Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Differentiation question

2010 · Shift 0 · Q40
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Differentiation
  5. /2010 · Shift 0 · Q40

Differentiation question

2010 · Shift 0 · Q40

JEE MainMathematicsDifferentiationMCQ+4 / −1
Let f:(−1,1)→Rf:\left( { - 1,1} \right) \to Rf:(−1,1)→R be a differentiable function with f(0)=−1f\left( 0 \right) = - 1f(0)=−1 and f′(0)=1f'\left( 0 \right) = 1f′(0)=1. Let g(x)=[f(2f(x)+2)]2g\left( x \right) = {\left[ {f\left( {2f\left( x \right) + 2} \right)} \right]^2}g(x)=[f(2f(x)+2)]2. Then g′(0)=g'\left( 0 \right) =g′(0)=
  1. A
    −4-4−4
  2. B
    000
  3. C
    −2-2−2
  4. D
    444
View written solutionFree

Correct answer: A

  1. We are given: f(0)=−1, f′(0)=1f(0)=-1,\, f'(0)=1f(0)=−1,f′(0)=1 and g(x)=[f(2f(x)+2)]2.g(x)=\left[f\big(2f(x)+2\big)\right]^2.g(x)=[f(2f(x)+2)]2.

  2. Let h(x)=f(2f(x)+2).h(x)=f\big(2f(x)+2\big).h(x)=f(2f(x)+2). Then g(x)=[h(x)]2.g(x)=[h(x)]^2.g(x)=[h(x)]2.

  3. Differentiate g(x)g(x)g(x) using the chain rule: g′(x)=2h(x)h′(x).g'(x)=2h(x)h'(x).g′(x)=2h(x)h′(x). So first compute h(0)h(0)h(0) and h′(0)h'(0)h′(0).

  4. Compute h(0)h(0)h(0): h(0)=f(2f(0)+2)=f(2(−1)+2)=f(0)=−1.h(0)=f\big(2f(0)+2\big)=f\big(2(-1)+2\big)=f(0)=-1.h(0)=f(2f(0)+2)=f(2(−1)+2)=f(0)=−1.

  5. Now differentiate h(x)=f(2f(x)+2)h(x)=f(2f(x)+2)h(x)=f(2f(x)+2): h′(x)=f′(2f(x)+2)⋅ddx(2f(x)+2)h'(x)=f'\big(2f(x)+2\big)\cdot \frac{d}{dx}(2f(x)+2)h′(x)=f′(2f(x)+2)⋅dxd​(2f(x)+2) =f′(2f(x)+2)⋅2f′(x).=f'\big(2f(x)+2\big)\cdot 2f'(x).=f′(2f(x)+2)⋅2f′(x).

  6. Evaluate at x=0x=0x=0: h′(0)=f′(2f(0)+2)⋅2f′(0)h'(0)=f'\big(2f(0)+2\big)\cdot 2f'(0)h′(0)=f′(2f(0)+2)⋅2f′(0) =f′(0)⋅2⋅1=f'(0)\cdot 2\cdot 1=f′(0)⋅2⋅1 =1⋅2=2.=1\cdot 2=2.=1⋅2=2.

  7. Therefore, g′(0)=2h(0)h′(0)=2(−1)(2)=−4.g'(0)=2h(0)h'(0)=2(-1)(2)=-4.g′(0)=2h(0)h′(0)=2(−1)(2)=−4.

  8. Hence the correct option is: −4\boxed{-4}−4​ i.e. Option A.

PreviousNext

More from Differentiation

  • Let y be an implicit function of x defined by x2x−2xxcoty−1=0. Then y′(1) equals2009 · MCQ
  • If xm.yn=(x+y)m+n, then dxdy​ is2006 · MCQ
  • If x=ey+ey+ey+.....∞, x>0, then dxdy​ is2004 · MCQ
  • If f(x)=xn, then the value of f(1)−1!f′(1)​+2!f′′(1)​−3!f′′′(1)​+..........n!(−1)nfn(1)​…2003 · MCQ
  • Let f(x) be a polynomial function of second degree. If f(1)=f(−1) and a,b,c are in A.P, then f′(a),f′(b),f′(c) are in2003 · MCQ
  • If y=(x+1+x2​)n, then (1+x2)dx2d2y​+xdxdy​ is2002 · MCQ
  • Let f:R→R be a twice differentiable function such that (sinxcosy)(f(2x+2y)−f(2x−2y))=(cosxsiny)(f(2x+2y)+f(2x−2y)), for all x,y∈R. If f′(0)=21​, then…2025 · MCQ
  • Let f:R→R be a thrice differentiable odd function satisfying f′(x)≥0,f′(x)=f(x),f(0)=0,f′(0)=3. Then 9f(loge​3) is equal to ​ .2025 · Numerical