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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.
| 1. |
Let [x] represent the greatest integer less than or equal to`x`If [`sqrt(n^2+lambda)]=[n^2+1]+2`, where `lambda,n in N ,`then `lambda`can assume`(2n+4)d iffe r e n tv a l u s``(2n+5)d iffe r e n tv a l u s``(2n+3)d iffe r e n tv a l u s``(2n+6)d iffe r e n tv a l u s`A. (2n+4) different valuesB. (2n+3) different valuesC. (2n+5) different valuesD. (3n+6) different values |
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Answer» Correct Answer - C We have `[sqrt(n^(2)+1)]=n` `:.[sqrt(n^(2)+lambda)]=[sqrt(n^(2)+1)]+2` `implies[sqrt(n^(2)+lambda)]=n+2` `implies n+2 lesqrt(n^(2)+lambda) lt n+3` `implies (n+2)^(2)lelambda lt(n+3)^(2)` `implies 4n+4lelambda6n+9` `implies lambda=4n+4,4n+5,....,6n+8`. Hence, `lambda` can assume (2n+5) distinct values. |
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| 2. |
The solution set of equation `(x+2)^(2)+[x-2]^(2)=(x-2)^(2)+[x+2]^(2)`, where [.] represents the greatest integer function, isA. NB. ZC. QD. R |
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Answer» Correct Answer - B We have, `[x+n]=[x]+n` where `n in Z` and `x in R`. `:. (x+2)^(2)+[x-2]^(2)=(x-2)^(2)+[x+2]^(2)` `implies (x+2)^(2)-(x-2)^(2)=[x+2]^(2)-[x-2]^(2)` `implies(x+2)^(2)-{(x+2)-4}^(2)=[x+2]^(2)-[(x+2)-4]^(2)` `implies (x+2)^(2)-{(x+2)-4}^(2)=[x+2]^(2)-{[x+2]-4}^(2)` `implies 8(x+2)=8[x+2]` `implies(x+2)-[x+2]=0` `implies {x+2}=0` , where{x} denotes the fractional part of x `implies x+2 in Z implies x in Z` Hence, the solution set of the given equation is the set ofintegers. |
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| 3. |
The number of solutions of `|[x]-2x|=4, "where" [x]]` is the greatest integer less than or equal to x, isA. 2B. 4C. 1D. infinite |
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Answer» Correct Answer - B When `x in Z` CASE I In this case, we have [x]=x `:. |[x]-2x|=4 implies |[x]|=4 implies x=+-4` CASE II When `x in Z` In this case, we have `x=n+lambda"where" n in Z "and "0 lt lambda lt 1` `implies[x]=n` `:. |[x]-2x|=4` `implies|n-2(n+lambda)|=4` `implies n+2lambda=+-4 implies n=+-4-2lambda " "`...........(i) This is possible, when `lambda=(1)/(2)` Putting `lambda=(1)/(2)` in (i), we get `n=+-4-1` `implies n=3,-5 impliesx=3+(1)/(2),-5+(1)/(2) implies x=(7)/(2),-(9)/(2)` Hence, `x=+-4,(7)/(2),-(9)/(2)` |
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| 4. |
Number of solutions of the equation `x^(2)-2=[sinx]`, where [.] denotes the greatest integer function, isA. 3B. 4C. 2D. 1 |
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Answer» Correct Answer - C We have, `[sin x]=-1,0,1` So, we have the following cases. CASE I When [sin x]=-1 In this case, we have `x^(2)-2=-1 impliesx=+-1` `:. X=-1 ` is the solution in this case. CASE II When [sin x]=0 In this case, we have `x^(2)-2=0 impliesx=+-sqrt(2)` But, `[sinsqrt(2)]=0` and `[sin(-sqrt(2))]=-1`. `:. x=sqrt(2)` is the solution in this case. CASE III When [sinx]=1 In this case, we have `x^(2)-2=1 impliesx=+-sqrt(3)` But, `=[sin sqrt(3)]=0` and `[sin(-sqrt(2))=-1`. Therefore, there is no solution in this case. REMARK It can be easily seen from the graphs of `y=x^(2)-2` and y=[sin x] that the two curves intersect at x=-1 and `x=sqrt(2)` . Hence, the given equation has two solutions only. |
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| 5. |
The equation `(0.4)^(x-1)=(6.25)^(6x-5)` hasA. no solutionB. one solutionC. two solutionsD. more than two solutions |
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Answer» Correct Answer - B We have, `(0.4)^(x-1)=(6.25)^(6x-5)` `implies ((2)/(5))^(x-1)=((5)/(2))^(12x-10)` `implies((5)/(2))^(13x-11)=((5)/(2))^(0) implies13x-11=0 implies x=(11)/(13)` |
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| 6. |
The number of real solutions of the equation `sin(e^x)=2^x+2^(-x)` is |
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Answer» Correct Answer - A We known that `AM geGM` `:. (2^(x)+2^(-x))/(2)gesqrt(2^(x)xx2^(-x))` `implies 2^(x)+2^(-x)ge2` But, `sin(e^(x)) le1`. Thus, `sin(e^(x)) ne 2^(x)+2^(-x)` for any `x in R` Hence, the given equation has no solution. |
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| 7. |
The equation `2sin^(2)((x)/(2))cos^(2)x=x+(1)/(x),0 lt x le(pi)/(2)` hasA. one real solutionB. no real solutionC. infinitely many real solutionsD. None of these |
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Answer» Correct Answer - B We know that `2"sin"^(2)(x)/(2)cosxle2` and `x+(1)/(2)ge2` for `0 lt x lt (pi)/(2)` Thus, `2"sin"^(2)(x)/(2)cosx=x,x+(1)/(2)` holds good only when each side is eaul to 2. We observe that `x+(1)/(x)=2` for x=1 only. But, `2"sin"^(2)(x)/(2) cos x ne 2` for x=1. Hence, the given equation has no real solution. |
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| 8. |
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,-4)`C. `(4,oo)`D. `[-4,4]` |
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Answer» Correct Answer - B `||x-1|+a|=4` `implies|x-1|+a=+-4` `implies|x-1|=+-a` For real solutions, we must have `+-4-a gt 0` `implies 4-a ge0` and `-4-a ge0` `ale4` and `a le-4` `implies ale-4 implies ale4 implies a in(-oo,-4]` |
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| 9. |
The number of real roots of the equation `sqrt(1+sqrt(5)x+5x^(2))+sqrt(1-sqrt(5)x+5x^(2))=4` is |
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Answer» Correct Answer - C We have, `sqrt(1+sqrt(5)x+5x^(2))+sqrt(1-sqrt(5)x+5x^(2))=4 " "`………(i) Also, `(1+sqrt(5)x+5x^(2))-(1-sqrt(5)x+5x^(2))=2sqrt(5)x " "`........(ii) Dividing (ii) by (i), we get `:. sqrt(1+sqrt(5)x+5x^(2))-sqrt(1-sqrt(5)x+5x^(2))=(sqrt(5))/(2)x " "`.......(iii) Adding (i) and (iii), we get `sqrt(1+sqrt(5)x+5x^(2))=2+(sqrt(5))/(4)x` `implies 1+sqrt(5)x+5x^(2)=4+(5)/(16)x^(2)+sqrt(5)x` `implies x^(2)=(16)/(25) impliesx=+-(4)/(5)` |
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| 10. |
The equation `(x/(x+1))^2+(x/(x-1))^2=a(a-1)`hasFour real roots if `a >2`Four real roots if `a |
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Answer» Correct Answer - D We have, `((x)/(x+1))^(2)+((x)/(x-1))^(2)=a(a-1)` `implies((x)/(x+1)+(x)/(x-1))^(2)-(2x^(2))/(x^(2)-1)=a(a-1)` `implies ((2x^(2))/(x^(2)-1))^(2)-((2x^(2))/(x^(2)-1))=a(a-1)` `implies ((2x^(2))/(x^(2)-1))((2x^(2))/(x^(2)-1)-1)=a(a-1)` ` implies y(y-1)=a(a-1)," where "y=(2x^(2))/(x^(2)-1)` `implies y^(2)-a^(2)-y+a=0` `implies (y-a)(y+a-1)=0 impliesy=a,y=1-a` When y=a, we have `(2x^(2))/(x^(2)-1)=a impliesx=+-sqrt((a)/(a-2))` Clearly, `x in R`if `a in (-oo,0) cup(2,oo)` When y=1-a, we have `(2x^(2))/(x^(2)-1)=a impliesx=+-sqrt((a-1)/(a-2))` Clearly, `x in R` if `a in (-oo,-1)cup (1,oo)`. Thus,the given equation has four real rootsif `a gt2` or a lt-1` and exactly two real roots if `1 lt a lt 2`. Hence,all the options are true . |
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| 11. |
The solution set of the equation `(x)^(2)+[x]^(2)=(x-1)^(2)+[x+1]^(2)`, where (x) denotes the least integer greater than or equal to x and [x] denotes the greatest integer less than or equal to x, isA. RB. R-ZC. R-ND. None of these |
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Answer» Correct Answer - B We have, `(x-1)=(x)-1" and "[x+1]=[x]+1` `:.(x)^(2)+[x]^(2)=(x-1)^(2)+[x+1]^(2)` `implies(x)^(2)+[x]^(2)={(x)-1}^(2)+{(x]+1)^(2)` `implies(x)^(2)+[x]^(2)=(x)^(2)-2(x)+1+[x]^(2)+2[x]+1` `implies[x]-(x)+1=0` Now, two cases arise. CASE I When `c in Z` In this case,we have [x]=x and `(x)=x implies[x]-(x)+ne0` So, the equation [x]-(x)+=0 has no solution. CASE II When `x in Z` In this case, we have x=n+k, where `n in Z` and `0 lt k lt 1` `:. [x]=n` and (x)=n+1 `:. [x]-x+1=0` is true for all x Hence, the solution set of the given equation is R-Z. |
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| 12. |
The solution set of the equation `|(x+1)/(x)|+|x+1|=((x+1)^(2))/(|x|)`, isA. `{x:xge0}`B. `{x:x gt 0} cup{-1}`C. `{-1,1}`D. `{x:x ge1" or ",x le-1}` |
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Answer» Correct Answer - B We have, `|(x+1)/(x)|+|x+1|=((x+1)^(2))/(|x|)` `implies(|x+1|)/(|x|)+|x+1|=(|x+1|^(2))/(|x|)` `implies|x+1|{(1)/(|x|)+1-(|x+1|)/(|x|)}=0` `implies |x+1|=0" or ",(1)/(|x|)+1=(|x+1|)/(|x|)` `implies |x+1|=0" or ",1+|x|=|x+1|` `implies x=-1" or ",1+|x|=|x+1|` In order to solve the equation `1+|x|=|x+1|`, we consider the following cases : In this case, we have |x|=-x and |x+1|=-(x+1) `:. 1+|x|=|x+1|` `implies 1-x=-(x+1)`, which is absurd. CASE II When `-1 lex lt 0` In this case, we have |x|=-a and |x+1|=x+1 `:. 1+|x|=|x+1|implies1-x=x+1 implies x=0` But, `-1 le x lt 0` So, there is no solution in this case. CASE III When `x ge0` In this case, we have |x|=x and |x+1|=x+1 `:. 1+|x|=|x+1|` `implies 1+x=x+1`, which is true for all x. Clearly, the given equation is meaningful for `x ne 0` Hence, the solution set of the given equation is `{x:x gt 0}cup{-1}` |
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| 13. |
If `(sqrt(2))^(x)+(sqrt(3))^(x)=(sqrt(13))^(x//2)`, then the number of real values of x isA. 2B. 4C. 1D. None of these |
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Answer» Correct Answer - C We have, `(sqrt(2))^(x)+(sqrt(3))^(x)=(sqrt(13))^(x//2)` `implies 2^(x//2)+3^(x//2)=(sqrt(13))^(x//2)` `implies ((2)/(sqrt(13)))^(x//2)+((3)/(sqrt(13)))^(x//2)=((2)/(sqrt(13)))^(2)+((3)/(sqrt(13)))^(2)` `implies(x)/(2)=2 " "`[On comparison] `implies x=4`. |
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| 14. |
The equation `(sqrt(5)+2)^(|x|)+(sqrt(5)-2)^(|x|)=18`. hasA. only one solutionB. two solutionsC. no solutionD. any number of solutions |
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Answer» Correct Answer - B We have, `(sqrt(5)+2)^(|x|)+(sqrt(5)-2)^(|x|)=(sqrt(5)+2)^2)+(sqrt(5)-2)^2)` `implies |x|=2 implies x=+-2` |
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| 15. |
The equation `(x^2+x=1)^2+1=(x^2+x+1)(x^2-x-5)`for `x in (-2,3)`will have number of solutions.`1`b. `2`c. `3`d. 0A. 2B. 4C. 3D. None of these |
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Answer» Correct Answer - D We have, `(x^(2)+x+1)^(2)+1=(x^(2)+x+1)(x^(2)-x-5)` `implies (x^(2)+x+1)+(1)/((x^(2)+x+1))=x^(2)-x=5` We know that `x^(2)+x+1 gt0` for all `x in R`. Therefore, LHS`=(x^(2)+x+1)+(1)/((x^(2)+x+1)) ge2` But, RHS `=x^(2)-x-5 lt 1` for `x in (-23)`. Hence. the given equation has no solution. |
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| 16. |
The roots of the equation `2^(x+2).3^((3x)/(x-1))=9` are given byA. `log_(2)((2)/(3)),-2`B. 3,-3C. `-2,1-(log3)/(log 2)`D. `1-log((2)/(3)),2` |
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Answer» Correct Answer - C We have, `2^(x+2)xx3^((3x)/(x-1))=3^(2)` `rArr(x+2)log 2+(3x)/(x-1)log 3=2 log 3` `rArr (x+2)log 2+((3x)/(x-1)-2)log 3=0` `rArr (x+2)log 2+((x+2)/(x-1))log 3=0` `rArr (x+2){log 2 +(log 3)/(x-1)}=0` `rArr x+2=0" or ",log 2+(log 3)/(x-1)=0` `rArrx=-" or ",x=1-(log 3)/(log 2)` |
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| 17. |
The number of real roots of the equation `1+3^(x//2)=2^(x)`, is |
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Answer» Correct Answer - B We have, `1+3^(x//2)=2^(x)` `implies ((sqrt(3))/(2))^(x)+((1)/(2))^(x)=1` `implies(sqrt(3)/(2))^(x)+((1)/(2))^(x)=((sqrt(3))/(2))^(2)+((1)/(2))^(2) implies x=2` Thus, x=2 is a solution of the given equation. Now, let `y=((sqrt(3))/(2))^(x)+((1)/(2))^(x)` Clearly,`((sqrt(3))/(2))^(x)" and "((1)/(2))^(x)` are decreasing functions of x. Therefore, `y=((sqrt(3))/(2))^(x)+((1)/(2))^(x)` is a decreasing function of x. Cosequently, we have `y=((sqrt(3))/(2))^(x)+((1)/(2))^(x) lt 1` for `x gt 2` and, `y gt 1` for `x lt 2`. Hence, the given equation has just one solution. REMARK Any equation of the form `a^(x)+b^(x)=1`, where `a^(2)+b^(2)=1` has just one solution given by x=2. |
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| 18. |
If `3^(x+1)=6^(log_(2)3)`, then x is equal toA. 3B. 2C. `log_(3)2`D. `log_(2)3` |
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Answer» Correct Answer - D We have, `3^(x+1)=6^(log_(2)3)` `3^(x+1)=2^(log2^(3))implies3^(x)xx3=3^(log2^(3))xx3` `x=log_(2)3` |
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| 19. |
The number of real roots of the equation `x^(2)+x+3+2 sin x=0, x in [-pi,pi]`, isA. 2B. 3C. 4D. None of these |
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Answer» Correct Answer - D We have, `x^(2)+x+3+2 sin x=0` `implies x^(2)+x+3=-2 sin implies(x+(1)/(2))^(2)+(11)/(4)=-2 sin x` We observe that `LHS gt (11)/(4) gt 2` for all x whereas RHS lies between -2 and 2 Hence, the given equation has no solution. |
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| 20. |
The number of real roots of the equation `log_(1//3)(1+sqrt(x))+log_(1//3)(1+x)=2`, is |
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Answer» Correct Answer - A We have, `log_(1//3)(1+sqrt(x))+log_(1//3)(1+x)=2` `implies -log_(3)(1+sqrt(x))-log_(3)(1+x)=2` Clearly, LHS is meaningful for `x ge0`. Also, `LHS le0` for all `x ge0` and RHS=0. So, the given equation has no solution. |
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| 21. |
If `5^x+(2sqrt(3))^(2x)geq1 3^x` then the solution set forA. `[2,oo)`B. {2}C. `(-oo,2]`D. [0,2] |
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Answer» Correct Answer - C We have, `5^(x)+(2sqrt(3))^(2x)ge13^(x)` `implies ((5)/13)^(x)+((12)/(13))^(x)ge1` `implies a^(x) +b^2ge 1`,where `s(5)/(12)` and `b=(12)/(13) ` ` implies a^(x) =b^(x) ge 1 `, where `a^(2) +b^(2) =1` Let `f(x) =a^(x) +b^( x)` we observe that `f(2) = a^(2) + b^(2) =1` Also, `f(x) gt 1` for `x lt 2` and `f(x) lt 1 ` for `x gt 2` Thus, `f(x) ge 1`for `x le 2`. Hence, the solution set of the given inequation is `(-oo,2]`. |
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| 22. |
The number of solutions of the equation `log_(3)(3+sqrt(x))+log_(3)(1+x^(2))=0`, is |
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Answer» Correct Answer - A Clearly, the given equation is meaningful for `x ge0`. We observe that `3+sqrt(x) gt 3` and `1+x^(2) gt 1` for all `x gt 0` `implies log_(3)(3+sqrt(x)) gt 1` and `log_(3)(1+x^(2)) gt 0` for all ` x gt 0` `implies log_(3)(3+sqrt(x))+log_(3)(1+x^(2)) gt 0 ` for all ` xge0` Hence, the given equation has no solution. |
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| 23. |
The equation `5^(1+log_(5)cosx)=5/2`hasA. no solutionB. one solutionC. two solutionsD. more than two solutions |
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Answer» Correct Answer - D We have, `5^(1+log_(5)cosx)=(5)/(2)` ` implies1+log_(5)5+log_(5) coss=log_(5)5-log_(5)2` `implies log_(5)cos x=-log_(5)2` `implies log_(5) cos x=log_(5)2^(-1)` `implies cosx=2^(-1) implies cos x=(1)/(2) impliesx=2n pi+-(pi)/(3),n in Z` |
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| 24. |
The equation `(|x^2-4x|+3)/(x^2+|x-5|)=1` hasA. no real solutionB. exactly one real solutionsC. two real solutionsD. three real solutions |
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Answer» Correct Answer - D Let us consider the following cases: CASE I When `x le0` or ,`4lex le5` In this case, we have `|x^(2)-4x|=x^(2)-4x` and `|x-5=-(-5)`. So, the given equation reduces to `(x^(2)-4x+3)/(x^(2)-x+5)=1` `implies x^(2)-4x+3=x^(2)-x+5 implies-3x=2 implies x=-(2)/(3)` CASE II When `0 lt x lt 4` In this case, we have `|x^(2)-4x|=-(x^(2)-4x)` and `|x-5|=-(x-5)`. So, the given equation reduces to `-x^(2)+4x+3=x^(2)-x+5` `implies 2x^(2)-5x+2=0 implies(2x-1)(x-2)=0 implies x=(1)/(2),2` CASE III When ` x gt 5` In this case, we have `|x^(2)-4=x^(2)-4x` and `|x-5|=x-5` So, the equation reduces to `x^(2)-4x+3=x^(2)+x-5 implies-5x=-8 impliesx=(8)/(5)` But, it does not belong to `(5,oo)`. Hence, the given equation has three real solutions. |
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| 25. |
The number of real solutions of equation `2^(x/2)+(sqrt2+1)^x=(5+2sqrt2)^(x/2)` isA. 1B. 2C. 4D. infinite |
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Answer» Correct Answer - A We have, `2^(x//2)+(sqrt(2)+1)^(x)=(5+2sqrt(2))^(x//2)` `(sqrt(2))^(x)+(sqrt(2)+1)^(x)={sqrt((sqrt(2))^(2)+(sqrt(2)+1)^(2))}^(x)` `implies (sqrt(2))^(x)+(sqrt(2)+1)^(x)=(sqrt(5+2sqrt(2)))^(x)` `implies ((sqrt(2))/(sqrt(5+2sqrt(2))))^(x)+((sqrt(2)+1)/(sqrt(5+2sqrt(2))))^(x)=1` `implies ((sqrt(2))/(sqrt(5+2sqrt(2))))^(x)+((sqrt(2)+1)/(sqrt(5+2sqrt(2))))^(x)=((sqrt(2))/(sqrt(5+2sqrt(2))))^(2)+((sqrt(2)+1)/(sqrt(5+2sqrt(2))))^(x)` `implies x=2` |
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| 26. |
The number of real solutions of the equation `e^x=x` isA. 1B. 2C. 0D. None of these |
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Answer» Correct Answer - C It is evident from the graphs of y=`e^(x)` and y=x (see Fig.1) that they do not intersect. Hence, the equation `e^(x)`=x has no solution. ALITER We observe that `e^(x) gt x` for all `x in R` `:. E^(x)=x` has no solution. |
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| 27. |
Let `(x_0, y_0)` be the solution of the following equations In `(2x)^(In2)=(3y)^(ln3)`and `3^(lnx)=2^(lny)` Then `x_0` isA. `(1)/(6)`B. `(1)/(3)`C. `(1)/(2)`D. 6 |
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Answer» Correct Answer - C We have, `(2x)^("In"2)=(3y)^("In"3)"and",3^("In"x)=2^("In"y) ` `rArrlog(2x)^("In"2)=log(3y)^("In"3)" and "log(3^("In"x))=log(2^("In"y))` `rArr log2(log2+logx)=log3(log+logy)` and, `log3 logx=log2 logy` `rArr (log2)^(2)+(log2)(logx)=(log3)^(2)+((log3)^(2)logx)/(log2)` [On eliminating logy] `rArr ((log2)^(2)-(log3)^(2))/(log2)logx=(log3)^(2)-(log2)^(2)` `rArr logx=-log2 rArrx=(1)/(2)` |
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