InterviewSolution
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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.
| 351. |
Let `f:(0,oo)to R` be a continuous function such that `F(x)=int_(0)^(x^(2)) tf(t)dt. "If "F(x^(2))=x^(4)+x^(5),"then "sum_(r=1)^(12) f(r^(2))=`A. 216B. 219C. 222D. 225 |
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Answer» Correct Answer - B It is given that f is continuous. Therefore, the integral function f(x) is differentiable. Also, `F(x)=underset(0)overset(x)int t(f(t)dt` `F(x^(2))=underset(0)overset(x^(2))int t(f(t)dt` `x^(4)+x^(5)=underset(0)overset(x^(2))int t(f(t)dt` Differentiating with respect to x, we get `4x^(3)+5x^(4)=(2x)x^(2)f(x^(2))` `Rightarrow f(x^(2))=2+(5)/(2)x` `Rightarrow f(r^(2))=2+(5)/(2)r` `Rightarrow underset(r=1)overset(12)sum f(r^(2))=underset(r=1)overset(12)sum (2+(5)/(2)r)=24+(5)/(2)xx78=219` |
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| 352. |
Let f(x)=`{{:(,(tanx-cotx)/(x-(pi)/(4)),x ne (pi)/(4)),(,a,x=(pi)/(4)):}` The value of a so that f(x) is a continous at `x=pi//4` is.A. 2B. 4C. 3D. 1 |
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Answer» Correct Answer - B If f(x) is continous at `x=(pi)/(4)`, then `f((pi)/(4))=underset(x to pi//4)lim f(x)` `Rightarrow a=underset(x to pi//4)lim (tan x- cotx )/(x-(pi)/(4))` `Rightarrow a=underset(x to pi//4)lim (sin^(2)x-cos^(2)x)/((x-(pi)/(4))sin x cos x)` `Rightarrow a=underset(x to pi//4)lim (sin2((pi)/(4)-x))/(2((pi)/(4)-x))xx(1)/(sinx cos x)` `Rightarrow a=2xx1xx2=4` |
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| 353. |
If `f (x) = (2x + 3 sin x )/(3x + 2 sin x )`, x ne 0 is continous at x = 0 then f (0) = ?A. -1B. 0C. 1D. None of these |
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Answer» Correct Answer - C |
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| 354. |
If `f(x)={{:(,x^(m)sin((1)/(x)),x ne 0),(,0,x=0):}` is a continous at x=0, thenA. `m in (0,oo)`B. `m in (-oo,0)`C. `m in (1,oo)`D. `m in (-oo,1)` |
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Answer» Correct Answer - A If f(x) is a continous at x=0, then `underset (x to 0^(-))f(x)=underset(x to 0^(+))lim f(x)=f(0)=0` Now, `underset(x to 0^(-))limf(x)=underset(h to 0)lim f(0-h)=underset(h to 0)lim (-h)^(m) sin ((-1)/(h))` `Rightarrow underset(x to 0^(-))lim f(x)=underset(h to 0)lim (-h)^(m) sin ((1)/(h))=0" only when m"gt0 and underset(x to 0^(+))limf(x)=underset(h to 0)lim f(0+h)` `Rightarrow underset(x to 0^(+))lim f(x)=underset(x to 0) h^(m) sin ((1)/(h))=0,"only when m"gt0` Hence, f(x) is a continous at x=0, if `m gt 0` |
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| 355. |
If `f (x) {{:((1 - cos 8 x )/(x^(2)) "," x ge 0 ),(lambda ", " x lt 0 ) :}` is continous at x = 0 than `lambda` = ?A. 32B. 16C. 64D. 8 |
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Answer» Correct Answer - A |
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| 356. |
If `f(x)={{:(,((4^(x)-1)^(3))/(sin(x//4)log(1+x^(2)//3)),x ne 0),(,k,x=0):}` is a continous at x=0, then k=A. `12(log, 4)^(2)`B. `96(log, 2)^(3)`C. `(log, 4)^(3)`D. none of these |
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Answer» Correct Answer - B Since f(x) is a continous at x=0. Therefore, `underset(x to 0)lim f(x)=f(0)` `Rightarrow underset(x to 0)lim ((4^(x)-1)^(3))/("sin" (pi)/(4)log(1+x^(2)/(3)))=k` `Rightarrow underset(x to 0)lim (12((4^(x)-1^(3))/(x)))/((("sin"(pi)/(4))/(pi//4))("log"(1+x^(2)//3)/(x^(2)//3)))=k` `Rightarrow 12(log 4)^(3)=k Rightarrow k=96(log2)^(3)` |
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| 357. |
Determine the value ofthe constant `k`so that the function `f(x)={{:((sin2x)/(5x), if x!=0),(k ,if x=0):}`is continuous at `x=0`. |
| Answer» Correct Answer - ` k = 2/5` | |
| 358. |
If the function `f(x)= {(3ax +b", " x gt 1 ),(11 ", "x=1),(5 ax-2b", " x lt 1):}` continuous at x= 1 then ( a, b) =?A. (3,2)B. (2,3)C. (1,4)D. (4,1) |
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Answer» Correct Answer - A |
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| 359. |
The derivative of `cos^(-1)(2x^(2)-1)` w.r.t. `cos^(-1)` is |
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Answer» Correct Answer - `(-2)/(sqrt(1-x^2))` < < |
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| 360. |
Differentiate `tan^-1sqrt((1-x^2)/(1+x^2))` with respect to `cos^-1""x^2` |
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Answer» Correct Answer - `1/2` |
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| 361. |
`tan ^-1((2x)/(1-x^2))` |
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Answer» Correct Answer - `2/(1+x^2)` |
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| 362. |
Let `g(x)=((x-1)^(n))/(log cos^(m)(x-1)), 0 lt xlt 2,` m and n are integers, `m ne 0, n gt 0`, and let p the left h and derivative of `|x-1| at x=1, "If " lim_(x to 1^(+)) g(x)=p, "then"`A. `n=1, m=1`B. `n=1, m=-1`C. `n=2, m=2`D. `n gt 2, m=n` |
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Answer» Correct Answer - C We have `f(x)=|x-1|={{:(,x-1,x ge 1),(,1-x,x lt 1):}` `therefore ` p=Left hand derivative of f(x) at x=1 `Rightarrow p=underset(x to 1^(-))lim (f(x)-f(1))/(x-1) =underset(x to 1^(-))lim (1-x-0)/(x-1)=-1` Now, `underset(x to 1^(+))lim g(x)=p` `Rightarrow underset(h to 0)lim g(1+h)=-1` `Rightarrow underset(h to 0)lim (h^(n))/(m log cos h)=1` `Rightarrow (1)/(m) underset(h to 0)lim (h^(n-2))/(((tan h)/(h)))=1` `Rightarrow n=2 and "in that case "=(n)/(m)=1` `Rightarrow m=n=2` |
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| 363. |
Determine the value ofthe constant `k`so that the function `f(x)={k x^2 , if xlt=2 3 , if x >2`is continuous. |
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Answer» Since a polynomial funcation is continuous and a constant function is continuous , the given function is continuous for all x lt 2and for all x gt2 So, consider the point x=2 , we have f(2) = 4k . `lim_( xto 2+) f(x) =lim_( h to0) f( 2+h) =lim_( h to0) 3=3` And, ` lim_( x to 2-) f(x) =lim_( h to 0) f( 2-h) =lim_( h to 0) k( 2-h)^(2) = 4k` for continuity , we must have 4k=3 , or ` k = 3/4` |
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| 364. |
`logsqrt((1-cos x)/(1+cos x))` |
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Answer» Correct Answer - cosec x |
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| 365. |
`tan^-1((1+x)/(1-x))` |
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Answer» Correct Answer - `1/(1+x^2)` |
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| 366. |
Differentiate w.r.t. x the function `(sinx-cosx)^((sinx-cosx)),pi/4 < x < (3pi)/4` |
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Answer» `y = (sinx-cosx)^(sinx-cosx)` Taking log both sides, `logy = (sinx-cosx)log(sinx-cosx)` Now, differentiating both sides, `1/ydy/dx = log(sinx-cosx)(cosx+sinx)+ (sinx-cosx)*1/(cosx-sinx)*(cosx+sinx)` `1/ydy/dx = (cosx+sinx)(log(sinx-cosx)+1)` Putting value of `y` `dy/dx = (sinx-cosx)^(sinx-cosx) ((cosx+sinx)(log(sinx-cosx)+1))` |
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| 367. |
`logsqrt((1+x^2)/(1-x^2))` |
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Answer» Correct Answer - `(2x)/( 1-4^x)` |
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| 368. |
Differentiate `logtan(pi/4+x/2)`with respect to `x`: |
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Answer» Correct Answer - sec x |
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| 369. |
`x/(sqrt(1-x^2))` |
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Answer» Correct Answer - `(1- x^2)^(-3//2)` |
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| 370. |
Differentiate `tan^(-1){sqrt((1+cosx)/(1-cosx))}, 0 |
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Answer» Correct Answer - `1/2` |
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| 371. |
Differentiate `sin^(-1)(3x-4x^3)`with respect to `x`, if `1/2 |
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Answer» Correct Answer - `3/sqrt(1-x^2)` |
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| 372. |
Differentiate `cos^(-1)(4x^3-3x)`with respect to `x`, if `x in (1/2, 1)` |
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Answer» Correct Answer - `(-3)/(sqrt(1-x^2))` |
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| 373. |
Find `(dy)/(dx)`, if `y=sin^(-1)x+sin^(-1)sqrt(1-x^2),-1lt=xlt=1`. |
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Answer» Let `x = sintheta`, then `sin^-1x = theta` Then, our given equation becomes, `y = theta+sin^-1(sqrt(1-sin^2theta))` `y = theta+sin^-1(costheta)` `y = theta+sin^-1(sin(pi/2-theta))` `y = theta+pi/2 -theta` `y = pi/2` `:. dy/dx = 0` |
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| 374. |
If `x=acos^3theta,y=a sin^3theta` then find `(d^2y)/(dx^(2))` |
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Answer» `x=acos^3theta` `rArr(dx)/(dtheta)=a. 3cos^(2)theta(-sintheta)=-cos^2thetasintheta` `y=asin^3theta` `rArr(dy)/(dx)=a . Sin^2 theta. Costheta=3a sin^2thetacostheta` `:. (dy)/(dx)=(dy//dtheta)/(dx//dtheta)=(3asin^2thetacostheta)/(-3a cos^2thetasintheta)=-tan theta` and `(d^2y)/(dx^2)=d/(dx)((dy)/(dx))=d/(dx)(-tantheta)` `=-d/(dtheta)(tantheta). (dtheta)/(dx)` `=(-sec^2theta)/((-3a cos^2theta. sintheta))` `=(sec^4theta. cosectheta)/(3a)`. |
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| 375. |
If `e^y(x+1)=1,s howt h a t(d^2y)/(dx^2)=((dy)/(dx))^2dot` |
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Answer» `e^(y)(x+1)=1` `impliese^(y)=(1)/(x+1) " " ` ...(1) Differentiate both sides w.r.t. x `e^(y)(dy)/(dx)=(-1)/((x+1)^(2))` `implies(1)/(x+1)(dy)/(dx)= -(1)/((x+1)^(2)) " " ` From equation (1) `implies (dy)/(dx)=-(1)/(x+1)` `implies(d^(2)y)/(dx^(2))=(d)/(dx)(-(1)/(x+1)) = (1)/((x+1)^(2))=(-(1)/(x+1))^(2)=((dy)/(dx))^(2)` ` " " ` Hence Proved. |
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| 376. |
Differentiate `logsqrt((1+sinx)/(1-sinx))`with respect to `x`: |
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Answer» Correct Answer - sec x |
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| 377. |
Differentiate `e^x` with respect to `sqrt(x))` |
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Answer» Correct Answer - `2sqrt(x.)e^x` |
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| 378. |
If `y=e^acos^((-1)x),-1lt=x |
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Answer» `y=e^(acos^(-1)x)` `implies(dy)/(dx)=e^(acos^(-1)x)*(d)/(dx)(acos^(-1)x)` `implies (dy)/(dx)=y*((-a))/(sqrt(1-x^(2)))` `impliessqrt(1-x^(2))(dy)/(dx)= -ay` `implies(1-x^(2))((dy)/(dx))^(2)=a^(2)y^(2)` Differentiate both sides w.r.t.x `(1-x^(2))*2(dy)/(dx)*(d^(2)y)/(dx^(2))+((dy)/(dx))^(2)(-2x)=a^(2)*2y(dy)/(dx)` Divide both sides by `2*(dy)/(dx)` `(1-x^(2))(d^(2)y)/(dx^(2))-x(dy)/(dx)=a^(2)y` `(1-x^(2))(d^(2)y)/(dx^(2))-x(dy)/(dx)-a^(2)y=0 " " ` Hence Proved. |
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| 379. |
Differentiate `x^8` with respect to `x^4` |
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Answer» Correct Answer - `2x^4` |
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| 380. |
If `x=a(cost+tsint) `and `y=a(sint-tcost), 0 |
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Answer» Correct Answer - tan t |
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