 4x – 1 and g(x) = x2<\/sup> + 2<\/p>\n

(i) (gof) (x) = g (f(x))
\n= g (4x – 1), \u2235 f(x) = 4x – 1
\n= (4x – 1)2<\/sup> + 2, \u2235 g(x) = x2<\/sup> + 2
\n= 16x2<\/sup> – 8x + 1 + 2
\n= 16x2<\/sup> – 8x + 3<\/p>\n

(ii) (gof)\\(\\left(\\frac{a+1}{4}\\right)\\) = \\(g\\left(f\\left(\\frac{a+1}{4}\\right)\\right)\\)
\n= \\(g\\left(4\\left(\\frac{a+1}{4}\\right)-1\\right)\\)
\n= g(a)
\n= a2<\/sup> + 2<\/p>\n

(iii) (fof)(x) = f(f(x))
\n= f(4x – 1), \u2235 f(x) = 4x – 1
\n= 4(4x – 1) – 1
\n= 16x – 4 – 1 = 16x – 5<\/p>\n

\"Inter<\/p>\n

(iv) (fof)(0) = f(f(0))
\n= f(4 \u00d7 0 – 1)
\n= f(-1)
\n= 4(-1) – 1 = -5
\nNow go(fof)(0)
\n= g(-5) = (-5)2<\/sup> + 2 = 27<\/p>\n

Question 10.
\nIf f: [0, 3] \u2192 [0, 3] is defined by
\n\"Inter
\nthen show that f[0,3] \u2286 [0, 3] and find fof.
\nSolution:
\n0 \u2264 x \u2264 2 \u21d2 1 \u2264 1 + x \u2264 3 ——— (1)
\n2 < x \u2264 3 \u21d2 -3 \u2264 -x \u2264 -2
\n\u21d2 3 – 3 \u2264 3 – x \u2264 3 – 2
\n\u21d2 0 \u2264 3 – x < 1 ——- (2)
\nfrom (1) and (2).
\nf[0, 3] \u2286 [0, 3]
\nWhen 0 \u2264 x \u2264 1, we have
\n(fof) (x) = f(f(x))
\n= f(1 + x) = 1 + (1 + x) = 2 + x
\n[\u2235 1 \u2264 1 + x \u2264 2]
\nWhen 1 < x \u2264 2, we have
\n(fof) (x) = f(f(x))
\n= f(1 + x)
\n= 3 – (1 + x)
\n= 2 – x, [\u2235 2 < 1 + x \u2264 3]
\nWhen 2 < x \u2264 3, we have
\n(fof) (x) = f(f(x))
\n= f(3 – x)
\n= 1 + (3 – x)
\n= 4 – x, [\u2235 0 \u2264 3 – x < 1]
\n\"Inter<\/p>\n

Question 11.
\nIf f, g : R \u2192 R are defined
\n\"Inter
\nand \"Inter
\nthen find (fog)(\u03c0) + (gof)(e).
\nSolution:
\n(fog)(\u03c0) = f(g(\u03c0)) = f(0) = 0
\n(gof)(e) = g(f(e)) = g (1) = -1
\n\u2234 (fog)(\u03c0) + (gof) (e) = -1.<\/p>\n

Question 12.
\nLet A = {1, 2, 3), B = {a, b, c}, C = (p, q, r}
\nIf f: A \u2192 B, g: B \u2192 C are defined by
\nf = {(1, a), (2, c), (3, b)},
\ng = {(a, q), (b, r), (c, p)} then
\nshow that f-1<\/sup>og-1<\/sup> = (gof)-1<\/sup>
\nSolution:
\nGiven that
\nf = {(1, a), (2, c), (3, b)} and
\ng = {(a, q), (b, r), (c, p)}
\nthen g f = {(1, q), (2, p), (3, r)}
\n\u21d2 (gof)-1<\/sup> = {(q, 1), (p, 2), (r, 3)}
\n\u21d2 g-1<\/sup> = {(q, a), (r, b), (p, c)} and
\nf-1<\/sup> = {(a, 1), (c, 2), (b, 3)}
\nf-1<\/sup> g-1<\/sup> = {(q, 1), (r, 3), (p, 2)}
\n\u21d2 (gof)-1<\/sup> = f-1<\/sup>og-1<\/sup><\/p>\n

\"Inter<\/p>\n

Question 13.
\nIf f: Q \u2192 Q, is defined by f(x) = 5x + 4 for all x \u2208 Q, show that f is a bijection and find f-1<\/sup>.(A.P Mar. ’16, May ’12, ’05)
\nSolution:
\nLet x1<\/sub>, x2<\/sub> \u2208 Q
\nNow f(x1<\/sub>) = f(x2<\/sub>)
\n\u21d2 5x1<\/sub> + 4 = 5x2<\/sub> + 4
\n\u21d2 5x1<\/sub> = 5x2<\/sub>
\n\u21d2 x1<\/sub> = x2<\/sub>
\n\u2234 f is an injection.
\nLet y \u2208 Q then x = \\(\\frac{y-4}{5}\\) \u2208 Q and
\nf(x) = f\\(\\left(\\frac{y-4}{5}\\right)\\) = 5\\(\\left(\\frac{y-4}{5}\\right)\\) + 4 = y
\n\u2234 f is a surjection.
\nHence f is a bijection
\n\u2234 f-1<\/sup> : Q \u2192 Q is also bijection
\nWe have (fof-1<\/sup>)(x) = I(x)
\n\u21d2 f(f-1<\/sup>(x)) = x, \u2235 f(x) = 5x + 4
\n\u21d2 5f-1<\/sup>(x) + 4 = x
\n\u21d2 f-1<\/sup>(x) =\\(\\frac{x-4}{5}\\) for \u2200 x \u2208 Q<\/p>\n

Question 14.
\nFind the domains of the following real valued functions.<\/p>\n

i) f(x) = \\(\\frac{1}{6 x-x^{2}-5}\\)
\nSolution:
\nf(x) = \\(\\frac{1}{6 x-x^{2}-5}\\) = \\(\\frac{1}{(x-1)(5-x)}\\) \u2208 R
\n\u21d4 (x – 1) (5 – x) \u2260 0
\n\u21d4 x \u2260 1, 5
\n\u2234 Domain of f is R – {1, 5}<\/p>\n

ii) f(x) = \\(\\frac{1}{\\sqrt{x^{2}-a^{2}}}\\), (a > 0)\u00a0 (A.P.) (Mar. ’15)
\nSolution:
\nf(x) = \\(\\frac{1}{\\sqrt{x^{2}-a^{2}}}\\) \u2208 R
\n\u21d4 x2<\/sup> – a2<\/sup> > 0
\n\u21d4 (x + a)(x – a) > 0
\n\u21d4 x \u2208 (-\u221e, -a) \u222a (a, \u221e)
\n\u2234 Domain of f is (-\u221e, -a) \u222a (a, \u221e) = R – [-a, a]<\/p>\n

\"Inter<\/p>\n

iii) f(x) = \\(\\sqrt{(x+2)(x-3)}\\)
\nSolution:
\nf(x) = \\(\\sqrt{(x+2)(x-3)}\\) \u2208 R
\n\u21d4 (x + 2)(x – 3) \u2265 0
\n\u21d4 x \u2208 (-\u221e, -2) \u222a [3, \u221e)
\n\u2234 Domain of f is
\n(-\u221e, -2] \u222a [3, \u221e) = R – (-2, 3)<\/p>\n

iv) f(x) = \\(\\sqrt{(x-\\alpha)(\\beta-x)}\\) (0 < \u03b1 < \u03b2)
\nSolution:
\nf(x) = \\(\\sqrt{(x-\\alpha)(\\beta-x)}\\) \u2208R
\n\u21d4 (x – \u03b1) (\u03b2 – \u03b1) \u2265 0
\n\u21d4 \u03b1 \u2264 x \u2264 \u03b2 ; (\u2235 \u03b1 < \u03b2)
\n\u21d4 x \u2208 [\u03b1, \u03b2]
\n\u2234 Domain of f is [\u03b1, \u03b2]<\/p>\n

v) f(x) = \\(\\sqrt{2-x}\\) + \\(\\sqrt{1+x}\\)
\nSolution:
\nf(x) = \\(\\sqrt{2-x}\\) + \\(\\sqrt{1+x}\\) + x \u2208 R
\n\u21d4 2 – x \u2265 0 and 1 + x \u2265 0
\n\u21d4 2 \u2265 x and x \u2265 -1
\n\u21d4 -1 \u2264 x \u2264 2
\n\u21d4 x \u2208 [-1, 2]
\n\u2234 Domain of f is [-1, 2].<\/p>\n

vi) f(x) = \\(\\sqrt{x^{2}-1}\\) + \\(\\frac{1}{\\sqrt{x^{2}-3 x+2}}\\)
\nSolution:
\nf(x) = \\(\\sqrt{x^{2}-1}\\) + \\(\\frac{1}{\\sqrt{x^{2}-3 x+2}}\\) \u2208 R
\n\u21d4 x2<\/sup> – 1 \u2265 0 and x2<\/sup> – 3x + 2 > 0
\n\u21d4 (x + 1)(x – 1) \u2265 0 and (x – 1)(x – 2) > 0
\n\u21d4 x \u2208 (-\u221e, -1] \u222a [1, \u221e) and x \u2208 (-\u221e, 1)u(2, \u221e)
\n\u21d4 x \u2208 (R – (-1, 1)) \u2229 (R – [1, 2])
\n\u21d4 x \u2208 R – {(-1, 1) \u222a [1,2]}
\n\u21d4 x \u2208 R – (-1, 2]
\n\u21d4 x \u2208 (-\u221e, -1] \u222a (2, \u221e)
\n\u2234 Domain of f is (-\u221e, -1) \u222a (2, \u221e) = R – (-1, 2]<\/p>\n

vii) f(x) = \\(\\frac{1}{\\sqrt{|x|-x}}\\)
\nSolution:
\nf(x) = \\(\\frac{1}{\\sqrt{|x|-x}}\\) \u2208 R
\n\u21d4 |x| – x > 0
\n\u21d4 |x| > x
\n\u21d4 x \u2208 (-\u221e, 0)
\n\u2234 Domain of f is (-\u221e, 0)<\/p>\n

viii) f(x) = \\(\\sqrt{|x|-x}\\)
\nSolution:
\n\\(\\sqrt{|x|-x}\\) \u2208 R
\n\u21d4 |x| – x \u2265 0
\n\u21d4 |x| \u2265 x
\n\u21d4 x \u2208 R
\n\u2234 Domain of f is R<\/p>\n

Question 15.
\nIf f = ((4, 5), (5, 6), (6, -4)} and g = ((4, -4), (6, 5), (8, 5)} then find
\n(i) f + g
\n(ii) f – g
\n(iii) 2f + 4g
\n(iv) f + 4
\n(v) fg
\n(vi) \\(\\frac{\\mathbf{f}}{\\mathbf{g}}\\)
\n(vii) |f|
\n(viii) \\(\\sqrt{f}\\)
\n(ix) f2<\/sup>
\n(x) f3<\/sup>
\nSolution:
\nGiven that f = ((4, 5), (5, 6), (6, -4)}
\ng {(4, \u20144), (6, 5), (8, 5)}
\nDomain of f = {4, 5, 6} = A
\nDomain of g = (4, 6, 8} = B
\nDomain of f \u00b1 g = A \u2229 B = {4, 6}
\ni) f + g = {4, 5 + (-4), (6, -4 + 5)}
\n= {(4, 1), (6, 1)}<\/p>\n

ii) f – g = {(4, 5 – (-4)), (6, -4, -5)}
\n= {(4, 9), (6, -9)}<\/p>\n

iii) Domain of 2f = A = {4, 5, 6}
\nDomain of 4g = B = {4, 6, 8}
\nDomain of 2f + 4g = A \u2229 B = {4, 6}
\n\u2234 2f = {(4, 10), (5, 12), (6, -8)}
\n4g = {(4, \u201416), (6, 20), (8, 20)}
\n\u2234 2f + 4g = {(4, 10 + (-16), 6, -8 + 20)}
\n= {(4, -6), (6, 12)}<\/p>\n

iv) Domain of f + 4 = A= {4, 5, 6}
\nf + 4 ={4, 5 + 4), (5, 6 +4), (6, -4 + 4)}
\n= ((4, 9), (5, 10), (6, 0)}<\/p>\n

\"Inter<\/p>\n

v) Domain of fg = A \u2229 B = {4, 6}
\nfg = {(4, (5) (-4), (6, (-4) (5))}
\n= {(4, -20), (6, -20)}<\/p>\n

vi) Domain of \\(\\frac{f}{g}\\) = {4, 6}
\n\u2234 \\(\\frac{f}{g}\\) = {(4, \\(\\frac{-5}{4}\\)), (6, \\(\\frac{-4}{5}\\))}<\/p>\n

vii) Domain of |f| = {4, 5, 6}
\n\u2234 |f| = {(4, |5|), (5, |6|), (6, |-4|)}
\n= {(4, 5), (5, 6), (6, 4))<\/p>\n

viii) Domain of \\(\\sqrt{f}\\) = {4, 5}
\n\u2234 \\(\\sqrt{f}\\) = {(4, \\(\\sqrt{5}\\)), (5, \\(\\sqrt{6}\\))}<\/p>\n

ix) Domain of f2<\/sup> = {4, 5, 6} = A
\n\u2234 f2<\/sup> = ((4, (5)2<\/sup>) (5, (6)2<\/sup>, (6, (-4)2<\/sup>)}
\nf2<\/sup> = {(4, 25), (5, 36), (6, 16)}<\/p>\n

x) Domain of f3<\/sup> = A = {4, 5, 6}
\n\u2234 f3<\/sup> = {(4, 53<\/sup>), (5, 63<\/sup>), (6, (-4)3<\/sup>}
\n= {(4, 125) (5, 216) (6, -64)}<\/p>\n

Question 16.
\nFind the domains and ranges of the following real valued functions.
\ni) f(x) = \\(\\frac{2+x}{2-x}\\)
\nii) f(x) = \\(\\frac{x}{1+x^{2}}\\)
\niii) f(x) = \\(\\sqrt{9-x^{2}}\\)\u00a0 (A.P.)(Mar. ’15)
\nSolution:
\ni) f(x) = \\(\\frac{2+x}{2-x}\\) \u2208 R
\n\u21d4 2 – x \u2260 0 x \u21d4 x \u2208 R -{2}
\n\u2234 Domain of f is R – {2}
\nLet f(x) = \\(\\frac{y}{1}\\) = \\(\\frac{2+x}{2-x}\\).
\nApply componendo and dividendo rule
\n\u21d2 \\(\\frac{y+1}{y-1}\\) = \\(\\frac{(2+x)+(2-x)}{(2+x)-(2-x)}\\)
\n\u21d2 \\(\\frac{y+1}{y-1}\\) = \\(\\frac{4}{2 x}\\)
\n\u21d2 x = \\(\\frac{2(y-1)}{y+1}\\)
\nClearly x is not defined for y + 1 = 0
\n(i.e.,) y = -1
\n\u2234 Range of f is R – {-1}.<\/p>\n

ii) f(x) = \\(\\frac{x}{1+x^{2}}\\)
\nSolution:
\nf(x) = \\(\\frac{x}{1+x^{2}}\\) \u2208 R
\n\u2235 \u2200 x \u2208 R, x2<\/sup> + 1 \u2260 0
\nDomain of f is R
\nLet f(x) = y = \\(\\frac{x}{1+x^{2}}\\)
\n\u21d2 x2<\/sup>y – x + y = 0
\n\u21d2 x = \\(\\frac{-(-1) \\pm \\sqrt{1-4 y^{2}}}{y}\\) is a real number.
\niff 1 – 4y2<\/sup> \u2265 0; y \u2260 0
\n\u21d4 (1 – 2y)(1 + 2y) \u2265 0 and y \u2260 0
\n\u21d4 y \u2208 \\(\\left[-\\frac{1}{2} ; \\frac{1}{2}\\right]\\) – {0}
\nAlso x = 0 \u21d2 y = 0
\n\u2234 Range of f = \\(\\left[-\\frac{1}{2}, \\frac{1}{2}\\right]\\)<\/p>\n

iii) f(x) = \\(\\sqrt{9-x^{2}}\\)
\nSolution:
\nf(x) = \\(\\sqrt{9-x^{2}}\\) \u2208 R
\n\u21d4 9 – y2<\/sup> \u2265 0
\n\u21d4 x \u2208 [-3, 3]
\n\u2234 Domain of f is [-3, 3]
\nLet f(x) = y = \\(\\sqrt{9-x^{2}}\\)
\n\u21d2 x = \\(\\sqrt{9-y^{2}}\\) \u2208 R
\n\u21d4 9 – y2<\/sup> \u2265 0 \u21d4 (3 + y)(3 – y) \u2265 0
\n\u2234 -3 \u2264 y \u2264 3
\nBut f(x) attains only non negative values
\n\u2234 Range of f = [0, 3].<\/p>\n

\"Inter<\/p>\n

Question 17.
\nIf f(x) = x2<\/sup> and g(x) = |x|, find the following functions.
\ni) f + g
\nii) f – g
\niii) fg
\niv) 2f
\nv) f2<\/sup>
\nvi) f + 3
\nSolution:
\nGiven f(x) = x2<\/sup>
\n\"Inter
\nDomain of f = Domain of g = R
\nHence domain of all the functions (i) to (vi) is R
\n\"Inter
\niv) 2f(x) = 2 f(x) = 2x2<\/sup><\/p>\n

v) f2<\/sup>(x) = (f(x))2<\/sup> = (x2<\/sup>)2<\/sup> = x4<\/sup><\/p>\n

vi) (f + 3)(x) = f(x) + 3 = x2<\/sup> + 3.<\/p>\n

Question 18.
\nDetermine whether the following functions are even or odd.
\ni) f(x) = ax<\/sup> – a-x<\/sup> + sin x
\nSolution:
\nGiven f(x) = ax<\/sup> – a-x<\/sup> + sin x
\n\u2234 f(- x) = a-x<\/sup> – ax<\/sup> + sin (-x)
\n= a-x<\/sup> – ax<\/sup> – sin x
\n= – (ax<\/sup> – ax<\/sup> – sin x) = – f(x)
\n\u2234 f(x) is an odd function.<\/p>\n

ii) f(x) = x\\(\\left(\\frac{e^{x}-1}{e^{x}+1}\\right)\\)
\nSolution:
\nf(x) = x\\(\\left(\\frac{e^{x}-1}{e^{x}+1}\\right)\\)
\n\"Inter
\n\u2234 f is an even function.<\/p>\n

\"Inter<\/p>\n

iii) f(x) = log (x + \\(\\sqrt{x^{2}+1}\\))
\nSolution:
\nGiven f(x) = log (x + \\(\\sqrt{x^{2}+1}\\))
\nThen f(-x) = log (-x + \\(\\sqrt{(-x)^{2}+1}\\))
\n= log (\\(\\sqrt{x^{2}+1}\\) – x)
\n\"Inter
\n\u2234 f is an odd function.<\/p>\n

Question 19.
\nFind the domains of the following real valued functions.
\ni) f(x) = \\(\\frac{1}{\\sqrt{[x]^{2}-[x]-2}}\\)
\nSolution:
\nf(x) = \\(\\frac{1}{\\sqrt{[x]^{2}-[x]-2}}\\) \u2208 R
\n\u21d4 [x]2<\/sup> – [x] – 2 > 0
\n\u21d4 ([x] + 1) ([x] – 2) > 0
\n\u21d4 [x] < -1, (or) [x] > 2
\nBut [x] < -1 \u21d2 [x] = -2, -3, -4, ……..
\n\u21d2 x < -1 [x] > 2 \u21d2 [x] = 3, 4, 5, ……
\n\u21d2 x \u2265 3
\n\u2234 Domain of f = (-\u221e, -1) \u222a [3, \u221e]
\n= R – [-1, 3)<\/p>\n

ii) f(x) = log (x – [x])
\nf(x) = log (x – [x]) \u2208 R
\n\u21d4 x – [x] > 0
\n\u21d4 x > [x]
\n\u21d4 x is a non integer
\n\u2234 Domain of f is R – Z<\/p>\n

iii) f(x) = \\(\\sqrt{\\log _{10}\\left(\\frac{3-x}{x}\\right)}\\)
\nSolution:
\nf(x) = \\(\\sqrt{\\log _{10}\\left(\\frac{3-x}{x}\\right)}\\) \u2208 R
\n\u21d4 log10<\/sub>\\(\\left(\\frac{3-x}{x}\\right)\\) \u2265 0 and \\(\\frac{3-x}{x}\\) > 0
\n\u21d4 \\(\\frac{3-x}{x}\\) \u2265 10\u00b0 = 1 and 3 – x > 0, x > 0
\n\u21d4 3 – x \u2265 x and 0 < x < 3
\n\u21d4 x \u2264 \\(\\frac{3}{2}\\) and 0 < x < 3
\n\u21d4 x \u2208 (-\u221e, \\(\\frac{3}{2}\\)] \u2229 (0, 3) = (0, \\(\\frac{3}{2}\\)]
\n\u2234 Domain of f is (0, \\(\\frac{3}{2}\\)]<\/p>\n

iv) f(x) = \\(\\frac{\\sqrt{3+x}+\\sqrt{3-x}}{x}\\)
\nSolution:
\nf(x) = \\(\\frac{\\sqrt{3+x}+\\sqrt{3-x}}{x}\\) \u2208 R
\n\u21d4 3 + x \u2265 0, 3 – x \u2265 0 and x \u2260 0
\n\u21d4 x \u2265 -3, x \u2264 3 and x \u2260 0
\n\u21d4 -3 \u2264 x \u2264 3, and x \u2260 0
\n\u21d4 x \u2208 [-3, 3] and x \u2260 0
\n\u21d4 x \u2208 [-3, 3] – {0}
\n\u2234 Domain of f is [-3, 3] – {0}<\/p>\n

Question 20.
\nIf f : A \u2192 B and g : B \u2192 C are two injective functions then the mapping gof : A \u2192 C is an injection.
\nSolution:
\nf : A \u2192 B and g: B \u2192 C are one\u2014one.
\n\u2234 gof : A \u2192 C
\nTo prove that g o f is one \u2014 one function
\nLet a1<\/sub>, a2<\/sub> \u2208 A \u2234 f(a1<\/sub>), f(a2<\/sub>) \u2208 B and g(f(a1<\/sub>)), g(f(a2<\/sub>)) \u2208 C i.e., (gof) (a1<\/sub>), gof(a2<\/sub>) \u2208 C
\nNow (gof) (a1<\/sub>) = gof (a2<\/sub>)
\n\u21d2 g(f(a1<\/sub>)) = g(f(a2<\/sub>))
\n\u21d2 f(a1<\/sub>) = f(a2<\/sub>) (\u2235 g is one -one)
\n\u21d2 a1<\/sub> = a2<\/sub> (\u2235 f is one-one)
\nHence gof: A \u2192 C is a one-one function.
\nNote : The converse of the above theorem is not true.
\nIf f : A \u2192 B, g: B \u2192 C and go f is one-one then both f and g need not be one-one.
\nFor, consider
\nA = {1, 2}, B = {p, q, r), C = {s, t}
\nLet f = {(1, p), (2, q)} and
\ng = {(p, s), (q, t), (r, t)}
\nNow gof = {(1, s), (2, t)} \u21d2 gof is one\u2014one
\nfrom A to C
\nObserve that g: B \u2192 C is not one\u2014one.<\/p>\n

\"Inter<\/p>\n

Question 21.
\nIf f : A \u2192 B and g : B \u2192 C are two onto (surjective) functions then the mapping gof : A \u2192 C is a surjection. (May. ’08)
\nSolution:
\nf : A \u2192 B and g : B \u2192 C are onto.
\n\u2234 gof : A \u2192 C
\nTo prove that gof is onto. Let c be any element of C.
\nSince g : B \u2192 C is an onto function there exists an element b \u2208 B such that g(b) = c
\nSince f : A \u2192 B is an onto function, there exists an element a \u2208 A such that f(a) = b
\nNow g(b) = c \u21d2 g [f(a)] c = (gof) (a) = c
\nThus for any element c \u2208 C there is an element a \u2208 A such that (gof) (a) = c
\n\u2234 gof : A \u2192 C is a surjection<\/p>\n

Question 22.
\nIf f : A \u2192 B and g : B \u2192 C are two bijective functions then the mapping gof : A \u2192 C is a bijection. (Mar. ’16, May ’12)
\nSolution:
\nf and g are injections gof: A \u2192 C is an injection.
\nf and g are surjections= gof : A \u2192 C is a surjection.
\nHence it follows that if f and g are bijections,
\ngof is also a bijection.
\nNote : The converse of the above theorem is not true.<\/p>\n

Question 23.
\nIf f : A \u2192 B and g : B \u2192 C are such that gof is a surjection, then g is necessarily a surjection.
\nSolution:
\nLet c \u2208 C. Since gof is a surjection, from A to
\nC there exists an element a \u2208 A such that (gof) (a) = c, i.e., g(f(a)) = c
\nSince g: B \u2192 C and f(a) \u2208 B, \u2200 c \u2208 C there exists an element belonging to B.
\nHence g is a surjection.<\/p>\n

Question 24.
\nIf f : A \u2192 B and g : B \u2192 C and h : C \u2192 D are functions then ho(gof) = (hog)of (Mar. ’12, ’08)
\nSolution:
\nf : A \u2192 B and g : B \u2192 C gof : A \u2192 C
\nNow gof : A \u2192 C and h : C \u2192 D
\n\u21d2 ho(gof) : A \u2192 D
\nSimilarly (hog) of : A \u2192 D
\nThus ho (gof) and (hog) of both exist and have the same domain A and co-domain D.
\nLet a be an element of A.
\nNow [ho(gof)](a) = h[(gof)(a)] = h[g(f(a))]
\n= (hog) [f(a)] = [(hog)of] (a)
\n\u2234 ho(gof) = (hog)of
\nNote : Thus composition of mappings is associative.<\/p>\n

Question 25.
\nLet f : A \u2192 B, IA<\/sub> and IB<\/sub> are identity functions on A and B respectively. Then for IA<\/sub> = f = IB<\/sub> of Mar.\u201912, ’08
\nSolution:
\n\u2235 IA<\/sub> : A \u2192 A and f : A \u2192 B are functions foIA<\/sub> is a function from A to B. Hence foIA<\/sub> and f are definded on same domain A.
\ni) Let a \u2208 A then (foIA<\/sub>)(a) = f(IA<\/sub>(a))
\n= f(a)
\n[\u2235 IA<\/sub>(a) = a, \u2200 a \u2208 A]
\n\u2234 foIA<\/sub> = f ————- (1)<\/p>\n

ii) \u2235 f : A \u2192 B, IB<\/sub> : B \u2192 B are functions, IB<\/sub> of is a function from A to B
\n\u2234 The functions IB<\/sub> of and f are defined on the same domain A.
\nLet a \u2208 A, then (IB<\/sub> of)(a)
\n= IB<\/sub>(f (a)) = f(a)
\n\u2235 f : A \u2192 B, we have f(a) \u2208 B
\n\u2234 IB<\/sub> of = f ———– (2)
\nFrom(1) & (2) foIA<\/sub> = IBo<\/sub>f = f<\/p>\n

\"Inter<\/p>\n

Question 26.
\nLet A and B be two non-empty sets. If f : A \u2192 B is a bijection, then f-1<\/sup> : B \u2192 A is also a bijection.
\nSolution:
\nf : A \u2192 B is a bijection.
\n\u2234 f-1<\/sup> : B \u2192 A is a unique function.
\n(i) To prove that f-1<\/sup> is one\u2014one:
\nLet b1<\/sub> and b2<\/sub> be any two different elements of B. i.e., b1<\/sub> \u2260 b2<\/sub>.
\nThen, we have to prove that
\nf-1<\/sup>(b1<\/sub>) \u2260 F-1<\/sup>(b2<\/sub>)
\nLet f-1<\/sup>(b1<\/sub>) = a1<\/sub> and f-1<\/sup>(b2<\/sub>) = a2<\/sub>
\nsuch that a1<\/sub>, a2<\/sub> \u2208 A
\nThen b1<\/sub> = f(a1<\/sub>) and b2<\/sub> = f(a2<\/sub>)
\nNow b1<\/sub> \u2260 b2<\/sub> \u21d2 f(a1<\/sub>) \u2260 f(a2<\/sub>)
\n\u21d2 a1<\/sub> \u2260 a2<\/sub> (\u2235 f is a bijection)
\n\u21d2 f-1<\/sup>(b1<\/sub>) \u2260 f-1<\/sup>(b2<\/sub>)
\n\u2234 f-1<\/sup> is one\u2014one.<\/p>\n

(ii) To prove that f-1<\/sup> is onto:
\nLet \u2018a\u2019 be an element of A.
\nThen there exists an element b \u2208 B such that f(a) = b (or) f-1<\/sup>(b) = a
\n(or) a = f-1<\/sup>(b)
\nThus \u2018a\u2019 is the f-1<\/sup> – image of the element b \u2208 B
\nHence f-1<\/sup> is onto.
\n\u2234 f : B \u2192 A is a bijection.<\/p>\n

Question 27.
\nIf f : A \u2192 B is a bijection, then f-1<\/sup>of = IA<\/sub> and fof-1<\/sup> = IB<\/sub>
\n(A.P) (Mar. 15,12, ’07; May ’07, \u201806)
\nSolution:
\nf : A \u2192 B is a bijection \u21d2 f-1<\/sup> : B \u2192 A is also a bijection.
\nBy definition, fof-1<\/sup> : B \u2192 B and f-1<\/sup>of : A \u2192 A are bijection
\nAlso IA<\/sub> : A \u2192 A and IB<\/sub> : B \u2192 B
\nTo prove that f-1<\/sup>of = IA<\/sub>
\nLet a \u2208 A
\nSince f : A \u2192 B there exists a unique element
\nb \u2208 B such that f(a) = b
\na = f-1<\/sup> (b) (\u2235 f is a bijection)
\n\u2234 (f-1<\/sup>of) (a) = f-1<\/sup>[f(a)]
\n= f-1<\/sup>(b) = a = IA<\/sub> (a)
\nf-1<\/sup> of = IA<\/sub>
\nSimilarly it can be shown that fof-1<\/sup> = IB<\/sub><\/p>\n

Question 28.
\nIf f : A \u2192 B and g : B \u2192 A are two functions such that gof = IA<\/sub> and fog = IB<\/sub> then g = f-1<\/sup>.
\nSolution:
\n(i) To prove that f is one\u2014one:
\nLet a1<\/sub>, a2<\/sub> \u2208 A.
\nSince f: A \u2192 B, f(a1<\/sub>), f(a2<\/sub>) \u2208 B
\nNow f(a1<\/sub>) = f(a2<\/sub>)
\n\u21d2 g[f(a1<\/sub>)] = g[f(a2<\/sub>)]
\n\u21d2 (gof)(a1<\/sub>) = gof(a2<\/sub>)
\n\u21d2 IA<\/sub> (a1<\/sub>) = IA<\/sub> (a2<\/sub>)
\n\u2234 a1<\/sub> = a2<\/sub> = \u2234 f is one \u2014 one<\/p>\n

(ii) To prove that f is onto :
\nLet b be an element of B
\n\u2234 b = IB<\/sub> fog(b)
\n\u21d2 b = f{(g(b)} \u21d2 f{g(b)} = b
\ni.e., there exists a pre-image g(b) \u2208 A for b, under the mapping f. \u2234 f is onto
\nThus f is one-one, onto and hence
\nf-1<\/sup> : B \u2192 A exists and is also one \u2014 one and onto.<\/p>\n

(iii) To prove g = f-1<\/sup>:
\nNow g : B \u2192 A and f-1<\/sup> : B \u2192 A
\nLet a \u2208 A and b be the f – image of a, where b \u2208 B
\n\u2234 f(a) = b \u21d2 a =f-1<\/sup>(b)
\nNow g(b) = g[f(a)] (gof) (a)
\n= IA<\/sub>(a) = a = f-1<\/sup>(b)
\n\u2234 g = f-1<\/sup><\/p>\n

\"Inter<\/p>\n

Question 29.
\nIf f : A \u2192 B and g : B \u2192 C are bijective functions, then (gof)-1<\/sup> = f-1<\/sup>og-1<\/sup>(AP) (Mar. \u201816’14, \u201811; May \u201811)
\nSolution:
\nf : A \u2192 B, g : B \u2192 C are bijections
\n\u21d2 gof : A \u2192 C is a bijection
\nAlso g-1<\/sup> : C \u2192 B and f-1<\/sup> : B \u2192 A are bijections
\n\u21d2 f-1<\/sup>og-1<\/sup> : C \u2192 A is a bijection.
\nLet c be any element of C.
\nThen \u2203 an element b \u2208 B such that g(b) = c
\n\u21d2 b = g-1<\/sup>(c)
\nAlso \u2203 an element a A such that f(a) = b
\n\u21d2 a = f-1<\/sup>(b)
\nNow (gof) (a) = g(f(a) = g(b) = c
\n\u21d2 a = f-1<\/sup>(b)
\nNow (gof) (a) = g(f(a) = g(b) = c
\n\u21d2 a = (gof)-1<\/sup> (c) \u21d2 (gof) (c) = a
\nAlso (f-1<\/sup>og-1<\/sup>) (c) \u21d2 f-1<\/sup> (g-1<\/sup> (c)) = f-1<\/sup> (b) = a
\n\u2234 From (1) and (2);
\n(gof)-1<\/sup> (c) = (f-1<\/sup>og-1<\/sup>(c))
\n\u21d2 (gof)-1<\/sup> = f-1<\/sup>og-1<\/sup><\/p>\n","protected":false},"excerpt":{"rendered":"

Students get through Maths 1A Important Questions Inter 1st Year Maths 1A Functions Important Questions which are most likely to be asked in the exam. Intermediate 1st Year Maths 1A Functions Important Questions I. Question 1. If f: R – {0} \u2192 R is defined by f(x) = x + , then prove that (f(x))2 … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[14],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/21925"}],"collection":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/comments?post=21925"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/21925\/revisions"}],"predecessor-version":[{"id":21928,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/21925\/revisions\/21928"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=21925"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=21925"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=21925"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}