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51.

If the function `f(x)=2x^3-9a x^2+12 x^2x+1,w h e r ea >0,`attains its maximum and minimum at `pa n dq`, respectively, such that `p^2=q ,`then `a`equal to1 (b) 2(c) `1/2`(d) 3A. 1B. 2C. `1/2`D. 3

Answer» Correct Answer - 2
`f(x)=2x^(3)-9ax^(2)+12a^(2)x+1`
`therefore f(X)=6x^(2)-18ax+12a^(2)and f(x)=12x-18a`
For maximum /minimum , `6x^(2)-18ax+12a^(2)=0`
or `x^(2)-3ax+2a^(2)=0`
or `(x-a)(x-2a)=0`
i.e x=a or x =2a
Now `f'(a)=12a=18a=6alt0`
Therefore f(X) is maximum at x=a and minimum at x=2a
Thus p=a and q =2a
Given that `p^(2) = q or a^(2)=2a or a (a-2) =0 or a=2`
Discussion
52.

If `H(x_(0))`=0 for some x=`x_(0)`and `(d)/(dx)H(x)gt2cxH(x)` for all `xgex_(0)`where `cgt0` thenA. H(x) = 0 has root for `x gt x_(0)`B. H(x) = 0 has no root for `x gt x_(0)`C. H(x) is a constant functioD. none of these

Answer» Correct Answer - 2
Given that `(d)/(dx)H(x)gt2cxH(x)`
or `e^(-cx^(2))(d)/(dx)H(x)-e^(-cx^(2))2cxH(x)gt0`
or `(d)/(dx)H(x)e^(-cx^(2))gt0`
Thus `H(x)e^(-cx^(2))` is and increasing function
But `h(x_(0))gt0` for all `xgtx_(0)`
Hence H(x) connot be zero for any `xgtx_(0)`
Discussion
53.

consider `f(X) =(1)/(1+|x|)+(1)/(1+|x-1|)` Let `x_(1)` and `x_(2)` be point wher f(x) attains local minmum and global maximum respectively .If `k=f(x_(1))+f(x_(2))` then 6k-9=________.

Answer» Correct Answer - 8
`f(X)=(1)/(1+|x|)+(1)/(1+|x-1|)`
clearly f(X) is non differeentiable at x =0,1
Also `f(1//2+x)=f(1//2-x)`
Hence graph of f(x) is symmetrical about line x=`1//2` where derivation is zero
Local minimum =`f(1/2)=4/3`
Global maximum `=f(x)=f(1)=3/2`
`terefore k=4/3+3/2=17/6`
Discussion
54.

Let f(x) =`xsqrt(4ax-x^(2)),(agt0)` Then f(x) isA. increasing in (0,3a) decreasing in (3a,4a)B. increasing in (a,4a),decreasing in `(5a,oo)`C. increasing in `(0,4a)`D. none of these

Answer» Correct Answer - 1
f(X)=`xsqrt(4ax-x^(2))`
`therefore f(X) =sqrt(4ax-x^(2))+x(4a-2x)/2sqrt(4ax-x^(2))`
`=(2x(3a-x))/sqrt(4ax-x^(2))`
Now if `f(X)gt0` then
`2x(3a-x)gt0`
`2x(x-3a)lt0`
or x in (0,3a)
Thus f(x) increases in (0,3a) and decreases in (3a,4a)
Thus f(X) increases in (0,3a) and decreases (3a,4a)
Discussion
55.

If `ab=2a+3b, agt0, b gt0`, then the minimum value of ab isA. 12B. 24C. `(1)/(4)`D. none of these

Answer» Correct Answer - B
`ab=2a+3b rArr b=(2a)/(a-3)`
Now `z=ab=(2a^(2))/(a-3)`
`rArr" "(dz)/(da)=(2[(a-3)2a-a^(2)])/((a-3)^(2))=(2[a^(2)-6a])/((a-3)^(2))`
Put `(dz)/(da)=0, therefore a^(2)-6a=0, a=0,6`
Clearly a = 6 is point of minima
when `a=6, b=4 rArr (ab)_("min")=6xx4=24`
Discussion
56.

Let `f(x)`be defined as`f(x)={tan^(-1)alpha-5x^2,0

Answer» Correct Answer - `alpha lt -tan 1`
`f(x)={{:(tan^(-1) alpha-5x^(2),0ltxlt),(-6x,xge1):}`
f(1)=-6
for maximum at x=1
`underset(xrarr1)lim f(x)=tan^(-1)alpha -5lt-6`
or `tan^(-1)alphalt-1 or alpha-tan 1`
Discussion
57.

Find the values of x where function `f(X)m = (sin x + cosx)(e^(x))` in `(0,2pi)` has point of inflection

Answer» Correct Answer - `x=pi//4,5pi//4`
We have f(x) =`(sinx+cosx)^(e^(x)`
`therefore f(x) =(sinx+cosx)e^(x)+e^(x)(cosx-sinx)`
`rarr f(x) =2e^(x)cosx`
`f'(x) =2(e^(x) cosx-=e^(x)sinx)`
`=2e^(x)(cosx-sinx)`
If f'(x) =0 then cos x - sinx =0
`therefore tanx=1 rarr x =(pi)/(4),(5pi)/(4)`
Discussion
58.

If `f(x)=x^3-x^2+100x+2002 ,t h e n``f(1000)>f(1001)``f(1/(2000))>f(1/(2001))``f(x-1)>f(x-2)``f(2x-3)>f(2x)`A. `f(1000)ltf(1001)`B. `f((1)/(2000))gtf((1)/(2001))`C. `f(x-1)gtf(x-2)`D. `f(2x-3)gtf(2x)`

Answer» Correct Answer - 2,3
`f(X)=x^(3)-x^(2)+100x+2002`
`f(x)=3x^(2)-2x+100gt0 forall x in R` thus
Therefore fX() is increasing (strictly) Thus
`f((1)/(2000))gtf((1)/(2001))`
Also `f(X-1)gtf(x-2)asx-1gtx-2forallx`
Discussion
59.

The fraction exceeds its `p^(th) ` power by the greatest number possible, where `p geq2` isA. `((1)/(p))^(1//(p-1))`B. `((1)/(p))^(p-1)`C. `p^(1//p-1)`D. none of these

Answer» Correct Answer - A
Let `y=x-x^(p)`, where x is the fraction
`rArr" "(dy)/(dx)=1-px^(p-1)`
For maximum of minimum, `(dy)/(dx)=0`
`rArr" "1-px^(p-1)=0 rArrx=((1)/(p))^(1//(p-1))`
Now, `(d^(2)y)/(dx^(2))=-p(p-1)x^(p-2)`
`therefore(d^(2)y)/(dx^(2)):|_(x=((1)/(p))^(1//(p-1)))=-p((1)/(p))^((p-2)//(p-1))lt0`
`therefore "y is maximum at x"=((1)/(p))^(1//(p-1))`
Discussion
60.

If `f^(prime)(x)=|x|-{x},`where {x} denotes the fractional part of `x ,`then `f(x)`is decreasing in`(-1/2,0)`(b) `(-1/2,2)``(-1/2,2)`(d) `(1/2,oo)`A. `((-1)/(2),0)`B. `((-1)/(2),2)`C. `((-1)/(2),2)`D. `((1)/(2),oo)`

Answer» Correct Answer - 1
`f(X) =|x|-{x}-|x|-(x-[x])=|x|-x+[x]`
For x `in (-1//2,0)`
`f(X) =-x-x-1=-2x-1`
Also for `-1/2ltxlt0 or 0lt-2xlt1 or -1lt-2xlt0`
or `f(x) lt0,f(X)` decreases in `(-1//2,0)`
Similary we can chech for other given option say for `x in (-1//2,2)`
`f(x)={{:((ix-x-x),(-1)/(2)ltxlt0),(x-x+0,0lex1),(x-x+1,1lexlt2):}`
Here f(X) decreases only in `(-1//2,0)` otherwise f(X) in other intervals is constant
Discussion
61.

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 - 2
`f(x)=(1)/(e^(x)+2e^(-x))=(e^(x))/(e^(2x)+2)`
`therefore f(x)=((e^(2x)+2)e^(x)-2e^(2x)e^(x))/(e^(2x)+2)^(2)`
So the maximum of f(x) is
`therefore 0lt1/3lt(1)/(2(sqrt(2))`
for some c in R we get
`f(c ) =1/3`
Discussion
62.

If `lim_(xrarra) f(x)=lim_(xrarra) [f(x)]` ([.] denotes the greates integer function) and f(x) is non-constant continuous function, thenA. `underset(xrarra)(lim)` f(x) is an integerB. `underset(xrarra)(lim)` f(x) is non-integerC. f(x) has local maximum at x = aD. f(x) has local minimum at x = a

Answer» Correct Answer - A::D
We have `underset(xrarra)(lim)f(x)=underset(xrarra)(lim)[f(x)]`.
The can occur only when `underset(xrarra)(lim)f(x)` is an integer.
`rArr" "f(a^(+))gt f(a) and f(a^(-))gt f(a)`
`rArr" "x = a` must be point of local minima.
Discussion
63.

Let `f(x)=cospix+10x+3x^2+x^3,-2lt=xlt=3.`The absolute minimum value of `f(x)`is0 (b) `-15`(c) `3-2pi`none of these

Answer» Correct Answer - 2
f(x) =`-pisin pi x +10+ 6x+3x^(2)`
`=3(x+1)^(2)+7 pi sin pi x gt 0 for all x `
Thus f(X) is increasing in `-2lexle3`
So absolute minimum =f -(2)=1 -20+12-8
Discussion
64.

If `m`is the minimum value of `f(x , y)=x^2-4x+y^2+6y`when `xa n dy`are subjected to the restrictions `0lt=xlt=1a n d0lt=ylt=1,`then the value of `|m|`is________

Answer» Correct Answer - 3
we have `f(x,y)=x^(2)+y^(2)-4x+6y`
Let `(x,y)=cos , theta sin theta)`.Then `theta in [0,pi//2]`and
`f(x,y)=f(theta)=cos^(2)theta+sin^(2)theta-4cos theta +6 sin`
`f(theta)= cos theta +4 sin theta gtforall0foralltheta in [0,pi//2]`
Therefore f`(theta)` is strictly increasing in `[0,pi//2]` thus
`f(theta)+_(min)=f(0)=1-4+0=-3`
Discussion