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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.
| 401. |
The function `f: R-{0} -> R` given by `f(x)=1/x-2/[e^2x-1]` can be made continuous at x=0 by defining f(0) as |
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Answer» `f(0) = lim_(x-> o-) f(x) = lim_(x->0+) f(x) = lim_(x->0) f(x)` `f(0)= lim_(x->0) f(x) = lim_(x->0) [1/x - 2/(e^(2x) - 1)]` `= lim_(x->0) (e^(2x)-1-2x)/(x(e^(2x) -1))` `= lim_(x->0) (1 + 2x + (2x)^2/(2!) + ..... -1-2x)/(x(1 + 2x + (2x)^2/(2!) +.....-1))` `= lim_(x->0) ((4x^2)/(2!) + (8x^3)/(3!) + ....)/(x^2(2 + (4x)/(2!) + (8x^2)/(3!)+ ....)` `f(0) = (4/(2!))/2 = 4/(2 xx2)` `f(0) = 1` option D is correct |
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| 402. |
`f(x)=1/sqrt(|[|x|-1]|-5)` |
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Answer» `[|x|-1]-5>0` `[|x|-1]>5` `[|x|-1]<-5` `[|x|-1]>5` `|x|-1>=6` `|x|>=7` `x>=7` `x in (-oo,-7] uu[7,oo)`. |
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| 403. |
Find the domain of the function :`f(x)=sin^(-1)((log)_2x)` |
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Answer» Correct Answer - `[1//2,2]` `f(x)=sin^(-1)(log_(2)x)` Since the domain of `sin^(-1)x " is " [-1,1],f(x)=sin^(-1)(log_(2)x)` is defined if `-1 le log_(2)x le 1` or ` 2^(-1) le x le 2^(1)` or `(1)/(2) le x le 2` or domain`=[(1)/(2),2]` |
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| 404. |
Find the domain of the function :`f(x)=sqrt(4^x+8^((2/3)(2x-2))-13-2^(2(x-1)))` |
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Answer» Correct Answer - `[2,oo)` `4^(x)=8^((2)/(3)(x-2))-13-2^(2(x-1)) ge 0` or `4^(x)+(4^(x))/(16)-(4^(x))/(4) ge 13` or `4^(x) ge 4^(2)" or " x in [2,oo)` |
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| 405. |
The decimal equivalent ot the binay number 10011.1 is |
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Answer» `(10011.1)_2 = 1*2^4+0*2^3+0*2^2+1*2^1+1*2^0+1*2^-1` `=16+0+0+2+1+1/2 = 19.5` `:. (10011.1)_2 = (19.5)_10` `:. (10011.1)_2` is `19.5` in decimal representation. |
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| 406. |
Find the domain and range of `f(x)=sin^(-1)x+tan^(-1)x+sec^(-1)x.` |
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Answer» Correct Answer - Domain: Range: `{pi//4,3pi//4}` We have `f(x)=sin^(-1)x +tan^(-1)x +sec^(-1)x` Domain of `sin^(-1)x " is " [-1,1].` Domain of `tan^(-1)x ` is R. Domain of `sec^(-1)x " is " R-(-1,1)` So, domain `f(x) " is " {-1,1}` `f(-1)=sin^(-1)(-1)+tan^(-1)(-1)+sec^(-1)(-1)` ` = -pi//2-pi//4+pi=pi//4` ` f(1)=sin^(-1)(1)+tan^(-1)(1)+sec^(-1)(-1)` `= pi//2+pi//4+0=3pi//4` Hence range is `{pi//4,3pi//4}` |
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| 407. |
Find the range of the function`f(x)=cot^(-1)(log)_(0. 5)(x^4-2x^2+3)` |
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Answer» Correct Answer - `[3pi//4,pi)` `y=(x^(2)-1)^(2)+2 ge 2` or `log_(0.5)(x^(4)-2x^(2)+3) le -1` or `cot^(-1)log_(0.5)(x^(4)-2x^(2)+3) in [(3pi)/(4),pi)` |
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| 408. |
Show that zero is the identity for addition on R and1 is the identity for multiplication on R. But there is no identity elementfor the operations `-: RxxR->R`and `-:: R_*xxR_*->R_*dot` |
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Answer» We know that a+0=a=0+a and ` axx1 =a=1xxa AA a in R ` `implies ` 0 is the additive identity and 1 is the multiplicative identity in R. Now ,there is no element e in R such that `a-e =a=e-a AA a in R ` `:.` For `- : R xx R to R `, there is no identity element in R. Similarly, for `div : R xx R to R `, there is no identity element in R. Hence Proved. |
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| 409. |
Show that the function `f: R->R` given by `f(x)=x^3`is injective. |
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Answer» Here, `f(x) = x^3` `:. f(x_1) = x_1^3` `f(x_2) = x_2^3` Let `f(x_1) = f(x_2)` Then, `x_1^3 = x_2^3` `=>x_1 = x_2` So, if `f(x_1) = f(x_2),` then `x_1 = x_2`. `:.` Given function is injective(one-one). |
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| 410. |
Let`A = {1, 2, 3}` Then number of relations containing `(1, 2)" and "(1, 3)`which are reflexive and symmetric but not transitive is (A) 1 (B) 2 (C) 3 (D) 4 |
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Answer» Here, `A={1,2,3}` `R_1={(1,1),(2,2),(3,3)(1,2)(1,3)(2,1)(3,1)(3,2)(2,3)}` We will work with the relations that contains `(1,2),(3,1)`. Relation R is reflexive as `(1,1)(2,2)(3,3) in R` Relation R is symmetric as `(1,2),(2,1) in R` and `(1,3)(3,1) in R`. Relation R is not transitive since `(3,1)(1,2) in R` but `(3,2) !in R`. Therefore the total number of relation containing `(1,2)(1,3)` which are reflexive ,symmetric but not transitive is `1`. However if we add the pair `(3,2)` and `(2,3)` to relation R then it will become transitive.Therefore, the correct answer is 1 (A). |
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| 411. |
If A={a,b,c,d}, then on A . (i) write the identity relation `I_(A)` . (ii) write a reflexive relation which is not the identity relation. |
| Answer» Correct Answer - (i) {(a,a),(b,b),(c,c),(d,d)} `" "` {(a,a),(b,b),(c,c),(d,d),(a,c),(b,d)} | |
| 412. |
(i) If A= {x,y,z}, B=(1,2,3} and R= {(x,2),(y,3),(z,1),(z,2), then find `R^(-1)`. (ii) If R is a relations such that R ={(4,5),(1,4),(4,6),(7,6),(3,7)}, then find `R^(-1) oR^(-1)` |
| Answer» Correct Answer - (i) `R^(-1)={(2,x),(3,y),(1,z),(2,z)}" " (ii){(5,1),(6,1),(6,3)}` | |
| 413. |
Give an example of a relation.Which is(i) Symmetric but neither reflexive nor transitive.(ii) Transitive but neither reflexive nor symmetric.(iii) Reflexive and symmetric but not transitive.(iv) Reflexive and transitive but not symmetric.(v) Symm |
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Answer» Correct Answer - On A={1,2,3} (i) {(1,1),(1,2),(2,3),(2,2),(3,3)}`" "`(ii) {(1,2),(2,1),(1,3),(3,1)} (iii) {(1,1),(2,2),(1,2),(2,1)}`" "`(iv) {(1,1),(2,2),(3,3),(1,3)} |
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| 414. |
Give an example of a relation. Which is(i) Symmetric but neither reflexive nor transitive.(ii) Transitive but neither reflexive nor symmetric.(iii) Reflexive and symmetric but not transitive.(iv) Reflexive and transitive but not symmetric.(v) Symmetric and transitive but not reflexive. |
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Answer» (i) Relation that is symmetric but neither reflexive nor transitive. `R = {(x,y),(y,x)}` (ii) Realtion that is transitive but neither reflexive nor symmetric. `R = {(x,y),(y,z),(x,z)}` (iii) Relation that is reflexive and symmetric but not transitive. `R = {(x,x),(y,y),(x,y),(y,x)}` (iv) Relation that is reflexive and transitive but not symmetric. `R = {(x,x),(y,y),(z,z),(x,y),(y,z),(x,z)}` (v) Relation that is symmetric and transitive but not reflexive. `R = {(x,y),(y,x),(y,z),(z,y),(x,z),(z,x)}` |
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| 415. |
Show that the Signumfunction `f: R->R`, given by`f(x)={1, if x >0 0, if x=0-1, if x |
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Answer» `f: R to R and f(x) = " " f(x) = {{:(1","if x gt 0),(0","if x =0),(-1","if x lt 0","):}` `because " " f(1) = f(2) = 1 ` ` therefore f ` is not one-one. Again `2 in R` and there does not exist `x in R` for which `f(x) = 2`. `therefore f` is not onto. Therefore, `f` is neither one-one nor onto. |
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| 416. |
Find the equation of the plane passing through the point (1, 3, 5) and perpendicular to each of the planes x + 2y + 3z = 7 and 3x + 3y + z = 0, Also find distance of new plane from (2,3,1). |
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Answer» `p_1=x+2y+3z=7``p_2=3x+3y+z=0``n_1=(1,2,3)``n_2=(3,3,1)`since normal vector is also perpendicular`n2=[[1,j,k],[1,2,3],[3,3,1]] ` on solving, `n_2=-7hat i +8 hat j_3hat` `vec n_3=vec n_1+vec n_2`on solving for `n_3` using (`vec r- vec a)*vec n``-7(x-1)+(y-3)8+(z-5)-3=0` `n3=-7x-8y+3z+2=0`(equation of plane)now the distance;`d=Abs(-5/sqrt(122))`hence solved. |
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| 417. |
Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is `2R/sqrt(3)` . Also find maximum volume. |
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Answer» `h^2=R^2-r^2` Volume of cylinder `r^2=R^2-h^2` `pir^2(2h)=2pir^2h` `2pi(R^2-h^2)h` `(dv)/(dh)=0` `d/dx[2pi(R^2-h^2)h]=0` `(R^2-h^2)+h*(-2h)=0` `R^2-3h^2=0` `h=R/sqrt3` Height of the cylinder at maximum `H=2h=(2R)/sqrt3` `V_(max)=2piR^2h=2pi(R^2-h^2)h` `=2pi(R^2-R^2/3)R/sqrt3` `=2pi(2R)^2/3*R/sqrt3` `(4piR^3)/(3sqrt3)` |
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| 418. |
If `f: R ->R`is defined by `f(x) = x^2- 3x + 2`, find `f(f(x))`. |
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Answer» `f: R rarr R` and `f(x)=x^2-3x+2` `therefore f(f(x))=f(x^2-3x+2)` `(x^2-3x +2)^2-3(x^2-3x+2 )+2` `x^4+9x^2+4- 6x^3+4x^2-12 x - 3x^2+9x - 6 +2` `=x^4-6x^3+10 x^2-3x` |
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| 419. |
Let `f: X rarr Y` be a function defined by f(x) = a sin ( x +`pi/4`) + c. If f is both one-one and onto, then find the set X and Y |
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Answer» `f(x)=a sin(x+pi/4)+bcosx+c` `=(a sinx)/sqrt2+(acosx)/sqrt2+bcosx+c` `=(asinx)/sqrt2+cosx(a/sqrt2+b)+c` `r=(a/sqrt2)^2+(a/sqrt2+b)^2=rsin(x+theta)+c` `Y in[c-r,c+r]` f(x) is one-one `r^2=a^2/2+a^2/2+b^2+sqrt2ab` `r=sqrt(a^2+b^2+sqrt2ab)` option a is correct. |
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| 420. |
Show that the function `f: R rarr { x in R: -1 lt x lt 1 } ` defined by `f(x)=(x)/(1+|x|), x in R` is one- one and onto function . |
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Answer» Let x, `y in R` and , f(x)=f(y) `rArr x/(1+|x|)=y/(1+|y|)` If x is positve and y is negative then `x gt y ` `rArr x-y gt 0 and 2xy lt 0 ` `therefore x/(1+x)=y/(1-y)` `rArr y+ xy =x -xy` `rArr 2xy = x-y` which is impossible then `f(x)=f(y)rArr (x)/(1+x)=(y)/(1+y)` `rArr x+y` If x and y both are negative then `f(x)=f(y)rArr x/(1-x)=y/(1-y)` `rArr x-xy = y-xy` `rArr x=y` Therefore , f is one - one Let y `in ` R be such that `-1 lt y lt 1` If y is negative then `x=y/(1+y) in R` is such that `f(x)=f((y)/(1+y))=((y)/(1+y))/(1+|(y)/(1+y)|)=((y)/(1+y))/(1-y/(1+y))=y` If y is positive then `x=y/(1-y) in R` is such that `f(x)=f(y/(1-y))=((y)/(1-y))/(1+((y)/(1-y)))=((y)/(1-y))/(1-y/(1+y))=y` `therefore` f is onto Therefore f is one - one onto . |
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| 421. |
If `f(x)=1/(1+e^(-1 /x)) ,x!=0 ` and 0,x=0 then at x=0 (A) right hand limit of f(x) exists but not left-hand limit (B) left-hand limit of f(x) exists but not right- hand limit (C) both limits exists but are not equal (D) both limits exist and are equal |
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Answer» LHL=`lim_(x->0-)(1/(1+e^(-1/x)))` `=lim_(h-.0)(1/(1+e^(1/h)))` `=0` RHL=`lim_(x->0+)1/(1+e^(-1/x))` `=lim_(h->0)1/(1+e^(-1/h)` `=1/(1+0)` LHL`!=`RHL option c is correct. |
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| 422. |
Find the values of x for which expression `sqrt(1-sqrt(1-sqrt(1-x^(2))))` is meaningful. |
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Answer» `sqrt(1-sqrt(1-sqrt(1-x^(2))))` is meaningful if `1-sqrt(1-sqrt(1-x^(2))) ge 0` or `sqrt(1-sqrt(1-x^(2))) le 1` or `1-sqrt(1-x^(2)) le 1` or `sqrt(1-x^(2)) ge 0` or `1-x^(2) ge 0` or `x^(2) le 1 " or "x in [-1,1]` |
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| 423. |
The coordinate of a moving point at time t are given by x = a ( 2t + sin 2t) , y = a ( 1-cos 2t). Prove that acceleration is constant. |
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Answer» x=a(2t+sin2t) `dx/dt=2a+a-cos2t*2` `d^x/dt=0+2a-sin2x2i` `=-4asin2thati` `dy/dt=asin2t*2` `d^y/dt^2=2a*2cos2t` `=4acos2thatj` `|veca|=sqrt(a_x^2+a_y^2)` `=sqrt(16a^2sin^2t+16a^2cos^2 2t)` `=sqrt(16a^2)` =4a. |
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| 424. |
If the tangent at a point `P` with parameter `t`, on the curve `x=4t^2+3`, `y=8t^3-1` `t in R` meets the curve again at a point Q, then the coordinates of Q are |
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Answer» `x=4t^2+3` dif. with respect to t `y=8t^2-1` difff. wih respect to t `dy/dx=(dy/dt)*(dx/dt)=(24t^2)/(8t)=3t` `(y-y_1)=(8t)*(x-x_1)` `9-(8t_1^3-1)-3t_1{n-t_1^2-3}` `8t^3-1-8t_1^3+1=3t_1{4t^2+3-4t_1^2-3}` `8(t-t_1){t^2+t_1^2+tt_1}-12t_1{t-t-1}(t_1+t)` `8t^2-4t_1^2-4t_t_1=0` `2t^2-t_1^2-tt_1=0` `(t^2-t_1)+t^2-tt_1=0` `(t-t_1)(t+t_1)+t(t-t_1)=0` `t+t_1+t=0` `t=-t_1/2` `x=4*(t_1/2)^2+3=3+t_1^2` `y=8*(-t_1/2)^3-1` `=t_1^3-1` option d is correct. |
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| 425. |
`[5cos theta+3cos(theta+pi/3)+3]` lies between (A) 4 and 10 (B) -4 and 10 (C) -10 and 4 (D) None of the above |
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Answer» `3cos(theta+pi/3)=3{costhetacospi/3-sinthetasinpi/3}` `=3costheta*1/2-3sinthetasqrt3/2` `=3/2costheta-3sqrt3/2sintheta` `5costheta+3cos(theta+pi/3)+3` `=5costheta+3/2costheta=3sqrt3/2sintheta+3` `=1/2(13costheta-3sqrt3sintheta)+3` `=sqrt(169+27)/2*cos(theta+phi)+3` `=sqrt196/2cos(theta+phi)+3` `=7cos(theta+phi)+3` `-4<=7cos(theta+phi)+3<=10` option c is correct. |
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| 426. |
If `g(x)=x^2+x-2a n d1/2gof(x)=2x^2-5x+2,`then which is not a possible `f(x)?``2x-3`(b) `-2x+2``x-3`(d) None of theseA. `2x-3`B. `-2x+2`C. `x-3`D. None of these |
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Answer» Correct Answer - C `(1)/(2)(gof)(x)=2x^(2)-5x+2` ` or (1)/(2)g[f(x)]=2x^(2)-5x+2` ` :. [{f(x)}^(2)+{f(x)}-2]=2[2x^(2)-5x+2]` ` or f(x)^(2)+f(x)-(4x^(2)-10x+6)=0` ` :. f(x)=(-1+-sqrt(1+4(4x^(2)-10x+6)))/(2)` `=(-1+-sqrt((16x^(2)-40x+25)))/(2)=(-1+-(4x-5))/(2)` `=2x-3 or -2x+2` |
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| 427. |
The domain of `f(x)=cos^(-1) ( 2 /(2+sin x))` contained in `[0,2pi]` |
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Answer» `-1<=2/(2+sinx)<=1` `1<=2+sinx<=3` `1/3<=1/(2+sinx)<=1` `2/3<=2/(2+sinx)<=2`. |
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| 428. |
`Det[[a^2,bc,ac+c^2],[a^2+ab,b^2,ac],[ab,b^2,c^2]]=4a^2b^2c^2` |
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Answer» `L.H.S. = |[a^2,bc,ac+c^2],[a^2+ab,b^2,ac],[ab,b^2+bc,c^2]|` Applying `R_1->R_1-R_2-R_3` `= |[-2ab,-2b^2,0],[a^2+ab,b^2,ac],[ab,b^2+bc,c^2]|` `=-2 |[ab,b^2,0],[a^2+ab,b^2,ac],[ab,b^2+bc,c^2]|` Now applying `R_2->R_2-R_1,R_3 ->R_3-R_1` `= -2 |[ab,b^2,0],[a^2,0,ac],[0,bc,c^2]|` `=-2abc |[a,b,0],[a,0,c],[0,b,c]|` `=-2a^2b^2c^2 |[1,1,0],[1,0,1],[0,1,1]|` Now expanding the determinant, `=-2a^2b^2c^2(1(-1)-1(1)+0)` `=-2a^2b^2c^2(-2)` `=4a^2b^2c^2=R.H.S.` |
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| 429. |
In `Delta`ABC if `2c cos^2(A/2)+ 2a Cos^2(C/2)` - 3b = 0. Prove that a, b, c are in AP |
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Answer» Given equation is , `2c cos^2(A/2)+2a cos^2(C/2) - 3b = 0` We know, `cos2x = 2cos^2x - 1 => 2cos^2x = 1+cos2x` `:. c(1+cosA)+a(1+cosC) - 3b = 0` `=>c(1+(b^2+c^2-a^2)/(2bc))+a(1+(a^2+b^2-c^2)/(2ab)) - 3b = 0` `=>c((2bc+b^2+c^2-a^2)/(2bc))+a((2ab+a^2+b^2-c^2)/(2ab)) - 3b = 0` `=>1/(2b)(2bc+b^2+c^2-a^2+2ab+a^2+b^2-c^2) - 3b =0` `=>1/(2b)(2b^2+2b(c+a)) - 3b = 0` `=>a+b+c - 3b = 0` `=>a+c = 2b` `=> b = (a+c)/2` So, `a,b and c` are in `A.P.` |
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| 430. |
If the graph of `y=f(x)`is symmetrical about the lines `x=1a n dx=2,`then which of the following is true?`f(x+1)=f(x)`(b) `f(x+3)=f(x)``f(x+2)=f(x)`(d) None of theseA. `f(x+1)=f(x)`B. `f(x+3)=f(x)`C. `f(x+2)=f(x)`D. none of these |
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Answer» Correct Answer - C From the given data, `f(1-x)=f(1+x) " (1)" ` ` and f(2-x)=f(2+x) " (2)" ` In (2), replacing x by `1 +x`, we have `f(1-x)=f(3+x)` ` or f(1+x)=f(3+x) " [From (1)]" ` ` or f(x) = f(2+x)` |
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| 431. |
If `f(x)={x^2 sin((pi x)/2), |x|=1` then `f(x)` isA. an even functionB. an odd functionC. a periodic functionD. None of these |
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Answer» Correct Answer - B `f(-x)={((-x)^(2)"sin"(pi (-x))/(2)",",|-x| lt 1),((-x)|-x|",",|-x| ge 1):},` `={(-x^(2)"sin"(pi x)/(2)",",|x| lt 1),(-x|x|",",|x| ge 1):}`, `= -f(x)` |
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| 432. |
If `f(x)=x^m n ,n in N ,`is an even function, then `m`iseven integer(b) odd integerany integer (d) `f(x)-e v e ni snotpos s i b l e`A. even integerB. odd integerC. any integerD. `f(x)`-even is not possible |
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Answer» Correct Answer - A Given `f(x)=sqrt(n)(x^(m)),n in N,` in an even function where `m in I`. So, `f(x)=f(-x)` `or root(n)(x^(m))=root(n)((-x)^(m))` ` or x^(m)=(-x)^(m)` i.e., m is an integer ` or m=2k, k in I` |
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| 433. |
Prove that `x^3 + x^2 + x` is factor of `(x+1)^n - x^n -1` where n is odd integer greater than 3, but not a multiple of 3. |
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Answer» `x^3+x^2+x=(x+1)-x^2-1` `x[1+x+x^2]` `1+w+w^2=0` `x(x-w)(x-w^2)` When x=0 `(0+1)^n-(0)^n-1=0` when x=w `(1+w)^n-w^n-1` `=(-w^2)^n-w^n-1` `=(-w)^n-(w)^n-(w^3)^n` `1+w+w^2=0` `1+w^2+w^(2n)=0` Only when n is odd n is not multiple of 3 `(-w)^n-(w^2)^n-1` n is odd. |
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| 434. |
The minimum number of times a fair coin needs to be tossed, so that the probability of getting at least two heads is at least 0.96 is : |
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Answer» `p=1/2,q=1-1/2=1/2` `P(r)=nC_r*p^r*q^(n-r)` r success in n trials Required probability=1-P(1)-P(0) `P(1)=nC_1*(1/2)^11*(1/2)^(n-1)` `P(1)=nC_1*(1/2)^n` `P(0)=nC_0*(1/2)^0*(1/2)^n` `=(1/2)^n` `P=1-n*1/(2^n)-1/(2^n)` `=1-1/(2^n)(n+1)>=0.96` `=11-0.96>=1/(2^n)*(n+1)` `0.04>=(n+1)/(2^n)` `2^n/(n+1)>=25` Which is possible when `n>=8` `n=8`. |
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| 435. |
Let m and N be two 3x3 matrices such that MN=NM. Further if `M!=N^2` and `M^2=N^4` then which of the following are correct. |
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Answer» `M!=N^2` `M^2=N^4` `M^2-N^4=0` `M^2-(N^2)^2=0` `(M-N^2)(M+N^2)=0` `M-N^2!=0` `M+N^2=0` `|M+N^2|=0` `|M^2+MN^2|=0` `|A|=0` `|A|U=0` It is possible for infinite matrices. Here,(M+MN^2)U=zero matrix. Option A and B is correct. |
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| 436. |
Let `AB` be a chord of the circle `x^2+y^2=r^2` subtending a right angle at the center. Then the locus of the centroid of the `Delta PAB` as `P` moves on the circle is(1) A parabola(2) A circle(3) An ellipse(4) A pair of straight lines |
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Answer» `A{rcos(theta+90^0),rsin(theta+90^0)}` `=A(-rsintheta,rcostheta)` `theta=0` `B(r,0),A(0,r)` Contract of`/_PAB is (alpha,beta)` `alpha=(r+0+rcostheta)/3=3alpha-r=rcostheta-(1)` `beta=(0+r+rsintheta)/3=3beta-r=rsintheta-(2)``(3alpha-r)^2+(3beta-r)^2=r^2cos^2theta+r^2sin^2theta=r^2` `(alpha-r/3)^2+(beta-r/3)^2=(r/3)^2` `(r/3,r/3),(r/3)` option b is correct. |
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| 437. |
If `f(x)`is an invertible function and `g(x)=2f(x)+5,`then the value of `g^(-1)(x)i s``2f^(-1)(x)-5`(b) `1/(2f^(-1)(x)+5)``1/2f^(-1)(x)+5`(d) `f^(-1)((x-5)/2)`A. `2f^(-1)(x)-5`B. `(1)/(2f^(-1)(x)+5)`C. `(1)/(2) f^(-1)(x)+5`D. `f^(-1)((x-5)/(2))` |
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Answer» Correct Answer - D Replacing x by `g^(-1)(x),` we get `x=2f(g^(-1)(x))+5` ` :. f(g^(-1)(x))=(x-5)/(2)` ` :. g^(-1)(x)=f^(-1)((x-5)/(2))` |
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| 438. |
The equation of the diameter of the circle `x^2+y^2+2x-4y-11=0` which bisects the chords intercepted on line 2x-y+3=0 is:a. x+y-7=0b. 2x-y-5=0c. x+2y-3=0d. none of these |
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Answer» `x^2+y^2+2x-4y-11=0` `x^2+y^2+2gx+2ffy+c=0` `g=c` `f=-2` Centre(-g,-f)=(-1,2) `/_OAD cong /_OAB` OA=OB AD=DB OD=OD `m_1*m_2=-1` `m_2=-1/2` `y=-1/2x+c` `Z=-1/2*(-1)+c` c=3/2 `y=(3-x)/2` `2y=3-x` `x+2y=3` option 3 is correct. |
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| 439. |
If `f: N to N, and x_(2) gt x_(1) implies f(x_(2)) gt f(x) AA x_(1), x_(2) in N and f(f(n))=3n AA n in N," then " f(2)=` _______. |
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Answer» Correct Answer - 3 `f(3n)=f(f(f(n))) = 3f(n) AA n in N` Putting `n=1, f(3)=3f(1)` If `f(f(1))=3 " giving " 1=3` which is absurd. Therefore, `f(1) ne 1.` Thus, `3=f(f(1)) gt f(1) gt 1` So, `f(1)=2` `f(2)=f(f(1)=3` |
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| 440. |
The number of integral values of `a`for which `f(x)="log"((log)_(1/3)((log)_7(sinx+a)))`is defined for every real value of `x`is ________ |
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Answer» Correct Answer - 3 `log_(1//3) log_(7)(sinx+a0 gt 0` or `0 lt (sinx +a) lt 7 AA x in R ` [a should be less than the minimum value of `7-sinx` and a must be greater than the maximum value of ` 1-sinx`] `implies 1-sinx lt a lt 7-sinx AA x in R` ` 2 lt a lt 6` |
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| 441. |
Two trains starting simultaneously from P&Q towards Q and P resp at 8 am. They cross each other at 12 noon. Train starting from Q, Thereafter takes 6 hours to reach P. On a particular day, the train starting from P, reduced its speed and arrived at Q, 200 minutes late. At what time did the trains cross each other on that day? |
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Answer» `4V_1=6V_2` `V_2=(V_1)/3` `V_1t=V_2 4` `V_1=8/3h` lime(Normal)=`4+8/3=20/3h` Specific lime=`20/3+200/60=10h` `d=V_1*20/3=V*10` `V_1=(2V_2/3)` `d=Vt+V_2t` `20/3V_1=4/3V_1t` `t=5 hours` Multiple time=1P.M. |
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| 442. |
Let `(x_0,y_0)` be the solution of the following equations: `(2x)^ln2 = (3y)^ln3 and 3^(lnx) = 2^(lny)`Then value of `x_0` is: |
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Answer» `(2x)^(ln2) = (3y)^ln3` Taking `ln` both sides, `=>(ln2) (ln2x) = (ln3)(ln3y)` `=>(ln2) (ln2 +lnx) = (ln3)(ln3+lny)->(1)` Now, we will take the second equation, `3^(ln x) = 2^(ln y)` Taking `ln` both sides, `=>(ln x)(ln 3) = (ln y)(ln 2)` `=> (ln y) = ((ln x)(ln 3))/(ln 2)` Putting value of `ln y` in (1), `(ln2) (ln2 +lnx) = (ln3)(ln3+((ln x)(ln 3))/(ln 2))` `=>(ln x)(ln 2- ((ln 3)^2/ln2)) = (ln3)^2-(ln2)^2` `=>(ln x)/(ln 2)((ln 2)^2- (ln 3)^2) = (ln3)^2-(ln2)^2` `=>(ln x)/(ln 2) = -1` `=>(ln x) = ln (2)^-1` `=> x = 2^-1 => x = 1/2`, which is the required value of `x_0`. |
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| 443. |
If `z_1 and bar z_1` represent adjacent vertices of a regular polygon of n sides where centre is origin and if `(Im(z))/(Re(z)) = sqrt(2) - 1`, then n is equal to: (A) 8 (B) 16 (C) 24 (D) 32 |
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Answer» Please refer to the diagram in the video. From the diagram, `tan theta = y/x` `tan theta = sqrt2-1` Now, `(2pi)/n = 2theta` `=>tan((2pi)/n) = tan2theta` `=>tan((2pi)/n) = (2tantheta)/(1-tan^2theta)` `=>tan((2pi)/n) = (2(sqrt2-1))/(1-(sqrt2-1)^2)` `=>tan((2pi)/n) = (2(sqrt2-1))/(1-(2+1-2sqrt2))` `=>tan((2pi)/n) = (2(sqrt2-1))/(2(sqrt2-1))` `=>tan((2pi)/n) = 1 = tan(pi/4)` `=>(2pi)/n = pi/4` `=> n = 8` So, option `A` is the correct option. |
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| 444. |
The function `f(x)=(x+1)/(x^3+1)`can be written as the sum of an even function `g(x)`and an odd function `h(x)`. Then the value of `|g(0)|`is___________ |
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Answer» `g(x)=(f(x)+f(-x))/(2)` `=(1)/(2)[(x+1)/(x^(3)+1)+(1-x)/(1-x^(3))]` `=(1)/(2)[(1)/(x^(2)-x+1)+(1)/(1+x+x^(2))]` `=(1)/(2)[(2(x^(2)+1))/((x^(2)+1)^(2)-x^(2))]` `=(x^(2)+1)/(x^(4)+x^(2)+1)` ` :. g(0)=1` |
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| 445. |
If `f(x)=sqrt(4-x^2)+sqrt(x^2-1)`, then the maximum value of `(f(x))^2`is ____________ |
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Answer» Correct Answer - 6 Let `x^(2)=4cos^(2) theta+sin^(2) theta.` Then `(4-x^(2))=3 sin^(2) theta and (x^(2)-1)=3 cos^(2)theta` ` :. f(x)=sqrt(3)|sin theta|+sqrt(3)|cos theta|` `or y_(min)=sqrt(3) and y_(max)=sqrt(3)((1)/(sqrt(2))+(1)/(sqrt(2)))=sqrt(6)` Hence, range of `f(x) " is " [sqrt(3),sqrt(6)].` Hence, maximum value of `(f(x))^(2) ` is 6. |
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| 446. |
If `f(x+2a)=f(x-2a),t h e np rov et h a tf(x)i sp e r iod i cdot` |
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Answer» Correct Answer - 4a ` f(x+2a)=f(x-2a)` Replacing `x` by `x+2a`, we get `f(x)=f(x+4a)` Therefore, `f(x)` is periodic with period 4a. |
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| 447. |
If `f(x) and g(x)`are periodic functions with periods 7 and 11, respectively, then the period of `f(x)=f(x)g(x/5)-g(x)f(x/3)`isA. 177B. 222C. 433D. 1155 |
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Answer» Correct Answer - D The period of `f(x)` is 7. So, the period of `f((x)/(3)) " is " (7)/(1//3)=21.` The period of g(x) is 11. So, the period of `g((x)/(5)) " is " (11)/(1//5) =55.` Hence, `T_(1)="Period of " f(x)g((x)/(5))=7xx 55=385` and `T_(2)="period of " g(x) f((x)/(3)) =11xx 21 =231` ` :. " Period of " F(x)=LCM {T_(1),T_(2)}` `=LCM{385,231}` `=7xx11xx3xx5` `=1155` |
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| 448. |
The period of the function`f(x)=c^((sin^2x) +sin^2 (x+pi/3)+cosxcos(x+pi/3))`is (where `c`is constant)A. 1B. `(pi)/(2)`C. `pi`D. None of these |
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Answer» Correct Answer - D `sin^(2)x+sin^(2)(x+(pi)/(3))+cosxcos(x+(pi)/(3))` `=sin^(2)x+((sinx)/(2)+(sqrt(3)cosx)/(2))^(2)+cosx((sinx)/(2)-(sqrt(3)cosx)/(2))` `=sin^(2)x+(sin^(2)x)/(4)+(3cos^(2)x)/(4)+(cos^(2)x)/(2)` `=(5sin^(2)x)/(4)+(5cos^(2)x)/(4)` `=(5)/(4)` Hence, `f(x)=c^(5//4)=` constant, which is periodic but has no fixed period. |
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| 449. |
Find the value of `1//x`for the given values of `xdot``x >3`(ii) `x |
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Answer» (i) We have `3 lt x lt oo` `implies (1)/(3) gt (1)/(x) gt (1) /(to oo)` `implies 0 lt (1)/(x) lt (1)/(3)` (ii) We have ` -oo lt x lt -2 ` `implies (1)/(to -oo)gt (1)/(x) gt (1)/(-2)` `implies 0 gt (1)/(x) gt -(1)/(2)` `implies -(1)/(2) lt (1)/(x) lt 0` (iii) `x in (-1,3) -{0}` `implies x in (-1,0) cup (0,3)` For `x in (-1,0), ` we have `(1)/(-1) gt (1)/(x) gt (1)/(to 0^(-))` `implies -1 gt (1)/(x) gt -oo` `implies -oo lt (1)/(x) lt -1 " " `(1) For `x in (0,3), ` we have `(1)/(to 0^(+)) gt (1)/(x) gt (1)/(3)` `implies oo gt (1)/(x) gt (1)/(3)` `implies (1)/(3) lt (1)/(x) lt oo " " ` (2) From (1) and (2) , `(1)/(x) in (-oo, -1) cup ((1)/(3), oo)` |
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| 450. |
Fifteen coupens are numbered `1,2,3,...15` respectively. Seven coupons are selected at random one at a time with replacement The Probability that the largest number appearing on a selected coupon is 9 is : |
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Answer» 1,2,3,4,5,6,7,8,9 `=9^7-8^7` `P=(9^7-8^7)/15^7` |
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