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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 99151. |
The nth term of an arithmetic sequence is an = 5 – 6n. Finditssumofnterms? |
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Answer» nth term = 5 - 6n 1st term = 5 - 6 x 1 = 5 - 6 = -1 \(S=\frac{n}{2}[f+l]=\frac{n}{2}[-1+5-6n]\) = \(\frac{n}{2}[4-6n]=n(2-3n)=2n-3n^2\) |
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| 99152. |
Consider the multiples of 7 in between 100 and 500. a. What are the first and last numbers? b. How many terms are there in the sequence? |
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Answer» a. First term = 100 – 2 + 7 = 105 Last term = 500 – 3 = 497 b. Common difference = xn = dn + (f - d) 497 = 7n + (105 - 7) \(n=\frac{(497-98)}{7}=57\) So, number of terms = 57 |
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| 99153. |
Write the terms of the sequence 5 × (1 + 6), 10 × (2 + 6), 15 × (3 + 6), 20 × (4 + 6) in the form : first term 5 × 1(1 + 6), second term 5 × 2(2 + 6). Write its algebraic expression |
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Answer» 5 x (1 + 6), 10 x (2 + 6), 15 x (3 + 6), 20 x (4 + 6), … this sequence can be written as 5 x 1(1 + 6), 5 x 2 (2 + 6), 5 x 3 (3 + 6), 5 x 4 (4 + 6). Algebraic expression = 5 x n (n + 6) i.e. xn = 5n2 + 30n |
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| 99154. |
Find the sum of first 24 terms of the list of numbers whose nth term is given by an = 3 + 2n. |
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Answer» xn = 3 + 2n Now, put n = 1,2,3,.... x1 = 3 + 2 x 1 = 5 x2 = 3 + 2 x 2 = 7 x3 = 3 + 2 x 3 = 9 Thus, the terms of the AP are 5, 7, 9 Here, f = 5 and d = 2 \(S_{24}=\frac{24}{2}[2\times5+(24-1)\times2]\) = 12/10 + 46l = 672 |
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| 99155. |
Look at the sequence 1 + (1 + 5), 2 + (2 + 5), 3 + (3 + 5) ……. a. Write next two terms b. Write its algbraic form. |
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Answer» a. Next two terms 4 + (4 + 5), 5 + (5 + 5), b. Algebraic expression xn = n + (n + 5) = n + n + 5 xn = 2n + 5 |
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| 99156. |
For the arithmetic sequence 6, 12, 18,……… a. What is the common difference? b. Find the 10th term? |
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Answer» a. Common difference = 6 b. 10th term = f + 9d = 6 + (9 × 6) = 6 + 54 = 60 |
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| 99157. |
a. Write the arithmetic sequence with first term 2 and common difference 3. b. Check whether 100 is a term in this sequence. c. Check whether the difference of any two terms of this sequence will be 2015. d. Find the position of the term 125 in this sequence. |
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Answer» a. f = 2 d. = 3 Sequence is 2, 5, 8, 11, ………. b. If (100 – 2) is not a multiple of common difference 3, then 100 is not a term of the arithmetic sequence. c. 2015 is not a multiple of common difference 3, so 2015 will not be the difference of any two terms of this sequence. d. xn = 3n – 1 3n-1 = 125 3n = 126 n = 42 |
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| 99158. |
Consider circles, points on its circumference and chords as shown in the figure. Mark two points on the circle and draw a chord. Mark one more point and draw three chords. Continue this process by adding one more point each time.a. Write the number of chords in each figure as a sequence.b. Write the algebraic expression of this sequence.c. Find the number of chords in the 10th figure. |
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Answer» a. No. of chords in figure 1 = 1 No. of chords in figure 2 = 1 + 2 = 3 No. of chords in figure 3 = 1+ 2 + 3 = 6 Sequence of number of chords = 1, 3, 6, 10, … b. No. of chords in figure n = 1 + 2 + 3 + .... + n = \(\frac{n(n+1)}{2}\) c. No. of chords in the 10th figure = \(\frac{10\times11}{2}\) = 55 |
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| 99159. |
Write the next two lines of the pattern above. Calculate the first and last terms of the 20th line. |
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Answer» 4, 7, 13, 22, 34, 49,……. (4, 4 + 3, 7 + 6, 13 + 9, 22 + 12, 34 + 15), Generally it written as 4 + 3(1 + 2 + 3 + ………..) First term in the fifth row = 4 + 3(1 + 2 + 3 + 4) = 4 + 30 = 34 First term in the sixth row = 4 + 3(1 + 2 + 3 + 4 + 5) = 4 + 45 = 49 Common differences in each row = 3 Fifth row 34, 37, 40, 43, 46,.. Sixth row 49, 52, 55, 58, 61, 64… 4 7 10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 55 58 61 64 …………………………. First term in the 20th row = 4 + 3(1 + 2 + 3 + 4 +……………. +19) = \(4+\frac{3\times19\times20}{2}=574\) Last term in the 20th row = 574 + 19 x 3 = 574 + 57 = 631 |
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| 99160. |
Consider the arithmetic sequence 171, 167, 163,……….. (i) Is ‘0’ is a term of this sequence? Why? (ii) How many positive terms are in this sequence? |
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Answer» Arithmetic senes : 171, 167, 163 ……………. i. Here common difference d = x2 - x1 = 167 - 171 = -4 Obtaining 0 as a term of this sequence 0 - 171 = -171 is a multiple of common difference -4. -171/-4 = 171/4 Quotient = 42, Remainder = 3 Remainder is not zero, so -171 is not a multiple of common difference -4. ∴ 0 will not be a term of the sequence ii. Remainder = 3, ∴ 0 + 3 = 3 is a term of the sequence. ∴ 3 is the last positive term of the sequence. ∴ Sum of positive terms of the sequence = \(\frac{x_n-x_1}{d}+1=\frac{3-171}{-4}+1\) = (-168/-4) + 1 = 42 + 1 = 43 |
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| 99161. |
The difference between 12th and 8th term of an arithmetic sequence is 20. Find the common difference. |
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Answer» 12th term = f + 11d 8th term = f + 7d Difference = f + 11d - (f + 7d) = 20 That is, 4d = 20, d = \(\frac{20}{4}=5\) Common difference (d) = 5 |
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| 99162. |
A pattern is formed using sticks of equal length as shown below:a. Write the sequence of number of sticks used in each figure.b. Write the sequence of number of squares and rectangles in each figure,c. Write the algebraic expression in the above two sequences.d. Find the number of sticks and squares in the 10 figure. |
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Answer» a. No of sticks in figure 1 = 1 + 3 = 4 No of sticks in figure 2 = 1 + 3 + 3 = 1+2 × 3 =7 No of sticks in the figure 3 = 1 + 3 × 3 = 10 No of sticks in the figure 4 = 1 + 4 × 3 = 13 Sequence of number of sticks = 4, 7, 10, 13, … (i) b. No of squares and rectangles in the figure 1 = 1 No of squares and rectangles in the figure 2 = 2 + 1 = 3 No of squares and rectangles in the figure 3 =3 + 2+ 1 = 6 No ofsquares and rectangles in the figure 4 = 4 + 3 + 2 + 1 = 10 Sequence of squares and rectangles = 1, 3, 6, 10, …. c. No of sticks in the nth figure 1 = 1+ n × 3 = 3n + 1 No of squares and rectangles in the nth figure = 1 + 2 + 3..... +n = \(\frac{n(n+1)}{2}\) d. No of sticks in the 10th figure = 3 × 10 + 1 = 31 No of squares and rectangles in the 10th figure = \(\frac{10\times11}{2}\) = 55 = 55 |
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| 99163. |
The tenth term of an arithmetic sequence is 65 and its 15th term is 80. Is 200 a term of this sequence? |
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Answer» To get the 15th term from the 10th term, we must add the common difference 5 times. 5 times of common difference = 80 – 65 = 15 Common difference (d) = 15/5 = 3 Dividing 65 by 3 we get the remainder as 2. Dividing 200 by 3 we get the remainder as 2. Therefore 200 is a term of this sequence. |
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| 99164. |
Consider an arithmetic sequence with common difference 6 and 7th term 52. Find the 15th term of the arithmetic sequence. Is it possible, to get a difference of 100 between any two terms of this sequence? |
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Answer» 15th term can be obtained by adding 8 times the common difference to the 7th term. x15 = x7 + 8d = 52 + 8 × 6 = 100 The difference between any two terms of an Arithmetic sequence will be a multiple of com-mon difference. 100 can’t be the difference be-tween any two terms of this sequence, since it is not a multiple of 6. |
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| 99165. |
If ten times tenth term of an arithmetic sequence is equal to fifteen times fifteenth term, find 25th term. Calculate the product of first 25th terms. |
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Answer» 10 × (f + 9d) = 15 × ( f + 14 d) ⇒ 10 f + 90 d = 15 f + 210 d ⇒ 5 f + 120 d = 0 ⇒ f + 24 d = 0 25th term = f + 24 d = 0 25th term is 0. Product of 25 terms is 0. |
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| 99166. |
If `f(x) = lim_(n->oo) sum_(r=0)^n (tan(x/2^(r+1)) + tan^3 (x/2^(r+1)))/(1- tan^2 (x/2^(r+1)))` then `lim_(x->0) f(x)/x` is |
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Answer» Correct Answer - 1 We have `tanA-tan A/2=(2tanA/2)/(1-tan^(2)A/2)-tanA/2` `=(tanA/2+tan^(3)A/2)/(1-tan^(2)A/2)` `f(x)=lim_(nrarroo)sum_(r=0)^(n)(tan x/(2^(r))=tanx/(2^(r+1)))=tanximplieslim_(xrarr0)(f(x))/x=1` |
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| 99167. |
`lim_(nrarroo)(1/(n+1)+1/(n+2)+…….+1/(6n))=kln6`, then find the value of `k` |
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Answer» Correct Answer - 1 `lim_(nrarroo)(1/(n+1)+1/(n+2)+…+1/(6n))` `=lim_(nrarroo)sum_(r=1)^(5n)1/(n+r)=lim_(rarroo)sum_(r=1)^(5n)1/n. 1/((1+r/n))=int_(0)^(5)1/(1+x)dx` `=[ln(1+x)]_(0)^(5)=ln6=ln1=ln6` |
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| 99168. |
If `e^(x)=(sqrt(1+z)-sqrt(1-z))/(sqrt(1+z)+sqrt(1-z))` and `tan (y/2)=sqrt((1-z)/(1+z))` then the value of `(dy)/(dx)` at `z=1` is equal to toA. `-2`B. `-1`C. 0D. 1 |
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Answer» Correct Answer - A `"Using"z=costheta," "atz=1rArrtheta=0^(@)` Now, `e^(x)=((costheta)/(2)-"sin"(theta)/(2))/((costheta)/(2)+"sin"(theta)/(2))=((pi)/(4)-(theta)/(2))` `:.x=lntan((pi)/(4)-(theta)/(2))` `(dx)/(d theta)=(1)/("tan"((pi)/(4)-(theta)/(2)))sec^(2)((pi)/(4)-(theta)/(2)).((-1)/(2))rArr(dx)/(d theta)=-sectheta` Also, `"tan"(y)/(2)="tan"(theta)/(2)rArr(y)/(2)=tan^(-1)("tan"(theta)/(2))rArr(y)/(2)=(theta)/(2)rArr(dy)/(d theta)=1` Hence `(dy)/(dx)=-costheta" at "theta=0^(@)rArr(dy)/(dx)=-1`. |
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| 99169. |
If `S_(n)=(3)/(1^(2)+2^(2))+(7)/(1^(2)+2^(2)+3^(2))+. . . . . . .. . . . . .` upto n terms, then value of `Lim_(ntooo) S_(n)` is equal toA. `(1)/(2)`B. 1C. 5D. 6 |
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Answer» Correct Answer - A `S_(n)=underset(r=1)overset(n)(sum)((2r+1)/(1^(2)+2^(2)+ . . . . . +r^(2)))=underset(r=1)overset(n)(sum)(6(2r+1))/(r(r+1)(2r+1))=6underset(r=1)overset(n)(sum)((1)/(r)-(1)/(r+1))=(6n)/(n+1)` `:." "underset(ntooo)LimS_(n)=underset(ntooo)Lim(n(6))/(n(1+(1)/(n)))=6` |
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| 99170. |
Given `lim_(n->oo)((3nC_n)/(2nC_n))^(1/n) =a/b` where a and b are relatively prime, find the value of `(a + b)`. |
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Answer» Correct Answer - 5 `A/B=lim_(nrarroo)(((2n+1)(2n+2)(2n+3)……(2n+n))/((n+1)(n+2)(n+3)…(n+n)))^(1//n)` `In(A/B)=lim_(nrarroo)1/n sum_(r=1)^(n)In((2+r/n)/(1+r//n))` `int(0)^(1)In(((2+x))/((1+x)))dx="ln" 27/16` `A=27` & `B=16` |
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| 99171. |
Let `f(x)= cosec 2x + cosec 2^2 x+ cosec 2^3 x+........+ cosec 2^n ,x in (0,pi/2) and g(x)=f(x)+cot 2^n x` . If `H(x)={ (cosx)^(g(x))+(sec)^(cisecx) if x lt 0 and p if x=0 and (e^x+e^(-x)-2cosx)/(x sin x) If x lt o` .Find the value of p, if possible to make the functieIf `H(x)` continuous at `x = 0`.A. `(1)/(2)`B. 1C. 2D. 0 |
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Answer» Correct Answer - A `f(x)="cosec"2x+"cosec"2^(2)x+ . . . . . . .+"cosec"2^(n)x` now cosec `2x=(1)/(sin2x)=(sin(2x-x))/(sinxsin2x)=cotx-cot2x` `|||ly" cosec"2^(2)x=cot2x-cot2^(2)x` `"cosec"2^(3)x=cot2^(2)x-cot2^(2)x-cot2^(3)x` `vdots` ` ul("cosec"2^(n)x=cot2^(n-1)x-cot2^(n)x)` `:.f(x)=cotx-cot2^(n)x` `:.g(x)=f(x)+cot2^(n)x=cotx` now `H(0+h)=underset(hto0)Lim((cos""h)^(cot""h)+(sec""h)^("cosec h"))=e^(overset(Lim coth (cosh-1))overset(hto0" ")" "+=e^(overset(Lim cosech(sech-1))overset(hto0" ")` =1+1=2 . . . . (1) `H(0-h)=underset(hto0)Lim(e^(-h)+e^(h)-2cosh)/(hsin""h)=underset(hto0)Lim{(e^(h)+e^(-h)-2)/(h^(2))+(2(1-cosh))/(h^(2))}=2` From (1) and (2) h(x) will be cont. if p=2 |
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| 99172. |
If `underset(0)overset(oo)(f)(dx)/((x+sqrt(1+x^(2)))^(5))=(m)/(n)` where m and relatively prime prime, then the value of (m+n) isA. 31B. 30C. 29D. 28 |
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Answer» Correct Answer - A Let `x+sqrt(1+x^(2))=t` `+x^(2)=t^(2)+x^(2)-2txrArr1=t^(2)-2txrArrx=(1)/(2)(t-(1)/(t))rArrdx=(1)/(2)(1+(1)/(t^(2)))dt` Now, `overset(oo)underset(1)(f)(1)/(t^(5))(1)/(2)(1+(t)/(t^(2)))dt=(1)/(2)overset(oo)underset(1)(f)((1)/(t^(5))+(1)/(t^(4)))dt` `I=(-1)/(2)((1)/(4t^(4))+(1)/(6t^(6)))underset(1)overset(oo)| ` `I=(5)/(24)=(m)/(n)` (given) So, (m+n)=29. |
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| 99173. |
The value of `Lim_(xto0) (5x^(2)+[x^(2)+1])^(((1)/(x^(2)+sin^(2)x)))` is, [Note : [y] denotes greatest interger less than or equal to y.]A. `(5)/(2)`B. `e^((5)/(2))`C. `(25)/(2)`D. `e^((25)/(2))` |
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Answer» Correct Answer - A `underset(xto0)Lim(5x^(2)+[x^(2)+1])^(((1)/(x^(2)+sin^(2)x)))=e^(L)` where `L=underset(xto0)Lim(1)/(x^(2)+sin^(2)x)(5x^(2)+[x^(2)+1]-1)=underset(xto0)Lim(5x^(2))/(x^(2)+sin^(2)x)=(5)/(2)` `:." Limit"=e^((5)/(2))`. |
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| 99174. |
42. If \( A=\{x \in N: 3 |
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Answer» Correct option is (b) A \(\cap \) B = {4, 6, 8, 10} A = {x \(\in N \) : 3 < x < 12} ∴ A = {4, 5, 6, 7, 8, 9, 10, 11} and B = {x \(\in N \) : x is even & x < 15} ∴ B = {2, 4, 6, 8, 10, 12, 14} Then A \(\cup\) B = {2, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14} A \(\cap \) B = {4, 6, 8, 10} A/B = A - B = \(\phi\) B/A = B - A = {2, 12, 14} |
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| 99175. |
The principal value of tan-1(-√3) is(A) 2π/3(B) 4π/3(C) -(π/3)(D) None of these |
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Answer» Correct option: (C) -(π/3) |
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| 99176. |
The police were able to trace the criminal because he left his ________ on the door handle. A) fingerprints B) finger-nails C) fingertips D) ringer ends E) finger-marks |
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Answer» Correct option is A) fingerprints |
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| 99177. |
If y = ax, then d2y/dx2 is equal to which of the following(A) ax loga (B) ax . (loga)2 (C) (ax)2 · loga (D) None of these |
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Answer» Correct option: (B) ax . (loga)2 |
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| 99178. |
The price of this TV set is not listed in the ________. A) catalogue B) guidebook C) directory D) record E) schedule |
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Answer» Correct option is A) catalogue |
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| 99179. |
What is BCNF (Boyce-Codd Normal Form)? |
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Answer» A relation schema R is in BCNF if it is in 3NF and satisfies an additional constraint that for every FD X A, X must be a candidate key. |
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| 99180. |
The rate of change of area of a cricle with respect to its radius r at r = 6 cm is(A) 12 π (B) 11 π(C) 10 π(D) 8 π |
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Answer» Correct option: (A) 12 π |
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| 99181. |
A: What ____ ?B: It’s November 22, 1984.A) the date isB) dayC) is the dateD) day is today |
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Answer» Correct option is C) is the date |
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| 99182. |
Write the first three terms of the sequence an = (-1)n-1 5n+1 |
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Answer» y2/4 - x2/9 = 1 ⇒ a2 = 4, b2 = 9 ⇒ a = 4, b = 3 c = √(a2 + b2) = √(4 + 9) = √13 The eccentricity e = c/a = √13/2 |
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| 99183. |
Find the slope of the line x/3 + y/2 = 1 |
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Answer» 2x + 3y = 6, m = - a/b = - 2/3 |
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| 99184. |
Write the negative of “For every real number x, x is less than x + 1.” |
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Answer» There exist real number x, x is not less than x + 1. |
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| 99185. |
If U = {x : x ≤ 10, x ∈N} A = {x : x ∈ N and x is prime}, B = {x : x ∈ N. x is event} find A ∩ B im roaster form. |
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Answer» U = { 1, 2, 3, 4, …………….. 10} A = {2, 3, 5, 7} B = {2, 4, 6, 8, 10}, A ∩ B = {2} |
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| 99186. |
Differentiate sin-1 (3x - 4x3) w.r.t x |
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Answer» assuming the given function as y= sin^-1( 3x- 4x^3) Let x =sinu => dx/du = cosu then y = sin^-1(3sinu-4sin^3u)=sin^-1(sin 3u) = 3u dy/du= 3 So dy/dx = (dy/du)/(dx/du)=3/cosu= 3/[(1-sin^2u)^1/2]=3/[(1-x^2)^1/2] let y = sin-1(3x - 4x3) let x = sinθ Then 3x - 4x3 = 3sinθ - 4sin3θ = sin3θ Therefore, y = sin-1(3x - 4x3) =sin-1(sin3θ) = 3θ Now, \(\frac{dy}{dx}\) = \(\frac{dy}{dθ}\) . \(\frac{dθ}{dx}\) = 3 \(\frac{dθ}{dx}\) (∵ y = 3θ) ∵ x = sinθ ⇒ \(\frac{dx}{dθ}\) = cosθ (By differentiating w.r.t θ) ⇒\(\frac{dθ}{dx}\) = \(\frac{1}{cosθ}\) = \(\frac{1}{\sqrt{1-sin^2θ}}\) = \(\frac{1}{\sqrt{1-x^2}}\) (∴ siinθ = x) Therefore \(\frac{dy}{dx}\) = \(3\times\frac{1}{\sqrt{1-x^2}}\) (∴\(\frac{dθ}{dx}\) = \(\frac{1}{\sqrt{1-x^2}}\)) ∵ \(\frac{d}{dx}\) sin-1(3x - 4x3) = \(\frac{3}{\sqrt{1-x^2}}\) (∵ = y = sin-1(3x - 4x3)) |
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| 99187. |
Sixteen players `P_(1),P_(2),P_(3)….., P_(16)` play in tournament. If they grouped into eight pair then the probability that `P_(4)` and `P_(9)` are in different groups, is equal toA. `7/15`B. `14/15`C. `2/15`D. `4/15` |
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Answer» Correct Answer - B Total ways `=(16!)/((2!)^(8)8!)` Number of ways in which `P_(4)` and `P_(9)` are in same groups `=(14!)/((2!)^(7)7!)` Number of ways in which they are in different groups `=(16!)/((2!)^(8)8!)-(14!)/((2!)^(7)7!)` `=(14!)/((2)^(7)7!)((15.16)/(2.(8))-1)=(14.14!)/((2)^(7)7!)` Probability `=((14.14!)/((2)^(7).7!))/((16!)/((2)^(8)8!))=(14.8.2)/(15.16)=14/15` |
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| 99188. |
The number of all subsets of a set containing `2n+1` elements which contains more than `n` elements isA. `2^(n)`B. `2^(2n)`C. `2^(n+1)`D. `2^(2n-1)` |
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Answer» Correct Answer - B `.^(2n+1)C_(n+1)+ .^(2n+1)C_(n+2)+…+ .^(2n+1)C_(2n+1)` `=1/2((.^2n+1)C_(0)+ .^(2n+1)C_(1)+ .^(2n+1)C_(3)+…+ .^(2n+1)C_(2n+1))=2 .^(2n)` |
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| 99189. |
Find the equation of circle passing through points (5,3) (1,5) (3,-1) |
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Answer» Let the required equation of the circle is x2 + y2 + 2gx + 2fg + c = 0 But this eqn passes through the points (5,3) 10g + 6f + C + 34 = 0 (1,5) 2g + 10 f + C + 26 = 0 (3,-1) 6g – 2f + C + 10 = 0 Solving 2,3 and 4 we get 2 – 3 ⇒ 2g – f + 2 = 0 3 – 4 ⇒ g – 3f – 4 = 0 Again solving we get y = -2, f = -2, c = -2 This the required egn of the circle is x2+ y2 – 4x – 4y -2 = 0 |
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| 99190. |
The value of `|(.^(10)C_(4).^(10)C_(5).^(11)C_(m)),(.^(11)C_(6).^(11)C_(7).^(12)C_(m+2)),(.^(12)C_(8).^(12)C_(9).^(13)C_(m+4))|` is equal to zero when `m` isA. `6`B. `4`C. `5`D. `7` |
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Answer» Correct Answer - C `Delta=|(.^(10)C_(4) .^(10)C_(5) .^(11)C_(m)),(.^(11)C_(6) .^(11)C_(7) .^(12)C_(m+2)), (.^(12)C_(8) .^(12)C_(9) .^(13)C_(m+4))|` `C_(2)rarrC_(2)+C_(1)` `Delta=|(.^(10)C_(4) .^(11)C_(5) .^(12)C_(m+2)),(.^(11)C_(6) .^(12)C_(7) .^(12)C_(m+2)), (.^(12)C_(8) .^(13)C_(9) .^(13)C_(m+4))|` `Delta=0` for `m=5(C_(2)-=C_(3))` |
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| 99191. |
Find the equations of the hyperbola whose foci are (±5, 0) the transverse axis is of length 8. |
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Answer» Foci are S(±5,0) ∴ ae = 5 Length of transverse axis = 2a = 8, a = 4 e = 5/4 b2 = a2 (e2 – 1) = 16(25/6 - 1) = 9 Equation of the hyperbola is x2/16 – y2/9 = 1 9x2 – 16y2 = 144. |
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| 99192. |
EXERCISE 16(A)Express each of the following statements in percentage form :(i) 13 out of 20(ii) 21 eggs out of 30 are \( g \) |
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Answer» (i) 13/20 x 100 = 13 x 5 = 65% (ii) 21/30 x 100 = 7x 10 = 70% \(\therefore\) 70% one good |
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| 99193. |
If `A` is square matrix of order `2xx2` such that `A^(2)=I,B=[{:(1,sqrt(2)),(0,1):}]` and `C=ABA` thenA. `C^(2009)=A[{:(1,2009sqrt(2)),(0,1):}]`B. `C^(2009)=A[{:(1,2009+sqrt(2)),(0,1):}]`C. `|C|^(2009)=1`D. `C^(2009)=A[{:(1,2009sqrt(2)),(0,1):}]` |
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Answer» Correct Answer - C::D `(ABA)^(2)=(ABA)(ABA)=AB(AA)BA=AB^(2)A` continuing in the manner `C^(2008)=AB^(2009)A` `B^(2)=[{:(1,2sqrt(2)),(0,1):}],B^(3)=[{:(1,3sqrt(2)),(0,1):}]` Continuing in this manner `B^(2009)=[{:(1,2009sqrt(2)),(0,1):}]` |
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| 99194. |
31. If I lend a sum of money at 6 per cent, the insterest for a certain time exceeds the loan by Rs 100 ; but if I lend it at 3 per cent, for a fourth of the time, the loan exceeds its interest by Rs 425 . How much do I lend? |
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Answer» Let the amount of loan be Rs.x and its interest exceeds the loan amount by Rs.100 in y years at 6% rate. So x+100= xy6/100 ........(1) By 2nd condition x - xy3/400=425 Or, 8x - xy6/100=3400........(2) Combining (1) and (2) we get 8x -x-100=3400 =>7x=3500 =>x = 500 |
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| 99195. |
If `f(x)=sin^(-1)[e^(x)]+sin^(-1)[e^(-x)]` then which of the following hold(s) good ? [Note: where [k] denotes the larest integer than or equal to k.]A. f(x) has isolated point remonable discontinuity at x=0B. domain of f(x) is (-In 2, In 2)C. Range of f(x) contains exactly 2 elements.D. Rang of `f(x)={(pi)/(2)}` |
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Answer» Correct Answer - A For domain of f(x), `e^(x)in(0,2)ande^(-x)in(0,2)` `:.x in(-ln2,ln2)` `"Now, "f(x)={:[(pi,,,x-0),((pi)/(2)",",,,x in(-ln2,ln2)-{0}):}` |
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| 99196. |
If `x=(1)/(1^(2))+(1)/(3^(2))+(1)/(5^(2))+` . . . `y=(1)/(1^(2))+(3)/(2^(2))+(1)/(3^(2))+(3)/(4^(2))+` . . . . `z=(1)/(1^(2))-(1)/(2^(2))+(1)/(3^(2))-(1)/(4^(2))+` . . .thenA. `(x)/(1)=(4y)/(3),(4z)/(3)`B. `(z)/(2),(x)/(3),(y)/(6)` are in A.P.C. `(y)/(6),(x)/(3),(z)/(2)` are in A.P.D. `6y,3x,2z` are in H.P. |
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Answer» Correct Answer - B::C `y-x=3(x-z)` `4x=3z+yimplies(y)/(6),(x)/(3),(z)/(2)` are in A.P. |
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| 99197. |
If `a f(x+1)+b f(1/(x+1))=x,x !=-1,a != b,`then `f(2)` is equal toA. `f(2)=(2a+b)/(2(a^(2)-b^(2))`B. `f(1)=0`C. `f((1)/(2))=(a+2b)/(a^(2)-b^(2))`D. `f(2)=(2a+b)/((a^(2)-b^(2)))` |
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Answer» Correct Answer - A::B `af(x+1)+bf((1)/(x+1))=(x+1)-1` ..(i) `af((1)/(x+1))+bf(x+1)=(1)/(x+1)=-1` . . .(ii) `(1)xxa-(2)xxbimplies(a^(2)-b^(2))f(x+1)` `=a(x+1)-a-(b)/(x+1)+b` putting `x=1,f(2)=(2a+b)/(2(a^(2)-b^(2)))` |
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| 99198. |
Cosider `f(x)=1-e^((1)/(x)-1)` Q. If S is the range of f(x), then S isA. `(-infty,infty)`B. `(-infty,1)`C. `(-infty,1)`D. `(-infty,1)-{(1-(1)/(e))}` |
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Answer» Correct Answer - B `e^((1)/(x)-1)=1-yge0` i.e., `(1)/(x)-1=l n(1-y)` i.e., `(1)/(x)=1+l n(1-y)=l n (1-y)` `becausex(1)/(l ne(1-y))` `because(e(1-y)ne1` i.e., `yne1-(1)/(e)` `becauseS=(-infty,1)-{1-(1)/(e)}` |
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| 99199. |
If (2, 0), (0, 1), (4, 5) and (0, c) are concyclic and then find c. |
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Answer» x2 + y2 + 2gx + 2fy + c1 = 0 Satisfies (2, 0), (0, 1) (4, 5) we get 4 + 0 + 4g + c1 = 0 –– (i) 0 + 1 + 2g. 0 + 2f + c1 = 0 –– (ii) 16 + 25 + 8g + 10f + c1 = 0 –– (iii) (ii) – (i) we get – 3 – 4g + 2f = 0 4g – 2f = – 3 –– (iv) (ii) – (iii) we get – 40 – 8g – 8f = 0 (or) g + f = – 5 –– (v) Solving(iv) and (v) we get g = -13/6, f = -17/6 Substituting g and f value in equation (i) we get 4 + 4 (-13/6) + c1 = 0 c1 = 14/3 Now equation x2 + y2 – (13x/3) - (17y/3) + (14/3) = 0 Now circle passes through (0, c) then c2 - (17c/3) + (14/3) = 0 3c2 – 17c + 14 = 0 ⇒ (3c – 14) (c – 1) = 0 (or) c = 1 or 14/3 |
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| 99200. |
in the given figure `y=x^(2)+bx+c` is a quadratic polynomial which meets x-axis at A and B and y-axis at C Q is vertex of quadratic polynomial and foot of perpendicular from Q to x-axis and y-axis are P and R respectively. Then answer the following questions. Q. If `("area of" DeltaABC)/("Area of rectangle" OPQR)=(8)/(3)` satisfies the relation `b^(2)=kc` then `k` will beA. `9`B. `(2)/(9)`C. `(9)/(2)`D. `2` |
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Answer» Correct Answer - B `((1)/(2)(beta-alpha)c)/(((b^(2)-4c)/(4))(-(b)/(2)))=(8)/(3)` squaring and using `(beta-alpha)^(2)=b^(2)-4c` `9c^(2)=4b^(4)-16b^(2)c` `4b^(4)-16b^(2)c-9c^(2)=0` `4b^(4)-18b^(2)c+2b^(2)c-9c^(2)=0` `(2b^(2)+c)(2b^(2)-9c)=0impliescgt0because2b^(2)+cne0` `(b^(2))/(c)=(9)/(2)implies(b^(2))/(c)=4.5` |
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