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| 251. |
Find e local maximum and local minima, of the function `f(x)= sin x-cos x`,`0 |
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Answer» Here, `f(x) = sinx-cosx` `f(x) = sqrt2(1/sqrt2sinx-1/sqrt2cosx)` `= sqrt2(cos(pi/4) sinx-sin(pi/4) cosx)` `=sqrt2(sin(x-pi/4))` So, `f(x)` will be maximum when `sin(x-pi/4) = 1` `=> sin(x-pi/4) = sinpi/2=> x-pi/4 = pi/2=>x = (3pi/4)` `f(x)` will be minimum when `sin(x-pi/4) = -1` `=> sin(x-pi/4) = sin-pi/2=> x-pi/4 = -pi/2=>x = (-pi/4)` `:. f(x)_(max) = sqrt2` `f(x)_(min) = -sqrt2` |
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| 252. |
The equation of tangents to the curve `y=cos(x+y), -2pi |
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Answer» `x+2y=0` `m=-1/2` `y=cos(x+y)` differentiate with respect to x `dy/dx=-sin(x+y)(1+y^2)` `-1/2=-sin(x+y)(1-1/2)` `-1=-sin(x+y)` `sin(x+y)=1` `x+y=pi/2` `y=cospi/2=0` `sinx=1` `at x=-3/2pi,pi/2` `(-3/2pi,0)` and `(pi/2,0)` `y-0=-1/2(x+3/2pi)` `y=-x/2-3/4pi` `4y+2x+3pi=0` `y-0=-1/2(x-pi/2)` `y=-x/2+pi/4` `4y+2x-pi=0`. |
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| 253. |
The range of `f(x)= sin^(-1)x +cos^(-1)x+ tan^(-1) x` is |
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Answer» `-1<=x<=1` `x in R` `sin^(-1)x+cos^(-1)x=pi/2` `f(x)=pi/2+tan^(-1)x` `R(f(x))=[pi/2-pi/4,pi/22+pi/4]` `=[pi/4,3/4pi]` Option 2 is correct. |
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| 254. |
The equation `||x-2|+a|=4`can have four distinct real solutions for `x`if `a`belongs to the interval`(-oo,-4)`(b) `(-oo,0)``(4,oo)`(d) none of theseA. `(-oo, -4)`B. `(-oo,0]`C. `[4,oo)`D. none of these |
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Answer» Correct Answer - A `|x-2|+a=+-4` `or |x-2|= +-4-a` For four real roots, `4-a gt 0 and -4 -a gt 0` ` or a in (-oo, -4)` |
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| 255. |
Let M be a 2 x 2 symmetric matrix with integer entries. Then M is invertible if (a)The first column of M is the transpose of the second row of M (b)The second row of Mis the transpose of the first olumn of M (c) M is a diagonal matrix with non-zero entries in the main diagonal (d)The product of entries in the main diagonal of Mis not the square of an integer |
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Answer» `M->2*2` symmetric matrix `M=M^T` Suppose`M=[[a,b],[c,d]]` `[[a,b],[c,d]]=[[a,c],[b,d]]` `b=c` Our matrix M=`[[a,b],[b,d]]` `[[a],[b]]^T=[[b,d]]` `[[a,b]]=[[b,d]]` `a=b,b=d` `a=b=d` `M=[[a,a],[a,a]]` `|M|=0` `[[b,d]]^T=[[a],[b]]` `a=b=-d` c)`[[a,0],[0,d]]` `|M|=>ad-0` `|M|!=0` Therefore,M is invertible matrix d)`M=[[a,b],[b,d]]` `|M|=ad-b^2` `|M|!=0` M is an invertible matricx Option C and D is correct. |
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| 256. |
If `z=x+iy` and `w=(1-iz)/(z-i)`, then `|w|=1` implies that in the complex plane(A)`z` lies on imaginary axis (B) `z` lies on real axis (C)`z` lies on unit circle (D) None of these |
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Answer» `w = (1-iz)/(z-i)` `=>|w| = |(1-iz)|/|(z-i)|` `=>|w|^2 = |(1-iz)|^2/|(z-i)|^2` `=>|w|^2 = |(1-i(x+iy))|^2/(|(x+iy)-i|^2` `=>1^2 = |(1+y) - ix|^2/|x+(y-1)i|^2` `=>x^2+(y-1)^2 = x^2+(1+y)^2` `=>y-1 = 1+y` `=> y = 0` `:. z = x+i(0) = x` `=> z= x` So, `z` lies on real axis. |
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| 257. |
Let `f:R to R ` be a function satisfying `f(2-x)=f(2+x) and f(20-x)=f(x) AA x in R.` For this function `f`, answer the following. The graph of `y= f(x)` is not symmetrial aboutA. symmetrical about `x=2`B. symmetrical about `x=10`C. symmetrical about `x=8`D. None of these |
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Answer» Correct Answer - C `f(2-x) =f(2+x) " (1)" ` Replace x by `2-x. " Then " f(x)=f(4-x) " (2)" ` Also, given `f(20-x)=f(x) " (3)" ` From (1) and (2), `f(4-x)=f(20-x).` Replace x by `4-x. " Then " f(x)=f(x+16).` Hence, the period of `f(x)` is 16. `f(2-x)=f(2+x)` Hence, `y=f(x)` is symmetrical about `x=2`. Also, `f(20-x)=f(x)` `or f(20-(10+x))=f(10+x)` `or f(10-x)=f(10+x)` Hence, `y=f(x)` is symmetrical about `x=10`. |
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| 258. |
Let `f:R to R ` be a function satisfying `f(2-x)=f(2+x) and f(20-x)=f(x) AA x in R.` For this function `f`, answer the following. If `f(2) ne f(6),` then theA. fundamental period of `f(x) " is " 1`B. fundamental period of `f(x) " may be " 1`C. period of `f(x)` cannot be 1D. fundamental period of `f(x) " is " 8` |
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Answer» Correct Answer - C `f(2-x) =f(2+x) " (1)" ` Replace x by `2-x. " Then " f(x)=f(4-x) " (2)" ` Also, given `f(20-x)=f(x) " (3)" ` From (1) and (2), `f(4-x)=f(20-x).` Replace x by `4-x. " Then " f(x)=f(x+16).` Hence, the period of `f(x)` is 16. If 1 is a period, then `f(x)=f(x+1) AA x in R` `or f(2)=f(3)=f(4)=f(5)=f(6)` which contradicts the given hypotheses that `f(2) ne f(6).` Therefore, 1 cannot be period of `f(x)`. |
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| 259. |
If f(x) = `(1)/(1 -x) x ne 1` and g(x) = `(x-1)/(x) , x ne0`, then the value of g[f(x)] is :A. `-x`B. xC. 2xD. None of these |
| Answer» Correct Answer - b | |
| 260. |
`7^(log_(3)5)+3^(log_(5)7)-5^(log_(3)7)-7^(log_(5)3)` |
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Answer» We know, `log_ab = log_xb/log_xa` So, `7^(log_(3)5) =7^(log_(7)5/(log_(7)3) ` `= (7^(log_(7)5))^(1/log_(7)3)` (As `a^(log_(a)b) = b`) `=5^((1/log_(7)3)` (As `a^(log_(a)b) = b`) `= 5^(log_(3)7)` (As `a^(1/log_(b)c) = a^(log_(c)b)`) So,`7^(log_(3)5) = 5^(log_(3)7)`->(1) Similarly, we can show that, `3^(log_(5)7) = 7^(log_(5)3)`->(2) So, using (1) and (2), we can write our expression as, `5^(log_(3)7)+7^(log_(5)3)- 5^(log_(3)7)-7^(log_(5)3) = 0` |
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| 261. |
Factorize `1+x^4+x^8` |
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Answer» `1 + x^4 + x^8` `= 1+ x^4 + x^8 + x^4 - x^4` using `x^8 = x^4*x^4= (x^4)^2` `= 1 + 2x^4 + x^8 - x^4` using `(a+b)^2 = a^2 + 2ab + b^2` `(1)^2 + 2*1*(x^4) + (x^4)^2 - x^4` `= (x^4+1)^2 - x^4` `= (x^4 + 1)^2 - (x^2)^2` `= (x^4 + 1 - x^2)(x^4+1+x^2)` `= (x^4-x^2+1)(x^4+2x^2+1-x^2` `= (x^4-x^2+1)((x^2+1)^2-x^2)` `= (x^4 - x^2+1)(x^2-x+1)(x^2+x+1)` answer |
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| 262. |
Factorize `x^4+4` |
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Answer» `=> x^4 +4` `(x^2)^2 + (2)^2` `= (x^2)^2 + (2)^2 + 2(x^2)2 - 2(x^2)2` `= (x^2 +2)^2 - (2x)^2 ` using `(a+b)^2 = a^2 + b^2` now by using `a^2 - b^2 = (a+b)(a-b) ` `= (x^2 +2+2x)(x^2+2-2x)` answer |
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| 263. |
Time period of f(x) = |sin x| + |cos x| is : |
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Answer» `f_1(x)=T_1,f_2(x)=T_2` `f_1(x)+f_2(x)=L.C.M of (T_1,t_2)` `T(|sinx|)=pi` `T(|cosx|)=pi` `T(|sinx|+|cosx|)=L.C.M(pi,pi)=pi`. |
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| 264. |
The value of `int_-pi^pi cos^2x/[1+a^x].dx`,a>0 is |
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Answer» Here, `I =f(x) = int_(-pi)^(pi)cos^2x/(1+a^x)dx->(1)` `I = f(-x) = int_(-pi)^(pi) cos^2(-x)/(1+a^-x)dx` `I= int_(-pi)^(pi) cos^2 x/(1+1/a^x)dx` `I= int_(-pi)^(pi) (a^x cos^2x)/(1+a^x)dx->(2)` Adding (1) and (2), `2I = int_(-pi)^(pi) (cos^2 x)(1+a^x)/(1+a^x)` `=> 2I = int_(-pi)^(pi) (cos^2 x)->(3)` Now, as `cos^2 x` is an even function, `:. int_(-pi)^(pi) (cos^2 x) =2 int_(0)^(pi) (cos^2 x)` So, (3) becomes, `:. 2I = 2 int_(0)^(pi) (cos^2 x)` Now, using `cos2x = 2cos^2x -1` in our expression, `=> I = int_(0)^(pi) 1/2 (1+cos2 x)` `=> I =1/2(x+ 2sin(2x)/2)_(0)^(pi)` `=>1/2(pi-0) = pi/2` `:. I = pi/2` |
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| 265. |
Find the vector equation of line passing through point (1,2,-4) and perpendicular to two lines `(x-8)/3 = (y+19)/-16 = (z-10)/7 and (x-15)/3 = (y-25)/8 =(z-5)/-5` |
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Answer» `L_3_|_L_1,L_3_|_L_1` `P_3_|_P_1,P_3_|_P_2` `P_3=vecP_1*vecP_2` `=hati(24)-hatj36+72hatk` `vecP_3=24hati+36hatj+72hatk` `vecP_3=12(2hati+3hatj+6hatk)` `A(1,2,-4)` `vecr-vecA=lambdavecP_3` `(xhati+yhatj+zhatk)-(hati+2jatj-4hatk)=lambda(2hati+3hatj+6hatk)` `(x-1)/2=(y-2)/3=(z+4)/6` |
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| 266. |
State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly.(i) If `P = {m , n} " and " Q = {n , m}, " then " P xxQ = {(m , n), (n , m)}`. |
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Answer» (i) false `p*q = {m,m),(m,n),(n,m),(n,n)}` (ii) true (iii) true |
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| 267. |
Let f be the subset of `Z xxZ`defined by `f = {(a b , a + b) : a , b in Z}`. Is f a function from Z to Z? Justify your answer. |
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Answer» `f(ab) = a+b` `f(12) = f(2*6) = 2+6=8` `f(4*3) = 3+4 = 7` `f(1*12) = 1+12 = 13` so this relation is not a function |
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| 268. |
The domain of the function f given by `f(x)=(x^(2)+2x+1)/(x^(2)-x-6)`A. `R-{3,-2}`B. `R-{-3,2}`C. `R-[-3,-2]`D. `R-[-3,-2]` |
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Answer» Correct Answer - A We have, `f(x)=(x^(2)+2x+1)/(x^(2)-x-6)` f(x) is defined , if `x^(2)-x6=0` `rArr x^(2)-3x+2x-6=0` `rArrx(x-3)+2(x-3)=0` `rArr (x-3)(x+2)=0` `:. x=-3,-2` `:.` Domain of f=R-{3,-2}` |
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| 269. |
Find the domain and range of the following real functions:(i) `f(x)=-|x|` (ii) `f(x)=sqrt(9-x^2)` |
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Answer» 1)f(x)=-|x| Domain={R}=R Range={y}=negative part of R 2)y=f(x) `y^2=9-x^2` `x^2+y^2=9` Domain={x}=(-3,3) Range={y}=(-3.3) |
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| 270. |
Find the domain of the function `f(x)`defined by `f(x)=sqrt(4-x)+1/(sqrt(x^2-1))`.A. `(-oo,-1)uu(1,4]`B. `(-oo,-1]uu(1,4]`C. `(-oo,-1)uu[1,4]`D. `(-oo,-1)uu[1,4)` |
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Answer» Correct Answer - A We have, `f(x)=sqrt(4-x)+(1)/(sqrt(x^(2)-1))` f(x) is defined, if `4-xge0orx^(2)-1gt0` `x-4le0 or (x-1)(x-1)gt0` `xle4 or xlt-1 and xgt1` `:.` Domain of `f=(-oo,-1)uu(1,4]` |
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| 271. |
Let `f`be a function defined on [0,2]. The prove that the domain of function `g(x)i sf(9x^2-1)dot` |
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Answer» Correct Answer - `[-(1)/(sqrt(3)),-(1)/(3)] cup [(1)/(3),(1)/(sqrt(3))]` g is meaningful if `0 le 9x^(2) -1 le 2 hArr 1 le 9x^(2) le 3` i.e., ` x in [-(1)/(sqrt(3)),-(1)/(3)] cup [(1)/(3),(1)/(sqrt(3))]` |
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| 272. |
Find the rang of y = cos(2sinx) |
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Answer» `-1<=sinx<=1` `-2<=2sinx<=2` `y=cos(2sinx)` `y in [cos2,1]`. |
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| 273. |
`f: RvecR ,f(x^2=x+3)+2f(x^2-3x+5)=6x^2-10 x+17AAx in R ,`then find the function `f(x)dot` |
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Answer» Correct Answer - `f(x)=2x-3` Obviously, `f` is a linear polynomial. Let `f(x)=ax+b.` Hence, `f(x^(2)+x+3)+2f(x^(2)-3x+5)=6x^(2)-10x+17` or ` [a(x^(2)+x+3)+b]+2[a(x^(2)-3x+5)+b]=6x^(2)-10x+17` or `a+2a=6 " (1) " ` and `a-6a= -10 " (2)" ` or `a=2 " " ` (Comparing coeff. of `x^(2)` and coeff. of `x` both sides) Again, `3a+b+10a+2b=17` or `6+b+20+2b=17` ` :. b= -3` or `f(x)=2x-3` |
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| 274. |
If `f(x)`is an even function and satisfies the relation `x^2dotf(x)-2f(1/x)=g(x),w h e r eg(x)`is an odd function, then find the value of `f(5)dot` |
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Answer» `x^(2)f(x)-2f((1)/(x))=g(x) " and " 2f((1)/(x))-4x^(2)f(x)=2x^(2)g((1)/(x))` or `-3x^(2)f(x)=g(x)+2x^(2)g((1)/(x))` or ` f(x)=((g(x)+2x^(2)g((1)/(x)))/(3x^(2)))` Since g(x) is odd `f(-x)= -f(x).` But given that `f(x)` is even. ` :. f(x)=0` ` :. f(5)=0` |
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| 275. |
If `f(x)`satisfies the relation `f(x)+f(x+4)=f(x+2)+f(x+6)`for all`x ,`then prove that `f(x)`is periodic and find its period. |
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Answer» Given `f(x)+f(x+4)=f(x+2)+f(x+6) " (1) " ` Replace x by `x+2`. Then `f(x+2)+f(x+6)=f(x+4)+f(x+8) " (2)" ` From (1) and (2), we have `f(x)=f(x+8).` Hence, `f(x)` is periodic with period 8. |
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| 276. |
Show that the relation `R`on the set `A`of points in a plane, given by `R={(P , Q):`Distance of the point `P`from the origin is same as the distance of thepoint `Q`from the origin}, is an equivalence relation.Further show that the set of all points related to a point `P!=(0, 0)`is the circle passing through `P`with origin as centre. |
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Answer» `R = {(P, Q):` distance of point `P` from the origin is the same as the distance of point `Q` from the origin} Clearly, `(P, P) in R` since the distance of point `P` from the origin is always the same as the distance of the same point `P` from the origin. `:.` `R` is reflexive. Now,Let `(P, Q) in R.` `=>` The distance of point `P` from the origin is the same as the distance of point `Q` from the origin.`=>` The distance of point `Q` from the origin is the same as the distance of point `P` from the origin. `=> (Q, P) in R`. `:.` `R` is symmetric. Now, Let `(P, Q), (Q, S) in R` The distance of points `P` and `Q` from the origin is the same and also, the distance of points `Q` and `S` from the origin is the same. `:.` The distance of points P and S from the origin is the same. `:. (P, S) in R`. `:. R` is transitive. As `R` is reflexive, summetric and transitive, `R` is an equivalence relation. The set of all points related to `P != (0, 0)` will be those points whose distance from the origin is the same as the distance of point `P` from the origin. In other words, if `O (0, 0)` is the origin and `OP = r`, then the set of all points related to ` P` is at a distance of `r` from the origin. Hence, this set of points forms a circle with the centre as the origin and radius `r` and this circle passes through point `P`. |
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| 277. |
If a vertex of an equilateral triangle is the origin and the side opposite to it has the equation x+y=1, then orthocentre of the triangle is : |
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Answer» `ax+by+c=0` `(x1,y1)` `(x-x1)/a=(y-y1)/b=-(ax1+by1+c)/(a^2+b^2)` `x/1=y/1=-(-1)/(1^2+1^2)` `x=1/2,y=1/2` `G=(((2xx1/2+1xx0)/(2+1),2xx1/2+1xx0)/(2+1))` `G=(1/3,1/3)` |
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| 278. |
Show that the relation R in the set `A={1,2,3,4,5}` given by ` R={(a,b):|a-b|` is even }, is an equivalence relation. |
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Answer» `A= {1,2,3,4,5}` `R={(a,b):|a-b|` is even } It is clear that for any clement `a in A,` we have `|a - a| = 0` (which is even). Therefore, R is reflexive. Let `(a,b) in R`. `implies |a-b|` is even, ` implies |-(a-b)|=|b-a|` is also even `implies (b,a) in R` Therefore, R is symmetric. Now, let `(a,b) in R ` and `(b,c) in R.` ` implies |a-b|` is even and `|b-c|` is even `implies (a-b)` is even and `(b-c)` is even ` " " ` (assuming that `a gt b gt c`) `implies (a-c)=(a-b)+(b-c)` is even ` " " ` [Sum of two even integers is even] ` implies |a-c|` is even `implies (a,c) in R` Therefore, R is transitive. Hence, R is an equivalence relation. |
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| 279. |
Examine each of the following relations given below and state in each case, giving reasons whether it is a function or not ? (i) `R={(4,1),(5,1),(6,7)}` (ii) `R={(2,3),(2,5),(3,3),(6,6)}` (iii) `R={(1,2),(2,3),(3,4),(4,5),(5,6),(6,7)}` (iv) `R={(1,1),(2,1),(3,1),(4,1),(5,1)}` |
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Answer» (i) Since 4, 5, 6 are the elements of domain of R having their unique images, this relation R is a function. (ii) Since the same first element 2 corresponds to two different images 3 and 5, this relation is not a function. (iii) Since every element has one and only one image, this relation is a function. (iv) Since every element has one and only one image, this relation is a function. |
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| 280. |
Show that the relation R in the set A of points in a plane given by `R = {(P , Q) :`distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set o |
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Answer» `R {(P,Q):` distance of point P from the origin is the same as the distance of point Q from the origin} Clearly, `(P,P) in R` since the distance of point P from the origin is always the same as the distance of the same point P from the origin. Therefore, R is reflexive. Now, let `(P,Q) in R`. ` implies` The distance of point P from the origin is the same as the distance of point Q from the origin. `implies` The distance of point Q from the origin is the same as the distance of point P from the origin. `implies (Q,P) in R` Therefore , R is symmetric. Now, let `(P,Q),(Q,S) in R.` `implies ` The distance of point P and Q from the origin is the same and also, the distance of point Q and S from the origin is the same. `rArr` the distance of points P and S from the origin is the same. ` implies (P,S) in R` Therefore, R is transitive. Therefore, R is an equivalence relation. |
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| 281. |
Given a non-empty set X, consider P(X) which is theset of all subsets of X. Define the relation R in P(X) as follows:For subsets A, B in P(X), ARB if and only if A B. Is R an equivalence relation on P(X)?Justify you answer |
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Answer» Since every set is a subset of itself, ARA for all ` A in P(X).` Thererfore, R is reflexive. Let ` ARB implies A subset B.` This cannot be implied to `B subset A`. For instant, if `A={1,2}` and `B={1,2,3},` then it cannot be implied that B is related to A. Therefore, R is not symmetric. Further if ARB and BRC, then ` A subset B` and `B subset C`. `implies A subset C` `implies ARC` Therefore, R is transitive. Hence, R is not an equivalence relation since it is not symmetric. |
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| 282. |
Show that the relation R defined in the set A of all triangles as `R={(T_(1),T_(2)):T_(1)` is similar to `T_(2)`}, is equivalence relation. |
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Answer» `R={(T_(1),T_(2)):T_(1)` is similar to `T_(2)`} R is reflexive since every triangle is similar to itself. Further, if `(T_(1),T_(2)) in R,` then `T_(1)` is similar to `T_(2)`. `implies T_(2)` is similar to `T_(2)` `implies (T_(2),T_(1)) in R`. Therefore, R is symmetric. Now, let `(T_(1),T_(2)),(T_(2),T_(3)) in R`. `implies T_(1)` is similar to `T_(2)` and `T_(2)` is similar to `T_(3)` `implies T_(1)` is similar to `T_(3)` `implies (T_(1), T_(3)) in R` Therefore, R is transitive. Thus, R is an equivalence relation. |
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| 283. |
Relation " parallel" in the set of all straight lines in a plane is :A. only reflexiveB. only symmetricC. only transitiveD. equivalence relation |
| Answer» Correct Answer - D | |
| 284. |
Relation " similar" in triangles in a plane is :A. reflexive, symmetric , transitiveB. reflexive, transitive but not symmetricC. symmetric , transitive but not reflexiveD. none of the above |
| Answer» Correct Answer - A | |
| 285. |
If n(A)=3 and n(B)=4 , then no. of of one-one function from A to B is :A. 12B. 24C. 36D. none of these |
| Answer» Correct Answer - B | |
| 286. |
If `f(x)=(x-1)/(x+1)` then `f(2x)` is equal toA. `(1+f(x))/(3+f(x))`B. `(1+3f(x))/(3+f(x))`C. `(3+f(x))/(1+f(x))`D. `(1+3f(x))/(3-f(x))` |
| Answer» Correct Answer - B | |
| 287. |
If R and S are two non-empty relations on set A, then incorrect statement is :A. R and S are reflexive , then `R cap S ` is also reflexive .B. R and S are symmetric , then `R cup S ` is also symmetricC. R and S are transitive , then `R cap S ` is also transitive .D. R and S are transitive , then `R cup S ` is also transitive. |
| Answer» Correct Answer - B | |
| 288. |
The number of proper sub sets of the set `{3, 4, 5, 6, 7}` is |
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Answer» Let `A = {3,4,5,6,7}` Number of proper subsets of a set containing `n` elements can be given as `2^n-1`. Here, `A` contains `5` elements. `:. n = 5` So,number of proper subsets of set `A` will be `=2^5 - 1 = 31.` |
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| 289. |
Let `A" "=" "{-1," "0," "1," "2}`, `B" "=" "{-4," "-2," "0," "2}`and `f,g:" "A" "->" "B`befunctions defined by `f(x)=x^2-x ,""""x in A`and `g(x)=2|x-(1/2)|-1, x in A`. Are f and g equal? Justify your answer. (Hint: One may note that two functio |
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Answer» Here A={-1, 0, 1,2} and B={-4, -2, 0 ,2} In f,g:`A rarr B,f(x) =x^2-x` and `g(x)=2|x -1/2|-1, x in A` Now `f(-1)=(-1)^2-(-1)=1+1=2` and `g(-1)=2 |-1-1/2|-1=3-1 =2` `therefore f(1-)=g(-1)` `f(0) =0^2-0=0` and g(0) =2|0-1/2|-1 = 1-2=0 `therefore f(0)=g(0)` and `g(1)=2|1-1/2|-1=2 (1/2)-1=0` `therefore f(1)=g(1)` `f(2)=2^2-2=4-2=2` `and g(2)=|2-1/2|-1=3-1=2` `therefore f(2)=g(2)` Therefore , `AA a in A,` `f(a)=g(a)` `rArr ` f and g are equal |
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| 290. |
A relation `R={(x,y):x,y in A and x lt y}` is defined on set A={1,2,3,4,5}. The relation R is :A. reflexiveB. symmetricC. transitiveD. equivalence |
| Answer» Correct Answer - C | |
| 291. |
If`f:R to A ,` where A =[-1,1] , is defined as `f(x)=cos x, ` thn find f isA. intoB. one-oneC. ontoD. none of these |
| Answer» Correct Answer - C | |
| 292. |
let `f(x)={x^2+ax+b, 0ltxlt2 and 3x+2 , 2lexlt4 and 2ax+5b , 4lexlt8 `. if f(x) is continuous on close interval`(0,8) `.find value of` a and b.` |
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Answer» Here, value of `f(x)` is changing at `2` and `4`. So, `f(x)` to be continuous, `f(2^-) = f(2^+)` `=>2^2+2a+b = 3(2)+2` `=>4+2a+b = 8` `=>2a+b = 4->(1)` Also, `f(x)` to be continuous, `f(4^-) = f(4^+)` `=>3(4) + 2 = 2a(4) +5b` `=>14 = 8a+5b` `=> 8a+5b = 14->(2)` Multiplying (1) with `4` and then subtracting (1) from (2), `8a+4b-8a-5b = 16 - 14` `=> b = -2` Putting `b = -2` in (1), `2a-2 = 4 => a = 3` So, for `a = 3` and `b = -2`, `f(x)` will be continuous. |
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| 293. |
If the function `f: Rsetminus``{0}vec`given by`f(x)=1/x-2/(e^(2x)-1)`is continuous at `x=0,`then find the value of `f(0)` |
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Answer» `F(x)=1/x-2/(e^2x-1)` `lim_(x-6^+)F(x)=lim_(x+6^-)F(x)=F(0)` `lim_(x-0)F(x)=lim_(x+0)(1/11-2/(e^(2x)-1))` `=lim_(x-0) (e^(2x)-1-2x)/(x(e^2x-1))` `=(0/0)L`hospital `=lim_(x-0) (e^(2x)xx2-0-2)/(x(e^(2x)xx0-0)+e^(2x))` `=lim_(x-0)(2e^(2x)-2)/(2xe^2x+e^2x-1)` `(0/0)L`hospital `=lim_(x-0) (2e^(2x)xx2-0)/2(xe^(2x)xx2+e^2x)+e^2xxx2-0` `=lim_(x-0)(4e^2x)/4xe^2x+4e^2x` `=4/4` option(4)`=1` |
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| 294. |
Let `f(x) = x+f(x-1)` where `xepsilonR`. If `F(0)=1` find `f(100)` |
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Answer» Given `f(x)=x+f(x-1) " and " f(0)=1` Put `x=1`. Then, `f(1)=1+f(0)=2` Put `x=2`. Then, `f(2)=2+f(1)=4` Put `x=3`. Then, `f(3)=3+f(2)=7` Thus, `f(0),f(1),f(2), …" form a series " 1,2,4,7, …. ` Let `S=1+2+4+7+ … +f(n-1)` `S=1+2+4+ ... +f(n-2)+f(n-1)` Subtracting , we get `0=(1+1+2+3+...+n" terms")-f(n-1)` ` :. f(n-1)=(n(n+1))/(2)` ` :. f(100)=5051` |
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| 295. |
The function f is defined as : `f(x)={{:(1","x gt0),(0"," x =0),(-1","x lt 0):}` The range of f is :A. {1,0}B. {0,-1}C. {1,-1}D. {1,0,-1} |
| Answer» Correct Answer - D | |
| 296. |
Given a non -empty set X, let `*:" "P(X)" "xx" "P(X) ->P(X)`be defined as A * B = (A B) `uu`(B A), `AA`A, B ` in `P(X)`A" "*" "B" "=" "(A" "-" "B) uu(B" "-" "A), AA""""A ," "B in P(X)`. Show that theempty set `varphi`is the identity for the |
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Answer» Operation on set P(X) is `A ** B =(A-B) cup (B-A) AA A ,B in p(X)` Let `A in P(X)` `therefore A ** phi =(A - phi ) cup (phi -A)` and `phi ** A =(phi-A)cup (A- phi)` `therefore A ** =A = phi ** A` `rArr phi` is the identity element in the binary peration .operation Again `A in P(X)` will be invertible if and only if exists an element B in P(X) such that `A ** B =Phi = B**A ` Now `Lambda ** Lambda=(A-A)cup (A-A)` `= phi cup phi = phi ` `therefore` All elements A of P(X) are invertible and `A^(-1)=A` |
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| 297. |
The function f(x) is defined for all real x. If `f(a+b)=f(ab) AA a " and " b " and " f(-(1)/(2))=-(1)/(2)` then find the value of `f(1005).` |
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Answer» We have `(f(a+b)=f(ab) AA a " and " b` Put ` a=0` ` :. f(b)=f(0)=k ` (say) Putting `b=0` in given relation, `f(a)=f(0)=k` Thus `f(a)=f(b)=k AA a,b` Hence, `f(x)` is a constant function. ` :. f(1005)=f(-(1)/(2))` `=-(1)/(2)` |
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| 298. |
If `f: N to N ` defined as `f(x)=x^(2)AA x in N, ` then f is :A. many-oneB. one-oneC. ontoD. none of these |
| Answer» Correct Answer - B | |
| 299. |
If `(f(x))^2+f((1-x)/(1+x))=64 x AA in D_f` thenA. `4x^(2//3)((1+x)/(1-x))^(1//3)`B. `x^(1//3)((1-x)/(1+x))^(1//3)`C. `x^(1//3)((1-x)/(1+x))^(1//3)`D. `x((1+x)/(1-x))^(1//3)` |
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Answer» Correct Answer - A `(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)` |
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| 300. |
If `(f(x))^(2)xxf((1-x)/(1+x))=64x AAx in D_(f),` then The domain of `f(x)` isA. `[0,oo)`B. `R-{1}`C. `(-oo,oo)`D. None of these |
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Answer» Correct Answer - B `(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)` |
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