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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. |
A differentiable function `f(x)`has a relative minimum at `x=0.`Then the function `f=f(x)+a x+b`has a relative minimum at `x=0`forall `a`and all`b`(b) all `b`if `a=0`all `b >0`(d) all `a >0`A. all a and all bB. all b if a =0C. all b `gt` 0D. all a `gt` 0 |
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Answer» Correct Answer - 2 Since f(x) has a relative minimum at x =0 f(X)=0 and `f(0)gt0` (assuming `f(X)ne 0)` If the funciton y =f(X) ax+b has a relative minimum at x=0 then `(dy)/(dx)=0 at x =0 or f(X) + a =0 for x =0` F(0)+a=0 or a =0 or a=0 Now `(d^(2)y)/(dx^(2))=f'(x)or (d^(2)y)/(dx^(2))_(x=0)=f'(0)gt0 [therefore f'(0)gt0]` Hence y has a relative minimum at x= 0 if a =0 and b can attain anyreal value. |
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| 2. |
If `g(x)=max(y^(2)-xy)(0le yle1)`, then the minimum value of g(x) (for real x) isA. `(1)/(4)`B. `3-sqrt3`C. `3+sqrt8`D. `(1)/(2)` |
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Answer» Correct Answer - B `y^(2)-xy=(y-(x)/(2))^(2)-(x^(2))/(4) rArr y^(2)-xy` is decreasing for `yle (x)/(2)` and increasing for `yge(x)/(2)` Thus the largest value of `y^(2)-xy` must be at `y=0,(x)/(2)` or 1. The values are `0,(x^(2))/(4),1-x` for `x in (0,1)` And for `x in (0,1)g(x)=max((x^(2))/(4),1-x)` Also `x^(2)+4x=4=0 rArr x = -2 pm sqrt8` `rArr" "g(x)=1-x " for "x le sqrt8-2 and (x^(2))/(4)` for `x ge sqrt8-2` `rArr" g(x) is minimum at "x=sqrt8-2` and minimum value is `3-sqrt8` |
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| 3. |
If the equation `4x^(3)+5x+k=0(k in R)` has a negative real root thenA. k=0B. `-ooltklt0`C. `0ltkltoo`D. `-ooltkltoo` |
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Answer» Correct Answer - 3 Let (x) =`4x^(3)+5x+k` `therefore f(x)=12x^(2)+5gt0 for all x in R` `therefore` f(X) is stictly increasing on R ltrbgt For negative root of the equation f(X) =0 graph of y =f(x) must intersect y axis at positive side |
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| 4. |
An extremum of the function `f(x)=(2-x)/picospi(x+3)+1/(pi^2)sinpi(x+3),0x |
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Answer» Correct Answer - 1,3 `f(x)=(2-x)/(pi)cospi(x+3)+(1)/(pi^(2))sinpi(x+3)` `f(X)=-(1)/(pi)cos pi(x+3)-(2-x)sinpi(x+3)` `=(x-2)sinpi(x+3)=0(let)` `therefore x=2,1,3` `f(x)=sinpi(x+3)+pi(x-2)cospi(x+3)` `f(X)=-pilt0f(2)=0f(3)=pigt0` Therefore x=1 is a maximum and x =3 is a minimum hence x=2 is the point of infleciton |
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| 5. |
Discuss the extremum of `f(x)=x(x^2-4)^(-1/3)` |
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Answer» Correct Answer - Point of local maxima : x=`-2sqrt(3)` Point of local minima : `x=2sqrt(3)` `f(x)=x(x^(2)-4)^(-1//3` `f(x)=(x^(2)-4)^(-1//3-1)/(3)(x^(2)-4)-4//3(2x)x` `=(1)/(x^(2)-4)^(1//3)-(2x^(2))/(3(x^(2)-4)^(4//3)` `=3(x^(2)-4)-2x^(2)/(3(x^(2-4))^(4//3)` `=(x^(2-12))/(3(x^(2)-4)^(4//3)` Sign scheme of f(X) is as follows: Thus `x=2 sqrt(3)` is point opf minima and x =`-sqrt(3)` is point of maxima `f_(min)=f(2 sqrt(3))=2 sqrt(3)(12-4)A^(-1//3)=sqrt(3)` Here maxima value is less than minimum value Thus is because f(X) discontinous x=2 Since f(X) is unbounded function both the extreme values are local |
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| 6. |
consider P(x) to be a polynomial of degree 5 having extremum at x=-1,1 and `lim_(xrarr0) p(x)/(x^(3))-2=4` Then the value of P(x)_____. |
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Answer» Correct Answer - 2.4 Give `underset(xrarr0)lim p(x)/(x^(3))-2=4` `therefore underset(xrarr0)lim(p(x))/(x^(3))=6` consider P(x) =`ax^(5)+bx^(4)+6x^(3)` Now `P(-1)=0 rarr5a-4b=-018` and `P(1)=0 rarr 5a+4b=-18` Hence P(x) =`(-18)/(5)x^(5)+6x^(3)` or P(1)`=12/3` |
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| 7. |
The funciton f(x)=`x^(2)+(lambda)/(x)`has aA. minimum at x =2 if `lambda` =16B. maximum at x =2 if `lambda` =16C. maximum for no real value of `lambda`D. point of inflectin at x=1 if `lambda`=-1 |
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Answer» Correct Answer - 1,3,4 `f(x)=2x-(lambda)/(x^(2)),f(x)=0` `rarr x=((lambda)/(2))^(1//3)` if `lambda=16,x=2` Now `f(x)=2+(2lambda)/(x^(3))` Thus if `lambda=16 f(x)gt0` i.e f(X) has a minimum x=2 Also `f{((lambda)/(2))^(1//3)}=2+(2lambda)/(lambda//2)=2+4gt0` Hence f(x) has maixmum for no real value of `lambda` When `lambda =-1 f(x)=0` if x=1 so f(x) has a point of inflection at x=1 |
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| 8. |
Separate the interval of convaity of y =x `log_(e)x-(x^(2))/(2)+(1)/(2)` |
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Answer» Correct Answer - Concave upward in (0,1) Concave downward in `(1,oo)` `y=1+log_(e)x-x` `rarr y'=(1)/(x)-1=(1-x)/(x)` `y' gt for x in (0,1)` and `y' lt 0 for x in (1,oo)` `therefore ` graph is concave upward for x in (0,1) and concave downward for x in (1,o)` |
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| 9. |
If a function `f(x)`has `f^(prime)(a)=0a n df^(a)=0,`then`x=a`is a maximum for `f(x)``x=a`is a minimum for `f(x)`it is difficult to say `(a)a n d(b)``f(x)`isnecessarily a constant function.A. x=a is a maximum for f(x)B. x=a is a minimum for(x)C. it is difficult to say (a) and (b)D. f(x) is necessarily a constant function |
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Answer» Correct Answer - 3 consider function `x^(3)-x^(4),x^(4)` |
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| 10. |
if `f(x)=4x^3-x^2-2x+1` and `g(x)={min{f(t): 0 |
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Answer» Correct Answer - 4 `f(X)=12x^(2)-2x-2` `=2(3x+1)(2x-1)` `thereforeg(X)={{:(f(x),0lex1//2),(f(1//2),1//2lexle1):}` `rarr g(1/4)+g(3/4)+g(5/4)` `=f(1/4)+f(1/2)+g(5/4)` `=5/2` |
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| 11. |
If `f(x)=alog|x|+b x^2+x`has its extremum values at `x=-1a n dx=2,`then`a=2,b=-1``a=2,b=-1//2``a=-2,b=1//2`(d) none of theseA. a=2,b=-1B. a=2,b=`-1//2`C. a=-2,b=`1//2`D. none of these |
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Answer» Correct Answer - 2 We have `f(X) =a log|x|+bx^(2) +x` or `f(X) =(a)/(x)+2bx+1` since f(X) attains its extremum values at x=1,2 f(-1) =0 and f(X) =0 or -a-2b+1 and `(a)/(2)`+4b+1=0 or a =2 and b =-`1//2` |
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| 12. |
Consider a polynomial y = P(x) of the least degree passing through A(-1,1) and whose graph has two points of inflection B(1,2) and C with abscissa 0 at which the curve is inclined to the positive axis of abscissa at an angle of `sec^(-1)sqrt(2)` The value of P(0) isA. 1B. 0C. `3/4`D. `1/2` |
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Answer» Correct Answer - 4 Since two points of inflection occur at x=1 and x=0 p(1)=P(0)=0 `therefore P(x)=a(x^(2)-x)` or `P(x)=a(x^(3)/(3)-x^(2)/(3))` also given `(dy)/(dx)_(x=0)=sec^(-1)sqrt(2)=tan^(-1)1` Hence P(0)=1 so b =1 thus `P(x)=a(x^(3)/(3)-x^(2)/(2))+1` `therefore P(x) =a(x^(4)/(12)-x^(3)/(6))=x+c` As P(1)=2 ,we have `a((1)/(12)-(1)/(6))+1+c=1` or `(a)/(12)+c=0` solving (1) and (2) we have `a =6 and c=1/2` `P(x) =6(x^(4)/(12)-x^(3)/(6))+x+1/2` `P(2)=5/2 land p(x)=1/2` `P(x) =6(x^(3)/(3))-(x^(2)/(2))+1=(x-1)^(2)(2x+1)` |
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| 13. |
Consider a polynomial y = P(x) of the least degree passing through A(-1,1) and whose graph has two points of inflection B(1,2) and C with abscissa 0 at which the curve is inclined to the positive axis of abscissa at an angle of `sec^(-1)sqrt(2)` The value of P(2) iusA. -1B. `(-3)/(2)`C. `(5)/(2)`D. `(7)/(2)` |
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Answer» Correct Answer - 3 Since two points of inflection occur at x=1 and x=0 p(1)=P(0)=0 `therefore P(x)=a(x^(2)-x)` or `P(x)=a(x^(3)/(3)-x^(2)/(3))` also given `(dy)/(dx)_(x=0)=sec^(-1)sqrt(2)=tan^(-1)1` Hence P(0)=1 so b =1 thus `P(x)=a(x^(3))/(3)-(x^(2))/(2)+1` `therefore P(x) =a(x^(4))/(12)-(x^(3))/(6)=x+c` As P(1)=2 ,we have `a(1)/(12)-(1)/(6)+1+c=1` or `(a)/(12)+c=0` solving (1) and (2) we have `a =6 and c=1/2` `P(x) =6(x^(4))/(12)-(x^(3))/(6)+x+1/2` `P(2)=5/2 and p(x)=1/2` `P(x) =6(x^(3))/(3)-(x^(2))/(2)+1=(x-1)^(2)(2x+1)` |
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| 14. |
If x = -1 and x = 2 are extreme points of f(x) = `alpha log|x| + beta x^2 + x`, thenA. `alpha =-6,beta=1/3`B. `alpha =-6,beta=-1/2`C. `alpha =2,beta=-1/2`D. `alpha = 2,beta=1/2` |
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Answer» Correct Answer - 3 `f(X)=(alpha)/(x)+2betax+1` `rarr 2betax^(2)+x+alpha`=0 has roots -1 and 2 `rarr alpha=2,beta =-1/2` |
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| 15. |
The least integral value of `x`where `f(x)=(log)_(1/2)(x^2-2x-3)`is monotonically decreasing is________ |
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Answer» Correct Answer - 4 `x^(2)-2x3gt0` or `(x-3)(x+1)gt0` i.e `xlt1 or xgt3` Now `f(X)=log_(1//2)(x^(2)-2x-3)` `=log_(e)(x^(2)-2x-3)/(log_(e)(1//2))` `f(X)=(2x-2)/(log_(e)(1//2))(x^(2)-2x-3)` For f(x) to decreasing `f(X)lt0` or `(x-1)/(log_(e)(1//2)(x-3)(x+1))lt0` or `xgt1` From 1 and 2 x `gt`3 |
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| 16. |
Let f be function `f(x)=cosx-(1-X^(2)/(2))`.ThenA. f(x) is an increasing function in `(0,oo)`B. f(X) is a decresing function in `(-oo,oo)`C. f(X) is an increasing function in `(-oo,oo)`D. f(x) is a decreasing function in `(-oo,0)` |
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Answer» Correct Answer - 1,4 f(x) =`cos x -(1-(x^(2))/(2))` `therefore f(x)=-sin x+x` `sinx lt x if x gt 0 and sinx gt x if x lt 0` |
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| 17. |
The set of value(s) of `a`for which the function`f(x)=(a x^3)/3+(a+2)x^2+(a-1)x+2`possesses a negative point of inflection is`(-oo,-2)uu(0,oo)`(b) `{-4/5}``(-2,0)`(d) empty setA. `(-oo,-2)cup(0,oo)`B. `{-4//5}`C. `(-2,0)`D. empty set |
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Answer» Correct Answer - 1 `f(x) = ax^(2)+2(a+2)x+(a-1)` `f'(x) =2ax+2(a+2)=0` Thus `x=(+2)/(a)` which is the point of inflection Given that we must have `(a+2)/(a) lt or a in (-oo,-2)cup(0,oo)` |
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| 18. |
Show that `f(x) = 2x + cot^-1 x + log(sqrt(1+x^2)-x)` is increasing in `R`A. increases in `[0,oo)`B. idecreases in `[0,oo)`C. neither increases nor decreases in `[0,oo)`D. increases in `(-oo,oo)` |
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Answer» Correct Answer - 1,4 We have f(x)=`2x+cot^(-1)x+logsqrt(1+x^(2)-x)` `therefore f(X)=2(1)/(1+x^(2))+(1)/(sqrt(1+x^(2))-x)(x)/(sqrt(1+x^(2))-1)` `=(1+2x^(2))/(1+x^(2))-(1)/sqrt(1+x^(2))=(1+32x^(2))/(1+x^(2))=-sqrt(1+x^(2))/(1+x^(2))` `=(x^(2)+sqrt(1+x^(2))sqrt(1+x^(2))-1)/(1+x^(2))gt0` for all x Hence f(X) is an increasing function in `(-oo,oo)` and in particular in `(0,oo)` |
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| 19. |
Let `f(x)=-sin^3x+3sin^2x+5on[0,pi/2]`. Find the local maximum and local minimum of `f(x)dot` |
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Answer» Correct Answer - Strictly increasing f(x) =-`sin^(3) x + 3 sin^(2) x+5` `therefore f(X) =-3 cos x sin^(2) + 6 sinxcos x` Now `sin x -2 lt 0 forall x in [0,(pi)/(2)]` sinx cos `x ge 0 forall x in [0,(pi)/(2)]` Thus f(x) is a strictly incresing function `forall x in [0,pi//2]` Hence f(x) is minimum when x=0 and maximum when `x =pi//2` `f_(min)=f(0)=5` `f_(max)=f(pi//2)=7` |
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| 20. |
The maximum value of the function `f(x)=((1+x)^(0. 6))/(1+x^(0. 6))`in the interval `[0,1]`is`2^(0. 4)`(b) `2^(-0. 4)`1(d) `2^(0. 6)`A. `2^(0.4)`B. `2^(-0.4)`C. 1D. `2^(0.6)` |
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Answer» Correct Answer - 3 `f(x) =0.6(1+x)^(0.4)(1+x^(0.6))-0.6x^(0.4)(1+x)^(0.4)/(1+x^(0.6))^(2)` `=0.6(1+x^(0.6))-x^(0.4(1+x^(1)))/(1+x^(0.6^(2))(1+x^(0.4)))` `=0.6 (x^(0.4)-1)/(1+x^(0.6^(2)))(1+x)^(0.4)x^(0.4)lt0 forallx in (0,1)` Hence f(X) is decreasing thus `f(x)_(max) =f(0)=1` |
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| 21. |
A tangent is drawn to the ellipse `(x^2)/(27)+y^2=1`at `(3sqrt(3)costheta(0,pi/2)dot`Then findthe value of `theta`such that the sum of intercepts on the axes made by thistangent is minimum.A. `(pi)/(3)`B. `(pi)/(6)`C. `(pi)/(8)`D. `(pi)/(4)` |
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Answer» Correct Answer - 2 Equation of the tangenth to the ellipese `(x^(2))/(27)+y^(2) =1 at 3sqrt(3 cos theta sin theta),theta in (0,pi//2)` is `(sqrt(3)cos theta)/(9)+y sin theta=1` `therefore` sum of the intercepts =S =`3sqrt(3)theta+cosec theta` For minimum values of s,`(ds)/(d theta)=0` `3(sqrt(3)sin theta)/(cos^(2)theta)-(cos theta)/(sin^(2)theta)=0` or `3sqrt(3)sin^(3) theta - cos^(3) theta =0` or `tan theta =(1)sqrt(3)=tan pi//6 rarr theta -pi//6` |
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| 22. |
Let `f(x)=2x-sin x and g(x)=3sqrt(x)`. ThenA. range of gof is RB. gof is one-oneC. both f and g are one-oneD. both f and g are onto |
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Answer» Correct Answer - 1,2,3,4 f(X)=`2x -sin x or f(x)=2 - cos x gt 0 forall x` .Hence f(X) is strictly increasing .Hence the function is one in and onto `g(x)=x^(1//3)or g(x)=(1)/(3)x^(-2//3)gt0 forallx` Hence g(x) si strictly increasing and hence one one onto also gof is one-one `gof(x)=(2x-sin x )^(1//3)` has range R as the range of 32x - sin x is R. |
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| 23. |
A function is matched below against an interval where it is supposed tobe increasing. Which of the following parts is incorrectly matched?Interval,Function[2, `oo)`, `2x^3-3x^2-12 x+6``(-oo,oo)`, `x^3=3x^2+3x+3``(-oo-4)`, `x^3+6x^2+6``(-oo,1/3)`, `3x^2-2x+1`A. `[2,oo),2x^(3)-3x^(2)-12x+6`B. `[-oo,oo),x^(3)-3x^(2)-3x+3`C. `[-oo,-4),x^(3)-6x^(2)+6`D. `[-oo,(1)/(3)],3x^(2)-2x+1` |
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Answer» Correct Answer - 4 f(x)=`3x^(2)-2x+1` `therefore f(X)=6x-2` f is increasing .Thus `f(x)ge0i.e` `6x-2ge or x ge1/3` |
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| 24. |
Find the range of f(x) =`(sinx)/(x)+(x)/(tanx) in (0,(pi)/(2))` |
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Answer» Correct Answer - `((2)/(pi),2)` It can be proved that both the functions `(sinx)/(x) and (x)/(tx)` are decreassing in `(0,(pi)/(2))` `therefor` e Range of the function is `(underset(xrarrpi//2)limf(x),underset(xrarr0)limf(x))=((2)/(pi),2)` |
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| 25. |
The least value of `a`for which the equation `4/(sinx)+1/(1-sinx)=a`has at least one solution in the interval `(0,pi/2)`9 (b)4 (c) 8(d) 1A. 9B. 4C. 8D. 1 |
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Answer» Correct Answer - 3 Since a =`((4)/(sinx)+(1)/(1-sinx))`,a is atleast `therefore (da)/(dx)=[(-4)/(sin^(2)x)+(1)/(1-sinx)^(2)]cosx=0` we have to find the value of x in the interval `(0,pi//2)` Thus cos `x ne 0` and the other factor when equated to zero gives sinx =`2//3` Now Put sin x=`2/3 and cos^(2) x=1-4/9=5/9` `therefore (d^(2)a)/(dx^(2)=0+[(8)/(8//27)+2xx27]5/9=81xx8/9=45gt0` Thus a is minimum and its value is `(4)/(2//3)+(1)/(1-2//3)=6+3=9` |
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| 26. |
The vertices of a triangle are (0,0), `(x ,cosx),`and `(sin^3x ,0),w h e r e0A. `3sqrt(3)/(32)`B. `(sqrt(3))/(32)`C. `(4)/(32)`D. `6 sqrt(3)/(32)` |
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Answer» Correct Answer - 1 f(x)=`(sin^(3)xcosx)/(2)` `therefore f(x)=(3 sin^(2)xcos^(23)x-sin^(4)x)/(2)` `f(x) =0 rarr 3 sin^(2)x cos^(2)x - sin^(4)x=0` or `3 cos^(2)x-sin^(2)x=0` or `4cos^(2)x-1=0` or `cosx=1/2` or `x=(pi)/(3)` which is the point of maxima `therefore f_(max)=(sqrt(3)//2)^(3)(1//2)/(2)=(3sqrt(3))/(32)` |
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| 27. |
If `f(x)=int_0^x (sint)/(t)dt,xgt0,` thenA. f(x) has a local maxima at x =`npi(n=2k,kin I^(+))`B. f(x) has a local minimum at x =`npi(n=2k,kin I^(+))`C. f(x) has neither maxima nor minima at x =`npi(n in I^(+))`D. f(x) has local maxima at x =`npi(n=2k-1,k in I^(+))` |
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Answer» Correct Answer - 2,4 `f(x) =(sinx )/(x)` `for f(x)=0,(sinx)/(x)=0 or x =npi,(n in I ,n ne0)` `f(x)=(xcosx-sinx)/(x^(2))` `f(npi)=(cosnpi)/(npi)lt0if n=2k -1and gt0nif n =2k,kin I^(+)` Hence f(X) has local maxima at `x = npi` , where n=2k -I and local minima at `x =npi, n =2k` where `k in I^(+)` |
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| 28. |
Find the number of solution of the equation `x^(3)+2x+cosx+tanx=0 in (-(pi)/(2),(pi)/(2))` |
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Answer» Correct Answer - one solution Let f(x) =`x^(3)+2x+cos x +tan x` `therefore f(x) =3x^(2)+(2-sinx)+sec^(2)xgt0` So f(x) is incresing funciton `therefore f(x) =1 and f(-1)lt0` Therefore equation has one solution in `((pi)/(2),(pi)/(2))` |
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| 29. |
The length of the longest interval in which the function `3sinx-4sin^3x`is increasing is`pi/3`(b) `pi/2`(c) `(3pi)/2`(d) `pi`A. `(pi)/(3)`B. `(pi)/(2)`C. `(pi)/(2)`D. `(pi)` |
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Answer» Correct Answer - 1 3 `sin x -4 sin^(3)x` = sin 3x which increases for `3x in -(pi)/(2),(pi)/(2) or x in -(pi)/(6),(pi)/(6)` whose length is `(pi)/(3)` |
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| 30. |
If `f(x)={sin^(-1)(sinx),xgt0` `(pi)/(2),x=0,then cos^(-1)(cosx),xlt0`A. x=0is apoint of maximaB. x=0 is a point of minimaC. x=0 is a point of intersectionnD. none of these |
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Answer» Correct Answer - 1 `f(x) =pi//2f(0^(+))=0f(0^(-))=0` Hence x =0 is the point of maxima |
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| 31. |
The function `f(x)=tan^(-1)(sinx+cosx)`is an increasing function in`(-pi/2,pi/4)`(b) `(0,pi/2)``(-pi/2,pi/2)`(d) `(pi/4,pi/2)`A. `(0,pi//2)`B. `(0,(pi)/(2))`C. `(-(pi)/(2),(pi)/(2))`D. `((pi)/(4),(pi)/(2))` |
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Answer» Correct Answer - 1 Here fX() =`tan^(-1) (sinx+cosx)` `therefore f(X)=(1)/(1+(sinx+cosx)^(2)(cosx-sinx)` `=(cosx-sinx)/(2+sin 2 x)` `for -(pi)/(2)ltxlt(pi)/(4),cos x gtsinx` Hence y = f(X) is increasing in `(-(pi)/(2),(pi)/(4))` |
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| 32. |
If `f(x)=|x-1|+|x+4|x-9|+….+|x-2500| AA x in R`, then all the values of x where f(x) has minimum values lie inA. (600, 700)B. (576, 678)C. (625, 678)D. none of these |
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Answer» Correct Answer - C `f(x)=|x-1|+|x+4|+|x-9|+ …. +[x-2500]` is non-differentiable at `x=1^(2), x=2^(2), … , x=50^(2)`. f(x) is minimum at the middle term of the above series which are `x=25^(2) and x=26^(2)`. |
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| 33. |
Let `f(x)=(x^(2)+2)/([x]),1 le x le3`, where [.] is the greatest integer function. Then the least value of f(x) isA. 2B. 3C. `3//2`D. 1 |
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Answer» Correct Answer - B `f(x)={{:(x^(2)+2",",1lexlt2),((x^(2)+2)/(2)",",2lexlt3),((x^(2)+2)/(3),x=3):}` `therefore" Least value of f(x) in " [1,2]" is "3` `" Least value of f(x) in "[2,3]" is "3` `f(3)=(11)/(3)` `therefore" Least value of f(x) is 3"` |
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| 34. |
If the function `f(x)=(Ksinx+2cosx)/(sinx+cosx)`is strictly increasing for all values of `x ,`then`K1``K2`A. `klt1`B. `kle2`C. `kge3`D. none of these |
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Answer» Correct Answer - 4 Since `f(x) = (K sin x+ 2 cosx)/(sin x+ cos x)` is strictly increasing `f(x) gt 0 for all x` or `(k-2)/(sinx _++ cosx)^(2)gt 0 for all x` or `k-2gt 0 or kgt2` |
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| 35. |
P(x) be a polynomial of degree 3 satisfying P(-1) =10 , P(1) =-6 and p(x) has maxima at x = -1 and p(x) has minima at x=1 then The value of P(1) isA. -12B. -10C. 15D. 21 |
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Answer» Correct Answer - 1 P(x) is polynomial of degree 3 Given p(x) has minima at x=1 `therefore p(x) =a(x-1)` `therefore p(x)=a(x^(2)/(2)-x)+b` Given p(x) has maximum at x =-1 `therefore p(-1)=a(x^(2)/(2)-x)+b` Given p(x) has maxima at x =-1 `therefore b=-3/2a` `therefore p(x)=a(x^(3)/(6))-(x^(2)/(3)x)+C` Given p(-1)=10 `therefore a(-1/6-1/2+3/21)+C=10` `therefore 5a+6c=60` `therefore p(1)=a(1/6-1/2-3/2)+c=-6` solving (1) and (2) we get a=6 an c=5 `therefore p(x)=x^(3)-3x^(2)-9x+5` `therefore p(2)=8-12-18+5=-17` `p(x)=3x^(2)-6x-9` `therefore p(1)=3-6=-12` |
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| 36. |
P(x) be a polynomial of degree 3 satisfying P(-1) =10 , P(1) =-6 and p(x) has maxima at x = -1 and p(x) has minima at x=1 then The value of P(2) isA. -15B. -16C. -17D. -22 |
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Answer» Correct Answer - 3 P(x) is polynomial of degree 3 Given p(x) has minima at x=1 `therefore p(x) =a(x-1)` `therefore p(x)=a(x^(2)/(2)-x)+b` Given p(x) has maximum at x =-1 `therefore p(-1)=a(x^(2)/(2)-x)+b` Given p(x) has maxima at x =-1 `therefore b=-3/2a` `therefore p(x)=a(x^(3)/(6)-(x^(2)/(3)x)+C` Given p(-1)=10 `therefore a(-1/6-1/2+3/21)+C=10` `therefore 5a+6c=60` `therefore p(1)=a(1/6-1/2-3/2)+c=-6` solving (1) alnd (2) we get a=6 an c=5 `therefore p(x)=x^(3)-3x^(2)-9x+5` `therefore p(2)=8-12-18+5=-17` p(x)=`3x^(2)-6x-9` `therefore p(1)=3-6=-12` |
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| 37. |
If f(x) is strictly increasing function then h(x) is non monotonic function givenA. ` a in (0,3)`B. `a in (-2,2)`C. `(3,oo)`D. `a in (-oo,0)cup (3,oo)` |
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Answer» Correct Answer - 4 h(x)is non monotonic function if `3af(x)^(2)-2af(x)+1` changes sign for which `Dgt0 or 4a^(2)-12agt0` `a in(-oo,0)cup(3,oo)` |
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| 38. |
If `f(x)=x^3+b x^2+c x+d`and `0A. f(x) is strictly increasing functionB. f(x) has local maximaC. f(x) is a strictly decreasing functionD. f(x) is bounded |
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Answer» Correct Answer - 1 `f(x) =x^(3)+bx^(2)+cx+d,0ltb^(2)ltc` `f(x) =3x^(2)+2bx+c` Discriminant =`4b^(2)-12c=4(b^(2)-3c)ltc` `therefore f(X) gt 0 forall x in R` Thus f(X) is strictly increasing `forall x inR` |
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| 39. |
If `f(x)={{:(3-x^(2)",",xle2),(sqrt(a+14)-|x-48|",",xgt2):}` and if f(x) has a local maxima at x = 2, then greatest value of a isA. 2013B. 2012C. 2011D. 2010 |
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Answer» Correct Answer - C Local maximum at x = 2 `rArr" "underset(hrarr0)(lim)f(2+h)lef(2)` `rArr" "underset(hrarr0)(lim)(sqrt(a+14)-|2+h-48|)le3-2^(2)` `rArr" "sqrt(a+14)le45 rArr a le 2011` |
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| 40. |
Let f(x) =`x^(3)-3(7-a)X^(2)-3(9-a^(2))x+2` The values of parameter a if f(x) has a positive point of local maxima areA. `pi`B. `-oo,-3cup(3,(29)/(7))`C. `-oo,(58)/(14)`D. none of these |
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Answer» Correct Answer - 2 `f(x)=x^(3)-3(7-a)x^(2)-3(9-a^(2))x+2` f(X)=`3x^(2)-6(7-a)x-3(9-a^(2))` for real root `Dge0` or `49+a^(2)-14a+9-a^(2)ge0 or ale58/14` when point of minma is negative point of maxima is also negative Hence equation f(x) =`3xA^(2)-6(7-a)x-3(9-a^(2))` =0 has both roots negative sum or roots =`2(7-a)lta or a gt 7 ` which is not possible as form (1) `ale58/14` When point of maxima is positive point of minima is also positive ltrbgt Hence equation `f(x) =3x^(2)-6(7-a)x-3(9-a^(2))=0` has both roots positive sum roots =`2(7-a)gt0 or alt7` Also product of roots is positive or `-(9-a^(2))gt0 or a^(32)gt9 or a in (-oo,-3)cup(3,oo)` From (1),(2) and (3) ain `(-oo,-3)cup(3,58//14)` For points of extrema of opposite sign equation (1) has roots of opposite sign Thus a in (-3,3). |
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| 41. |
If `f(x)=(t+3x-x^2)/(x-4),`where `t`is a parameter that has minimum and maximum, then the range of valuesof `t`is`(0,4)`(b) `(0,oo)``(-oo,4)`(d) `(4,oo)`A. `(0,4)`B. `(0,oo)`C. `-(oo,4)`D. `(4,oo)` |
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Answer» Correct Answer - 3 `f(x)=(t+3x-x^(2))/(x-4),f(x)=(x-4)(3x-2x)-(t+3x-x^(2))/(x-4)^(2)` for maximum or minimum f(X)=0 `-2x^(2)+11x-12-t-3x+x^(2)=0` `-x^(2)+8x-(12-t)=0` For one maxima and minima `64-4(12+t)gt0` `16-12-tgt0, i.e 4gtt or tlt4` |
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| 42. |
Leg `f(x)=x^4-4x^3+6x^2-4x+1.`Then,`f`increase on `[1,oo]``f`decreases on `[1,oo]``f`has a minimum at `x=1``f`has neither maximum nor minimumA. f increases on `[1,oo]`B. f decreases on `[1,oo]`C. f has a minimum at x=1D. f has neither maximum nor minimum |
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Answer» Correct Answer - 1,3 f(x)=`4(x^(3)=3x^(2)+3x-1)=4(x-1)^(3)gt0 for x gt1` Hence f increases in `[1,oo)` Moreover f(x)`lt0 for x lt 1` Hence f has a minimum at x =1 |
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| 43. |
Let `f: R rarr R ` be a fucntion such that `f(x) = ax +3 sin x +4 cos x` Then f(x) is invertible ifA. `klt1`B. `kgt1`C. `klt2`D. `kgt2` |
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Answer» Correct Answer - 4 f(X) =a+3 cos x -4 sinx `=a+5 cos (x+alpha)` where `cos alpha =3/5` for invertible f(X) must be monotonic .Thus `f(x) ge0 forall x or f(X) leoforallx` i.e `a + 5 cos (x+alpha)ge 0 or a+5 cos(x+alpha)le0` i.e `age-5 cos(x+alpha) or a le5 cos (x+alpha)` i.e `agt5 or ale-5` |
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| 44. |
If `alpha`is an integer satisfying `|alpha|lt=4-|[x]|,`where `x`is a real number for which `2xtan^(-1)x`is greater than or equal to `ln(1+x^2),`then the number of maximum possible values of `a`(where [.] represents the greatest integer function) is_____ |
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Answer» Correct Answer - 9 (9) Let y =2x `tan^(-1)x-ln(1+x^(2))` `y=2 tan^(-1)x+(2x)/(1+x^(2))-(2x)/(1+x^(2))` `therefore y gt 0 forall lx in R^(+) ,y lt0 forall x in R^(-)` Therefore 4-|[x]| takes the values 0,1,2,3,4 `|alpha|le4-|[x]|` is satisfied by`alpha=0 pm1,pm 2,pm 3,pm 4` Therefore number of values of `alpha` is 9 |
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| 45. |
Let `f(x)`be an increasing function defined on `(0,oo)`. If `f(2a^2+a+1)>f(3a^2-4a+1),`then the possible integers in the range of `a`is/are`1`(b) 2 (c)3 (d) 4A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - 2,3,4 Since f is defined on `(0,oo) ,2a^(2)+a+1gt0` which is true as `Dlt0` Also `3a^(2)-4a+1gt0` `(3a-1)(a-1)gt0 i.e alt1//3 or agt1` As f is increasing `f(2a^(2)+a+1)gtf(3a^(2)-4a+1)` or `2a^(2)+a+1gt3a^(2)-4a+1` or `0gta^(2)-5a` or `0gta^(2)-5a` or `a(a-5)lt0 or a in (0,5)` from (1) and (2) we get `ain (0,1//3)cup(1,5)` Therefore possible integers are {2,3,4} |
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| 46. |
Let `f:R rarr(0,oo) and g:R rarr` R be twice differntiable function such that f' and g' ar continous fucntion on R. Suppose `f(x2)=g(2)=0,f'(2)ne0and g'(2)ne.If lim_(xrarr2) (f(X)g(x))/(f(x)g(x))=1` thenA. f has a local minimum at x=2B. f has a local maximum at x=2C. `f'(2)gtf(x)`D. `f(X)-f'(x)=0 for at least one x in R |
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Answer» Correct Answer - 1,4 `underset(xrarr2)lim (f(x)g(x))/(f(x)g(x))=1` `rarr underset(xrarr2)lim(f(x)g(x)+g(x)f(x))/(f(x)g(x)+f(x)g(x))` `(f(2)g(2)+g(2)f(2))/(f(2)g(2)+f(2)g(2))=1` `rarr f(2)=f(2)` Hence option 4 is correct As `f(2)=f(2) in (0,oo)` `rarr f(2)gt0` `rarr` f has local min at x=2 hence option 1 is correct |
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| 47. |
The value of `a`for which the function `f(x)=(4a-3)(x+log5)+2(a-7)cotx/2sin^2x/2`does not possess critical points is`(-oo,-4/3)`(b) `(-oo,-1)``[1,oo)`(d) `(2,oo)`A. `(-,oo,-4//3)`B. `(-oo,-1)`C. `[1,oo)`D. `(2,oo)` |
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Answer» Correct Answer - 1,4 we have `f(x)=(4a-3)(x+log5)+2(a-7)cot (x)/(2)sin^(2)(x)/(2)` `=(4a-3)(x+log5)+(a-7)sinx)` or `f(X)=(4a-3)+(a-7)cosx` If f(X) does ot have critical ponts then f(x) =0 does not have any solution in R Now `f(x)=0 cos x =(4a-3)/(7-a)` or `|(4a-3)/(7-a)|le1` or `-1le(4a-3)/(7-a)le1` or `a-7le4a-3le7-a` Thus f(x) =0m has solution in R if `-4//3leale2` `ain(-oo,-4//3)cup(2,oo)` |
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| 48. |
The function `f(x)=(ln(pi+x))/(ln(e+x))`isincreasing in `(0,oo)`decreasing in `(0,oo)`increasing in `(0,pi/e),`decreasing in `(pi/e ,oo)`decreasing in `(0,pi/e),`increasing in `(pi/e ,oo)`A. increasing in `(0,oo)`B. decreasing oin `(0,oo)`C. increasing in `(0,pi//e),decreasing in(pi//e,oo)`D. decreasing in `(0,pi//e)` increasing in `(pi//e,oo)` |
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Answer» Correct Answer - 2 `f(X)=(log(e+x)xx(1)/(pi+x)-log(pi+x)(1)/(e+x))/(log(e+x))^(2)` `=(log(e+x)xx(e+x)-(pi+x)log(pi+x))/(pi+x)(e+x)(log(e+x))/6(2)` Since log function is an increasing function and `eltpi` `log(e+x)ltlog(pi+x)` Thus `(e+x)log(e+x)lt(e+x)log(pi+x)lt(log(pi+x)` for all `x gt 0` Thus f(x)=0 ltrbgt Therefore f(X) decrease on `(0,oo)` |
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| 49. |
A rectangular sheet of fixed perimeter with sides having their lengthsin the ratio `8: 15`is converted into anopen rectangular boxby folding after removing squares of equal area from all four corners.If the total area of removed squares is 100, the resulting box has maximumvolume. Then the length of the sides of the rectangular sheet are24 (b) 32(c) 45 (d)60A. 24B. 32C. 45D. 60 |
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Answer» Correct Answer - 1,3 Let the sides of rectangle be 15k and 8k and side of square be x then (15k -2x) (8k-2x) x is volume `v=2(2x^(3))-23kx^(2)+60k^(2)x)` `(dv)/(dx)|_(x=5)=0 or 6x^(2)-46kx+60k^(2)|_(x=5)=0` or `6k^(2)-23k+15-=0` or `k=3, k=5/6` only k=3 is permissiable So the sides are 45 nd 24 |
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| 50. |
Let `f(x)={x+2,-1lelt0` `1,x=0 (x)/(2),0ltxle1`A. a point of minimaB. a point of maximaC. both points of minima and maximaD. neither a pointof minimanor that of maxima |
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Answer» Correct Answer - 4 `f(0)gtf(0^(+)) land f(0)ltf(0^(-))` Hence x=0 is neither a maximum nor a minimum |
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