InterviewSolution
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.
| 2851. |
The value of int_(0)^(pi//2) (cos3x+1)/(2cos x-1)dx is |
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Answer» 2 `=underset(0)overset(pi//2)int (cos 3x-cos""(3pi)/(4))/(2(cosx-cos""(pi)/(3)))dx` `=underset(0)overset(pi//2)int ((4cos^(3)x-3cosx)-(4cos^(3)""(pi)/(4)-3cos""(pi)/(3)))/(2(cos x-cos""(pi)/(3)))dx` `=2underset(0)overset(pi//2)int ((cos^(3)x-cos^(3)""(pi)/(3))/(cos x-cos""(pi)/(3)))dx` `=2underset(0)overset(pi//2)int (cos^(2)x+cos^(2)""(pi)/(3)+cos x cos ""(pi)/(3))dx-(3)/(2)underset(0)overset(pi//2)int 1dx` `=underset(0)overset(pi//2)int (1+cos2x+(1)/(2)+cosx)dx-(3pi)/(4)` `=(3pi)/(4)+1-(3pi)/(4)=1` |
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| 2852. |
int_(0)^(1)x e^(x^(2)) dx |
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Answer» Solution :`" LET I=" int_(0)^(1) xe^(x^(2))dx` `"Let "x^(2)=trArr 2x=(DT)/(dx) rArr XD x= (dt)/(2)` `x=0rArr t=0 ` `"and " x=1 rArrt=1` ` :, I=(1)/(2) int_(0)^(1) E^(t) dt=(1)/(2)[e^(t)]_(0)^(1)` `=(1)/(2)(e^(1)-e^(0))=(1)/(2)(e-1)` |
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| 2853. |
Determine for which values of x, the following functions are increasing or decreasing : f(x)=x^(3)+(1)/(x^(3)), x ne0 |
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| 2854. |
If a,b,c are the sides of a triangle then the range of (a^(2)+b^(2)+c^(2))/(ab+bc+ca) is |
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Answer» [1,2) |
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| 2855. |
(x+(1)/(x))^(x) +x^((1+(1)/(x))) |
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Answer» SOLUTION :`"Let " y = (x+(1)/(x))^(x) +x^((1+(1)/(x)))` `"Let" u=(x+(1)/(x))^(x) " and "V = x^(1+(1)/(x))` `therefore y=u + v rArr (dy)/(dx) = (DU)/(dx) + (dv)/(dx)"" ...(1)` `"Now", u = (x+(1)/(x))^(x)` `rArr "LOG " u = "log" (x+(1)/(x))^(x) = x" log"(x+(1)/(x))` `rArr (1)/(u) (du)/(dx) = x* (d)/(dx) "log" (x+(1)/(x)) + "log"(x+(1)/(x))(d)/(dx) x` `rArr (du)/(dx) = u [(x)/(x+(1)/(x))(1-(1)/x^(2))+"log" (x+(1)/(x))]` `=(x + (1)/(x))^(x) [(x^(2)-1)/(x^(2)+1) + "log" (x+(1)/(x))]` ` "and "v = x^((1+(1)/(x))` `rArr "log " v = "log" {x^((1+(1)/(x)))} = (1+(1)/(x))"log"x` `rArr (1)/(v) (dv)/(dx) = (1+(1)/(x))(d)/(dx) "log"x + "log"x (d)/(dx) (1+(1)/(x))` `rArr (dv)/(dx) = v[(1+(1)/(x))*(1)/(x)+"log"x (-(1)/(x^(2)))]` `rArr (dv)/(dx) = x^((1+(1)/(x)))*(1)/(x^(2))[x+1-"log"x]` `therefore "From equation (1)"` `(dy)/(dx) = (x+(1)/(x))^(x)[(x^(2)-1)/(x^(2)+1)+ "log" (x+(1)/(x))] + x^((1+(1)/(x))) * (1)/(x^(2))[x+1-"log"x]` |
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| 2856. |
If the lines (x-1)/(-3)=(y-2)/(2k)=(z-3)/2 and (x-1)/(3k)=(y-1)/1=(z-6)/(-3) are perpendicular, find the value of k. |
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| 2857. |
If there is an error of 1/(10)% in the measurement of the radius of a sphere, then the percentage erro in the calculation of the volume of the sphere is |
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Answer» `3//10` |
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| 2858. |
Reflexion of the point (alpha, beta, gamma) in XY plane is |
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Answer» 1.`(alpha, beta, 0)` |
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| 2859. |
Thevalue(S) of alpha in [0,2pi] forwhichvectorvec(a)= hat(i) + 3hat(j) + (sin 2 alpha) hat(k)makes an obtuse angle with the z-axis and thevectors vec(b) = (tan alpha)hat(i) - hat(j) + 2 sqrt(sin(alpha)/(2))hat(k) andvec(c) = (tan alpha)hat(i) (tan alpha)hat(j)- 3 sqrt(cosec(alpha)/(2))hat(k) are |
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Answer» `TAN^(-1) 3` |
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| 2860. |
A coin is tossed successively three times. The probability of getting exactly one head or 2 heads, is |
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Answer» `1/4` |
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| 2861. |
Find the area enclosed by y=log_(e)(x+e) and x=log_(e)((1)/(y)) and the x-axis. |
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| 2862. |
For a group of 50 male workers, the mean and S.D. of their daily wages are Rs. 630 and Rs. 90 respectively. For a group of 40 female workers, these are Rs. 540 and Rs. 60 respectively. The S.D. of these 90 workers is |
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Answer» 60 |
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| 2863. |
If p rarr (q vv r) is false, then the truth values of p, q, r are respectively |
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Answer» T,F,F |
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| 2864. |
Total number of species which use d -orbital in their bonding SO_(2),SO_(3),SO_(4)^(-2),CIO_(4)^(oplus),XeO_(2)F_(2),XeO_(3) |
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| 2865. |
If y = sin ^(3) x + cos^(6) x find (dy)/(dx). |
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| 2866. |
If (dy)/(dx)+(y)/(sqrt(x^(2)+a^(2)))=3x,y(0)=a^(2) then the value of (y(sqrt(3a)))/(a^(2)).(8-3sqrt3)/(2-sqrt3) is equal to 19k, then find k. |
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| 2867. |
Let f: Rto R be by f(x)=(x-1)(x-2)(x-5). Define F(x)=int_(0)^(x)f(t)dt,xgt0 Then which of the following options is/are correct ? |
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Answer» `F(x) ne 0` for all `x in (0,5)` Since , `F(x)-int_(0)^(x) f(t)dt,x gt 0` So, `F'(x)=(x-2)(x-5)` Accroding to wavy curve method `(-""+""-""+)/(1"2"5)` F'(x) chages , it's sign form negative to positive at `x =1` and 5, So, F(x) has minima at ` x=1` and 5 and as F'(x) changes, it's sign from positive to negative at` x=2` so F(x) has maxima at ` x=2` ` because F(2)=int_(0)^(2) f(t)dt= int_(0)^(2)(t^2-8t^2+17t-10)dt` ` =[(t^4)/(4)-8(t^3)/(3)+17(t^2)/(2)-10T]_(0)^(2)` ` =4-(64)/(3)+34-20=38-(124)/(3)=-(10)/(3)` `because` At the point of maxima `x=2`, the functional value `F(2)=-(10)/(3)`, negative for the internal `x in (0,5)`, so `F(x)ne 0` for any value of `x in (0,5)`, Hence , OPTIONS (a),(b) and (d) are correct. |
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| 2868. |
Evaluate P (AnnB)if 2P(A)=P(B)=5/13and P(A/B)=2/5. |
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Answer» <P> SOLUTION :P(A/B)=2/5`rArr (P(ANNB))/(P(B))=2/5`rArr(P(AnnB))=2/5xxP(B)`2/5xx5/13=2/13 P(A)=5/26 THUS,`P(AnnB)` `=`P(A)+P(B)-P(AnnB)` =5/26+5/13-2/13=11/26 |
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| 2869. |
Integrate the following functions x/(9-4x^2) |
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Answer» Solution :Let t = `9-4x^2`. Then `dt = -8X DX GT x dx = -1/8 dt` THEREFORE`int x/(9-4x^2) dx = int 1/t xx -1/8 dt` =`-1/8 log|t|+C` =-`1/8 log|9-4x^2| +c` |
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| 2870. |
Find the area enclosed by y^2=x^3,x=0,y=1 |
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Answer» SOLUTION :Given CURVE is `y^2=x^3` `impliesx=y^(2//3)` It PASS through the origin So the REQUIRED AREA `=int_0^1xdy= int_0^1 y^(2/3)dy=[y^(2/3)/(5//3)]_0^1=3/5` |
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| 2871. |
Evalute the following integrals int (1)/(x^(2)) tan^(-1) xdx |
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| 2872. |
The value of int_(0)^(4) 3^(sqrt(2x+1) )dx is |
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Answer» `(6)/(LOG 3) (13- (4)/(log 3) )` |
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| 2873. |
The value of sum_(r = 1)^15 r^2 ((""^15C_r)/(""^15C_(r - 1)) ) of is equal to |
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Answer» 1240 |
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| 2874. |
Let the circle (x - 1)^(2) + (y - 1)^(2) = 25 cuts a rectangular with transverse axis along y = x at four points A, B, C and D having co-ordinates (x_(1) , y_(1)) : i = 1,2,3,4 respectively O being the centre of the hyperbola, Now match the entries from the following two columns |
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| 2875. |
Integrate the following functions 1/sqrt(7-6x-x^2) |
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Answer» SOLUTION :`7-6x-x^2 = 7-(x^2+6x)` =`7-(x^2+6x+9-9)` `16-(x+3)^2` therefore` INT 1/sqrt(7-6x-x^2) dx` =`int 1/sqrt(16-(x+3)^2) dx` =`sin^-1((x+3)/4)+C` |
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| 2876. |
The sum of the series 1/1.2 - 1/2.3 + 1/3.4 … upto infty is equal to |
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Answer» `log_(e) 2 - 1` |
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| 2877. |
The numbers of terms of the A.P 3, 7, 11, 15, …… to be taken so that the sum is 406 is |
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Answer» 5 |
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| 2878. |
The area bounded by the curves y=|x|-1 and y= -|x|+1 is |
| Answer» ANSWER :D | |
| 2879. |
One compartment of a purse contains three 25 paise coins and 2 one rupee coins and the other compartment contains two 25 ps. Coins and 3 one rupee coins. The probability of drawing a rupee from the purse is |
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Answer» `1//5` |
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| 2880. |
Let S denotes the set consisting of four functions and S = { [x], sin^(-1) x, |x|,{x}} where , {x} denotes fractional part and[x] denotes greatest integer function , Let A, B , C aresubsets of S. Suppose A : consists of odd functions (s) B : consists of discontinuous function (s) and C: consists of non-decreasing function(s) or increasing function (s). If f(x) in A nn C, g(x) in B nnC, h (x) in B" but notC and" l(x) inneither A nor B nor C . Then, answer thefollowing. The range ofg(f(x)) i |
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Answer» `{-1,0,1}` |
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| 2881. |
Let S denotes the set consisting of four functions and S = { [x], sin^(-1) x, |x|,{x}} where , {x} denotes fractional part and[x] denotes greatest integer function , Let A, B , C aresubsets of S. Suppose A : consists of odd functions (s) B : consists of discontinuous function (s) and C: consists of non-decreasing function(s) or increasing function (s). If f(x) in A nn C, g(x) in B nnC, h (x) in B" but notC and" l(x) inneither A nor B nor C . Then, answer thefollowing. The range of f(h(x)) is |
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Answer» `(0, pi/2)` |
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| 2882. |
Let S denotes the set consisting of four functions and S = { [x], sin^(-1) x, |x|,{x}} where , {x} denotes fractional part and[x] denotes greatest integer function , Let A, B , C aresubsets of S. Suppose A : consists of odd functions (s) B : consists of discontinuous function (s) and C: consists of non-decreasing function(s) or increasing function (s). If f(x) in A nn C, g(x) in B nnC, h (x) in B" but notC and" l(x) inneither A nor B nor C . Then, answer thefollowing. The function l (x) is |
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Answer» periodic |
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| 2883. |
If x+y+z=180^(@) then (sinx+siny+sinz)/("cos"x/2."cos"y/2."cos"z/2)= |
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Answer» 0 |
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| 2884. |
Integrate the following intcos(2-7x)dx |
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Answer» SOLUTION :`INT(COS(2-7x)dx=intcosthetacdot(-1/7d theta)` put 2-7x=`theta` then `-7x=d"theta`or `dx=(-1/7)d"theta` `(-1/7)sintheta+C=(-1/7sin(2-7x)+C` |
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| 2885. |
Evaluate int(1-cos2x)/(1+cos2x)dx. |
| Answer» SOLUTION :`INT(1-cos2x)/(1+cos2x)=int(2sin^2x)/(2cos^2x)dx=inttan^2xdx=int(sec^2x-1)dx=tanx-x+C` | |
| 2887. |
' lim_ (x to 0) (a^(x)-b^(x))/(e^(x)-1) is equal to |
| Answer» Answer :A | |
| 2888. |
int_(0)^((pi)/(2))(f(sinx))/(f(cosx)+f(sinx))dx is : |
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Answer» `pi` |
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| 2889. |
If P(A)=(6)/(11), P(B)=(5)/(11) amd P(A cup B)=(7)/(11), find (i) P(A cup B) (ii) P(A|B) (iii) P(B|A) |
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| 2890. |
Statement :1 If A nad B are two fixed points and P is any point such that PA + PB =k then locus of P is an ellipse. Because Statemet :2 In any ellipse with foci at A and B is PA + PB =k. |
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Answer» Stateme-1 is True, Statement-2 is True, Statemetn-2 is correct explanation for Statement-1 |
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| 2891. |
A particle is thrown with 12 m/s in positive X direction while acceleration is 2 m//s^2 in negative X direction then distance travelled by particle between t=5 to t=7 sec is |
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Answer» 0 m |
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| 2892. |
Solve the following equations, using inverse of a matrix : {:(2x+y+z=1),(2x-4y-2x=3),(3y-5z=9):} |
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| 2893. |
Evaluate the integral underset(0)overset(pi//2)int tan^(5)x cos^(8)x dx |
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| 2894. |
Obtain the general solution of the following differential equations.dy/dz=sqrt(1-y^2)/sqrt(1-z^2) |
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Answer» SOLUTION :`dy/dz=SQRT(1-y^2)/sqrt(1-z^2)` `RARR dy/sqrt(1-y^2)=dz/sqrt(1-z^2)` `rArr intdy/sqrt(1-y^2)=intdz/sqrt(1-z^2)`rArr SIN^(-1)y=sin^(-1)z+C` |
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| 2895. |
Integrationof certainirrational expressions I=int(dx)/(x(2+3sqrt((x-1)/(x)))). |
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| 2896. |
State whether the points A(3, 2), B(-4,-3) lie on the same side or opposite sides of the line 2x - 3y +4 = 0. |
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| 2898. |
Let O be a point inside DeltaABC such that angleOAB = angleOBC = angle OCA = theta Area of DeltaABC is equal to |
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Answer» `((a^(2) + b^(2) + c^(2))/(4)) tan theta` Applying sine rule in `DeltaAOB`, we have `(OA)/(sin angleABO) = (AB)/(sin angle AOB)` or `OA = (c sin angleABO)/(sin angleAOB) = (c sin (B - theta))/(sin B)`...(i) `[ :' angle ABO = B - theta, angle AOB = 180^(@) - theta - angleABO = 180^(@) -B]` Again in `DeltaAOC`, we have `(OA)/(sin angleACO) = (AC)/(sin angleAOC)` `rArr OA = (b sin angleACO)/(sin angleAOC) = (b sin theta)/(sin A)` `[ :' angleOAC = A - theta, angleAOC = 180^(@) - theta - angleOAC = 180^(@)]` From Eqs. (i) and (ii), we have `(c sin (B - theta))/(sin B) = (b sin theta)/(sin A)` or `c sin A (B - theta) = b sin theta sin B` `= b sin theta sin (A +C)` or `2R sin C sin A (sin B cos theta - cos B sin theta)` `= 2R sin B sin theta (sin A cos C + cos A sin C)` Dividing both SIDES by `2R sin theta sin A sin B sin C`, we get `cot theta - cot B = cot C + cot A` or `cot theta = cot A + cot B + cot C` Squaring both sides, we have `cot^(2) theta = cot^(2) A + cot^(2) B + cot^(2)C + 2(cotA cot B + cot B cot C + cot C cot A)` or `cosec^(2) theta - 1 = (cosec^(2) A -1) + (cosec^(2) B -1) + (cosec^(2) C -1) + 2` [SINCE in `DeltaABC, cot A cot B + cot B cot C + cot C cot A = 1`] or `cosec^(2) theta = cosec^(2) A + cosec^(2) B + cosec^(2)C` Area of triangle ABC, `Delta = Delta_(1) + Delta_(2) + Delta_(3)` `=(1)/(2) [a OB + b OC + c OA] sin theta` `=(1)/(4) tan theta [2 a OB cos theta + 2B OC cos theta+ 2c OA cos theta]` `=(1)/(4) tan theta [(a^(2) + x^(2) -y^(2)) + (b^(2) + y^(2) - z^(2)) + (c^(2) + z^(2) - x^(2)]` `= (1)/(4) tan theta [a^(2) + b^(2) + c^(2)]` |
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| 2899. |
If n(A)=n then n{(x,y,z),x,y,z in A ,x ne y, y ne z ,zne x}= |
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Answer» `n^(3)` |
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| 2900. |
The vector equation of the line of intersection of the planes vecr=vecb+lamda_1 (vecb-veca) + mu_1 (veca-vecc) and vecr=vecc+lamda_2(vecb-vecc)+mu_2(veca+vecb) veca,vecb,vecc being non - coplanar vectors, is |
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Answer» `VECR = vecb+mu_1 (veca+vecc)` |
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