Saved Bookmarks
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.
| 301. |
Which of the following statements is not correct for the R by aRb if and only if b lives within one kilometer from a (a) R is reflexive (b) R is symmetric (c) R is not anti-symmetric (d) None of the above |
|
Answer» aRb aRa `(a,a) in R` R is reflexive `(a,b) in R` `(b,a) in R` R is symmetric. `a!=b` Option C is correct. |
|
| 302. |
If `a_(0)=x,a_(n+1)=f(a_(n)), " where " n=0,1,2, …,` then answer the following questions. If `f:R to R ` is given by `f(x)=3+4x and a_(n)=A+Bx,` then which of the following is not true?A. `A+B+1=2^(2n+1)`B. `|A-B|=1`C. `underset (h to oo)(lim)(A)/(B)= -1`D. None of these |
|
Answer» Correct Answer - C Given `a_(n+1)=f(a_(n))` Now, `a_(1)=f(a_(0))=f(x)` `or a_(2)=f(a_(1))=f(f(a_(0)))=fof(x)` `or a_(n)=(fofofof …f(x))/("n times")` Since `a_(1)=g(x)=3+4x,` we have `a_(2)=f{g(x)}=g(3+4x)=3+4(3+4x)=(4^(2)-1)+4^(2)x` `a_(3)=g{g^(2)(x)}=g(15+4^(2)x)=3+4(15+4^(2)x)=63+4^(3)x` `=(4^(3)-1)+4^(3)x` Similarly, we get `a_(n)=(4^(n)-1)+4^(n)x` `or A=4^(n)-1 and B=4^(n)` `or A+B+1=2^(2n+1),|A-B|=1, and ` `underset(n to oo)(lim)(4^(n)-1)/(4^(n))=underset(n to oo)(lim)(1-(1)/(4^(n)))=1` |
|
| 303. |
If `(f(x))^(2)xxf((1-x)/(1+x))=64x AAx in D_(f),` then The value of `f(9//7)` isA. `8(7//9)^(2//3)`B. `4(9//7)^(1//3)`C. `-8(9//7)^(2//3)`D. None of these |
|
Answer» Correct Answer - C `(f(x))^(2)f((1-x)/(1+x))=64x " (1)" ` Putting `(1-x)/(1+x)=y or x=(1-y)/(1+y),` We get `{f((1-y)/(1+y))}^(2)*f(y)=64((1-y)/(1+y))` ` or f(x)*{f((1-x)/(1+x))}^(2)=64((1-x)/(1+x)) " (2)" ` Squaring (1) and dividing by (2), `(f(x)^(4){f((1-x)/(1+x))}^(2))/(f(x){f((1-x)/(1+x))}^(2))=((64x)^(2))/(64((1-x)/(1+x)))` `or {f(x)}^(3)=64x^(2)((1+x)/(1-x))` ` or f(x)=4x^(2//3)((1+x)/(1-x))^(1//3)` ` :. f(9//7)= -4(9//7)^(2//3)(2)` |
|
| 304. |
`f(x)={(x-1",",-1 le x le 0),(x^(2)",",0le x le 1):} and g(x)=sinx` Consider the functions `h_(1)(x)=f(|g(x)|) and h_(2)(x)=|f(g(x))|.` Which of the following is not true about `h_(1)(x)`?A. It is a periodic function with period `pi`.B. The range is [0, 1].C. The domain is R.D. None of these |
|
Answer» Correct Answer - D `|g(x)|=|sinx|, x in R` `f(|g(x)|)={(|sinx|-1",",-1 le |sinx| lt 0),((|sinx|)^(2)",", 0le (|sinx|) le 1):}=sin^(2)x,x in R` `f(g(x))={(sinx-1",",-1 le sinx lt 0),(sin^(2)x",", 0le sinx le 1):}` `={(sinx-1",",(2n-1) pi lt x lt 2n pi),(sin^(2)x",", 2n pi le x le (2n+1)pi):},n in Z.` or `|f(g(x))|={(sinx-1",",(2n-1) pi lt x lt 2n pi),(sin^(2)x",", 2n pi le x le (2n+1)pi):},n in Z.` Clearly, `h_(1)(x)=f(|g(x)|)=sin^(2)x` has period `pi`, range [0, 1], and domain R`. |
|
| 305. |
`f(x)={(x-1",",-1 le x le 0),(x^(2)",",0le x le 1):} and g(x)=sinx` Consider the functions `h_(1)(x)=f(|g(x)|) and h_(2)(x)=|f(g(x))|.` If for `h_(1)(x) and h_(2)(x)` are identical functions, then which of the following is not true?A. Domain of `h_(1)(x) and h_(2)(x)" is " x in [2n pi,(2n+1)pi],n in Z.`B. Range of `h_(1)(x) and h_(2)(x)" is " [0,1]`C. Period of `h_(1)(x) and h_(2)(x) " is " pi`D. None of these |
|
Answer» Correct Answer - C `|g(x)|=|sinx|, x in R` `f(|g(x)|)={(|sinx|-1",",-1 le |sinx| lt 0),((|sinx|)^(2)",", 0le (|sinx|) le 1):}=sin^(2)x,x in R` `f(g(x))={(sinx-1",",-1 le sinx lt 0),(sin^(2)x",", 0le sinx le 1):}` `={(sinx-1",",(2n-1) pi lt x lt 2n pi),(sin^(2)x",", 2n pi le x le (2n+1)pi):},n in Z.` or `|f(g(x))|={(sinx-1",",(2n-1) pi lt x lt 2n pi),(sin^(2)x",", 2n pi le x le (2n+1)pi):},n in Z.` For `h_(1)(x)-=h_(2)(x)=sin^(2)x, x in [2n pi,(2n+1) pi], n in Z,` and has range [0, 1] for the common domain. Also, the period is `2pi` (from the graph). |
|
| 306. |
If `a_(0)=x,a_(n+1)=f(a_(n)), " where " n=0,1,2, …,` then answer the following questions. If `f(x)=(1)/(1-x),` then which of the following is not true?A. `a_(n)=(1)/(1-x) " if " n=3k+1`B. `a_(n)=(x-1)/(x) " if " n=3k+2`C. `a_(n)=x " if " n=3k`D. None of these |
|
Answer» Correct Answer - D Given `a_(n+1)=f(a_(n))` Now, `a_(1)=f(a_(0))=f(x)` `or a_(2)=f(a_(1))=f(f(a_(0)))=fof(x)` `or a_(n)=(fofofof …f(x))/("n times")` Now, if `f(x)=(1)/(1-x), fof(x)=(1)/(1-(1)/(1-x))=(x-1)/(x)` or `fofof(x)=((1)/(1-x)-1)/((1)/(1-x))=x` `or a_(n)=(fofof ... of(x))/("n times")=(1)/(1-x) " if " n=3k+1` `=(x-1)/(x) " if " n=3k+2` `=x " if " n=3k` |
|
| 307. |
If `a_0 = x,a_(n+1)= f(a_n)`, where `n = 0, 1, 2, ...,` then answer thefollowing questions. If `f (x) = msqrt(a-x^m),x lt0,m leq 2,m in N`,thenA. `a_(n)=x, n=2k+1,` where k is an integerB. `a_(n)=f(x) " if " n=2k,` where k is an integerC. The inverse of `a_(n)` exists for any value of n and mD. None of these |
|
Answer» Correct Answer - D Given `a_(n+1)=f(a_(n))` Now, `a_(1)=f(a_(0))=f(x)` `or a_(2)=f(a_(1))=f(f(a_(0)))=fof(x)` `or a_(n)=(fofofof …f(x))/("n times")` `a_(1)=f(x)=(a-x^(m))^(1//m)` `or a_(2)=f(f(x))=[a-{(a-x^(m))^(1//m)}^(m)]^(1//m)=x` `or a_(3)=f(f(f(x)))=f(x)` Obviously, the inverse does not exist when m is even and n is odd. |
|
| 308. |
If `x in [0,5]`, then what is the probability that `x^2 - 3x + 2 > 0` |
|
Answer» `x^2-3x+2 ge 0` `=>(x-1)(x-2) ge 0` It means, `x ge 2` and `x le 1` If we draw it on the number line, from points `0` to `5`, we can see that `4/5` of the part is covered.So, the required probability is `4/5`. Please refer to video to see the number line. |
|
| 309. |
Consider a triangle OAB on the xy- plane in which O is taken as origin of reference and position vector of A and B are `vec a`and`vec b` respectively. A line AC parallel to OB is drawn a from A. D is the mid point of OA. Now a line DC meets AB at M. Area of `DeltaABC` is 2 times the area of `DeltaOAB` On the basis of above information, answer the following questions1. Position vector of point C is2. Position vector of point M is |
|
Answer» `vec(OM)=vec(OA)+vec(AM)` `vec(OM)=veca+t(vecb-veca)` `vec(OM)=vec(OD)+vec(DM)` `vec(OM)=veca/2+S(veca/2+2vecb)` `veca+tvecb-tveca=veca/2+sveca/2+2svecb` `1-t=(1+5)/2,t=25` `1=5s` `s=1/5` `vec(OM)=veca+2/5(vecb-veca)` `=veca+2/5vecb-2/5veca` `=(3veca+2vecb)/5` |
|
| 310. |
If `f(x)=sinx+cosa x`is a periodic function, show that `a`is a rational number |
|
Answer» Period of `sinx=2 pi =(2pi)/(I)` and period of `cos ax=(2 pi)/(|a|)` ` :. " Period of " sinx +cos ax= LCM " of " (2pi)/(I) " and " (2pi)/(|a|)` `=(LCM" of " 2pi " and " 2pi)/(HCF " of " 1 and a)` `=(2pi)/(lambda)` where ` lambda` is the HCF of 1 and `a,(1)/(lambda) " and " (|a|)/(lambda)` should both be integers. Suppose `(1)/(lambda)=p " and "(|a|)/(lambda)=q.` Then, `((|a|)/(lambda))/((1)/(lambda)) =(q)/(p), " where " p, q in Z` i.e., `|a|=(p)/(q)` Hence, a is the rational number. |
|
| 311. |
If `5cosx+12cosy=13`, then the maximum value of `5sinx+12siny` is (A) `12` (B) `sqrt(120)` (C) `sqrt(20)` (D) 13 |
|
Answer» `(5cosx+12cosy)=13` `(5cosx+12cosy)^2=13^2=169` `25cos^2x+120cosxcosy+144cos^2y=169-(1)` `5sinx+12siny=A` `25sin^2x+120sinxsiny+144sin^y=A^2` adding eqution 1 and 2 25+144+120(cosxcosy+sinxsiny) `=A^2+169` `120(cosxcosy+sinxsiny)=A^2` `120cos(x-y)=A^2` `A^2` is a max when`cos(x-y)=1` `A_(max)^2=120` `A_(max)=sqrt120` |
|
| 312. |
The area of the region bounded by the curves `y=sqrt[[1+sinx]/cosx]` and `y=sqrt[[1-sinx]/cosx]` bounded by the lines x=0 and `x=pi/4` is |
|
Answer» `sinx=(2tanx/2)/(1+tan^2x/2)` `cosx=(1-tan^2x/2)/(1+tan^2x/2)` `y=sqrt((1+((2tanx/2)/(1+tan^2x/2))/((1-tan^2x/2)/(1+tan^2x/2)))` `y=(1+tanx/2)/(sqrt(1-tan^2x/2)` `y=(1-tanx/2)/sqrt(1-tan^2x)` `I=int_0^(pi/4)sqrt((1+sinx)/(cosx))-sqrt((1-sinx)/(cosx))dx` `I=int_0^(pi/4)(2tan(x/2))/sqrt(1-tan^2x/2)dx` `I=int_0^(sqrt2-1)(2t2dt)/(sqrt(1-t^2)(1+t^2))` `I=int_0^(sqrt2-1)(4dt)/((1+t^2)sqrt(1-t^2))` option b is corect. |
|
| 313. |
If `a`, `b`, `c` `p`, `q`, `r` six complex numbers such that `p/a+q/b+r/c=1+i` and `a/p+b/q+c/r=0` then value of `p^2/a^2+q^2/b^2+r^2/c^2=` |
|
Answer» `p^2/a^2+q^2/b^2+r^2/c^2 = (p/a+q/b+r/c)^2-2((pq)/(ab)+(qr)/(bc)+(rp)/(ca))` `=(1+i)^2-2((pqc)/(abc)+(qra)/(abc)+(rpb)/(cab))` `=(1+i)^2-2/(abc)*(pqr)(c/r+a/p+b/q)` As, `(a/p+b/q+c/r) = 0`, our equation becomes, `p^2/a^2+q^2/b^2+r^2/c^2 = (1+i)^2` `=1+i^2+2i = 1-1+2i = 2i` `:. p^2/a^2+q^2/b^2+r^2/c^2 = 2i` |
|
| 314. |
If the domain of `y=f(x)i s[-3,2],`then find the domain of `g(x)=f(|[x]|),w h e r[]`denotes the greatest integer function. |
|
Answer» Correct Answer - [-2, 3) Here, `f(x)` is defined for `x in [-3,2].` For `g(x)=f(|[x]|)` to be defined, we must have `-3 le |[x]| le 2` or `0le |[x]| le 2 " " ["As" |x| ge 0 " for all " x]` or ` -2 le [x] le 2 " " ["As " |x| le a implies-a le x le a]` or `-2 le x lt 3 " " ` [By the definition of greatest integral function] Hence, domain of `g(x)` is `[-2,3)`. |
|
| 315. |
Are f and g both necessarily onto, if `gof`is onto? |
|
Answer» Let `f: A->B` and `g:B->C` are two functions such that `gof: A->C` We are given `gof` is onto. Let `A = {a_1,a_2,a_3}` Let `C = {c_1,c_2,c_3}` As `gof` is onto there will be mapping between each elements of set `C` with every element of set `A`. Now, Let `B = {b_1,b_2,b_3,b_4}` As set `B` contains more elements than `A` and `C` so it will not have mapping of its all elements with elements of set `A` and set `C`. That means it is not neccesary that `f` and `g` are also onto function. |
|
| 316. |
Let N be the set of natural numbers and f : N->N be a function given by f(x)=x+1 for `x in N`.Which one of the following is correct?a. f is one-one and ontob. f is one-one but not ontoc. f is only ontod. f is neither one-one nor onto |
|
Answer» A function, `f(x)` is a one-one function if`f(a) = f(b)`, then, `a =b` Here, `f(x) = x+1` `f(a) = a+1` `f(2) = b+1` If, `f(a) = f(2)` Then, `a+1 = b+1 => a = b` So, `f(x)` is a one-one function. For a function `f(x)` to be an onto function, it should cover all elements of `x`. Here, as `x in N`. So, `x = (1,2,3,4...) `. `f(x) in N`. `f(x) = (2,3,4,5...)`. As `f(x)` does not have value `1` that is present in `x`, so, `f(x)` is not an onto function. So, option `B` is the correct answer. |
|
| 317. |
If `f(x)=log((1+x)/(1-x))`, `f(2x/(1+x^2))=2f(x)` |
|
Answer» `f(y)=log((1+y)/(1-y))` `y=(2x)/(1+x^2)` `f((2x)/(1+x^2))=log((1+((2x)/(1+x^2)))/(1-(2x)/(1+x^2)))` `=log((1+x^2+2x)/(1+x^2-2x))` `=log((1+x^2)/(1-x^2))` `=2log((1+x)/(1-x))`. |
|
| 318. |
Let `f(x) =(alphax)/(x+1)` Then the value of `alpha` for which `f(f(x) = x` is |
|
Answer» Correct Answer - -1 `f(x)=(alpha x)/(x+1), x ne -1` Now `f(f(x))=x` `implies(alpha((alpha x)/(x+1)))/((alpha x)/(x+1)+1)=x` `implies(alpha^(2)x)/((alpha +1)x+1)=x` `implies (alpha+1)x^(2)+(1-alpha^(2))x=0 " ...(1)" ` Now this is true for all real x. `implies alpha +1=0 " and " 1-alpha^(2)=0` `implies alpha = -1 ` (common value) |
|
| 319. |
Let `f: A to B` and `g: B to C` be two functions. Then; if gof is onto then g is onto; if gof is one one then f is one-one and if gof is onto and g is one one then f is onto and if gof is one one and f is onto then g is one one. |
|
Answer» Let `z in D,` then there is some `y in C` such that `g(y)=z` ` :.` Now if `y notin B`, then there is no x for which `f(x)=y` ` :. gof` is not onto. |
|
| 320. |
The number of integer value(s) of n when `3^512`-1 is divided by `2^n` is/are |
|
Answer» `3^512 - 1 = (3^256)^2 - 1` `= (3^256 - 1)(3^256 + 1)` `= ((3^128)^2 - 1^2)(3^256 + 1)` `= (3^128 -1)(3^128 +1)(3^256 +1)` `3^512 - 1 = (3^64 - 1)(3^64 + 1)(3^128 + 1)(3^256+1)` `= (3^32 - 1)(3^32+1)` `= (3^16 - 1)(3^16+1)` `= (3^8 - 1)(3^8+1)` `= (3^4-1)(3^4+1)` `= (3^2-1)(3^2+1)` `3^512 - 1= (3-1)(3+1)3^2` `3^512 - 1= 2(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)(3^64+1)(3^128+1)(3^256+1)` `3^256 = (2+1)^256` `= .^256C_0(2)^256 (1)^0 + .^256C_1(2)^256(1)^1.......+1` `= 2(even) + 2(odd)` `= = 20^11` option c is correct answer |
|
| 321. |
Let`f"":""N rarr Y`be a function defined as `f""(x)""=""4x""+""3`, where `Y""=""{y in N"":""y""=""4x""+""3`for some `x in N}`. Show that f is invertibleand its inverse is(1) `g(y)=(3y+4)/3`(2) `g(y)=4+(y+3)/4`(3) `g(y)=(y+3)/4`(4) `g(y)=(y-3)/4` |
|
Answer» `F(x)=4x+3` `F(x)-3=4x` `x=(F(x)-3)/4` `F^-1(x)=(x-3)/4` `g(y)/N=(y-3)/4` `y=4x+3` `g(y)=(y-3)/4`is the answer |
|
| 322. |
Let `S_n` denote the sum of n terms of an AP whose first term is a. If common difference d is given by `d=Sn-kS_(n-1)+S_(n-2)` , then k is : |
|
Answer» `d = a_n - a_(n-1)` `a_n = s_n - s_(n-1)` `= n/2 [2a + (n-1)d] - (n-1)/2[2a + (n-2)d]` `= (2an + n(n-1)d - (n-1)[2a+ (n-1)d])/2` `= (2a+ (n-1)d-d +nd)/2` `= a + nd-d` `= a + (n-1)d` `k=2` so, option B is correct answer |
|
| 323. |
If `f(x)=log[(1+x)/(1-x)],`then prove that `f[(2x)/(1+x^2)]=2f(x)dot` |
|
Answer» `f(x)=log[(1+x)/(1-x)]` or `f((2x)/(1+x^(2)))=log[(1+(2x)/(1+x^(2)))/(1-(2x)/(1+x^(2)))]=log[(x^(2)+1+2x)/(x^(2)+1-2x)]` `=log[(1+x)/(1-x)]^(2)=2 log [(1+x)/(1-x)]=2f(x)` |
|
| 324. |
Let `f(x) = 1 + |x|,x < -1 [x], x >= -1, where [*]` denotes the greatest integer function.Then `f { f (- 2.3)}` is equal to |
|
Answer» Correct Answer - 3 `f(x)={(1+|x|",",x lt -1),([x]",",x ge -1):}` `f(-2.3)=1+|-2.3|=1+2.3=3.3` Now, `f(f(-2.3))=f(3.3)=[3.3]=3` |
|
| 325. |
If `f(x) = -1 +|x-1|, -1 le x le 3 " and " g(x)=2-|x+1|, -2 le x le 2,` then find `fog(x) " and " gof(x).` |
|
Answer» `f(g(x))= -1+|g(x)-1|,-1le g(x) le 3,-2 le x le 2` `= -1 +|2-|x+1|-1|,-1le2-|x+1|le3, -2 le x le 2` `= -1+|1-|x+1||,-1 le2-|x+1| le3,-2 le x le 2` Now `-1 le 2-|x+1|le 3` `implies -3 le -|x+1| le 1` `implies -1 le |x+1|le3` `implies 0le |x+1|le 3` ` implies -3 le x+1 le 3` `implies -4 le x le 2 ` Also ` -2 le x le 2` ` :. fog(x) = -1+|1-|x+1||,-2 le x le2` `g(f(x))=2-|f(x)+1|,-2le f(x) le 2, -1 le x le 3` `=2-| -1+|x-1|+1|,-2 le-1+|x-1| le 2, -1 le x le 3` `= 2-|x-1|,-2 le -1 +|x-1| le 2, -1 le x le 3 ` Now `-2 le -1+|x-1| le 2` `implies -1 le |x-1| le 3` `implies -3 le x-1 le 3` `implies -2 le x le 4` Also `-1 le x le 3` ` :. g(f(x))=2 - |x-1|,-1 le x le 3` |
|
| 326. |
`f(x)={x+1, x |
|
Answer» `f(g(x))={(g(x)+1",", g(x) lt0),({g(x)}^(2)",",g(x) ge0):}` `f(x)={(x^(3)+1",",x^(3) lt0", " x lt1),(2x-1+1",",2x-1 lt 0", "x ge1),((x^(3))^(2)",",x^(3) ge0", " x lt1),((2x-1)^(2)",",2x-1 ge 0", "x ge1):}` `={(x^(3)+1",", x lt0),(x^(6)",",0le x lt1),((2x-1)^(2)",", x ge1):}` Clearly, domain of function is R. For `x lt 0, x^(3) +1 in (-oo,1).` For `0le x lt 1, x^(6) in [0,1).` For ` x ge 1, (2x-1)^(2) in [1, oo).` Hence, the range is R. |
|
| 327. |
The function `f(x)` is defined in `[0,1]`. Find the domain of `f(tanx).` |
|
Answer» Here, `f(x)` is defined in `[0,1].` So, `x in [0,1],` i.e., the only values of x that we can subsitute lies in `[0,1].` For `f(tanx)` to be defined, we must have `0 le tan x le 1 " [As x is replaced by tan x]" ` i.e., `n pi le x le n pi +(pi)/(4), n in Z " [in general]" ` Thus, the domain of `f(tanx)` is `underset(n in Z)(cup)[n pi,n pi+(pi)/(4)]` |
|
| 328. |
Suppose that `g(x)=1+sqrt(x) " and " f(g(x))=3+2sqrt(x)+x.` Then find the function `f(x)`. |
|
Answer» `g(x)=1+sqrt(x) " and " f(g(x))=3+2sqrt(x)+x " " `(1) ` :. f(1+sqrt(x))=3+2sqrt(x)+x` Put `1+sqrt(x)=y " or " x=(x-y)^(2).` Then, `f(y)=3+2(y-1)+(y-1)^(2)=2+y^(2).` ` :. f(x)=2+x^(2)` |
|
| 329. |
Which of the following functions from Z to itself are bijections? aA. `f(x)=x^(3)`B. `f(x)=x+2`C. `f(x)=2x+1`D. `f(x)=x^(2) +1` |
|
Answer» Correct Answer - B Here, `f(x) =x +2impliesf(x_(1))=f(x_(2))` `x_(1)+2=x_(2)+2implies x_(1)=x_(2)` Let `" " y=x+2` `x=y-2 in Z, AA y in x` Hence, f(x) is one-one and onto. |
|
| 330. |
If `f:R to R` be defined by `f(x)=3x^(2)-5` and `g: R to R ` by `g(x)= (x)/(x^(2)+1).` Then, gof isA. `(3x^(2)-5)/(9x^(4)-30x^(2)+26)`B. `(3x^(2)-5)/(9x^(4)-6x^(2)+26)`C. `(3x^(2))/(x^(4)+2x^(2)-4)`D. `(3x^(2))/(9x^(4)+30x^(2)-2)` |
|
Answer» Correct Answer - A Given that , `f(x) = 3x^(2)-5` and `g(x) = (x)/(x^(2)+1)` `gof= g{f(x)}=g(3x^(2)-5) ` `=(3x^(2)-5)/((3x^(2)-5)^(2)+1)=(3x^(2)-5)/(9x^(4)-30x^(2)+25 +1)` `=(3x^(2)-5)/(9x^(4)-30x^(2)+26)` |
|
| 331. |
If `f: RvecR`is given by `f(x)=(x^2-4)/(x^2+1)`, identify the type of function. |
|
Answer» Correct Answer - many-one, into `f(x)=f(-x), " So, "f` is many-one. Also, `f(x)=1-(5)/(x^(2)+1) in [-4,1) " So, "f` is into. |
|
| 332. |
If`f: NvecZf(n)={(n-1)/2,w h e nni sod d-n/2,i d e n t ifyt h ew h e nni se v e n` |
|
Answer» Correct Answer - bijective When n is even, let `f(2m_(1))=f(2m_(2))` or `-(2m_(1))/(2)=-(2m_(2))/(2)` or `m_(1)=m_(2)` When n is odd, let `f(2m_(1)+1)=f(2m_(2)+1)` or `(2m_(1)+1-1)/(2)=(2m_(2)+1-1)/(2) " or "m_(1)=m_(2)` Therefore, `f(x)` is one-one. Also, when n is even, `-(n)/(2)=-(2m)/(2)= -m.` When n is odd, `(n-1)/(2)=(2m+1-1)/(2)=m.` Hence, the range of the function is Z. Therefore, function is onto. |
|
| 333. |
Find the range of `tan^(-1)((2x)/(1+x^2))` |
|
Answer» First, we must get the range of `(2x)/(1+x^(2))=y` We have `yx^(2)-2x+y=0` Since x is real, `D ge 0, " i.e., "4-4y^(2) ge 0 " or " -1 le y le 1.` So, ` "tan"^(-1)(y) in [-(pi)/(4),(pi)/(4)]` (As `tan x` is an increasing function) |
|
| 334. |
Find the range of `f(x)=|sinx|+|cosx|, x in R`. |
|
Answer» `f(x)=|sinx|+|cosx|, x in R`. Clearly, `f(x) gt 0`. Also, `f^(2)(x)=sin^(2)x+cos^(2)x+|2 sinx cos x|=1+|sin2x|` Now, `0 le |sin zx|le1` ` :. 1 le 1 +|sin zx| le 2` ` :. 1 le f^(2)(x) le 2` or ` 1 le f(x) le sqrt(2)` |
|
| 335. |
Find the range of `f(x)=sin^2x-sinx+1.` |
|
Answer» `f(x)=sin^(2)x-sinx +1=(sinx - (1)/(2))^(2)+(3)/(4)` Now, ` -1 le sinx le 1 " or " -(3)/(2) le sinx-(1)/(2) le (1)/(2)` or ` o le (sinx -(1)/(2))^(2) le (9)/(4) " or " (3)/(4) le (sinx -(1)/(2))^(2) l+ (3)/(4) le 3` Hence, the range is `[(3)/(4),3]`. |
|
| 336. |
The equation of the curve passing through the point `(1,pi/4)` and having a slope of tangent at any point (x,y) as `y/x - cos^2(y/x)` is |
|
Answer» Here, slope is given as `y/x-cos^2(y/x)` `:. dy/dx = y/x-cos^2(y/x)->(1)` Let, `y/x = v => y = vx => dy/dx = v+x(dv)/dx` so, (1) becomes, `=>v+x(dv)/dx = v- cos^2v` `=>(dv)/cos^2v = -dx/x` Integrating both sides, `int (dv)/cos^2v = - int dx/x` `=>int sec^2v (dv) = - int dx/x` `=>tanv = -ln x+c` `=>tan(y/x) = -ln x+c` As, the curve is passing through `(1,pi/4)`, it will satisfy the above equation. `:. tan (pi/4) = -ln (1) +c` `=> c = 1` So, the rfequired equation becomes, `tan(y/x) = -ln x+1` `=> tan(y/x) - 1= -ln x` `=> x = e^(1-tan(y/x))`, which is the required equation. |
|
| 337. |
Find the range of `f(x)=sin^(-1)x+tan^(-1)x+cos^(-1)xdot` |
|
Answer» Clearly, the domain of the function is `[-1,1].` Also `tan^(-1)x in [-(pi)/(4),(pi)/(4)] " for " x in [-1,1]` Now, `sin^(-1)x+cos^(-1)x =(pi)/(2)" for " x in [-1,1]` Thus, `f(x)=tan^(-1)x+(pi)/(2), " where " x in [-1,1]` Hence, the range is `[-(pi)/(4)+(pi)/(2),(pi)/(4)+(pi)/(2)]-=[(pi)/(4),(3pi)/(4)]` |
|
| 338. |
Find the matrix A satisfying the matrix equation: `[[2,1],[3,2]]` A`[[-3,2],[5,-3]]`=`[[1,0],[0,1]]` |
|
Answer» `[[2,1],[3,2]]A[[-3,2],[5,-3]] = [[1,0],[0,1]]` Let `B = [[2,1],[3,2]]` `C = [[-3,2],[5,-3]]` Then, `BAC = I` `=>B^-1BAC^-1C = B^-1C^-1I` `=>A = B^-1C^-1 = (CB)^-1` Now, `CB = [[-3,2],[5,-3]][[2,1],[3,2]] = [[0,1],[1,-1]]` Now, `(CB)^-1 = (adj(CB))/|CB|` `=>adj(CB) = [[-1,-1],[-1,0]]` `=>|CB| = 0-1 = -1` `:. (CB)^-1 = [[1,1],[1,0]]` `:. A = (CB)^-1 = [[1,1],[1,0]]` |
|
| 339. |
Suppose that `f(x)` is a function of the form `f(x)=(ax^(8)+bx^(6)+cx^(4)+dx^(2)+15x+1)/(x), (x ne 0)." If " f(5)=2`, then the value of `f(-5)` is _________. |
|
Answer» Correct Answer - 28 `f(x)=(ax^(8)+bx^(6)+cx^(4)+dx^(2)+15x+1)/(x)` `=underset("odd function")(underbrace(ax^(7)+bx^(5)+cx^(3)+dx+(1)/(x)+15))` Now, `f(x)+f(-x)=30` `or f(-5)=30-f(5)=28` |
|
| 340. |
If a and b are positive integers with no common factor,show that `[a/b]+[(2a)/b]+[(3a)/b]. . . . . [((b-1)a)/b]=((a-1)(b-1))/2` when [.] shows the greatest integer function |
|
Answer» Let `S = [a/b]+[(2a)/b]+[(3a)/b]+...+[((b-1)a)/b]->(1)` We can write it from end as, `S = [((b-1)a)/b]+ [((b-2)a)/b]+...[(2a)/b]+[a/b]->(2)` Adding (1) and (2), `2S = ([a/b]+[((b-1)a)/b])+([(2a)/b]+[((b-2)a)/b])+...+([((b-1)a)/b+[a/b]])` Now, `[a/b]+[((b-1)a)/b] = [a/b]+[a-a/b] = a-1` Similarly, `[(2a)/b]+[((b-2)a)/b] = [(2a)/b]+[a-(2a)/b] = a-1` So, each term will be equal to `a-1.` As, there are `b-1` terms, `:. 2S = (a-1)(b-1)` `=>S = ((a-1)(b-1))/2.` |
|
| 341. |
Find the domain of the following functions (a) `f(x)=(1)/(sqrt(x-2)) " (b) " f(x)=(1)/(x^(3)-x)` (c ) `f(x)= root(3)(x^(2)-2)` |
|
Answer» Correct Answer - (a) `(2,oo) " (b) " R-{-1,0,1} " (c ) "R` (a) `f(x)=(1)/(sqrt(x-2))` is defined if `x-2 gt 0 " or " x gt 2.` Therefore, domain is `(2,oo)`. (b) `f(x)=(1)/(x^(3)-x)` is not defined if `x^(3)-x=0 " or " x(x-1)(x+1)=0" or " x=-1,0,1.` Therefore, domain is `R-{-1,0,1}.` (c ) `f(x)=root(3)(x^(2)-2)`. We know that cube roots are defined for any real value. So, `x^(2)-2` can take any real value. So, x can take any real value. Therefore, domain is set R. |
|
| 342. |
Find the range of the following functions. (a) `f(x)=5-7x " (b) "f(x)=5-x^(2)` (c ) `f(x)=(x^(2))/(x^(2)+1)` |
|
Answer» Correct Answer - (a) `R " (b) " (-oo,5] " (c ) "[0,1)` (a) `f(x)=5-7x" or " y=5-7x. "So, "x=(5-y)/(7).` Clearly, for any real value fo y there exists a real number x. So, range is set R. (b) `f(x)=5-x^(2)" or "y=5-x^(2). " So "x^(2)=5-y.` Since, `x^(2) ge 0, 5-y ge 0. ` So, `y le 5`. Therefore, range is `(-oo,5]`. (c ) `f(x)=(x^(2))/(x^(2)+1)=y.` ` :. x^(2)=yx^(2)+yimpliesx^(2)=(y)/(1-y).` Since `x^(2) ge 0` for all real `x,(y)/(1-y) ge " or " (y)/(y-1) le 0.` So, `0le y lt 1.` Therefore, range is `[0,1)` |
|
| 343. |
Find the domain and range of `f(x)=(2-5x)/(3x-4)`. |
|
Answer» Correct Answer - Domain: R - {4/3}, Range: R - {-5/3} `f(x)=y=(2-5x)/(3x-4)` `implies 3yx-4y=2-5x` `impliesx=(4y+2)/(3y+5)` Hence ` x in R-{4//3} " and " y in R-{-5//3}` |
|
| 344. |
Find the domain of `f(x)=sqrt(sinx)+sqrt(16-x^2)` |
|
Answer» Correct Answer - `[-4,-pi] cup [0,pi]` `f(x)=sqrt(sinx)+sqrt(16-x^(2))` or ` sinx ge 0 " and " 16-x^(2) ge 0` or `2n pi le x le (2n+1)pi " and " -4 le x le 4` Therefore, domain is `[-4,-pi] cup [0,pi].` |
|
| 345. |
Find the domain and range of `f(x)=sqrt(4-16x^(2))`. |
|
Answer» Correct Answer - Domain: [-1/2, 1/2], Range: [0, 2] `f(x)=y=sqrt(4-16x^(2))` We must have `4-16x^(2) ge 0 implies x^(2) le 1//4` `implies x in [-1//2,1//2]` Hence domain is `[-1//2,1//2]` Also, minimum value of function is 0 and maximum value 2 when `16x^(2)=4` Hence range is `[0,2]` |
|
| 346. |
Let `f: R->R` where, `f(x)=(x^2+4x+7)/(x^2+x+1)` Is `f(x)` one-one? |
|
Answer» `f(x) = (x^(2)+4x+7)/(x^(2)+x+1)=1+(3(x+2))/(x^(2)+x+1)` Let `f(x_(1))=f(x_(2))` `implies 1+(3(x_(1)+2))/(x_(1)^(2)+x_(1)+1)=1+(3(x_(2)+2))/(x_(2)^(2)+x_(2)+1)` `implies x_(1)x_(2)^(2)+x_(1)x_(2)+x_(1)+2x_(2)^(2)+2x_(2)+2` `=x_(1)^(2)x_(2)+x_(1)x_(2)+x_(2)+2x_(1)^(2)+2x_(1)+2` `implies (x_(1)-x_(2))(2x_(1)+2x_(2)+x_(1)x_(2)+1)=0` Let us consider `2x_(1)+2x_(2)+x_(1)x_(2)+1=0` `implies x_(2)=(1+2x_(1))/(2+x_(1))` This relation is satisfied by infinite number of pairs `(x_(1), x_(2))," where " x_(1) ne x_(2), e.g., (0,-1//2),(1,-1) ` etc. Hence f(x) is many-one. |
|
| 347. |
Find the range of `f(x)sqrt(x-1)+sqrt(5-1)` |
|
Answer» Correct Answer - `[2,2sqrt(2)]` Let `y=sqrt(x-1)=sqrt(5-x)` or `y^(2)=x-1+5-x+2 sqrt((x-1)(5-x))` or `y^(2)=4+2 sqrt(-x^(2)-5+6x)` or `y^(2)=4+2 sqrt(4-(x-3)^(2))` then `y^(2)` has minimum value 4 [when `4-(x-3)^(2)=0`] and maximum value 8 when `x=3`. Therefore, `y in [2,2sqrt(2)]`. |
|
| 348. |
Find the domain and range of `f(x)=sqrt(3-2x-x^2)` |
|
Answer» Correct Answer - Domain: [-3, 1], Range: [0, 2] `f(x)=sqrt(3-2x-x^(2))` is defined if `3-2x-x^(2) ge 0` or `x^(2)+2x-3 le 0` or `(x-1)(x+3) le 0` or ` x in [-3,1]` Also, `f(x)=sqrt(4-(x+1)^(2))` which has maximum value when `x+1=0.` Hence, the range is `[0,2].` |
|
| 349. |
Evaluate `|[x,y,x+y],[y,x+y,x],[x+y,x,y]|` |
|
Answer» `C_1->C_1+C_2` `|[2(x+y),y,x+y],[2(x+y),x+y,x],[2(x+y),x,y]|` `R_1->R_1-R_3,R_2->r_2-R_3` `2(x+y)|[0,y-x,x],[0,y,x-y],[1,x,y]|` `2(x+y)((y-x)(x-y)-xy)` `2(x+y)(xy-x^2-y^2+xy-xy)` `2(x+y)(xy-x^2-y^2)`. |
|
| 350. |
Find the value of p for which curves `x^2=9p(9-y)` and `x^2=p(y+1)` cut each other at right angles. |
|
Answer» Equations of the given curves, `x^2 = 9p(9-y)->(1)` `x^2 = p(y+1)->(2)` From (1) and (2), `9p(9-y) = p(y+1)` `=>81p-9py = py+p` `=>10py = 80p=> y =8` `:. x^2 = (8+1)p = 9p` `=>x = 3sqrtp` So, point of intersection of these curve is `(3sqrtp,8)`. Now, we have to calculate their slopes by differentiating their equations. For first curve, Differentiating its equation, `2x = 9p(-dy/dx)=>dy/dx = -(2x)/(9p)` At point `(3sqrtp, 8)`, `dy/dx = m_1 = -(2sqrtp)/(3p)` For second curve, Differentiating its equation, `2x = p(dy/dx) =>dy/dx = (2x)/p` At point `(3sqrtp, 8)`, `dy/dx = m_2 =(6sqrtp)/(p)` As these two curves are at right angles, multiplication of their slopes should be equal to `-1`. `:. -(2sqrtp)/(3p)**(6sqrtp)/(p) = -1` `=>-12 = -3p => p = 4` |
|