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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.
| 101. |
Differentiate (log x) with repect to tan x. |
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Answer» Correct Answer - ` cot x ` |
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| 102. |
differentiate: log(logx) |
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Answer» Let `y="log"("log"x)` `implies(dy)/(dx)=(d)/(dx)"log"("log"x)` `=(1)/("log"x)(d)/(dx)("log"x)` `=(1)/(x"log"x)=(x log x)^(-1)` `implies(d^(2)y)/(dx^(2))=(d)/(dx)(x log x)^(-1)` `= -1* (x logx)^(-2)(d)/(dx)(x logx)` `=([x*(d)/(dx)logx+"log"x(d)/(dx)x])/((xlogx)^(2))` `=([x*(1)/(x)+"log"x])/((xlogx)^(2))` `=((1+"log"x))/((xlogx)^(2))` |
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| 103. |
Differentiate `(log x)^(cos x)` with respect to x. |
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Answer» `"Let " y = ("log "x)^("cos "x)` `rArr "log " y = "log "{("log "x)^("cos "x)}` `= "cos "x * "log "("log "x)` Diffenentiate both sides w.r.t.x `(1)/(y) (dy)/(dx) = "cos "x * (d)/(dx)"log" ("log " x) + "log"("log " x)(d)/(dx)"cos"x` `rArr (dy)/(dx) = y {("cos"x)/("log"x)*(d)/(dx)("log"x)+"log"("log" x)(-"sin"x)}` `rArr (dy)/(dx) =("log" x)^("cos"x){("cos"x)/(x*"log"x)-"sin"x * "log" x ("log"x)}` |
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| 104. |
Differentiate log x with respect to sin x |
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Answer» Correct Answer - `1/ x cot x ` |
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| 105. |
Differentiate `log(secx+tanx)`with respect to `x`: |
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Answer» Let y=log (sec x + tan x) `rArr (dy)/(dx)log (sec x +tan x) ` `1/(sec x+tan x ).(d)/(dx)(sec x tan x )` `1/(sec x+tan x ).(d)/(dx)(sec x tan x +sec ^2 x )` `=(sec x(tan x+ sec x))/(secx + tan x )=sec x.` |
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| 106. |
Differentiate the following w.r.t. x:`log(cose^x)` |
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Answer» `"Let"y = "log" ("cos" e^(x))` `(dy)/(dx) = (d)/(dx) " log" ("cos" e^(x))` `= (1)/("cos"e^(x)) (d)/(dx) "cos"e^(x)` `= (1)/("cos"e^(x))* (-"sin"e^(x))*(d)/(dx)e^(x)` `= - e^(x)*"tan"e^(x)` |
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| 107. |
If `y=A e^(m x)+B e^(n x)`, show that `(d^2y)/(dx^2)-(m+n)(dy)/(dx)+m n y=0`. |
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Answer» `y=Ae^(mx)+Be^(nx)` `implies(dy)/(dx)=Ame^(mx)+"Bne"^(nx)` `implies(d^(2)y)/(dx^(2))=Am^(2)e^(mx)+"Bn"^(2)e^(nx)` `L.H.S.=(d^(2)y)/(dx^(2))-(m+n)(dy)/(dx)+mny` `=[Am^(2)e^(mx)+Bn^(2)e^(nx)]-(m+n)[Ame^(mx)+"Bn"e^(nx)]+mn[Ae^(mx)+Be^(nx)]` `=Am^(2)e^(mx)+Bn^(2)e^(nx)-"Am"^(2)e^(mx)-"Bmn"e^(nx)-"Amn"e^(mx)-"Bn"^(2)e^(nx)+"Amn"e^(mx)+"Bmn"e^(nx)` `=0=R.H.S. " " `Hence Proved. |
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| 108. |
x=a cos t, y=b sin t |
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Answer» Correct Answer - `-b/a cot t` |
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| 109. |
log cos x |
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Answer» Correct Answer - `-tan x` |
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| 110. |
`sin^@` |
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Answer» Correct Answer - `pi/(180)cos( (pi x)/(180))` |
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| 111. |
`sqrt(sin x)` |
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Answer» Correct Answer - `(cos x)/(2 sqrt(x))` |
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| 112. |
If `y = 3 cos (log x) + 4 sin (log x)`, show that `x^2y_2+xy_1+y=0` |
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Answer» `y=3" cos"(log x)+4" sin"(log x) " "`...(1) `impliesy_(1)=(d)/(dx)[3" cos"(log x) +4 sin (log x)]` `= -(3" sin"(log x))/(x)+(4" cos"(log x))/(x) ` `impliesxy_(1)= -3 "sin"(log x)+4cos(log x)` Again differentiate both sides w.r.t.x `x*y_(2)+y_(1)*1=(3cos(logx))/(x)-(4sin(log x))/(x)` `implies x^(2)y_(2)+xy_(1)= -[3cos(logx)+4sin(logx)]= -y` ` " " `from equation (1) `implies x^(2)y_(2)+xy_(1)+y=0 " " ` Hence proved. |
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| 113. |
`log(x^2+3)` |
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Answer» Correct Answer - `(2x)/(x^2+3)` |
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| 114. |
If `y=cos^(-1)x`, find `(d^2y)/(dx^2)`in terms of `y`alone. |
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Answer» `y="cos"^(-1)x impliesx=cos y` `implies(dy)/(dx)= -sin y` `implies(dy)/(dx)= -(1)/("sin"y)= -"cosec " y` `implies (d^(2)y)/(dx^(2))=(d)/(dx)(-"cosec " y)` `="cosec "y" cot "y*(dy)/(dx)` `="cosec "y" cot"y(-"cosec "y)` `= -"cosec"^(2)y" cot "y` |
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| 115. |
`e^(x^2+1)` |
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Answer» Correct Answer - `2x.e^(x^2+1)` |
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| 116. |
Differentiate`sin^(-1)((2x)/(1+x^2))`with respectto `tan^(-1)((2x)/(1-x^2)),`if `-1A. 1B. 2C. -1D. 2 |
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Answer» Correct Answer - A |
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| 117. |
If `y=5" cos"x-3" sin"x`, prove that `(d^(2)y)/(dx^(2))+y=0.` |
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Answer» `y=5" cos"x-3" sin"x` `implies(dy)/(dx)=(d)/(dx)(5" cos"x-3" sin"x)= -5sinx-3cos x` `implies(d^(2)y)/(dx^(2))=(d)/(dx)(-5" sin"x-3 cos x)` `= -5 cos x+3 sin x ` `= -(5 cos x-3 sin x)= -y` `implies(d^(2)y)/(dx^(2))+y=0 " " `Hence Proved. |
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| 118. |
`cot ^2 x` |
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Answer» Correct Answer - `-2 cos x .cosec^2 x` |
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| 119. |
`e^(x/a)` |
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Answer» Correct Answer - `1/a.e^(x//a)` |
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| 120. |
Find the derivatiive of `cos(tan x^3)` |
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Answer» Let y `=cos (tan x^3) ` `rArr (dy)/(dx)=d/(dx)cos (tan x^3)` `=-sin (tan x^3)d/(dx)(tan x^3)` `=-sin (tan x^3).sec^2 x^3.d/dx x^3` `=-sin (tan x^3).sec^2 x^3.(3x^2)` `=-3x^2sex^2x^3.sin(tan x^3).` |
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| 121. |
the derivative of `sin(log x)` is |
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Answer» Let `y= "sin" (log x)` `implies(dy)/(dx)=(d)/(dx)"sin"("log"x)` `=(1)/("log"x)(d)/(dx)("log"x)` `=cos(log x)(d)/(dx)log x=("cos"(log x))/(x)` `implies(d^(2)y)/(dx^(2))=(d)/(dx)[("cos"(log x))/(x)]` `=(x(d)/(dx)"cos"(log x)-"cos"(log x)(d)/(dx)x)/(x^(2))` `=(x{-"sin"(log x)}*(1)/(x)-"cos"(log x))/(x^(2))` `= -[("sin"(log x) +cos(log x))/(x^(2))]` |
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| 122. |
`e^(5x)` |
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Answer» Correct Answer - `5.e^(5x)` |
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| 123. |
If `x^(2//3)+y^(2//3)=a^(2//3)`, find `(dy)/(dx)` |
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Answer» Correct Answer - `(-y^(1//3))/(x^(1//3))` |
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| 124. |
Let `f(x)={{:((alphacotx)/(x)+(beta)/(x^(2))",",0ltxle1),((1)/(3)",",x=0):}` If f(x) is continuous at x =0, then the value of `alpha^(2)+beta^(2)` isA. 1B. 2C. 5D. 9 |
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Answer» Correct Answer - B `underset(xrarr0)(lim)f(x)=(1)/(3)` `rArr" "underset(xrarr0)(lim)(x.alphacotx+beta)/(x^(2))=(1)/(3)` `rArr" "underset(xrarr0)(lim)(xalpha+betatanx)/(x^(2).tanx)=(1)/(3)` `rArr" "underset(xrarr0)(lim)(alphax+beta(x+(x^(3))/(3)+...oo))/(x^(3)((tanx)/(x)))=(1)/(3)` `rArr" "underset(xrarr0)(lim)((alpha+beta)x+((beta)/(3))x^(3)+...oo)/(x^(3))=(1)/(3)` So, `" "alpha+beta=0` Also,`" "beta=1rArr alpha=-1` |
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| 125. |
If y= `y=sqrt((1-sin 2x )/(1+sin 2 x))` prove that `(dy)/(dx)+sec^2(pi/4-x)=0.` |
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Answer» `y= sqrt((1-sin 2x)/(1+sin 2x))` `=sqrt((cos^2 x+sin^2x-2sin x cos x)/(cos^2+sin^2x+2 sin x cos x ))` `=sqrt(((cos x-sin x)^2)/(cos x +sin x )^2)` `=(cos x - sin x)/(cosx+sinx)` `=(1-tanx)/(1+tan x)=(tan"(pi)/(4)tanx)/(1+tanpi/2.tanx)=tan (pi/(4)=x)` `rArr dy/dx=d/dxtan ((pi)/4-x)` `=sec^2((pi)/(4)-x)d/dx(pi/4-x)` `=-sec^2(pi/4-x)` `rArr (dy)/(dx)+sec^2(pi/4-x)=0.` |
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| 126. |
If `y=sin^(-1)((1-x^2)/(1+x^2))`, then `(dy)/(dx)=``-2/(1+x^2)`(b) `2/(1+x^2)`(c) `1/(2-x^2)`(d) `2/(2-x^2)`A. `2/(1+x^2)`B. `(-2)/(1+x^2)`C. `(-1)/(1+x^2)`D. None of these |
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Answer» Correct Answer - B |
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| 127. |
Find the 2nd derivative if `x^3` log x with respect to x. |
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Answer» Correct Answer - `x(5+6logx)` |
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| 128. |
Find the 2nd dervative of `e^(ax+b)` with respect to x. |
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Answer» Correct Answer - `a^2. e^(ax+b)` |
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| 129. |
If `y=tan^(-1)x^3` then find `(d^2y)/(dx^(2))`. |
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Answer» Correct Answer - `(6x(1-2x^6))/((1+x^6)^2)` |
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| 130. |
`cos^(-1)(sqrt(1+cos x)/2)` |
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Answer» Correct Answer - `1/2` |
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| 131. |
If `y= tan^-1((1-x)/(1+x))+cot^-1((1-x)/(1+x))` then `dy/dx` ?A. -1B. 1C. 0D. None of these |
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Answer» Correct Answer - C |
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| 132. |
Differentiate `sqrt(((x-1)(x-2))/((x-3)(x-4)(x-5)))`with respect to `x` |
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Answer» `"Let "y =sqrt(((x-1)(x-2))/((x-3)(x-4)(x-5)))` `rArr "log " y = "log" sqrt(((x-1)(x-2))/((x-3)(x-4)(x-5)))` `=(1)/(2)["log"(x-1) +"log" (x-2) - "log" (x-3)- "log" (x-4)- "log" (x-5)]` Differentiate both sides w.r.t.x. `(1)/(y)(dy)/(dx) = (1)/(2){(1)/(x-1)+ (1)/(x-2) - (1)/(x-3) - (1)/(x-4) - (1)/(x-5)}` `rArr (dy)/(dx) = (1)/(2)sqrt(((x-1)(x-2))/((x-3)(x-4)(x-5))){(1)/(x-1)+ (1)/(x-2) - (1)/(x-3) - (1)/(x-4) - (1)/(x-5)}` |
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| 133. |
If `x=at^2,y=2` at then find `(d^2y)/(dx^2)`. |
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Answer» Correct Answer - `(-1)/(2at^3)` |
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| 134. |
`tan^(-1)((sin x)/(1+cos x))` |
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Answer» Correct Answer - `1/2` |
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| 135. |
If `x=a cos^3 t " and "y=a sin ^3 t "then find " dy/dx` |
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Answer» `x=a cos ^3 t ` `dx/dt=a.d/dt(cos t)^3=-3a cos^2 t. d/dt(cos t )` and `y=a sin^3 t` `rArr dy/dt =a. d/dt(sin t )^3=a.3 sin ^2 t .d/dtsin t ` `= 3a sin ^2 t cos t` `Now dy/dx=(dy//dt)/(dx//dt)=(3asin^2 tcost )/(-3 a cos^2 t.sin t)=-tan t `. |
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| 136. |
`"cos" x *"cos" 2x* "cos" 3x` , find dy/dx |
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Answer» `"Let" y = "cos" x * "cos" 2x * "cos" 3x` `rArr "log" y ="log" ("cos" x "cos" 2x "cos" 3x)` `="log cos" x + "log cos" 2x + "log cos" 3x` Differentiate both sides W.r.t.x ` (1)/(y) (dy)/(dx) = (1)/("cos" x)(d)/(dx)("cos" x)+(1)/("cos" 2x)(d)/(dx)("cos" 2x)+(1)/("cos" 3x)(d)/(dx)"cos" 3x` `=-("sin" x)/("cos" x) + ((-"sin" 2x))/("cos" 2x)*(d)/(dx)(2x)+((-"sin" 3x))/("cos" 3x)(d)/(dx)3x` `rArr (dy)/(dx)= -y("tan" x + 2"tan" 2x + 3"tan" 3x)` `rArr (dy)/(dx) = "cos" x " cos" 2x " cos" 3x ("tan " x + 2"tan "2x + 3 "tan "3x)` |
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| 137. |
`(x+(1)/(x))^(x) +x^((1+(1)/(x)))` |
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Answer» `"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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| 138. |
If x=a (t+sin t) and y=a(1-cos t ) then find `dy/dx`. |
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Answer» `x=a (t + sin t)` `rArr dx/dy =a(1+cos t)` and `y=a(1-cos t)` `rArr dy/dx =a(0+ sin t)=a sin t ` `therefore dy/dx=(dy//dt)/(dx//dt)=(a sin t)/(a(1+cos t ))` `=(2 sin""t/2 cos""t/2)/(2 cos^2""t/2)=tan""t/2` |
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| 139. |
(i)` b tan ^(-1)(x/a+tan^(-1)"(x)/a)` (ii) `(sin ^(-1)x)^(2)-(cos^(-1)x)^2` |
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Answer» Correct Answer - `(b(2a^2+x)^2)/(a(a^2+x^2)[1+(x/a +tan^-1""x/a)^2] ` (ii) `pi/sqrt(1-x^2)` |
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| 140. |
Find derivative of `cos(logx+e^x)`, `x >0`w.r.t. to `x` |
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Answer» `"Let" y = "cos" ("log" x +e^(x))` `rArr (dy)/(dx) =(d)/(dx)"cos" ("log" x + e^(x))` `=-"sin"("log" x + e^(x))(d)/(dx)("log" x +e^(x))` `=-"sin"("log" x +e^(x))((1)/(x)+e^(x))` `=(-"sin" ("log" x + e^(x))(1 + x * e^(x)))/(x)` |
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| 141. |
The set of points of discontinuity of the function `f(x)=(1)/(log|x|),is`A. {0}B. {-1,1}C. {-1,0,1}D. none of these |
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Answer» Correct Answer - C Clearly, f(x) is not defined at x=0, 1,-1 At all points f(x) is continuosu. Hence, required set is `{-1,0,1}` |
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| 142. |
If `f(x)-{x^2}-({x})^2, `where (x) denotes the fractional part of x, thenA. `f (x) ` is continuous at ` x = 2 ` but not at ` x = - 2 `B. `f (x) ` is continuous at ` x = - 2 ` but not at ` x = 2 `C. ` f (x)` is continuous at ` x = 2 and x = - 2 `D. `f (x) ` is discontinuous at ` x = 2 and x = - 2 ` |
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Answer» Correct Answer - A We have, ` f (x) = {x^(2)} - ( {x}) ^(2)` ` therefore lim_( x to 2 ^(-)) f (x) = lim_(h to 0 ) f ( 2 - h ) ` ` rArr lim_( x to 2^(-)) f (x) = lim _( h to 0 ){ (2 - h ) ^(2)} - ({ 2- h }) ^(2)` ` rArr lim _( x to 2 ^(-)) f (x) = lim_( h to 0 )[ ( 2 - h ) ^(2) - [(2 - h ) ^(2)] - (( 2 - h ) - [(2 - h)]) ^(2)]` ` rArr lim_( x to 2 ^(-)) f (x) = lim_( h to 0 )[ ( 2 - h) ^(2) - 3 - (2 - h - 1) ^(2)] ` ` rArr lim_( x to 2 ^(-)) f (x) = lim_( h to 0 ) [ 4 - 4h + h ^(2) - 3 - (1 - 2h + h ^(2))] = 0 ` ` lim_( x to 2 ^(+)) f (x) = lim_( h to 0 ) f ( 2 + h )` `rArr lim_( x to 2 ^(+)) f (x) = lim_( h to 0 ) [ {(2 + h ^(2)) } - { 2+ h } ^(2)]` ` rArr lim_(x to 2 ^(+)) f (x) = lim_( x to 0 ) [ ( (2 + h ) ^(2) - [ (2 + h ) ^(2)] ) - ((2 + h ) - [ 2 + h ] ) ^(2)]` ` rArr lim_( x to 2 ^(+)) f (x) =lim_( h to 0 ) [ ( 4 + 4h + h ^(2) - 4) - ( 2 + h - 2 )^(2)]` ` rArr lim_(x to 2^(+)) f (x) = lim_( h to 0 )4h =0 ` and, ` f ( 2 ) = { 2 ^(2)} - ( {2}) ^(2) = 0 - 0 = 0 ` ` therefore lim_( x to 2^(-)) f (x) = lim_( x to 2 ^(+)) f (x) = f (2) ` So, f (x) is continuous at ` x = 2`. Now, ` " " lim_( x to - 2^( - )) f (x) = lim_( h to 0 ) f (-2 - h )` ` rArr lim _( x to - 2^( -)) f (x) = lim_( h to 0) [{(-2 - h )^(2) } - ({( - 2 - h )}) ^(2) ] ` ` rArr lim_(x to - 2 ^(-))f (x) = lim_( h to 0 ) [ (( - 2 - h ) ^(2) - [ (-2 - h ) ^(2)] ) - (( - 2 - h ) - [ - 2 - h ] ) ^(2)]` ` rArr lim_( x to - 2 ^(-)) f (x) = lim _( h to 0) [ (-2 - h )^(2) - 4 - ( - 2 - h + 3 ) ^(2)]` `rArr lim_( x to - 2 ^( - )) f (x) = lim _( h to 0) [ 4h + h ^(2) - ( 1- h ) ^(2)] = - 1 ` ` " " lim_( x to - 2 ^(+)) f (x) = lim_( h to 0) f (-2 + h )` `rArr lim_( x to - 2 ^(+)) f (x) = lim_( h to 0 ) [{( - 2 + h ) ^(2)} - { ( - 2 + h )} ^(2) ] ` ` rArr lim_( x to - 2 ^( + ) ) f (x) =lim _( hto 0 ) [(( - 2 + h ) ^(2) - [ (-2 + h ) ^(2) ] )- (-2 + h - [ - 2 + h ] ) ^(2) ]` ` rArr lim _( x to - 2 ^( + )) f (x) = lim _( h to 0 ) [ (( - 2 + h ) ^(2) - 3 ) - ( - 2 + h + 2) ^(2)]` ` rArr lim _( x to - 2 ^(+)) f (x) = lim _(h to 0) [(-2 + h ) ^(2) - 3 - h ^(2)] = 1 ` ` therefore lim _( x to - 2^( - ))f (x) ne lim_( x to - 2 ^( + )) f (x) ` So, f (x) is continuous at ` x = - 2 ` |
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| 143. |
`cot^(-1)sqrt(x)` |
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Answer» Correct Answer - `(-1)/(2sqrt(x)(1+x))` |
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| 144. |
`e^(ax).sin^-1bx` |
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Answer» Correct Answer - `(e^(tan^(-1_x)).cos e^(tan^(-1_x)))/(1+x^2)` |
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| 145. |
If `y=a^(x^(a^x..oo))` then prove that `dy/dx=(y^2 log y )/(x(1-y log x log y))` |
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Answer» `y=a^(x^(a^x..oo))` `y-a^(x^(y)` `rArr log y =log x^(x^(y))` `x^(y).loga` `y-a^(x^(y)` `rArr log y =log x^(x^(y))` `x^(y).loga` `rArr log(log y)=log x^y+log(log a)` `=ylog x+log +log (log a)` Differentiate both sides with respect to x `1/(y log y )dy/dx=y/x+log x. dy/dx +0` `rArr (1/(y log y )-log x ) dy/dx = y/x ` `rArr ((1-y log x log y ) )/ (y log y ) .dy /dx = y/x` `dy/dx = (y^2 log y )/(x(1-y log x log y)` |
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| 146. |
If `x=at^2` and y=2 at then find `dy/dx` |
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Answer» `x=at^2` `rArr dy/dx =a.d/dt(t^2)=a.2t=2at` and y=2 at `rArr dy/dx =2a.d/dt (t)=2a .(1)=2a` Now `dy/dx=(dy//dt)/(dx//dt)=(2a)/(2 at) =1/t.` |
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| 147. |
`sin^-1(cos x)+tan ^-1(cot x)` |
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Answer» Correct Answer - -2 |
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| 148. |
If `(x-a)^2+(y-b)^2=c^2`, for some `c > 0`, prove that `([1+((dy)/(dx))^2]^(3/2))/((d^2y)/(dx^2))`is a constant independent of a and b. |
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Answer» `(x-a)^(2)+(y-b)^(2)=c^(2) " "` …(1) `implies2(x-a)+2(y-b)(dy)/(dx)=0` `implies(dy)/(dx)= -(x-a)/(y-b) " " ` ...(2) `implies (d^(2)y)/(dx^(2))= -((y-b)*1-(x-a)(dy)/(dx))/((y-b)^(2))` `= -((y-b)-(x-a)[-(x-a)/(y-b)])/((y-b)^(2))` `= -((y-b)^(2)-(x-a)^(2))/((y-b)^(3))` `= -(c^(2))/((y-b)^(3))` ` " " `From equation (1) Now `([1+((dy)/(dx))^(2)]^(3/2))/((d^(2)y)/(dx^(2)))=([1+((x-a)/(y-b))^(2)]^(3/2))/((-c^(2))/((y-b)^(3)))` `=([((y-b)^(2)+(x-a)^(2))/((y-b)^(2))]^(3//2))/(-(c^(2))/((y-b)^(3)))` `= -[(c^(2))/((y-b)^(2))]^(3//2)*((y-b)^(3))/(c^(2))` `= -(c^(3))/((y-b)^(3))*((y-b)^(3))/(c^(2))= -c` which is a constant independent of a and b. ` " " `Hence Proved. |
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| 149. |
Differentiate the following w.r.t. x:`(logx)^x+x^(logx)` |
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Answer» `y = x^("log" x) + ("log" x)^(x)` `=u + v " " (say)` `implies (dy)/(dx) = (du)/(dx) + (dv)/(dx) " " ….(1)` `"Now",u=x^("log"x)` `rArr " log"u-"log"(x^("log"x))-"log"x*"log"x-("log"x)^(2)` `rArr (1)/(u)(du)/(dx) = (2"log"x)/(x)` `rArr (du)/(dx) = (2u "log"x)/(x) = (2"log"x)/(x).x^("log"x)` `"and "v = ("log"x)^(x)` `rArr "log"v = "log" ("log"x)^(x) = x"log" ("log"x)` `rArr (1)/(v) (dv)/(dx) = (x)/(x"log"x) + "log"("log"x)` `rArr (dv)/(dx) = v[(1)/("log"x) + "log"("log"x)]` `=("log"x)^(x) [(1)/("log"x) + "log"("log"x)]` From equation (1) `(dy)/(dx) = x^("log"x)*(2"log"x)/(x) + ("log"x)^(x)[(1)/("log"x) + "log"("log"x)]` |
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| 150. |
`y=sqrt(sin +cos x+sqrt(sinx+ cos x +sqrt(sin x +cos x +....oo))) " then find " dy/dx` |
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Answer» `y=sqrt(sin +cos x+sqrt(sinx+ cos x +sqrt(sin x +cos x +....oo))) ` `rArr sqrt(sin x +cos x+y )` `rArr y^2 =sin x+ cos x+y ` `y^2-y =sin x+cos x` Differentiate both sides w.r.t. x, `2y(dy)/(dx))-dy/dx = cos x -sin x ` `rArr dy /dx (2y -1) =cos x -sin x ` `rArr dy /dx =(cos x-sin x )/(2y -1)` |
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