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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.
| 351. |
If `f(x)=sqrt(x^2+a x+4)`is defined for all `x ,`then find the values of `adot` |
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Answer» Correct Answer - [-4, 4] `f(x)=sqrt(x^(2)+ax+4)` is defined for all x. Therefore, `x^(2)+ax+4 ge 0` for all `x` or `D=a^(2)-16 le 0` or `a in [-4,4]` |
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| 352. |
Solve the following system of equations `3x - 4y + 5z = 0, x + y - 2z = 0, 2x + 3y + z = 0` |
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Answer» D=`[[3,-4,5],[1,1,-2],[2,3,1]]` =`3(1-(-6)+4(5)+5*1` =46. `D!=0` `D_x=D_y=D_z=0` `x=D_x/D=0,y=0,z=0` `(x,y,z)=(0,0,0)`. |
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| 353. |
If `bar(DA) =bar(a); bar(AB)=bar(b)` and `bar(CB)=kbar(a)` where `k>0` and `X,Y` are the midpoint of `DB` and `AC` respectively such that `|bar(a)|=17` and `|bar(XY)|=4`, then k= |
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Answer» `X=(veca+vecb)/2` `Y=(veca+vecb+veca(1-k))/2` `Y=veca+vecb/a-(vecaK)/2` `|XY|=4` `XY=veca+vecb/2-(vecaK)/2-(veca+vecb)/2` `=veca/2-(vecak)/2` `=veca/2(1-k)` `|XY|=4=(|veca|(1-k))/2` `4=(17(1-k))/2` `1-k=8/17` `k=9/17`. |
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| 354. |
The sum of all the real roots of equation `x^4-3x^3-2x^2-3x+1=0` is |
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Answer» `x^4-3x^3-2x^2-3x+1 = 0` `(x^2-4x+1)(x^2+x+1) = 0` Now, roots of equation,`x^2-4x+1 = 0`, `x = (4+-sqrt(16-4))/2 = 2+-sqrt3` These are real roots. Now, roots of equation,`x^2+x+1 = 0`, `x = (-1+-sqrt(1-4))/2 = -(1+-sqrt(-3))/2` These are imaginary roots. `:.` Sum of real roots ` = (2+sqrt3)+(2-sqrt3) = 4` |
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| 355. |
If 0 |
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Answer» x=[x]+{x} [x]=x-{x} `[x/2]+[x/3]+[x/5]=(31x)/30` `x/2-{x/2}+x/3-{x/3}+x/5-{x/5}=(31x)/30` `x/2+x/3+x/5-[(x/2)+(x/3)+(x/5)]=(31x)/30` `(3x)/30=t` `t=0` x is multiple of 2,3,5 LCM of(2,3,5)=30 x=30m `0<30m<1000` n=1to33 33 possibles values of x. |
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| 356. |
Find domain of the function `10^x+10^y=10` |
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Answer» `10^y=10-10^x` `10^y=10(1-10^(x-1))` `log_10 10^y=log10(1-10^(x-1))` `y=log_10 10+log_10(1-10^(x-1))` `y=1+log_10(1-10^(x-1))` `log_10(1-10^(n-1))` `1-10^(x-1)>0` `1>10^(x-1)` `10>10x` `10^x<10^1` `x<1` Domain=`(-oo,1)`. |
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| 357. |
`lim_(n->oo) {1/1.3+1/3.5+1/5.7+.....+1/((2n+1)(2n+3))` is equal to |
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Answer» `lim_(n->oo)[1/2((3-1)/(1.3))+1/2((5-3)/(3.5))+...+1/2(((2n+3)-(2n+1))/((2n+1)(2n+3)))]` `lim_(h->oo)[1/2(1-1/3_+(1/3-1/5)+(1/5-1/7)+...+(1/(2n+1)-1/(2n+3))]` `lim_(h->oo)1/2(1-1/(2n+3))` `lim_(h->oo)1/2((2n+2)/(2n+3))` `lim_(h->oo)1/2((2n(1+1/n))/(h(2+3/n)))` `1/2(2(1+0))/(2+0)` `1/2`. |
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| 358. |
If ` tan (pi/12 - x) , tan (pi/12) , tan (pi/12 + x) ` in G.P. then sum of all the solutions in [0,314] is `k pi`. Find k |
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Answer» `tan^2pi/12=tan(pi/12-x)*tan(pi/12+x)` `tan^2pi/12=((tanpi/12+tanx)/(a-tanpi/12*tanx))*((tanpi/12-tanx)/(1+tanpi/12tanx))` `tan^(2pi/12)=(tam^2pi/12-tan^2x)/(1-tan^2pi/12*tan^2x)` `tan^2(pi/12)(1-tan^2pi/12*tan^2x)=tan^2pi/12-tan^2x` `tan^2pi/12-tan^4pi/12tan^2x=tan^2pi/12-tan^2x` `tan^2x-tan^4pi/12tan^2x=0` `tan^2x[1-tan^4pi/12]=0` `tan^2x=0` `tanx=0` `x=npi` `=pi[1+2+3+..+99]` `=pi(99*100)/2` `=4950pi` `=kpi` `k=4950`. |
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| 359. |
Number of solution of `tan(2x)= tan(6x)` in `(0,3pi)` is : |
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Answer» y=tan6x `[0-pi)=>2-1=` `[pi-2pi)=2` `(2pi-3pi)=2` `(0,3pi)=>5`. |
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| 360. |
If `x=2+2^(2/3)+2^(1/3)`, then the value of `x^3-6x^2+6x` is: |
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Answer» `x-2=2^(2/3)+2^(1/3)` `(x-2)^3=(2^(2/3)+2^(1/3))^3` `x^3-8-6x^2+12x=4+2+6(2^(2/3)+2^(1/3))` `x^3-6x^2+!2x-8=6+6(x-2)=6+6x-12` `x^3-6x^2+6x-8=-6` `x^3-6x^2+6x=2`. |
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| 361. |
Find value of ` Sin ( tan^(-1) x)` |
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Answer» Let `tan^-1x = theta` Then, `x = tan theta` `=>tan theta = x/1` `:. sin theta = x/sqrt(x^2+1^2)` `=>sin theta = x/sqrt(x^2+1)->(1)` Now, `sin(tan^-1x ) = sin theta` From (1), `sin(tan^-1x ) = x/sqrt(x^2+1)` |
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| 362. |
Two men A and B start with velocities v at the same time from the junction of two roads inclined at `45^@` to each other.If they travel by different roads,find the rate at which they are being separated. |
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Answer» With the given details, we can create a right angle triangle. Please refer to video for the figure. From the figure, `(OP)/(OQ) = tan45^@ = 1` `:. OP = OQ` Let `OP = OQ = k` Then, `PQ = sqrt(k^2+k^2) = ksqrt2` Now, in right angle triangle `OPQ`, `OP^2+OQ^2 = PQ^2` As we have to find the rate of separation, we will differentiate above equation. So, `2OP(d(OP))/dt+2OQ(d(OQ))/dt = 2PQ(d(PQ))/dt` As we are given velocities of two person as `v`, `:. (d(OP))/dt = (d(PQ))/dt = v` So, our equation becomes, `=>2kv+2k(d(OQ))/dt = 2sqrt2kv` `=>v+(d(OQ))/dt =sqrt2v` `=> (d(OQ))/dt = (sqrt2-1)v` So, rate of separation will be `(sqrt2-1)v`. |
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| 363. |
`int 1/(x( 6 ( log x)^2 + 7 log x + 2) dx` |
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Answer» `logx=t` `1/xdx=dt` `intdt/(6t^2+7t+2)=1/6intdt/(t^2+7/6t+2/6)` `1/6intdt/((t+7/12)^2-(1/12)^2)` `ln((t+1/2)/(t+2/3))+c` `ln((2t+1)/(3t+2))+c` `=log((2logx+1)/(3logx+2)*3/2)+c` `=log((6logx+3)/(6logx+4))+c`. |
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| 364. |
If the `4^(th)` term in the expansion of `( 2 + 3x/8)^10` has the maximum value, then the values of x for which this will be true is : |
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Answer» `T_4` in `(2+(3x)/8)^10 = C(10,3)(2^7)((3x)/8)^3` `T_3` in `(2+(3x)/8)^10 = C(10,2)(2^8)((3x)/8)^2` `T_5` in `(2+(3x)/8)^10 = C(10,4)(2^6)((3x)/8)^4` Now, we are given `T_4` has the maximum value. So,`T_4 gt t_3` `=> C(10,3)(2^7)((3x)/8)^3 gt C(10,2)(2^8)((3x)/8)^2` `=>C(10,3)((3x)/8) gt C(10,2)(2)` `=>(10**9**8)/(3**2**1)((3x)/8) gt (10**9)/(2**1)(2)` `45x gt 90=> x gt 2` Also, `T_5 lt T_4` `=> C(10,4)(2^6)((3x)/8)^4 lt C(10,3)(2^7)((3x)/8)^3` `=>C(10,4)((3x)/8) lt C(10,3)(2)` `=>(10**9**8**7)/(4**3**2**1)((3x)/8) lt (10**9**8)/(3**2**1)(2)` `=>630x/8 lt 240` `=>x lt 192/63 => x lt 64/21` So, `2 lt x lt 64/21`, is the value for which `x` will be true. |
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| 365. |
`int e^(cotx) (cos x - cosec x ) dx` is equal to : |
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Answer» `int e^(cotx) (cosx-cosecx) dx` Let `y = e^(cotx)sinx` `:. dy/dx = e^cotx cosx +sinx e^cotx(-cosec^2x)` `=> dy/dx = e^cotx cosx - sinx e^cotx(cosec x)` `=> dy/dx = e^cotx(cos-cosecx)` Now, integrating both sides, `y+c = int e^(cotx) (cosx-cosecx) dx` `=> int e^(cotx) (cosx-cosecx) dx = e^(cotx)sinx+c` So, option `D` is the correct option. |
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| 366. |
Solve `||x-1| -5| ge 2`. |
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Answer» `||x-1| -5| ge 2` i.e., `|x-1| -5| le -2 " or " |x-1| -5| ge 2` i.e., `|x-1| le 3 " or " |x-1| ge 7` i.e., `-3 le x-1 le 3" or " x-1 le -7 " or " x-1 ge 7` i.e., `-2 le x le 4 " or " x le -6 " or " x ge 8 ` i.e., `x in (-oo,-6]cup [-2,4] cup [8,oo)` |
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| 367. |
Consider a function `f:[0,pi/2]->R` given by `f(x)=sin x and g:[0,pi/2]->R` given by `g(x)=cos x.` Show that f and g are one-one, but `f+g` is not one-one. |
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Answer» one-one: `a_1!=a_2`, then `f(a_1)!xf(a_2)` `f(a_1)=f(a_2)`,then `(a_1=a_2)` `sqrt2(sinx/sqrt2+cosx/sqrt2)` `=sqrt2(sinxcos45+cosxsin45)` `=sqrt2(sin(A+B)` `=sqrt2sin(x+45)` `a_1!=a_2` `f(g)(a_1)=(f+g)(a_2)` this function is one-one. |
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| 368. |
Let `X={1,2,3,4,5,6,7,8,9}.` Let R be a relation in `X` given by `R_1={(x,y):x-y` is divisible by `3}` and `R_2` another on `X` given by `R={(x,y):(x,y)uu{1,4,7}} or {x,y} uu {2,5,8} or {x,y} uu {3,6,9}}` Show that `R_1=R_2.` |
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Answer» Here, `X = {1,2,3,4,5,6,7,8,9}` `R_1 = {(x,y): x-y` is divisible by `3}` As, `x-y` is divisible by `3`. `x-y = 3n` where `n in N` `=>x= y+3n` `y=1, x = 4` when `n=1` `y=1, x = 7` when `n=2` `y=4, x = 7` when `n=1` `y=4, x = 1` when `n=-1` `y=7, x = 4` when `n=-1` `y=7, x = 1` when `n=-2` `:. (x,y) = {(1,4),(1,7),(4,7),(4,1),(7,4),(7,1)}` `:.(x,y) sub {1,4,7}` Similarly, ` (x,y) = {(2,5),(2,8),(5,8),(5,2),(8,5),(8,2)}` `:.(x,y) sub {2,5,8}` Similarly, ` (x,y) = {(3,6),(3,9),(6,9),(6,3),(9,6),(9,3)}` `:.(x,y) sub {3,6,9}` `:. R_1 = {(x,y):(x,y) sub {1,4,7}} or {(x,y):(x,y) sub {2,5,8}} or {(x,y):(x,y) sub {3,6,9}}` We are given, `:. R_2 = {(x,y):(x,y) sub {1,4,7}} or {(x,y):(x,y) sub {2,5,8}} or {(x,y):(x,y) sub {3,6,9}}` `:. R_1 = R_2` |
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| 369. |
238. EXAMPLE If the normals at three points P, Q and R in a s be the focus, meet point O and prove that SP. SQ. SR a SO As in the previous question we know that the normals a the points (ami, am 1), (am am2) and (am3, 2am3) meet in the point (h k), if m1 1m2 1m3 0 2a -h m2m3 m3m1 m1m2 (2) |
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Answer» `y=mx-2am-am^3` `(am_1^2,-2am_1)(am_2^2,-2am_2)(am_3^2,-2am_3)` `am^3+2am-mh+k=0` `m_1+m_2+m_3=0` `m_1m_2+m_1m_3+m_2m_3=(2a-h)/a` `m_1m_2m_3=-k/a` =SP*SQ*SR `=(am_1^2+a)(am_2^2+a)(am_3^2+a)` `=a^3(m_1^2+1)(m_2^2+!)(m_3^2+1)` `=a^3(m_1^2*m_2^2m_3^2+m_1^2m_2^2+m_1^2m_3^2+n_1^2+m_3^2m_3^2+m_2+m_3^2+1)` `=a^3[(h^2-2ah+a^2)/a^2]` `a^3/a^2(h-a)^2` `a(OS)^2`. |
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| 370. |
If f(x) = `(x^(2))/(1 + x^(2))` , then the value of f{f(2)} is :A. `(9)/(41)`B. `(25)/(41)`C. `(16)/(25)`D. `(16)/(41)` |
| Answer» Correct Answer - d | |
| 371. |
Match the following lists: |
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Answer» Correct Answer - `a to r,s; b to p,q,r,s; c to s; d to p` a. `tan^(-1)((2x)/(1-x^(2))) in (-(pi)/(2),(pi)/(2))` `or 2 tan^(-1)x in (-(pi)/(2),(pi)/(2))` `or tan^(-1)x in (-(pi)/(4),(pi)/(4))` `or tan^(-1)x in (-1,1)` b. `f(x)=sin^(-1)(sinx) and g(x)=sin(sin^(-1)x)` `f(x)` is defined if `sin x in [-1,1]` which is true for all `x in R`. But g(x) is defined for only `x in [-1,1]`. Hence, `f(x)` and g(x) are identical if `x in [-1,1]`. c. `f(x)=log_(x^(2))25 and g(x)=log_(x)5` `f(x)` is defined ` AA x in R - {0,1} and g(x) ` is defined for `(0, oo) -{1}.` Hence, `f(x)` and g(x) are identical if `x in (0,1) cup (1,oo).` d. `f(x)=sec^(-1)x+"cosec"^(-1)x,g(x)=sin^(-1)x+cos^(-1)x` `f(x)` has domain `R -(-1,1) and g(x)` has domain [-1, 1] Hence, both the functions are identical only if `x=-1,1.` |
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| 372. |
If f(x) = x - `(1)/(x)` , then the value of f(x) + `f((1)/(x))` is : |
| Answer» Correct Answer - a | |
| 373. |
Let `E_(1)={x in R :x ne 1 and (x)/(x-1) gt 0} and E_(2)={x in E_(1):sin^(-1)(log_(e)((x)/(x-1))) " is real number"}` (Here, the inverse trigonometric function `sin^(-1)x` assumes values in `[-(pi)/(2),(pi)/(2)]`.) Let `f:E_(1) to R` be the function defined by `f(x)=log_(e)((x)/(x-1)) and g:E_(2) to R` be the function defined by `g(x)=sin^(-1)(log_(e)((x)/(x-1))) .` The correct option isA. `a to s, b to q, c to p, d to p`B. `a to r, b to r, c to u, d to t`C. `a to s, b to q, c to p, d to u`D. `a to s, b to r, c to u, d to t` |
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Answer» Correct Answer - A `E_(1):(x)/(x-1) gt 0` `implies x in (-oo ,0) cup (1,oo)` Also `E_(2) : -1 lt log_(e)((x)/(x-1)) le 1` `implies (1)/(e) le (x)/(x-1) le e` `implies (1)/(e) le 1 +(1)/(x-1) le e` `implies (1)/(e)-1 le (1)/(x-1) le e-1` `implies (x-1) in (-oo, (e)/(1-e)] cup [(1)/(e-1),oo)` `implies x in (-oo, (e)/(1-e)] cup [(1)/(e-1),oo)` Now, `(x)/(x-1) in (0, oo)-{1} AA x in E_(1)` `implies log_(e) ((x)/(x-1)) in (-oo,oo)-{0}` `implies sin^(-1)(log_(e)((x)/(x-1))) in [-(pi)/(2),(pi)/(2)]-{0}` `("considering "log_(e) ((x)/(x-1)) in [-1,1]-{0})` |
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| 374. |
If f(x) = ` log_(e) ((1-x)/(1+x))` , then `f((2x)/(1 + x^(2)))` is equal to :A. `[f(x)]^(2)`B. 2f(x)C. 4f(x)D. None of these |
| Answer» Correct Answer - b | |
| 375. |
`(sin 70^@ + cos 40^@)/(cos 70^@ + sin 40^@)` |
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Answer» `(sin70^@ +cos40^@)/(cos70^@+sin40^@)` We know, `sintheta = cos(90-theta)`. So, our expression becomes `(cos20^@+cos40^@)/(sin20^@+sin40^@)` As ,`cosA+cosB = 2cos((A+B)/2)cos((A-B)/2)` and `sinA+sinB = 2sin((A+B)/2)cos((A-B)/2)` Our expression becomes, `=(2cos((20^@+40^@)/2)cos((20^@-40^@)/2))/(2sin((20^@+40^@)/2)cos((20^@-40^@)/2))` `=cot30^@ = sqrt3` |
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| 376. |
Solve the differential equation`sinxdy/dx+3y=cosx` |
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Answer» `dy/dx+(3y)/sinx=cosx/sinx` `P(x)=3/sinx` `Q(x)=cotx` `IF=e^(intp(x)dx)=e^(3/sinx) dx=e^(3intcosecx)`dx `=e^(ln(1+cosecx+cotx)^(-3)` `y*IF=intIF*Q(x) dx` `=int-dt/t^3` `=-(t^(-4+1)/(-4+1))` `y*if=1/(3(cosecx+cotx)^3` `y=1/((cosecx+cotx)^3^3(cosecx+cotx)^3)` `y=1/3`. |
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| 377. |
A and C lie on a circle center O with radius `5sqrt2`. The point B inside the circle is such that angle `/_`ABC = 90, AB = 6, BC = 2. Find `sqrt(OB^2 -1)` |
| Answer» A and C lie on a circle with radius `5*sqrt2` as radius.There is a point B inside the circle making `/_ABC = 90^circ`We need to find, `sqrt(OB^2 - 1)`Let center of circle be O.`:. OA = 5*sqrt2`Given,AB = 6BC = 2`AC^2 = AB^2 + BC^2``= 36 + 4 = 40 rArr AC = 2*sqrt10`Let us draw a perpendicular from O on line AB cutting it at D and equal to length y (say).BD = x (say)Now we will draw lines parallel to OD and BD in order to form a rectangle OEBD. OB will be the diagonal of this rectangle.In `/_ OEC``OC^2 = OE^2 + EC^2``rArr 50 = x^2 + (2+y)^2``rArr 50 = x^2 + 4 + 4y + y^2``rArr 46 = x^2 + y^2 + 4y` .....(1)In `/_ OAD``OA^2 = AD^2 + OD^2``rArr 50 = y^2 + (6-x)^2``rArr 50 = x^2 + y^2 - 12y + 36``rArr 14 = x^2 + y^2 - 12x` ......(2)Solving (1) & (2)`x = 1 , y = 5``OB^2 = x^2 + y^2 = 26`Hence, `sqrt( OB^2 - 1) = sqrt(25) = 5` | |
| 378. |
If `f9x)=a x^7+b x^3+c x-5,a , b , c`are real constants, and `f(-7)=7,`then the range of `f(7)+17cosxi s``[-34 ,0]`(b) `[0, 34]``[-34 ,34]`(d) none of theseA. [-34, 0]B. [0, 34]C. [-34, 34]D. None of these |
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Answer» Correct Answer - A `f(7)+f(-7)= -10` `or f(7)= -17` `or f(7)+17 cosx= -17 +17 cosx` which has range `[-34,0]`. |
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| 379. |
If `f(a-x)=f(a+x) " and " f(b-x)=f(b+x)` for all real x, where `a, b (a gt b gt 0)` are constants, then prove that `f(x)` is a periodic function. |
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Answer» `f(x)=f(b+(x-b))` ` =f(b-(x-b))` `=f(2b-x)` `=f(a+(2b-x-a))` `=f(a-(2b-x-a))` `=f(2a-2b+x)` Hence, `f(x)` is periodic with period `2a-2b.` |
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| 380. |
If `547.527/0.0082=x`, then the value of `547527/82` is: |
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Answer» `547.527/0.0082=x` `x=5475270/82` `547527/82=x/10`. |
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| 381. |
An alloy of zinc and tin contains 37% zinc by weight. Find the weight of zinc which must be added to 400 kg of this alloy if the final percentage of zinc is to be 70. |
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Answer» Total weight of alloy `= 400 kg` Let we add `x` kg zinc to make its weight `70%` in the alloy. Then, total alloy `= 400+x` Total zinc after adding `x` kg` = 37**400/100+x` `:. (400+x)**70/100 = (37**4)+x` `=>2800+7x = 1480+10x` `=>3x = 1320=> x = 440kg` So, zinc of weight of `440` kg should be added to get the desired mixture. |
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| 382. |
If R is a relation on a finite set having n elements, then the number of relations on A is |
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Answer» let there be a relation such that (a,b) is a relation. now a has `n` options and b also has `n`options. o total no. of ordered pairs=`nxxn=n^2` now each ordered pair ha 2 options, either it willl be a realtion or not so=>no. of relations=>`2^(n^2)` |
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| 383. |
If `e^x + e^(f(x)) = e` then domain of f(x) is |
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Answer» `e^x+e^y=e` `e^y=e-e^x` taking log both side `loge^y=ln(e-e^x)` `yloge=log(e-e^x)` `y=f(x)=log(e-e^x)` `e-e^x>0` `e>e^x` `lne>lne^x` `1>x` `x in (-oo,1)` |
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| 384. |
The range of `x^2+4y^2+9z^2-6yz-3xz-2xy` is |
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Answer» `f=x^2+4y^2+9z^2-6yz-3xz-2xy` `f=x^2+(2y)^2+(3z)^2-(2y)(3z)-(x)(3z)-(2y)(x)` `f=1/2*(2a^2+2b^2+2c^2-2ab-2bc-2ca)` `f=1/2{(x-2y)^2+(2y-3z)^2+(x-3z)^2}` `f>=0` `f[0,oo)`. |
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| 385. |
`int (sin theta+cos theta)/sqrt(1+sin2 theta) d theta` |
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Answer» `int(sintheta+costheta)/sqrt(1+sin2theta)d theta` `int(sintheta+costheta)/sqrt(sin^2theta+cos^2theta+2sinthetacostheta)d theta` `int(sintheta+costheta)/sqrt((sintheta+costheta)^2) d theta` `int(sintheta+costheta)/(sintheta+costheta)d theta` `int1d theta` `theta+C`. |
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| 386. |
Write explicit functions of `y`defined by the following equations and also find the domains of definitions of the given implicit functions:`x+|y|=2y`(b) `e^y-e^(-y)=2x``10^x+10^y=10`(d) `x^2-sin^(-1)y=pi/2` |
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Answer» (i) `x+|y|=2y` If `y ge 0, ` we have ` x+y=2yimpliesx` `implies y=x,x ge 0` If `y lt 0, ` we have ` x-y=2yimplies y=(x)/(3)` `implies y=(x)/(3), x lt 0` Hence, `y={((x)/(3)", "x lt 0),(x", "x ge 0):}` (ii) `e^(y) -e^(-y)=2x` `implies e^(2y)-2xe^(y)-1=0 " " `(Multiplying by `e^(y)`) `implies e^(y)=(2x+-sqrt(4x^(2)+4))/(2)=x+-sqrt(x^(2)+1)` `implies e^(y)=x+sqrt(x^(2)+1)" (as "sqrt(x^(2)+1) gtx, " then " x-sqrt(x^(2)+1) lt 0, " which is not possible)" ` `implies y=log_(e)(x+sqrt(x^(2)+1))` (iii) `10^(x)+10^(y)=10` `implies 10^(y)=10-10^(x)` `implies y=log_(10)(10-10^(x))` |
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| 387. |
Set A has m distinct elements and set B has n distinct elements. Then how many different mappings from set A to set B can be formed? |
| Answer» Since each element(preimage) in set A can corresponds to any one of the n elements (images) in set B, the number of mappings can be formed is `n xx n xx n … m" times "=n^(m)`. | |
| 388. |
If A is set of different triangles in the plane and B is set of all positive real numbers. A relation R is defined from set A to set B such that every element of set A is associated with some number in set B which is measure of area of triangle. Is this relation as function? |
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Answer» Two or more different triangles may have same area. But area of each triangle is unique. This means each element (triangle) of set A is associated with unique value (Area) of set B. So, given relation is function. |
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| 389. |
If `f(x)=(a-n^(n))^(1//n)` where `a gt 0` and `n in N` , then `f[f(x)]` is equal to :A. xB. aC. `x^(n)`D. None of these |
| Answer» Correct Answer - a | |
| 390. |
A function `f : RrarrR` is defined as `f(x) = x^(2) +2`, then evaluate each of the following : (i) `f^(-1)(-6)` (ii) `f^(-1)(18)` |
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Answer» (i) `f^(-1)(-6)={x in R : f (x) =-6}` = `{x in R : x^(2) + 2 =- 6}` = `{x in R : x^(2) =- 8} = phi` Ans. (ii) `f^(-1)(18)= {x in R : f(x) = 18}` = `{x in : x^(2) + 2 = 18 } = {x in R : x^(2) = 16 }` = `{x in R : x = +- 4 } = {-4,4}`. Ans. |
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| 391. |
If `f:R -> R, g:R -> R` defined as `f(x) = sin x and g(x) = x^2`, then find the value of `(gof)(x) and (fog)(x) `and also prove that `gof != fog`. |
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Answer» (fog) (x) = f[(g(x)] = `f(x^(2)) = sin x^(2)`Ans. and (gof)(x)= g[f(x)] =g(sinx) = `(sinx)^(2)= sin^(2) x` Ans. Therefore, fog `ne` gof . Hence Proved. |
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| 392. |
If `f(x)=|x+1|` then the true statement from the following is :A. `f(x^(2))={f(x)}^(2)`B. `f(x+y)=f(x)+f(y)`C. `f(|x|)=|f(x)|`D. None of these |
| Answer» Correct Answer - d | |
| 393. |
Function `f: R to R and g : R to R ` are defined as `f(x)=sin x and g(x) =e^(x)` . Find (gof)(x) and (fog)(x). |
| Answer» Correct Answer - `(gof)(x)=e^(sinx), (fog)(x)=sine^(x)` | |
| 394. |
Let `f: R ->R`be defined as `f(x) = 10 x + 7`. Find the function `g: R ->R`such that`gof=fog=1_R` |
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Answer» In `F: R rarr R`, f(x) = 10 x +7 Let x,y `in`R and f(x)=f(y) `rArr 10 x +7 =10 y +7` `rArr 10x = 10 y ` `rArr x=y` `therefore` f is one - one Again let f(x) =y where `y in R` `rArr 10x +7 =y rArr x(y-7)/(10)i R` `therefore` f is onto Now `(fog )(x) = f((y-7)/(7))` `=10 ((y-7)/(10))+7=y` `therefore go f =I_R "and" fog =I_R` `therefore g: R rarr R ` is defined as g(y) `=(y-7)/(10)` |
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| 395. |
If `f(x)=3x+|x|`, then the value of `f(3x)+f(-x)-f(x)` is:A. `3(x+|x|)^(2)`B. `3(x+|x|)`C. `(x-|x|)^(3)`D. None of these |
| Answer» Correct Answer - b | |
| 396. |
If `A={x:(pi)/(6) lt x lt (pi)/(3)}` and `f(x) =cosx-x(1+x),` then `f(A)` is equal to :A. `[(pi),(6),(pi)/(3)]`B. `[(-pi)/(3),(-pi)/(6)]`C. `[(1)/(2)-(pi)/(3)(1+(pi)/(3)),(sqrt(3))/(2)-(pi)/(6)(1+(pi)/(6))]`D. None of these |
| Answer» Correct Answer - c | |
| 397. |
`int1/(y(1-ay))`dy |
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Answer» `int1/(y(1-ay))dy` `Let 1/(y(1-ay))=A/y+B/(1-ay)` `1=A(1-ay)+By` `-aA+B=0` `A=1` `B=a` `intdy/(y(1-ay))=intdy/y+int(ady)/(1-ay)` `=lny+a(-1/a)ln(1-ay)` `=lny-ln(1-ay)` `=ln(y/(1-ay))+c`. |
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| 398. |
Find the domain of the function :`f(x)=3/(4-x^2)+(log)_(10)(x^3-x)` |
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Answer» Correct Answer - `(-1,0) cup (1,2) cup (2,oo)` `f(x)=(1)/(4-x^(2))+log_(10)(x^(3)-x)` `f` is defined when `x ne +-2 " and " x^(3) -x gt 0` or `x ne +-2 " and "x(x^(2)-1) gt 0` or `x ne +-2, x in (-1,0) cup (1,oo)` or `x in (-1,0) cup (1,2) cup (2,oo)` |
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| 399. |
Find the domain of the function :`f(x)=(log)_((x-4))(x^2-11 x+24)` |
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Answer» Correct Answer - `(8,oo)` `f(x)=log_(x-4)(x^(2)-11x+24).` `f(x)` is defined if `x-4 gt 0 " and " ne 1 " and " x^(2)-11x+24 gt0` or `x gt 4 " and " ne 5 " and " (x-3)(x-8) gt 0` i.e., `x gt 4 " and " ne 5 " and " x lt 3 " or " x gt 8` or ` x gt 8` or domain`(y)=(8,oo)` |
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| 400. |
The interval (-1, 1) is the range ofA)`x/(1+|x|)` B) `cos x - sin x ` C) `x^4+4x^3+1` D) `|x|/(1+x)` |
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Answer» x=-3/4 `(3/4)/(1-(3/4))=3` Option A is correct. |
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