Maths 2A Important Questions<\/a> TS Inter Second Year Maths 2A Binomial Theorem Important Questions Long Answer Type to help strengthen their preparations for exams.<\/p>\nTS Inter Second Year Maths 2A Binomial Theorem Important Questions Long Answer Type<\/h2>\n
Question 1.
\nState and prove Binomial Theorem. [March ’09]
\nSolution:
\n<\/p>\n
<\/p>\n
<\/p>\n
<\/p>\n
Question 2.
\nFind the numerically greatest term in the expansion of (4 + 3x)15<\/sup> where x = \\(\\frac{7}{2}\\).
\nSolution:
\nGiven
\n(4 + 3x)15<\/sup>
\n\u21d2 415<\/sup> (1 + \\(\\frac{3 x}{4}\\))15<\/sup> = 415<\/sup> (1 + x)n<\/sup>
\nwhere n = 15; x = \\(\\frac{3 x}{4}\\)
\n\\(|x|=\\left|\\frac{3 x}{4}\\right|=\\left|\\frac{3}{4} \\cdot \\frac{7}{2}\\right|=\\frac{21}{8}\\)
\nNow, \\(\\frac{(n+1)|x|}{|x|+1}=\\frac{(15+1)\\left|\\frac{3 x}{4}\\right|}{\\left|\\frac{3 x}{4}\\right|+1}\\)
\n= \\(\\frac{16 \\cdot \\frac{21}{8}}{\\frac{21}{8}+1}=\\frac{336}{29}\\) = 11.5
\n\u2234 T12<\/sub> is numerically greatest.
\nr + 1 = 12
\n\u21d2 r = 11
\nThe general term in the expansion of (x + a)n<\/sup> is
\nTr + 1<\/sub> = \\({ }^n C_r\\) xn – r<\/sup> ar<\/sup>
\nT11 + 1<\/sub> = \\({ }^{15} \\mathrm{C}_{11}\\) (4)15 – 11<\/sup> (3x)11<\/sup>
\n12th term in this expansion is<\/p>\n<\/p>\n
Question 3.
\nFind the numerically reatest tenu in the expansion of (4a – 6b)13<\/sup> when a = 3, b = 5.
\nSolution:
\nGiven (4a – 6b)13<\/sup> = (4a)13<\/sup> (1 – \\(\\frac{6 b}{4 a}\\))13<\/sup>
\n= (4a)13<\/sup> (1 + x)n<\/sup>
\nwhere, x = \\(-\\frac{6 b}{4 a}\\), n = 13
\n|x| = \\(\\left|\\frac{-6 b}{4 a}\\right|=\\left|\\frac{-6 \\cdot 5}{4 \\cdot 3}\\right|=\\frac{5}{2}\\)
\nNow, \\(\\frac{(n+1)|x|}{|x|+1}=\\frac{(13+1) \\frac{5}{2}}{\\frac{5}{2}+1}\\)
\n= \\(\\frac{14 \\cdot 5}{5+2}=\\frac{70}{7}\\) = 10
\n\u2234 T10<\/sub> and T11<\/sub> are numerically greatest.
\nT10<\/sub>:
\nr + 1 = 10
\n\u21d2 r = 9
\nThe general term in the expansion of (1 + x)n<\/sup> is
\nTr + 1<\/sub> = \\({ }^{\\mathrm{n}} \\mathrm{C}_{\\mathrm{r}}\\) xn – r<\/sup> ar<\/sup>
\nT10<\/sub> in this expansion is
\nT9+1<\/sub> = \\({ }^{13} \\mathrm{C}_9\\) (4a)13-9<\/sup> (- 6b)9<\/sup>
\nT10 = \\({ }^{13} \\mathrm{C}_9\\) (4a)4<\/sup> (- 6b)9<\/sup>
\n= \\({ }^{13} \\mathrm{C}_9\\) (44<\/sup> . 34<\/sup>) ((- 6)9<\/sup> (5)9<\/sup>)
\n= \\({ }^{13} \\mathrm{C}_9\\) 124<\/sup> . 309<\/sup>
\n|T10<\/sub>| = \\({ }^{13} \\mathrm{C}_9\\) 124<\/sup> . 309<\/sup><\/p>\nT11<\/sub>:
\nr + 1 = 11
\n\u21d2 r = 10.
\nThe general term in the expansion of (1 – x)n<\/sup> is
\nTr + 1<\/sub> = \\({ }^{\\mathrm{n}} \\mathrm{C}_{\\mathrm{r}}\\) xn – r<\/sup> ar<\/sup>
\nT11<\/sub> in this expansion is
\nT10+1<\/sub> = \\({ }^{13} \\mathrm{C}_{10}\\) (4a)13-10<\/sup> . (- 6b)10<\/sup>
\n= \\({ }^{13} \\mathrm{C}_{10}\\) (4a)3<\/sup> (6b)10<\/sup>
\nT11<\/sub> = \\({ }^{13} \\mathrm{C}_{10}\\) (4 . 3)3<\/sup> (6 . 5)10<\/sup>
\n= \\({ }^{13} \\mathrm{C}_{10}\\) 123<\/sup> . 3010<\/sup>
\nT11<\/sub> = \\({ }^{13} \\mathrm{C}_{10}\\) 123<\/sup> . 3010<\/sup>
\n\u2234 |T10<\/sub>| = |T11<\/sub>|.<\/p>\n<\/p>\n
Question 4.
\nIf the coefficient of x10<\/sup> in the expansion of \\(\\left(a x^2+\\frac{1}{b x}\\right)^{11}\\) is equal to the coefficient of x-10<\/sup> in the expansion of \\(\\left(a x-\\frac{1}{b x^2}\\right)^{11}\\) find the relation between a and b where a and b are real numbers. [AP – May 2015, AP – Mar. 2019]
\nSolution:
\nCase – I :
\nGiven \\(\\left(a x^2+\\frac{1}{b x}\\right)^{11}\\)
\nHere x = ax2<\/sup>, a = \\(\\frac{1}{\\mathrm{bx}}\\); n = 11
\nNow, the general term in the expansion is
\nTr+1<\/sub> = \\({ }^n C_r\\) xn-r<\/sup> ar<\/sup>
\n= \\({ }^{11} \\mathrm{C}_{\\mathrm{r}}\\) a11-r<\/sup> x22-2r<\/sup> 1r<\/sup> b-r<\/sup> x-r<\/sup>
\n= \\({ }^{11} \\mathrm{C}_{\\mathrm{r}}\\) a11-r<\/sup> b-r<\/sup> x22-3r<\/sup> …………….(1)
\nTo find the coefficient of x10<\/sup>,
\nPut 22 – 3r = 10
\n3r = 12
\nr = 4
\nSubstituting r = 4 in equation (1) we get
\nT4+1<\/sub> = \\({ }^{11} C_4\\) a11-4<\/sup> b-4<\/sup> x22-12<\/sup>
\nT5<\/sub> = \\({ }^{11} C_4\\) a7<\/sup> b-4<\/sup> x10<\/sup>
\nT5<\/sub> = \\({ }^{11} C_4\\) a7<\/sup> b-4<\/sup> x10<\/sup>
\n\u2234 The coeff. of x10<\/sup> \u00a1n the expansion of \\(\\left(a x^2+\\frac{1}{b x}\\right)^{11}\\) is \\({ }^{11} C_4\\) a7<\/sup> b-4<\/sup><\/p>\nCase II:
\nGiven \\(\\left(a x-\\frac{1}{b x^2}\\right)^{11}\\)
\nHere x = ax, a = \\(\\left(\\frac{-1}{b x^2}\\right)\\), n = 11
\nNow, the general term in the expansion is
\nTr + 1<\/sub> = \\({ }^n C_r\\) xn-r<\/sup> ar<\/sup>
\n= \\({ }^{11} C_r(a x)^{11-r}\\left(\\frac{-1}{b x^2}\\right)^r\\)
\n= \\({ }^{11} \\mathrm{C}_{\\mathrm{r}}\\) a11-r<\/sup> x11-r<\/sup> (- 1)r<\/sup> b-r<\/sup> x-2r<\/sup>
\n= \\({ }^{11} \\mathrm{C}_{\\mathrm{r}}\\) (- 1)r<\/sup> b-r<\/sup> x11-3r<\/sup> ……………(2)
\nTo find the coefficient of x-10<\/sup>
\nput 11 – 3r = – 10
\n\u21d2 3r = 21
\n\u21d2 r = 7
\nSubstitute r = 7 in equation (2) we get
\nT7+1<\/sub> = \\({ }^{11} \\mathrm{C}_7\\) a11-7<\/sup> b7<\/sup> (- 1)7<\/sup> x11-21<\/sup>
\nT8<\/sub> = – \\({ }^{11} \\mathrm{C}_7\\) a4<\/sup> b-7<\/sup> x-10<\/sup>
\n\u2234 The coeff. of x-10<\/sup> in the expansion of \\(\\left(a x-\\frac{1}{b x^2}\\right)^{11}\\) is – \\({ }^{11} \\mathrm{C}_7\\) a4<\/sup> b-7<\/sup>
\nGiven that, these coefficients are equal
\n\\({ }^{11} \\mathrm{C}_7\\) a7<\/sup> b-4<\/sup> = – \\({ }^{11} \\mathrm{C}_7\\) a4<\/sup> b-7<\/sup>
\n\\({ }^{11} C_4 \\frac{a^7}{b^4}=-{ }^{11} C_4 \\frac{a^4}{b^7}\\)
\n\\(\\frac{a^7}{b^4}=\\frac{-a^4}{b^7}\\)
\n\u21d2 a3<\/sup>b3<\/sup> = – 1
\n\u21d2 (ab)3<\/sup> = (- 1)3<\/sup>
\n\u21d2 ab = – 1<\/p>\nQuestion 5.
\nIf n is a positive integer, then show that
\ni) C0<\/sub> + C1<\/sub> + C2<\/sub> + ……………… + Cn<\/sub> = 2
\nii) C0<\/sub> + C2<\/sub> + C4<\/sub> + ……………. = C1<\/sub> + C3<\/sub> + C5<\/sub> + ……….. = 2n-1<\/sup> [March ’97, ’90]
\nSolution:
\nWe know that
\n(1 + x)n<\/sup> = C0<\/sub> + C1<\/sub>x + C2<\/sub>x2<\/sup> + ……………… + Cn<\/sub>xn<\/sup> ………….(1)
\ni) Put x = 1 in equation (1)
\n\u21d2 (1 + 1)n<\/sup> = C0<\/sub> + C1<\/sub>(1) + C2<\/sub>(1)2<\/sup> + C3<\/sub>(1)3<\/sup> + C4<\/sub>(1)4<\/sup> + ……………… + Cn<\/sub>(1)n<\/sup>
\n2n<\/sup> = C0<\/sub> + C1<\/sub> + C2<\/sub> + C3<\/sub> + C4<\/sub> + …………. + Cn<\/sub>
\n\u2234 C0<\/sub> + C1<\/sub> + C2<\/sub> + C3<\/sub> + C4<\/sub> + …………. + Cn<\/sub> = 2n<\/sup><\/p>\nii) Put x = – 1 in equation (1), we get
\n(1 – 1)n<\/sup> = C0<\/sub> + C1<\/sub> (- 1) + C2<\/sub> (- 1)2<\/sup> + C3<\/sub> (- 1)3<\/sup> + ……………. + Cn<\/sub> (- 1)n<\/sup>
\no = C0<\/sub> – C1<\/sub> + C2<\/sub> – C3<\/sub> + C4<\/sub> – ………………
\nC0<\/sub> + C2<\/sub> + C4<\/sub> + …………………. = C1<\/sub> + C3<\/sub> + C5<\/sub> + ………….
\nSince a = b
\n\u21d2 a = b = \\(\\frac{a+b}{2}\\)
\nC0<\/sub> + C2<\/sub> + C4<\/sub> + …………………. = C1<\/sub> + C3<\/sub> + C5<\/sub> + ………….
\n= \\(\\frac{\\mathrm{C}_0+\\mathrm{C}_2+\\mathrm{C}_4+\\ldots \\ldots+\\mathrm{C}_1+\\mathrm{C}_3+\\mathrm{C}_5+\\ldots \\ldots}{2}\\)
\n= \\(\\frac{\\mathrm{C}_0+\\mathrm{C}_1+\\mathrm{C}_2+\\mathrm{C}_3+\\mathrm{C}_4+\\mathrm{C}_5+\\ldots \\ldots}{2}\\)
\n= \\(\\frac{2^{\\mathrm{n}}}{2}\\)
\n= 2n – 1<\/sup>
\n\u2234 C0<\/sub> + C2<\/sub> + C4<\/sub> + …………………. = C1<\/sub> + C3<\/sub> + C5<\/sub> + …………. = 2n – 1<\/sup>
\nC0<\/sub> + C2<\/sub> + C4<\/sub> + …………………. = 2n – 1<\/sup>
\nC1<\/sub> + C3<\/sub> + C5<\/sub> + …………………… = 2n – 1<\/sup><\/p>\n<\/p>\n
Question 6.
\nProve that for any real numbers a, d, a . C0<\/sub> + (a + d) . C1<\/sub> + (a + 2d) . C2<\/sub> + ……………… + (a + nd) . Cn<\/sub> = (2a + nd) 2n – 1<\/sup> [May \u201898]
\nSolution:
\nLet S = a . C0<\/sub> + (a + d) . C1<\/sub> + (a + 2d) . C2<\/sub> + ……………… + (a + nd) . Cn<\/sub> ………………(1)
\nBy writing the terms in R.H.S of (1) in reverse order has done we get
\nS = (a + nd)Cn<\/sub> + (a + (n – 1)d)Cn-1<\/sub> + (a + (n-2)d)Cn-2<\/sub> + …………. + aC0<\/sub>
\nS = (a + nd)C0<\/sub> + (a + (n – 1)d)C1<\/sub> + (a + (2n – 2)d)C2<\/sub> + ………….. + aCn<\/sub> ……………(2)
\nAdding (1) and (2) we get<\/p>\n<\/p>\n
2s = (2a + nd) (C0<\/sub> + C1<\/sub> + C2<\/sub> + …………. + Cn<\/sub>)
\n2s = (2a + nd)2n<\/sup>
\n\u21d2 S = (2a + nd)2n-1<\/sup>
\n\u2234 aC0<\/sub> + (a + d)C1<\/sub> + (a + 2d)C2<\/sub> + …………. + (a + nd)Cn<\/sub> = (2a + nd) 22n-1<\/sup>.<\/p>\nQuestion 7.
\nFor r = 0, 1, 2, ….., n, prove that C0<\/sub> . Cr<\/sub> + C1<\/sub> . Cr+1<\/sub> + C2<\/sub> . C