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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
If 2a+3b+6c = 0, then show that the equation `a x^2 + bx + c = 0` has atleast one real root between 0 to 1. |
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Answer» `"Let "f(x) =(ax^(2))/(3)+(bx^(2))/(2)+cx` `f(0) =0 " " and" "f(1) =(a)/(3)+(b)/(2) +c =2a +3b +6c =0` `"if "" " f(0) =f(1) " then " f(x) =0 " for some value of " x in (0,1)` `rArr " "ax^(2) +bx +c =0` for at least one `x in (0,1)` |
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| 2. |
Let f: RR be a continuous function defined by `f(x)""=1/(e^x+2e^(-x))`.Statement-1: `f(c)""=1/3,`for some `c in R`.Statement-2: `0"" |
| Answer» Correct Answer - D | |
| 3. |
`(sinx)^(logx)` |
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Answer» Let `y = (sinx)^logx` Taking logs both sides, `logy = logx logsin x ` Differentiating both sides w.r.t. x `1/y dy/dx = logx(1/sinx)(cosx)+ logsinx (1/x)` `=>1/ydy/dx = logxcotx+logsinx/x` `=> dy/dx = y( logxcotx+logsinx/x)` `=> dy/dx = (sinx)^logx( logxcotx+logsinx/x)`, which is the required solution. |
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| 4. |
The total cost C(x) in Rupees, associated with the production of x units of an item is given by `C(x)=0. 005 x^3-0. 02 x^2+30 x+5000`. Find the marginal cost when 3 units are produced, where by marginal cost we mean the instantaneous rate of change of total cost at any level of output. |
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Answer» `C(x)=0.005x^3-0.02x^2+30x+5000` `(d(cx))/dx=(d(0.005x^3))/dx-(d(0.02))/dx+(d(30))/dx+0` `=0.005(3x^2)-0.02(2x)+30xx1` `=0.005(3xx9)-0.02(2xx3)+30` `=0.135-0.012+30` `=30.015` `(dc)/(dx)=30.015` |
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| 5. |
Find the equation of the normal to curve `x^2=4y`which passes through the point (1, 2). |
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Answer» `h^2=4k` slope of normal`=-1/(dy/dx)=-2/h` equation of normal`(y-k)=-2/h(x-h)` `k=2+2/h(1-h)` `h^2/4=2+2/h(1-h)` h=2 k=1 equation of line (y-1)=-1(x-2) x+y=3 |
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| 6. |
Find all the tangents to the curve `y=cos(x+y),-2pilt=xlt=2pi`that are parallel to the line `x+2y=0.` |
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Answer» Correct Answer - `x + 2y = (pi)/(2) and x + 2y = (-3pi)/(2)` Given, `y = cos (x + y)` `rArr ((d y )/(dx)) = - sin (x + y ) * ( 1+ (dy)/(dx))" " ` … (i) Since, tangent is parallel to `x + 2y =0`, then slope `(dy)/(dx) = - (1)/(2)` From Eq. (i), `- (1)/(2) = - sin (x + y ) *( 1- (1)/(2))` `rArr sin (x + y ) = 1, ` which shows `cos ( x+ y ) =0` ` therefore " " y =0` `rArr " " x + y = (pi)/(2) or - ( 3pi)/(2)` `therefore " " x = (pi)/(2) or - ( 3pi)/(2)` Thus, required points are ` ((pi)/(2), 0 ) and ( - ( 3pi) /( 2), 0 )` `therfore ` Equation of tangents are `" " (y -0)/( x - pi//2) = - (1) /(2)` and ` " " ( y - 0 ) /( x + 3pi//2) = -(1) /(2) rArr 2y = - x + (pi) /(2)` and ` " " 2y = - x - ( 3pi) /(2)` `rArr " " x + 2y = (pi)/(2)` and `" "x + 2h = - ( 3pi ) /(2)` are the required equations of tangents. |
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| 7. |
Manufacturer can sell x items at a price of rupees `(5-x/(100))`each. The cost price of x items is Rs `(x/5+500)`. Find the number of items he should sell to earn maximum profit |
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Answer» profit `= x(5 - x/100) - (x/5 + 500)` `= 5x -x^2/100-x/5 - 500` `(d(P))/dt = 5 - x/50-1/5 = 0` `24/5 - x/50 = 0` `x = (50+24)/5 = 240` `(d^2P)/(dx^2) = -1/50 <0 ` profit is maximum at x `= 240` answer |
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| 8. |
Find the equation of tangents to the curve `y=cos(x+y),-2pilt=xlt=2pi`that are parallel to the line `x + 2y = 0`. |
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Answer» Equation of the line parallel to the tangent is, `x+2y = 0 => y = -1/2x` `:.` Slope`(m) = -1/2` So, slope of the tangent will be `-1/2`. Now, `y = cos(x+y)` `=>dy/dx = -sin(x+y)(1+dy/dx)` As `m = -1/2, :. dy/dx = -1/2` `=> -1/2 = -sin(x+y)(1-1/2)` `=> sin(x+y) = 1` `=>x+y = pi/2 or x+y = (-3pi)/2` As, `y = cos(x+y)` `:. y = cos(pi/2) = 0 or y = cos((-3pi)/2) = 0` `:. x = pi/2 or x = (-3pi)/2` So, Equation of the line, when `x = pi/2` `(y-0) = -1/2(x-pi/2)` `=> y = -1/2x+pi/4` `=>4y+2x-pi = 0` Equation of the line. when `x =(-3 pi)/2` `(y-0) = -1/2(x+(3pi)/2)` `=> y = -1/2x-(3pi)/4` `=>4y+2x+3pi = 0` |
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| 9. |
If`M(x_o,y_o)` is the point on the curve `3x^2-4y^2=72` which is nearest to the line `3x+2y+1=0`, then the value of `(x_o + y_o)` is equal to (A) 3 (B) `-3` (C) 9 (D) `-9` |
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Answer» `3x^2-4y^2=72` `m=-3/2` `6x-8y(dy)/(dx)=0` `(dy)/(dx)=(3x)/(4y)` `(3x_0)/(4y_0)=-3/2` `x_o=-2y_o` `8y_o^2=72` `y=pm3` `3x+2y+1=0` `3x+2y+c=0` `c=-3x_0-2y_o` `|c-1| is minimum=4y_0` |c-1| is minimum at` c=12 andy_o=3` `x_0+y_0=3+3*(-2)=-3` option b is correct. |
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| 10. |
`2/(x-f(pi/12))+3/(x-f(pi/4))+4/(x-f((5pi)/12))=0`A. `(f((pi)/(12)),f((pi)/(4)))`B. `(0,f((pi)/(12)))`C. `(f((5pi)/(12)),oo)`D. `(f((pi)/(4)),f((5pi)/(12)))` |
| Answer» Correct Answer - A::D | |
| 11. |
The point on the curve `x^2=2y`which is nearest to the point (0, 5) is(A) `(2sqrt(2),4)` (B) `(2sqrt(2),0)` (C) (0, 0) (D) (2, 2) |
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Answer» If we see the options, option - `(B)` do not satisfy the equation of the curve. So, it is not the answer. Now, we will calculate the distance of rest of three options from `(0,5)`. For Option - (A), `d = sqrt((2sqrt2-0)^2+(4-5)^2) = 3` For Option - (C), `d = sqrt((0-0)^2+(0-5)^2) = 5` For Option - (D), `d = sqrt((2-0)^2+(2-5)^2) = sqrt13` As, `d` in case of option - (A) is the minimum, so `(2sqrt2,4)` is the nearest point. |
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| 12. |
Let `f(x)=x^2+(1/x^2)` and `g(x)=x-1/x` `xinR-{-1,0,1}`. If `h(x)=(f(x)/g(x))` then the local minimum value of `h(x)` is: (1) 3 (2) `-3` (3) `-2sqrt(2)` (4) `2sqrt(2)`A. 3B. `-3`C. `-2sqrt(2)`D. `2sqrt(2)` |
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Answer» Correct Answer - D We have , `f(x)=x^(2)+(1)/(x^(2))andg(x) =x-(1)/(x) rArr h(x)=(f(x))/(g(x))` `:. h(x)=(x^(2)+(1)/(x^(2)))/(x-(1)/(x))=((x-(1)/(x))^(2)+2)/(x-(1)/(x))` `rArr h(x)=(x-(1)/(x))+(2)/(x-(1)/(2))` `x-(1)/(x)gt0, (x-(1)/(x))+(2)/(x-(1)/(x))in [ 2 sqrt(2),oo]` `x-(1)/(x)lt0, (x-(1)/(x))+(2)/(x-(1)/(x))in [oo, 2sqrt(2)]` `:.` Local minimum value is `2sqrt(2)` |
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| 13. |
Let `f(x)=(x^2-1)^n(x^2+x+1)`then `f(x)`has local extremum at`x=1`when`n=2`(b) `n=3`(c)4 (d) `n=6`A. n=2B. n=3C. n=4D. n=6 |
| Answer» Correct Answer - A::C::D | |
| 14. |
Prove that the curve `y = x^2` and `xy = k` intersect orthogonally if `8k^2 = 1`. |
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Answer» Given, equation of curves are `=2x=y^(2)` ………….(i) and `2xy=k`……………….(ii) `rArr y=k/(2x)` [From eq.(ii)] From Eq.(i), `2x=(k/(2x))^(2)` `rArr 8x^(3)=k^(2)` `rArr x^(3)=1/8k^(2)` `rArr x=1/2k^(2//3)` `rArr y=k/(2x) = k/(2.1/2k^(2//3)) = k^(1//3)` Thus, we get point of intersection of curves which is `(1/2k^(2//3), k^(1//3))` From Eqs. (i) and (ii), `2=2y(dy)/(dx)` and `2[x.(dy)/(dx) + y.1]=0` `rArr (dy)/(dx) =1/y` and `(dy)/(dx) =(-2y)/(2x) =-y/3` `rArr (dy)/(dx) = (-2y)/(2x)=-y/x` `rArr (dy)/(dx)_(1/2k^(2//3),k^(1//3)) = 1/k^(1//3)` [say `m_(1)`] and `(dx)/(dy)_(1/2k^(2//3),k^(1//3)) = k^(1//3)/(1/2k^(2//3)) = -2k^(-1//3)` [say `m_(2)`] Since, the curves intersect orthogonally. i.e., `m_(1).m_(2)=-1` `rArr 1/k^(1//3).(-2k^(-1//3))=-1` `rArr -2k^(-2//3) = -1` `rArr 2/(k^(2//3)) =1` `k^(2//3)=2` `therefore k^(2)=8` Which is the required condition. |
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| 15. |
Let `f,g` and `h` be real-valued functions defined on the interval `[0,1]` by `f(x)=e^(x^2)+e^(-x^2)` , `g(x)=x e^(x^2)+e^(-x^2)` and `h(x)=x^2 e^(x^2)+e^(-x^2)`. if `a,b` and `c` denote respectively, the absolute maximum of `f,g` and `h` on `[0,1]` thenA. `a=b and c ne b`B. `a=c and a ne b`C. `a ne b and c ne b`D. `a=b=c` |
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Answer» Correct Answer - D Given function, `f(x)=x^(x^(2))+e^(-x^(2)),g(x)=xe^(x^(2))+e^(-x^(2)) and h(x)=x^(2)e^(x^(2))+e-x^(2)` are stricctly increasing on [0,1]. Hence, at x=1, the given function, attains absolute maximum all equal to e+1/e. `rArr a=b=c` |
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| 16. |
The function `f(x)=2|x|+|x+2|=||x|2|-2|x||`has a local minimum or a local maximum at `x=``-2`(b) `-2/3`(c) 2(d) `2/3`A. `-2`B. `(-2)/(3)`C. `2`D. `(2)/(3)` |
| Answer» Correct Answer - A::B | |
| 17. |
If the curves `ay+x^2=7` and `x^3=y` cut orthogonally at `(1, 1)` then `a=` (A) 1 (B) `-6` (C) 6 (D) `1/6`A. 1B. 0C. `-6`D. 6 |
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Answer» Correct Answer - D We have, `ay=x^(2)=7 "and" x^(3)=y` On differentiating w.r.t. x in both equations, we get `a.(dy)/(dx)+2x=0` and `3x^(2)=(dy)/(dx)` `rArr (dy)/(dx)_(1,1)=-2/a=m_(1)` and `(dy)/(dx)_(1,1)=3.1=3=m_(2)` Since, the curves cut orthogonally at (1,1). `therefore m_(1).m_(2)=-1` `rArr (-2/a).3=-1` `therefore a=6` |
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| 18. |
Let `f(x)=x^(m/n)` for `x in R` where m and n are integers , m even and n odd and `0A. f(x) dereases in `(-oo,0]`B. f(x) increases on `[0,oo)`C. f(x) increases on `(-oo,0]`D. f(x) decreases on `[0,oo)` |
| Answer» Correct Answer - A::B | |
| 19. |
The number of values of x where the function `f(x)=cos x +cos (sqrt(2)x)` attains its maximum value is |
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Answer» `f(x) = cos x+ cos ( sqrt2 x)` `x=0 => f(x)_(max) = 2` `T(cos x) = 2 pi` `T ( cos sqrt2 x) = ( 2 pi)/ sqrt2 = sqrt 2 pi` so `2x = 2n pi` `x= n pi` `sqrt2 x = 2m pi` `x = sqrt2 m pi` So option 2 is correct Answer |
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| 20. |
Match the conditions/ expressions in Column I with statements in Column II. Let the functions defined in Column I have domain ` (-pi//2, pi//2)`. |
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Answer» `(d)/(dx) ( x + sin x ) = 1 + cos x = 2 cos ^(2)""(x)/(2) gt 0 "for " - (pi)/(2) lt x lt (pi)/(2)`. Therefore , ` x + sin x ` is increasing in the given interval. Therefore, `(A) to (p)` is the answer. Again `(d)/(dx) (sec)= secx tanx ` which is ` gt 0 " for " 0 lt x lt pi//2 ` and `" " lt 0 " for " (-pi)/(2) lt x lt 0` Therefore, `sec x ` is neither increasing nor decreasing in the given interval. Therefore, `(B) to (r) ` is the answer. |
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| 21. |
The length of the longest interval, in which the function `3sin x-4 sin^3x` is increasing isA. `(pi)/(3)`B. `(pi)/(2)`C. `(3pi)/(2)`D. `pi` |
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Answer» Correct Answer - A Let `f(x) = 3 sin x - 4 sin ^(3) x = sin 3x ` The longest interval in which `sin x ` is increasing is of length `pi`. So, the length of largest interval in which `f(x) = sin 3x ` is increasing is `(pi)/(3)`. |
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| 22. |
Let a `in R` and let f: `R to R` be given by f(x) `=x^(5) -5x+a.` thenA. f(x) has three real roots if a `gt4`B. f(x) has only one real root is `a gt 4`C. f(x) has three real roots if `a lt -4`D. f(x) has three real roots if `-4 lt a lt4` |
| Answer» Correct Answer - B::D | |
| 23. |
`"Let "f(x) ={ underset( x^(2) +8 " "." "x ge0)(-x^(2)" "." "x lt0).` Equation of tangent line touching both branches of `y=f(x)` isA. `y= 4x+1`B. `y=4x+4`C. `y=x+4`D. `y=x+1` |
| Answer» Correct Answer - B | |
| 24. |
A wire of length 2 units is cut into two parts which are bentrespectively to form a square of `s i d e=x`units and a circle of `r a d i u s=r`units. If the sum of the areas of the square andthe circle so formed is minimum, then :(1) `2x=(pi+4)r`(2) `(pi+4)x=pir`(3) `x=2r`(4) `2x=r`A. `(4-pi) x=pir`B. `x=2r`C. `2x=r`D. `2x=(pi+4)r` |
| Answer» Correct Answer - B | |
| 25. |
If `f(x)=(x^2-1)/(x^2+1)`. For every real number `x ,`then the minimum value of `fdot`does not exist because `f`is unboundedis not attained even through `f`is boundedis equal to 1is equal to `-1`A. does not exist because `f ` is unboundedB. is not attained even through `f` is boundedC. is 1D. is `-1` |
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Answer» Correct Answer - D Given, ` f(x) = (x ^(2) - 1)/( x ^(2) + 1) = 1 - ( 2)/( x ^(2) + 1)` `f(x)` will be minimum, when ` (2)/(x ^(2) + 1)` is maximum, i.e., when `x ^(2) + 1 ` is minimum. i.e., at ` x = 0` `therefore ` Minimum value of ` f (x) ` is `f (0) = -1 ` |
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| 26. |
Let `f(x)=5-|x-2| and g(x)=|x+1|, x in R`. If f(x)n artains maximum value at `alpha` ang g(x) attains minimum value of `beta`, then `lim_(xto-alpha beta) ((x-1)(x^(2)-5x+6))/(x^(2)-6x+8)` is equal toA. `1//2`B. `-3//2`C. `-1//2`D. `3//2` |
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Answer» Correct Answer - A Given function are f(x)=-5-|x-2| and g(x) |x+1|, where `x in R`. Clearly, maximum of f(x) occurred at `x=2, so alpha, 2`. And minimum of g(x) occurred at `x=- 1, so beta=-1` `rArr alpha beta=-2` Now, `underset(x to- alpha beta) lim ((x-1)(x^(2)-5x+6))/(x^(2)-6x+8)` `underset(x to-2) lim ((x-1)(x-3)(x-2))/((x-4)(x-2))" " [ :. alpha beta =-2]` `underset(x to-2) lim ((x-1)(x-3))/((x-4))=((2-1)(2-3))/((2-4))=(1xx(-1))/((-2))=(1)/(2)` |
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| 27. |
which of the following statement `is // are ` true ? (i) f(x) =sin x is increasing in interval `[(-pi)/(2),(pi)/(2)]` (ii) f(x) = sin x is increasing at all point of the interval `[(-pi)/(2),(pi)/(2)]` (3) f(x) = sin x is increasing in interval `((-pi)/(2),(pi)/(2)) UU ((3pi)/(2),(5pi)/(2))` (4) f(x)=sin x is increasing at all point of the interval `((-pi)/(2),(pi)/(2)) UU ((3pi)/(2),(5pi)/(2))` (5) f(x) = sin x is increasing in intervals `[(-pi)/(2),(pi)/(2)]& [(3pi)/(2),(5pi)/(2)]`A. all are correctB. all are falseC. (3) and (4) are correctD. (1),(4) & (5) are correct |
| Answer» Correct Answer - D | |
| 28. |
Let `g(x)=(f(x))^3-3(f(x))^2+4f(x)+5x+3sinx+4cosxAAx in Rdot`Then prove that `g`is increasing whenever is increasing.A. `phi` is increasing wherever f is increasingB. `phi` is increasing whenever f is decreasingC. `phi` is decreasing whenever f is decreasingD. `phi` is decreasing if `f(x)=-11` |
| Answer» Correct Answer - A::D | |
| 29. |
Let f(x) `={underset( 3 sin x " " x le 0)(X^(3) +X^(2) -10X " " X lt0).` Examine the behaviour of f(x) at =0 |
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Answer» f(x) is continuous at x=0 `f(x) ={underset( 3 cos x " "x gt 0)(3x^(2) +2x -10 x lt 0).` f(0) =3 and f(0) =- 10 thus f(x) is non- differentiable at `x=0" "rArr " "x =0 ` is a critical point. Also derivative change sigh from negative to positive , so x=0 is a point of local minima. |
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| 30. |
Find the critical points of the function f(x) `=4x^(3)-6x^(2) -24x+9 " if f(i) x in [0,3] (ii) x in [-3,3] (iii) x in [-1,2]` |
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Answer» ` f(x) =12(x^(2) -x-2)` `=12 (x-2) (x+1)` `f(x) =0 " "rArr " "x=- 1 " or " 2` `(i) if x in [0,3] x=2 " is critical point "` `(ii) " if " x in [-3,3] " then we have two critical points " x =- 1,2` `(iii) " if "Xin [- 1,2] " then no critical point as both " x=-1 " and " x=2 " become boundary points "` Note : Critical points are always interior points of an interval. |
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| 31. |
If x = -1 and x = 2 are extreme points of f(x) = `alpha log|x| + beta x^2 + x`, thenA. `alpha =2 , beta =-(1)/(2)`B. `alpha=2,beta=(1)/(2)`C. `alpha=-6 , beta =(1)/(2)`D. `alpha=-6, beta =-(1)/(2)` |
| Answer» Correct Answer - A | |
| 32. |
At what points on the curve `x^2+y^2-2x-4y+1=0`, the tangents are parallel to the `y-a xi s ?` |
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Answer» Given, equation of curve which is `x^(2) + y^(2)-2x-4y+1=0` `rArr 2x+2y(dy)/(dx)-2-4(dy)/(dx)=0` `rArr (dy)/(dx)(2y-4)=2-2x` `rArr (dy)/(dx) = (2(1-x))/(2(y-2))` Since, the tangents are parall to the Y-axis i.e., `tantheta = tan90^(@)=(dy)/(dx)` `therefore (1-x)/(y-2)=1/0` `rArr y-2=0` `rArr y=2` For y=2 from Eq.(i), we get `x^(2)+2^(2)-2x-4 xx 2+1=0` `rArr x^(2)-2x-3=0` `rArr x(x-3)+1(x-3)=0` `rArr (x+1)(x-3)=0` `therefore x=-1, x=3` So, the required points are `(-1,2)` and (3,2). |
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| 33. |
If the line `a x+b y+c=0`is a normal to the curve `x y=1,`then`a >0,b >0``a >0,b |
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Answer» Correct Answer - B::C Given, `" " xy = 1 rArr y = (1)/(x)` `rArr " " (dy)/(dx) = - (1)/(x^(2))` Thus, slope of normal = `x^(2)` ( which is always positive) and it is given `ax + by + c =0` is normal, whose slope `= -(a)/(b)`. `rArr " " - (a)/(b) gt 0 or (a)/(b) lt 0` Hence, a and b are of opposite sign. |
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| 34. |
statement:`e^pi` is bigger than `pi^e`statement 2:`f(x)=x^(1/x)` is an increasing function whenn `xe[e,oo)`A. Statement -1 is True Statement-2 is True : Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statement -2 is True : Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is True Statement-2 is TrueD. Statement-1 is False, Statement -2 is True |
| Answer» Correct Answer - C | |
| 35. |
`"Let "f(x) ={underset(-2x + log_(2) (b^(2)-2), x gt1)(x^(3) -x^(2) +10 x-5 , x le 1).` The set of values of b for which f(x) has greatest value at x= 1 is given by .A. ` 1 le b le 2`B. `b={1,2}`C. `b in (-oo,-1)`D. `[-sqrt(130), -sqrt(2)) uu(sqrt(2), sqrt(130)]` |
| Answer» Correct Answer - D | |
| 36. |
If `f(x)=alog|x|+b x^2+x`has extreme values at `x=-1`and at `x=2`, then find `a`and `b`.A. `a=2`B. `b=-1//2`C. `a=-2`D. `b=1//2` |
| Answer» Correct Answer - A::B | |
| 37. |
If the normals any point to the parabola `x^(2)=4y` cuts the line y = 2 in points whose abscissar are in A.P., them the slopes of the tangents at the 3 conormal points are in |
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Answer» `y^2=4ax----------` `y=mx-2am-am^3` `x^2=4ay` `x=my-2am-am^3` `a=1` `x=my-2m-m^3` `P(h,k)` `h=mk-2m-m^3` `m^3+2m-mk+h=0` `m^3+(2-k)m+h=0` Cubic equation`m_1,m_2,m_3` `m_1+m_2+m_3=0` `m_1^3+m_2^3+m_3^3=3m_1m_2m_3` `m_1,m_2,m_3->slope of normal` `-1/m_1,-1/m_2,-1/m_3`->slope of tangent at foot of normal `x=my-2m-m^3//y=2` `x=2m-2m-m^3` `x_1=-m_1^3,x_2=-m_2^3,x_3=-m_3^3` `x_1+x_3=2x_2` `-m_1^3-m_3^3=-2m_2^3` `m_1^3+m_3^3=2m_2^3` `2m_2^3+m_2^3=3m_1m_2m_3` `m_2^2=m_1m_3` `1/m_2^2=1/m_1xx1/m_3` `(-1/m_2)^2=-1/m_1xx1/m_3` `b^2=ac` `-1/m_1,-1/m_2,-1/m_3`are in GP option `(b)` |
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| 38. |
The greatest, the least values of the function ,`f(x) =2-sqrt(1+2x+x^(2)), x in [-2,1]` are respectivelyA. `2,1`B. `2, -1`C. `2,0`D. ` -2,3` |
| Answer» Correct Answer - C | |
| 39. |
On the ellipse `4x^2+9y^2=1,`the points at which the tangents are parallel to the line `8x=9y`are`(2/5,1/5)`(b) `(-2/5,1/5)``(-2/5,-1/5)`(d) `(2/5,-1/5)`A. `((2)/(5), (1)/(5))`B. `(-(2)/(5), (1)/(5))`C. `(-(2)/(5), - (1)/(5))`D. `((2)/(5), - (1)/(5))` |
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Answer» Correct Answer - B::D Given, ` 4x ^(2 ) + 9y^(2) = 1 " " `… (i) On differentiating w.r.t. x, we get `" " 8x + 18y (dy)/(dx) =0` `rArr " " (dy)/(dx) =- ( 8x)/( 18y ) =- ( 4x )/( 9y)` The tangent at point `(h, k)` will be parallel to ` 8x = 9y`, then `" " - ( 4h)/( 9k) = (8)/(9)` `rArr " " h = - 2k ` Point `(h, k)` also lies on the ellipse. `therefore " " 4h^(2) + 9k^(2) = 1" " `... (ii) On putting value of h in Eq. (ii), we get `" " 4 (-2 k)^(2) + 9k ^(2) = 1 ` `rArr " " 16 k ^(2) = 9 k^(2) = 1 ` `rArr " " 25 k^(2) = 1 ` `rArr " " k ^(2) = (1)/( 25)` `rArr " " k = pm (1)/(5)` Thus, the point, where the tangents are parallel to `8x = 9y ` are `(-(2)/(5), (1)/(5)) and ((2)/(5), - (1)/(5))`. Therefore, options (b) and (d) are the answers. |
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| 40. |
if f(x) = `tan^(1) x -(1//2) en x.` thenA. the greatest value of f(x) on `[ 1//sqrt(3) , sqrt(3)]` is `pi//6 +(1//4) en 3`B. the least value of f(x) on `[ 1//sqrt(3),sqrt(3) ] " is " pi//3 -(1//4) en 3`C. f(x) decreases on `(0,oo)`D. f(x) increases on `(-oo,0)` |
| Answer» Correct Answer - A::B::C | |
| 41. |
Let f(x) `=(40)/(3x^(4) +8x^(3) -18x^(2)+60).` Which of the following statement (s) about f(x) is (are ) correct ?A. f(x) has local minima at x=0B. f(x) has local maxima at x=0C. absolute maximum value of f(x) is not defined .D. f(x) is local maxima at x=- 3, x =1 |
| Answer» Correct Answer - A::C::D | |
| 42. |
Let `f(x)=(1+b^2)x^2+2b x+1`and let `m(b)`be the minimum value of `f(x)dot`As `b`varies, the range of `m(b)`is`[0,}`b. `(0,1/2)`c. `1/2,1`d. `(0,1]`A. `[0,1]`B. `(0,(1)/(2)]`C. `[(1)/(2),1]`D. `(0,1]` |
| Answer» Correct Answer - D | |
| 43. |
Find two positivenumbers `x`and `y`such that `x+y=60`and `x y^3`is maximum. |
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Answer» `x+y =60` `rArr " "x=60 -y" "rArr" "xy^(3) =(60 -y)y^(3)` `"Let "" "f(y) =(60-y) y^(3) " "," "y in (0,60)` Let maximizing f(y) let us find critical points `f(y) =3y^(2) (60-y) -y^(3) =0` `f(y) =y^(2) (180 -4y) =0` `rArr " "y =45` `f(45) lt 0 " and " f(45) gt 0.` Hence local maxima at y =45 So `x=15 " and " y=45` |
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| 44. |
A rectangle is inscribed in a semi-circle of radius `r` with one of its sides on diameter of semi-circle. Find the dimensions of rectangle so that the area is maximum. Find the area also. |
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Answer» `A(theta)=widthxxheight` `=(2rcostheta)xxrsintheta` `=2r^2sinthetacostheta` `=r2xx(2sinthetacostheta)` `=sin2theta` `A(theta)=x^2xxsin(2theta)` `sin(2theta)=1` `2theta=pi/4` `theta=pi/4` Maximum area = `r^2xx1=r^2` `width=2xcostheta=2rcospi/4` `=2rxx1/sqrt2` `=sqrt2r` `height =rsintheta=rxxsin(pi/4)` `=r/sqrt2` |
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| 45. |
Prove that the radius of the right circular cylinder ofgreatest curved surface area which can be inscribed in a given cone is halfof that of the cone.A. one third that of the coneB. `1//sqrt(2)` times that of the coneC. `2//3` that of the coneD. `1//2` that of the cone |
| Answer» Correct Answer - D | |
| 46. |
Show that the triangle of maximum area that can be inscribed in a given circle is an equilateral triangle. |
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Answer» OB=9 BD=`sqrt(a^2-x^2)` CD=`sqrt(a^2-x^2)` AD=a+x<*(a+x)br>Area=`1/2*(a+x)*2sqrt(a^2-x^2` A=`(a+x)sqrt(a^2-x^2` `(dA)/(dx)=sqrt(a^2-x^2)+(a+x)/(2sqrt(a^2-x^2)`-2x=0. `=2a^2-2x^2-2ax-2x^2=0` `a^2-ax-2x^2=0` `(a-2x)(a+x)=0` `a-2x=0 or a+x=0` `x=a/2,-x` BD=`sqrt(a^2-x^2)=sqrt3/2a` BC=`sqrt3a` `AB^2=3/4a^2+9/4a^2=3a^2` `AB=sqrt3a`. BC and AB are equal. |
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| 47. |
The largest area of a rectangle which has one side on the x-axis and the two vertices on the curve `y=e^(-x^2)` isA. `sqrt(2) e^(-1//2)`B. `2 e^(-1//2)`C. `e^(-1//2)`D. none of these |
| Answer» Correct Answer - A | |
| 48. |
In a regular triangular prism the distance from the centre of one base to one of the vertices of the other base is e. The altitude of the prism for which the volume is greatest, is :A. `(e) /(2)`B. `(e) /(sqrt(3))`C. `(e)/(3)`D. `(e)/(4)` |
| Answer» Correct Answer - B | |
| 49. |
Consider the following statements : `S_(1):` The function `y=(2x^(2)-1)/(x^(4))` is neither increasing nor decreasing `(S_(2):` if f(x) is strictly increasing real function defined on R and C is a real constant then number of Solutions of f(x) =c is always equal to one. `S_(3):` Let f(x) =x : `x in (0,1)` , f(x) does not has any point of local maxima`//`minima `S_(4):` f(x)={x} has maximum at x=6 (here{} denotes fractional part function) State , in order ff whether `S_(1),S_(2),S_(3),S_(4)` are true or falseA. TTFTB. FTFTC. TFTFD. TFFT |
| Answer» Correct Answer - C | |
| 50. |
the function `(|x-1|)/x^2` is monotonically decreasing at the pointA. `x=3`B. `x=1`C. `x=2`D. none of these |
| Answer» Correct Answer - A | |