380 + Interview Questions in CONTINUITY-AND-DIFFERENTIABILITY IN GENERAL AWARENESS Page 4 InterviewSolution

151.	If `y=sin^(- 1)x+sin^(- 1)sqrt(1-x^2)` then find `(dy)/(dx)`
Answer» `y="sin"^(-1)x+"sin"^(-1)sqrt(1-x^(2))` Let `x=sin theta` ` :. y=theta+"sin"^(-1)sqrt(1-"sin"^(2))theta=theta+"sin"^(-1)sqrt(cos^(2)theta)` `=theta+"sin"^(-1)(cos theta)=theta+"sin"^(-1)"sin"((pi)/(2)-theta)` `=theta +(pi)/(2)-theta=(pi)/(2)` `implies(dy)/(dx)=(d)/(dx)((pi)/(2))=0`

Discussion

152.	`"If "xsqrt(1+y)+ysqrt(1+x)=0," prove that "(dy)/(dx)=-(1)/((x+1)^(2)).`
Answer» `xsqrt(1+y)+ysqrt(1+x)=0` `impliesxsqrt(1+y)= -ysqrt(1+x)` `impliesx^(2)(1+y)=y^(2)(1+x)` `impliesx^(2)+x^(2)y=y^(2)+xy^(2)` `impliesx^(2)-y^(2)+x^(2)y-xy^(2)=0` `implies(x-y)(x+y)+xy(x-y)=0` `implies(x-y)(x+y+xy)=0` `impliesx+y+xy=0` `impliesy(1+x)= -x` `impliesy=(-x)/(1+x)` Differentiate both sides w.r.t. x `(dy)/(dx)=((1+x)(d)/(dx)(-x)-(-x)(d)/(dx)(1+x))/((1+x)^(2))` `=((1+x)(-1)+x(1))/((1+x)^(2))=(-1)/((1+x)^(2)) " " ` Hence proved.

Discussion

153.	Differentiate the function `(sin x)^x+ sin^1 x` with respect to x.
Answer» `"Let " y = ("sin"x)^(x) + "sin"^(-1) sqrt(x)` `"Let " u = ("sin"x)^(x) "and " v = "sin"^(-1) sqrt(x)` `therefore y = u+v` `rArr (dy)/(dx) = (du)/(dx) + (dv)/(dx) " " .....(1)` `"Now", u = ("sin" x)^(x)` `rArr "log"u = "log"("sin" x)^(x) = x "log " "sin"x` `rArr (1)/(u) (du)/(dx) = x (d)/(dx) "log sin" x + "log sin"x (d)/(dx)x` `rArr (du)/(dx) = u[x* ("cos"x)/("sin"x) + "log sin" x * 1]` `rArr (du)/(dx) = ("sin"x)^(x) [x "cot" x + "log sin"x]` `"and " v="sin"^(-1) sqrt(x)` `rArr (dv)/(dx) = (d)/(dx)"sin"^(-1) sqrt(x)` `= (1)/(sqrt(1-(sqrt(x))^(2))) (d)/(dx)sqrt(x) = (1)/(2sqrt(x)(1-x))` From equation (1) `(dy)/(dx) = ("sin" x)^(x)[x "cot"x + "log sin"x] + (1)/(2sqrt(x)(1-x))`

Discussion

154.	Differentiate w.r.t. x the function `x^x+x^a+a^x+a^a ,`for some fixed `a > 0 a n d x > 0`.
Answer» Let `y=x^(x)+x^(a)+a^(x)+a^(a)` `implies(dy)/(dx)=(d)/(dx)x^(x)+(d)/(dx)x^(a)+(d)/(dx)a^(x)+(d)/(dx)a^(a)` `implies(dy)/(dx)=(d)/(dx)x^(x)+ax^(a-1)+a^(x)loga+0 " " `...(1) Let `u=x^(x)` `implieslogu=logx^(x)=xlogx` `implies(1)/(u)(du)/(dx)=x(d)/(dx)logx+logx(d)/(dx)x` `implies(du)/(dx)=u(x(1)/(x)+logx1)=x^(x)(1+logx)` From equation (1) `(dy)/(dx)=x^(x)(1+logx)+ax^(a-1)+a^(x)loga`

Discussion

155.	`y = sin ^(-1)((1 - x^(2))/(1+ x^(2))) 0 lt x lt 1`
Answer» `y = sin ^(-1)((1-x^(2))/(1+ x^(2)))` Let x = tantheta `rArr theta^(-1) x` `rArr y = sin ^(-1)((1- tan^(2)theta)/(1+ tan^(2)theta))` `= sin ^(-1)((cos ^(2)theta-sin^(2)theta)/(cos^(2)theta+ sin^(2)theta))` `= sin ^(-1)(cos 2 theta) = sin ^(-1) sin((pi)/(2)-2 theta)` `= (pi)/(2) - 2 theta= (pi)/(2)-2tan^(-1) x`. `rArr (dy)/(dx)=(d)/(dx)((pi)/(2)-2tan^(-1)x.)` `= 0 - (2xx1)/(1+x^(2))=-(2)/(1+x^(2))`.

Discussion

156.	Differentiate `x^x^(2-3)+(x-3)^x^2`with respect to `x`:
Answer» Let `y=x^(x^(2)-3)+(x-3)^(x^(2))` Let `u=x^(x^(2)-3)" and "v=(x-3)^(x^(2))` ` :. y=u+v` `implies(dy)/(dx)=(du)/(dx)+(dv)/(dx) " " `...(1) Now, `u=x^(x^(2))-3` `implieslogu=logx^(x^(2)-3)=(x^(2)-3)logx` `implies(1)/(u)(du)/(dx)=(x^(2)-3)(d)/(dx)logx+logx(d)/(dx)(x^(2)-3)` `implies(du)/(dx)=u[(x^(2)-3)/(x)+2xlogx]` `=x^(x^(2)-3)[(x^(2)-3)/(x)+2xlogx]` and `v=(x-3)^(x^(2))` `implieslogv=log(x-3)^(x^(2))=x^(2)log(x-3)` `implies(1)/(v)(dv)/(dx)=x^(2)*(d)/(dx)log(x-3)+log(x-3)(d)/(dx)x^(2)` `implies(dv)/(dx)=v[(x^(2))/(x-3)+2x log(x-3)]` `(x-3)^(x^(2))[(x^(2))/(x-3)+2x log(x-3)]` ` :.`From equation (1) `(dy)/(dx)=x^(x^(2)-3)[(x^(2)-3)/(x)+2xlogx]+(x-3)^(x^(2))[(x^(2))/(x-3)+2xlog(x-3)]`

Discussion

157.	`x^("sin"x) + ("sin" x)^("cos"x)`
Answer» `"Let" y = x^("sin"x) + ("sin" x)^("cos"x)` `"Let" u = x^("sin"x) "and " v=("sin" x)^("cos"x)` `therefore y=u+v` ` rArr (dy)/(dx) = (du)/(dx) + (dv)/(dx) " ".....(1)` `"Now", u=x^("sin"x)` `rArr "log"u = "log"(x^("sin"x)) = "sin"x * "log"x` `rArr (1)/(u) (du)/(dx) = "sin"x * (d)/(dx)"log"x + "log"x * (d)/(dx)"sin"x` `rArr (du)/(dx) = u[("sin"x)/(x) + "log"x * "cos"x]` `rArr (du)/(dx) = x^("sin"x)[("sin"x)/(x) + "log"x * "cos"x]` `"and "v=("sin"x)^("cos"x)` ` rArr " log"v = "log" ("sin"x)^("cos"x)` `= "cos" x * "log"("sin"x)` `rArr (1)/(v) (dv)/(dx) = "cos"x (d)/(dx)"log"("sin"x) + "log"("sin"x)(d)/(dx)("cos"x)` `rArr (dv)/(dx) = v["cos"x * ("cos"x)/("sin"x) + "log" ("sin"x) * (-"sin"x)]` ` rArr (dv)/(dx) = ("sin"x)^("cos"x)["cos"* "cot"x + "sin"x "log"("sin"x)]` From equation (1) `(dy)/(dx) = x^("sin"x)[("sin"x)/(x) + "log"x * "cos"x] + ("sin"x)^("cos"x)["cos"x * "cot"x - "sin"x * "log"("sin"x)]`

Discussion

158.	If `=sqrt(x)+sqrt(y)=sqrt(a)` differentiate both sides with respect to x
Answer» `1/(2sqrtx)+1/(2sqrt(y))dy/dx =0` `rArr dy/dx =-sqrt(y)/sqrt(x).`

Discussion

159.	`y = cos ^(-1)((1 - x^(2))/(1+ x^(2))) 0 lt x lt 1`
Answer» `y = cos ^(-1)((1 - x^(2))/(1+ x^(2)))` Let `x tan theta` ltbgt `rArr theta = tan^(-1) x ` `rArr y = cos^(-1)((1 - tan^(2)theta)/(1 +tan^(2)theta))` `=cos^(-1)((cos^(2)theta-sin^(2)theta)/(cos^(2)theta + sin^(2) theta))` `= cos ^(-1)(cos2theta)= 2theta= 2tan^(-1)x` `rArr (dy)/(dx) = 2 (d)/(dx) tan^(-1)x = (2)/(1+ x^(2))`.

Discussion

160.	`x^(y) + y^(x) = 1`
Answer» `x^(y) + y^(x) = 1` `rArr u+v = 1 " where "u=x^(y) " and "v=y^(x)` `rArr (du)/(dx) + (dv)/(dx) = 0 " "....(1)` `"Now " u = x^(y)` `rArr "log"u = "log"(x^(y)) = y"log"x` `rArr (1)/(u) (du)/(dx) = y * (d)/(dx)"log"x + "log"x * (d)/(dx)y` `rArr (du)/(dx) = u[(y)/(x) + "log"x(dy)/(dx)]` `=x^(y)[(y)/(x) + "log"x(dy)/(dx)] = y * x^(y-1) + x^(y) "log"x(dy)/(dx)` `"and " v = y^(x)` `rArr " log"v = "log"y^(x) = x"log"y` `rArr (1)/(v) (dv)/(dx) = x (d)/(dx)"log" y + "log" y * (d)/(dx)x` `rArr (dv)/(dx) = v[(x)/(y)(dy)/(dx) + "log"y * 1]` `=y^(x)[(x)/(y)(dy)/(dx) + "log"y]` `= x * y^(x-1) (dy)/(dx) + y^(x) "log"y` From equation (1) `y * x^(x-1) + x^(y)"log" x (dy)/(dx) + x * y^(x-1)(dy)/(dx) + y^(x) "log" y = 0` `rArr (dy)/(dx) (x^(y)"log"x + x * y^(x-1))` `=-(yx^(y-1) + y^(x)"log"y)` `rArr (dy)/(dx) = -(y * x^(y-1) + y^(x) "log"y)/(x^(y) "log"x + x * y^(x-1))`

Discussion

161.	`e^(mx).cos n x`
Answer» Correct Answer - `e^(mx).(m cos nx -n sin nx)`

Discussion

162.	Differentiate `x^(xcosx)+(x^2+1)/(x^2-1)`with respect to `x`:
Answer» `"Let "y = x^(x"cos"x) + (x^(2) + 1)/(x^(2)-1)` `"Let "u = x^(x"cos"x) " and "v=(x^(2)+1)/(x^(2)-1)` `therefore y=u+v` `rArr (dy)/(dx) = (du)/(dx) + (dv)/(dx) " ".....(1)` `"Now", u=x^(x"cos"x)` `rArr "log" u = "log"(x^(x"cos"x)) = x"cos"x * "log"x` `rArr (1)/(u)(du)/(dx) = x "cos"x * (d)/(dx)"log"x + x"log"x * (d)/(dx)"cos"x + "cos"x * "log"x * (d)/(dx)x` `rArr (du)/(dx) = u(x"cos"x * (1)/(x)-x"log"x * "sin"x + "cos"x * "log"x)` `= x^(x"cos"x) ("cos"x-x"log"x"sin"x + "cos"x"log"x)` `"and "v = (x^(2) + 1)/(x^(2)-1)` `rArr (dv)/(dx) = ((x^(2)-1)(d)/(dx)(x^(2)+1) - (x^(2)+1)(d)/(dx)(x^(2)-1))/(x^(2)-1)^(2)` `= ((x^(2)-1) * 2x - (x^(2)+1) * 2x)/(x^(2)-1)^(2)` `=-(4x)/(x^(2)-1)^(2)` From equation (1) `(dy)/(dx) = x^(x"cos"x)("cos" x - x"log"x"sin"x + "cos"x"log"x)-(4x)/(x^(2)-1)^(2)`

Discussion

163.	`e^(-2x)sin 4 x`
Answer» Correct Answer - `2e^(-2x)(2cos 4x-sin4x)`

Discussion

164.	`y=tan^-1[(3x-x^3)/(1-3x^2)],-1/sqrt3ltxlt1/sqrt3`
Answer» `y = tan ^(-1)((3x-x^(3))/(1-3x^(2)))` Let `x + tantheta` `rArr theta= tan^(-1) x` `rArr y = tan^(-1)((3tan theta- tan^(3)theta)/(1-3 tan^(2)theta))` `= tan^(-1) (tan3theta) = 3theta = 3 tan^(-1)x ` `rArr (dy)/(dx) = 2 (d)/(dx) tan^(-1)x = (3)/(1+x^(2))`

Discussion

165.	(i) `tan^(-1)sqrt(x)` (ii) `tan ^(-1)(2x+1)`
Answer» Correct Answer - `(1)/(2sqrt(x)(1+x))` (ii) `(1)/(2x^2+2x+1)`

Discussion

166.	`(e^x+e^-x)/(e^x-e^(-x))`
Answer» Correct Answer - `(-4)/(e^x-e^(-x))^2`

Discussion

167.	`cos ^(-1)(x/a)`
Answer» Correct Answer - `(-1)/(sqrt(a^2-x^2))`

Discussion

168.	If the following function f(x) is continuous at x=0 , find the values of `a, b and c`. `f(x) ={ {:( (sqrt(x+bx^(2))-sqrt(x))/(bx^(3/2)) ," if " x gt 0),( " " c, " if " x =0),((sin (a+1)x+sin x)/(x) , " if " x lt 0):}`
Answer» we havef(0)=a ` lim_(x to 0-) f( 0-h)` `lim_(h to0) (sin (a +1) (-h) +sin(-h))/((-h)) = lim_( hto0)(-{sin (a+1) h +sin h})/(-h)` `lim_(h to o) (sin(a+1) +sinh)/h = lim_(hto0) (2sin (a/2 +1)h.cos(ah)/2)/h ` `2.lim_(h to0) {(sin (a/2=1)h)/((a/2+1)h).(a/2+1).cos""(ah)/2}` `2(a/2+1) .lim_(h to0) (sin (a/2=1)h)/((a/2+1)).lim_(hto0) cos""(ah)/2` ` (a + 2) xx 1 xx 1 = (a +2)` ` lim_(hto0) {([sqrt(h+bh^(2))-sqrth])/(bh^(3//2))xx([sqrt(h+bh^(2))+sqrth])/([sqrt(h +bh^(2))+sqrth])` ` lim_(hto0) ((h+bh^(2)-h))/(bh^(3//2) (sqrt(h+bh^(2))+sqrth)) =lim_(h to 0) (bh^(2))/(bh^(2)(sqrt(1+bh+1))` `lim_(hto0) 1/((sqrt(1+bh+1)))=1/2` since f (x) is continuous at x =0, we have `f(0) =lim_(xto0-)f(x) Rightarrow c=1/2 anda+2=1/2` `c = 1/2 and a = (-3)/2`

Discussion

169.	Graph of `y=sin^(-1)((2x)/(1+x^2))`
Answer» ` y = sin ^(-1) ((2x)/(1+x^(2)))` Let `x = tan theta` `rArr theta = tan ^(1) x` `rArr y = sin^(-1) ((2tan theta)/(1+ tan^(2) theta))` ` = sin^(-1)((2tantheta)/(sec^(2)theta)) = sin^(-1)(2sin theta costheta)` `= sin ^(-1) (sin 2 theta) = 2 theta = 2 tan^(-1) x ` `rArr (dy)/(dx) = 2 (d)/(dx) tan 6(-1) x = (2)/(1+ x^(2))`.

Discussion

170.	Differentiate the following w.r.t. x:`(xcosx)^x+(xsinx)^(1/x)`
Answer» Let `y = (xcosx)^x+(xsinx)^(1/x)` Let `u = (xcosx)^x and v = (xsinx)^(1/x)` Then, `y = u+v` `dy/dx = (du)/dx+(dv)/dx` Now, `u = (xcosx)^x` Taking log both sides, `log u = x(logx+logcosx)` Differentiating both sides, `1/u (du)/dx = (logx+cosx)1+x(1/x+1/cosx(-sinx))` `1/u (du)/dx = (logx+cosx)+1-xtanx` `(du)/dx = (xcosx)^x(log(xcosx)+1-xtanx)->(1)` Now, `v = (xsinx)^(1/x)` Taking log both sides, `log v = 1/x(logx+logsinx)` Differentiating both sides, `1/v (dv)/dx = 1/x^2(x(1/x+1/sinx(cosx))-(logx+logsinx)1)` `1/v(dv)/dx = 1/x^2(1+cotx -log(xsinx))` `(dv)/dx = (xsinx)^(1/x)1/x^2(1+cotx -log(xsinx))` So, `dy/dx = (du)/dx+(dv)/dx` `dy/dx = (xcosx)^x(log(xcosx)+1-xtanx)+(xsinx)^(1/x)((1+cotx -log(xsinx))/x^2)`

Discussion

171.	Differentiate the following w.r.t. x:`(xcosx)^x+(xsinx)^(1/x)`
Answer» `"Let" y = (x "cos" x)^(x) + (x"sin"x)^(1//x)` `"Let" u = (x "cos" x)^(x) " and " v = (x"sin"x)^(1//x)` `therefore y=u+v` `rArr (dy)/(dx) = (du)/(dx) + (dv)/(dx) " "....(1)` `"Now "u = (x"cos" x)^(x)` `rArr "log" u = "log"(x"cos"x)^(x) = x("log"x + "log cos"x)` `rArr (1)/(u)(du)/(dx) = x * (d)/(dx)("log" x + "log cos"x) + ("log"x + "log cos"x)(d)/(dx)x` `rArr (du)/(dx) = u[x((1)/(x)-("sin"x)/("cos"x)) + ("log"x + "log cos"x) * 1]` `rArr (du)/(dx) = (x"cos"x)^(x)[1-x"tan"x + "log"x + "log cos"x]` `"and "v = (x"sin"x)^(1//x)` `rArr "log" v = "log"(x"sin"x)^(1//x) = (1)/(x)("log"x + "log sin"x)` `rArr (1)/(v)(dv)/(dx) = (1)/(x)(d)/(dx)("log"x + "log sin"x) + ("log"x + "log sin"x)(d)/(dx)((1)/(x))` `rArr (dv)/(dx) =v[(1)/(x)((1)/(x) + ("cos"x)/("sin"x)) - (1)/(x^(2))("log"x + "log sin"x)]` `rArr (dv)/(dx) = (x"sin"x)^(1//x)[(1 + x"cot"x - "log"(x"sin"x))/(x^(2))]` From equation (1) `(dy)/(dx) = (x"cos"x)^(x)[1 - x"tan"x + "log"x + "log cos"x] + (x"sin"x)^(1//x)[1 - x"tan"x + "log"(x"sin"x)]`

Discussion

172.	`y=(log_(e)x)^(sinx)`
Answer» Correct Answer - `(log_(e)X)^(sinx)[(sinx)/(xlog_(e)x)+cosxcdotlog_(e)(log_(e)x)]`

Discussion

173.	Show that the function `f(x)=x^3/2`is not differentiable at x=0.
Answer» `f(x)=x^(3//2)` `R,f (0)=underset(hrarr0)lim f(0+h)-f(0))/(h)` `=underset(hrarr0)lim(h^(3//2)-0)/(h)=underset(hrarr0) limh^(1//2)=0` and L.f(0) L.f(0)`=underset(hrarr0)lim (f(0-h)-f(0))/((-h))` `=underset(hrarr0)lim(h^(3//2)-0)/(-h)` `=underset(hrarr0)lim((-h)^(3//2)-0)/(-h)` `=underset(hrarr)lim^(1//2)` which is imaginary . `therefore` L.f (0) does not exist Therefore f(x) is not differentiable at x=0.

Discussion

174.	If `x^4+y^4-a^2xy=0 " then find " dy/dx`
Answer» `x^4+y^4-a^2xy=0 ` `4x^3+4y^3 dy/dx` `4x^3+4y^3 dy/dx-a^2(xdy/dx+y.1)=0` `rArr 4x^3 +3y^3dy/dx-a^2xdy/dxa^2y=0` `rArr dy/dx(4y^3-a^3x)=a^2y-4x^3` `rArr dy/dx =(a^2y-4x^3)/(4y^3-a^2x)`

Discussion

175.	Show that the function f(x)=\|x-2\| is continuous but not differentiable at x=2.
Answer» `f(x)=abs(x-2)={:{(x-2 ", " x ge 2),(-(x-2)", "x lt2):}` `R.H.L=underset(xrarr2^+)(lim)f(x)=underset(hrarr0)limf(2+h)` `=underset(xrarr0)(lim)f(2+h-2)=underset(hrarr0)limf(2+h)` f(2)2-2=0 and `L.H.L=underset(xrarr2^-)(lim)f(x)=underset(hrarr0)limf(2-h)` `=underset(hrarr0)(lim)-(2+h-2)=underset(hrarr0)limh(2-h)` `therefore`R.H.L = L.H.L=f(2) `therefore` f(x) is not differentialble at x=2.

Discussion

176.	Show that the function `f(x)={:{(1+x ", " x le2","),(5-x", "x gt2):}` is not differentiable at x=2
Answer» R.f(2) `=underset(hrarr0)lim((f2+h)-f(2))/(h)` `=underset(hrarr0)lim({5-(2+h)}-(1+2))/h` `=underset(hrarr0)lim(-h)/h=underset(hrarr0)lim(-1)=-1` and L.f(2) `=underset(hrarr0)lim(f(2-h)-f(2))/(-h)` `=underset(hrarr0)lim({1+(2-h)}-(1+2))/(-h)` `=underset(hrarr0)lim(-h)/(-h)lim(1)=1` `therefore R.f(2) ne L.f(2)` `therefore` f(x) is not differentiable at x=2.

Discussion

177.	`y=(logx)^(x)`
Answer» Correct Answer - `(logx)^(x)[log(logx)+1/(logx)]`

Discussion

178.	`y=x^(tanx)`
Answer» Correct Answer - `x^(tanx)cdot(tanx/x+logxcdot sec^(2)x)`

Discussion

179.	`y=x^(logx)+(logx)^(x)`
Answer» Correct Answer - `x^(logx)cdot(2logx)/x+(logx)^(x)cdot{1/(logx)+log(logx)}`

Discussion

180.	`cos^(-1)((1-x)/(1+x))`
Answer» Correct Answer - `(1)/(sqrt(x)(1+x))`

Discussion

181.	`sin^(-1)"(x)/(1+x)`
Answer» Correct Answer - `(1)/((x+1)sqrt(2x+1))`

Discussion

182.	The points where the function `f(x) = [x] + \|1 - x\|, -1 < x < 3` where `[.]` denotes the greatest integer function is not differentiable, areA. `(-1,0,1,2,3)`B. `(-1,0,2)`C. `(0,1,2,3)`D. `(-1,0,1,2)`
Answer» Correct Answer - C We have `f(x)={{:(,-x,-1 lexlt0),(,1-x,0lexlt1),(,x,1lexlt2),(,1+x,2lexlt3),(,5,x=3):}` Clearly, f(x) is discontinuous at x=0,1,2 and 3. So, it is not differentiable at these points. At x=-1, we have `underset(x to 1^(+))lim f(x)=underset(x to 1^(+))lim-x=1=f(-1)` So, it is continuous at x=-1 Also, (RHD at x=1)=-1 (a finite number). Therefore, f(x) is differentiable at x=-1

Discussion

183.	Let `f(x)={((x-1)^2 sin(1/(x-1))-\|x\|,; x != 1), (-1,; x=1):}` then which one of the following is true?A. f is differential at x=0 but not at x=1B. f is differentiable at x=1 but not at x=0C. f is neither differentiable at x=0 nor at x=1D. f is differentiable at x=0 and at x=1
Answer» Correct Answer - A We observe that `underset(x to 1)lim (f(x)-f(1))/(x-1)` `underset(x to 1)lim ((x-1)sin ((1)/(x-1)))/(x-1)=underset(x to 1)lim sin ((1)/(x-1))` =An oscillating number between -1 and 1. `therefore underset(x to 1)lim (f(x)-f(0))/(x-0)` `underset(x to 1)lim ((x-1)sin ((1)/(x-1))-sin1)/(x)` `underset(x to 1)lim (2sin(x)/(2(x-1))cos{(2-x)/(2(x-1))})/({(x)/(2(x-1))}2(x-1))` `=-sin 1+cos 1` `Rightarrow` f(x) is differentiable at x=0.

Discussion

184.	If cos (x+y)=y sin x then find `dx/dy`
Answer» cos(x+y)= y sin x Differentiate both sides with respect to x `-sin (x+y) d/dx(x+y)=y.cos x+sin x dy/dx` `rArr -sin (x+y)[1+dy/dx]=y.cos x+sin xdy/dx` `rArr -sin(x+y)-sinx.dx/dy` `rArr -sin (x+y)[1+dy/dx]=y.cos x+sin xdy/dx` `rArr -sin(x+y)-sinx.dy/dx=cos x+sin x. dy/dx` `rArr -dy/dx[sin (x+y)+sin x]=ycosx+sin (x+y)` `rArr dy/dx =-(y cos c +sin (x+y))/(sin (x+y)+sin x)`

Discussion

185.	`y=(sinx)^(logx)`
Answer» Correct Answer - `(sinx)^(logx)[logxcdotcotx+1/xlogsinx]`

Discussion

186.	`y=x^(sinx)+a^(sinx)`
Answer» Correct Answer - `x^(sinx)[sinx/x+cosxcdotlogx]+a^(sinx)logacdotcosx`

Discussion

187.	Differentiate the functions with respect to x`sec(tan(sqrt(x)))`
Answer» Here =`sec(tax(sqrtx))`differentiate with respect to x=`delta/(deltax)(sec(tan(sqrtx))`=`sec(tan(sqrtx)tan(tan(sqrtx))sec^2sqrtx1/(2sqrtx)`=`1/(2sqrtx)sec^2sqrtx*sec(tan(sqrtx))tan(tan(sqrtx))`

Discussion

188.	`sin^(-1)(xsqrt(x)),0 le x le 1`
Answer» Let `y=sin^(-1)(xsqrt(x))` `implies(dy)/(dx)=(d)/(dx)sin^(-1)(xsqrt(x))` `=(1)/(sqrt(1-(xsqrt(x))^(2)))(d)/(dx)(xsqrt(x))` `=(1)/(sqrt(1-x^(3)))(d)/(dx)(x^(3//2))` `=((3)/(2)x^(1//2))/(sqrt(1-x^(3)))=(3sqrt(x))/(2sqrt(1-x^(3)))`

Discussion

189.	`(5x)^(3cos 2x)`
Answer» Let `y=(5x)^(3cos 2x)` `implies log y=log[(5x)^(3cos 2x)]` `=3cos2xlog(5x)` Differentiate both sides w.r.t x `(1)/(y)(dy)/(dx)=3[cos2x(d)/(dx)log(5x)+log(5x)(d)/(dx)cos2x]` `implies(dy)/(dx)=3y[cos2x(5)/(5x)+log(5x)(-2sin2x)]` `implies(dy)/(dx)=3*(5x)^(3cos2x)[(cos2x)/(x)-2sin2xlog(5x)]`

Discussion

190.	Differentiate w.r.t. x the function`sin^3x+cos^6x`
Answer» Let `y=sin^(3)x+cos^(6)x` `implies(dy)/(dx)=(d)/(dx)(sin^(3)x+cos^(6)x)` `=(d)/(dx)(sinx)^(3)+(d)/(dx)(cosx)^(6)` `=3 (sinx)^(2)(d)/(dx)(sinx)+6(cos x)^(5)(d)/(dx)(cos x)` `=3sin^(2)x cos x-6 cos^(5)xsinx` `=3sin x cos x(sinx-2cos^(4)x)`

Discussion

191.	Differentiate w.r.t. x the function. `(3x^2-9x+5)^9`
Answer» `(3x^2-9x+5)^9`diffenentiate with respect to x`delta/(deltax)(3x^2-9x+5)``9(3x^2-9x+5)^8(6x-9)``27(3x^2-9x+5)^8(2x-3)`

Discussion

192.	If the function `f(x)=(3x^2ax+a+3)/(x^2+x-2)` is continuous at `x=-2,` then the value of `f(-2)` is
Answer» Correct Answer - B Since the function is continuous at `x=-2`, then `f(-2)=underset(xrarr-2)(lim)f(x)` `" "=underset(xrarr-2)(lim)(3x^(2)+ax+a+3)/(x^(2)+x-2)` this limit will exist if `15-a=0` `rArr" "a=15" (i)"` So, `" "underset(xrarr-2)(lim)f(x)=underset(xrarr-2)(lim)(3x^(2)+15x+18)/(x^(2)+x-2)` `" "=underset(xrarr-2)(lim)(3(x+2)(x+3))/((x+2)(x-1))` `" "=underset(xrarr-2)(lim)(3(x+3))/((x-1))` `" "=(3)/(-3)=-1` Hence, `f(-2)=-1`

Discussion

193.	If `f(x)={(x^3+x^2-16x+20)/(x-2)^2,x ne 0` k, x=0 is continuous for all real values of x, find the value of K.
Answer» x=0 f(0)=k R.H.L `=underset(xrarr0^+)lim f(x)==underset(xrarr0)limf(0+h)` `=underset(hrarr0)lim(h^3+h^2-16h+20)/(h-2)^2` `=underset(hrarr0)lim((h-2)^2(h+5))/(h-2)^2` `=underset(hrarr0)lim((h-5))/(1)=0+5=5` `L.H.L =underset(xrarr0^-)limf(x)=underset(xrarr0)limf(0-h)` `=underset(hrarr0)lim((-h)^2+(-h)^2-16(-h)+20)/(-h-2)^2` `=underset(hrarr0)lim(-h^3+h^3+16h+20)/(h+2)^2` `=underset(hrarr0)lim((h+2)^2(-h+5))/((h+5)^2)` `=underset(hrarr0)lim((-h+5))/(1)=5` `therefore`(fx) is continuous at x=0 `therefore` R.H.L=f(0)=L.H.L `rArr` K=5.

Discussion

194.	Show that f(x)=cos x is continuous for all values of x.
Answer» Let x=a be any value . At x=a f(a)=cos a R.H.L `underset(xrarr0)limf(x)=underset(hrarr0)limf(a+h)` `=underset(hrarr0)lim cos (a+h)=cos a` `L.H.L=underset(xrarra^-)(lim)f(x)=underset(hrarr0)limf(0-h)` `=underset(hrarr0)limcos(a-h)=cosa` `therefore` R.H.L =f (a) =L.H.L and a is arbitrary value of x. `therefore` f(x) =cos x is continous for all values of x. Hence proved

Discussion

195.	Discuss the continuity of `f(x)=\|x\|+\|x-1\| ` at x=0 and x=1.
Answer» `f(x)=\|x\|={:{(x, xge 0),(-x,x lt 0):}` and `abs(x-1)={:{(x-1, xge 1),(-(x-1), xlt 1):}` `therefore f(x)=abs(x)+\|x-1\|={:{(1-2x, x lt 0),(1, 0 le x lt1),(2x-1, x ge1):}` At x=0 f(0)=1 `R.H.L=underset(xrarr0^+)lim=underset(xrarr)lim f(0+h)=underset(xrarr0)lim 1=1` `L.H.L=underset(xrarr0^-)lim=underset(xrarr0)lim f(0+h)` `=underset(hrarr0)lim{-(0-h)}=0` `therefore` R.H.L = f(0)=L.H.L `therefore` f(x) is constinuous at x=0 Hence proved.

Discussion

196.	Show that `f(x)=\|x\|` is continuous at x=0
Answer» `f(x)=\|x\|={:{(x, xge 0),(-x,x lt 0):}` Now f(0)=0 `R.H.L = underset(xrarr0^+)(limf(x))=underset(hrarr0)(lim)f(0+h)` `=underset(hrarr0)(lim)h=0` `L.H.L=underset(xrarr0^-)(lim)f(x)=underset(hrarr0)limf(0-h)` `=underset(hrarr0)(lim){-(0-h)}=0` `therefore` R.H.L = f(0)=L.H.L `therefore` f(x) is constinuous at x=0 Hence proved.

Discussion

197.	(i)`lim_(x->0) ((sqrt(1+x)-sqrt(1-x))/x)`(ii)`lim_(x->0)(x-sinx)/x^3`
Answer» `=lim_(x->0)(1-1+2sin^2(x/2))/(sin^2x)` `=lim_(x->0)((sin(x/2))/(x/2))^2((x/2)^2/(sinx/x))x^2` `=lim_(x->0)(2x^2/4*1/x^2)=1/2`.

Discussion

198.	`f(x)={x+sqrt2asinx,0ltxltpi/4 and 2xcotx+b , pi/4lexltpi/2 and acosx-bsinx , pi/2lexltpi `. Determine the value of a and b If function is continuous for interval `[0,pi]`
Answer» Here, value of `f(x)` is changing at `pi/2` and `pi/4`. So, `f(x)` to be continuous, `f((pi/4)^-) = f((pi/4)^+)` `=>pi/4+sqrt2 a1/sqrt2 = 2pi/4(1)+b` `=>pi/4 +a = pi/2+b` `=>a-b = pi/4->(1)` Also, `f(x)` to be continuous, `f((pi/2)^-) = f((pi/2)^+)` `=>2(0) + b = a(0) - b(1)` `=>2b = 0` `=> b = 0` Putting `b = 0` in (1), `=> a - 0 = pi/4` `=> a = pi/4` So, for `a = pi/4` and `b = 0`, `f(x)` will be continuous.

Discussion

199.	If (x) `{:{(x^(2),x ne "0,"),(4,x=0):}` then find whether f(x) is constinuosus at x=0
Answer» At x=0 f(0) = 4 R.H.L `=underset(xrarr0^+)limf(x)=underset(hrarr0)(limf(0+h)` `underset(hrarr0)lim(0+h)^(2)=0` `therefore " R.H.L " ne f(0)` `therefore f(x) ` is discontinous at x= 0.

Discussion

200.	Show that the function `f(x)=x^2+3x+5`,is continuos at x=1.
Answer» Atx = 1 `underset(xrarr1)R.H.L= underset(hrarr0)limf(x)=limf(1+h)` =`lim{(1+h)^2+3(1+h)+5}` `=(1+0)^(2)+3(1+0)+5=9` `L.H.L =underset(xrarr1)limf(x)=underset(hrarr0)limf(1-h)` `underset(hrarr0)lim{(1-h)^(2)+3(1-h)+5}` `=(1-0)^2+3(1-0)+5=9` and f(1)`=1^2+3(1)+5=9` Now R.H.L = f(1)=L.H.L `therefore f(x)=x^2+3x+5` is continuous at x=1

Discussion

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