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NCERT Exemplar · Class 12 Mathematics Continuity and Differentiability

106 questions · 106 still being checked

EXERCISE 5.3 71–80 (part 8 of 11)

  1. Exercise 71

    Find the points on the curve y=(cos⁡x−1)\displaystyle y=(\cos x-1) in [0,2π]\displaystyle [0,2 \pi], where the tangent is parallel to x\displaystyle x-axis.

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    NCERT’s answer
    \(\displaystyle (\pi,-2)\)
    Continuous on \(\displaystyle [0,2\pi]\), differentiable on \(\displaystyle (0,2\pi)\), and \[y(0)=\cos 0-1=0,\qquad y(2\pi)=\cos 2\pi-1=0 \;\Rightarrow\; y(0)=y(2\pi) \] By Rolle's theorem, \(\displaystyle \dfrac{dy}{dx}=0\) at some interior point \(\displaystyle c\in(0,2\pi)\) (at \(\displaystyle x=0,\,2\pi\) only a one-sided derivative exists). \[\frac{dy}{dx}=-\sin x \] \[-\sin c=0,\ \ c\in(0,2\pi) \;\Rightarrow\; c=\pi \] \[y(\pi)=\cos\pi-1=-2 \] NCERT_Solution_Class12_Maths_Exemplar_Ch5_Ex5-3_Q71 Answer: \(\displaystyle (\pi,-2)\).

  2. Exercise 72

    Using Rolle's theorem, find the point on the curve y=x(x−4),x∈[0,4]\displaystyle y=x(x-4), x \in[0,4], where the tangent is parallel to x\displaystyle x-axis.

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    NCERT’s answer
    \(\displaystyle (2, -4)\)
    A polynomial: continuous on \(\displaystyle [0,4]\), differentiable on \(\displaystyle (0,4)\). \[f(0)=0,\qquad f(4)=4\cdot 0=0 \;\Rightarrow\; f(0)=f(4) \] \[f'(x)=2x-4 \] \[f'(c)=0 \;\Rightarrow\; c=2\in(0,4) \] \[y(2)=2(2-4)=-4 \] NCERT_Solution_Class12_Maths_Exemplar_Ch5_Ex5-3_Q72 Answer: The point is \(\displaystyle (2,-4)\).
  3. Verify mean value theorem for each of the functions given Exercises $\displaystyle 73$ to 76.

    Exercise 73

    f(x)=14x−1\displaystyle f(x)=\frac{1}{4 x-1} in [1,4]\displaystyle [1,4].

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    \(\displaystyle f\) is continuous on \(\displaystyle [1,4]\) and differentiable on \(\displaystyle (1,4)\), since \(\displaystyle 4x-1\neq0\) there. \[f'(x)=-\frac{4}{(4x-1)^2} \] \[\frac{f(4)-f(1)}{4-1}=\frac{\frac1{15}-\frac13}{3}=-\frac{4}{45} \] \[f'(c)=-\frac{4}{45}\Rightarrow(4c-1)^2=45 \] \[4c-1=3\sqrt5\quad(4c-1>0) \] \[c=\frac{1+3\sqrt5}{4}\approx1.93\in(1,4) \] NCERT_Solution_Class12_Maths_Exemplar_Ch5_Ex5-3_Q73 Answer: Mean value theorem holds, with \(\displaystyle c=\frac{1+3\sqrt5}{4}\).

  4. Exercise 74

    f(x)=x3−2x2−x+3\displaystyle f(x)=x^3-2 x^2-x+3 in [0,1]\displaystyle [0,1].

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    \(\displaystyle f\) is a polynomial, so continuous on \(\displaystyle [0,1]\) and differentiable on \(\displaystyle (0,1)\). \[f(0)=3,\quad f(1)=1-2-1+3=1 \] \[\frac{f(1)-f(0)}{1-0}=-2 \] \[f'(x)=3x^2-4x-1 \] \[f'(c)=-2\Rightarrow3c^2-4c+1=0 \] \[(3c-1)(c-1)=0\Rightarrow c=\tfrac13\ \text{or}\ 1 \] Only \(\displaystyle c=\tfrac13\) lies in \(\displaystyle (0,1)\). NCERT_Solution_Class12_Maths_Exemplar_Ch5_Ex5-3_Q74 Answer: Verified, with \(\displaystyle c=\tfrac13\).

  5. Exercise 75

    f(x)=sin⁡x−sin⁡2x\displaystyle f(x)=\sin x-\sin 2 x in [0,π]\displaystyle [0, \pi].

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    \(\displaystyle f\) is continuous and differentiable everywhere, so on \(\displaystyle [0,\pi]\) and \(\displaystyle (0,\pi)\). \[f(0)=0,\quad f(\pi)=\sin\pi-\sin2\pi=0 \] \[\frac{f(\pi)-f(0)}{\pi-0}=0 \] \[f'(x)=\cos x-2\cos2x \] \[f'(c)=0\Rightarrow\cos c-2(2\cos^2c-1)=0 \] \[4\cos^2c-\cos c-2=0 \] \[\cos c=\frac{1\pm\sqrt{33}}{8} \] Both values lie in \(\displaystyle (-1,1)\) (about \(\displaystyle 0.84\) and \(\displaystyle -0.59\)), so each gives a \(\displaystyle c\) in \(\displaystyle (0,\pi)\). NCERT_Solution_Class12_Maths_Exemplar_Ch5_Ex5-3_Q75 Answer: Verified, with \(\displaystyle c=\cos^{-1}\frac{1+\sqrt{33}}{8}\approx0.57\) and \(\displaystyle c=\cos^{-1}\frac{1-\sqrt{33}}{8}\approx2.21\).

  6. Exercise 76

    f(x)=25−x2\displaystyle f(x)=\sqrt{25-x^2} in [1,5]\displaystyle [1,5].

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    \(\displaystyle f\) is continuous on \(\displaystyle [-5,5]\) and differentiable on \(\displaystyle (-5,5)\), so on \(\displaystyle [1,5]\) and \(\displaystyle (1,5)\). \[f(1)=\sqrt{24}=2\sqrt6,\quad f(5)=0 \] \[\frac{f(5)-f(1)}{5-1}=-\frac{\sqrt6}{2} \] \[f'(x)=-\frac{x}{\sqrt{25-x^2}} \] \[-\frac{c}{\sqrt{25-c^2}}=-\frac{\sqrt6}{2}\Rightarrow4c^2=6(25-c^2) \] \[c^2=15\Rightarrow c=\sqrt{15}\in(1,5) \] NCERT_Solution_Class12_Maths_Exemplar_Ch5_Ex5-3_Q76 Answer: Verified, with \(\displaystyle c=\sqrt{15}\).

  7. Exercise 77

    Find a point on the curve y=(x−3)2\displaystyle y=(x-3)^2, where the tangent is parallel to the chord joining the points (3,0)\displaystyle (3,0) and (4,1)\displaystyle (4, 1).

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    NCERT’s answer
    \(\displaystyle \left(\frac{7}{2}, \frac{1}{4}\right)\)
    Tangent slope must equal the chord slope. \[\text{slope of chord}=\frac{1-0}{4-3}=1 \] \[\frac{dy}{dx}=2(x-3) \] \[2(x-3)=1\Rightarrow x=\tfrac72 \] \[y=\left(\tfrac72-3\right)^2=\tfrac14 \] NCERT_Solution_Class12_Maths_Exemplar_Ch5_Ex5-3_Q77 Answer: The point is \(\displaystyle \left(\tfrac72,\tfrac14\right)\).

  8. Exercise 78

    Using mean value theorem, prove that there is a point on the curve y=2x2−5x+3\displaystyle y=2 x^2-5 x+3 between the points A(1,0)\displaystyle \mathrm{A}(1,0) and B(2,1)\displaystyle \mathrm{B}(2,1), where tangent is parallel to the chord AB. Also, find that point.

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    NCERT’s answer
    \(\displaystyle \left(\frac{3}{2}, 0\right)\)
    The curve is a polynomial, so continuous on \(\displaystyle [1,2]\) and differentiable on \(\displaystyle (1,2)\). \[f(x)=2x^2-5x+3,\quad f(1)=0,\quad f(2)=1 \] \[\frac{f(2)-f(1)}{2-1}=1=\text{slope of AB} \] By the mean value theorem there is \(\displaystyle c\in(1,2)\) with \(\displaystyle f'(c)=1\): the tangent there is parallel to AB. \[f'(x)=4x-5 \] \[4c-5=1\Rightarrow c=\tfrac32\in(1,2) \] \[f\left(\tfrac32\right)=2\cdot\tfrac94-\tfrac{15}{2}+3=0 \] NCERT_Solution_Class12_Maths_Exemplar_Ch5_Ex5-3_Q78 Answer: The point is \(\displaystyle \left(\tfrac32,0\right)\).

  9. Exercise 79

    Find the values of p\displaystyle p and q\displaystyle q so that f(x)={x2+3x+p, if x≤1qx+2, if x>1f(x)= \begin{cases}x^2+3 x+p & , \text { if } x \leq 1 \\ q x+2 & , \text { if } x>1\end{cases} is differentiable at x=1\displaystyle x=1.

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    NCERT’s answer
    \(\displaystyle p=3, q=5\)
    Differentiable at \(\displaystyle x=1\) requires continuity there. \[\lim_{x\to1^-}f(x)=1+3+p,\quad \lim_{x\to1^+}f(x)=q+2 \] \[4+p=q+2\Rightarrow q=p+2 \] The one-sided derivatives must also agree. \[f'(1^-)=\left.(2x+3)\right|_{x=1}=5 \] \[f'(1^+)=q \] \[q=5\Rightarrow p=3 \] Answer: \(\displaystyle p=3,\ q=5\).

  10. Exercise 80

    If xm.yn=(x+y)m+n\displaystyle x^m . y^n=(x+y)^{m+n}, prove that
    (i)
    dydx=yx\displaystyle \frac{d y}{d x}=\frac{y}{x} and
    (ii)
    d2ydx2=0\displaystyle \frac{d^2 y}{d x^2}=0.

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    Take logarithms and differentiate with respect to \(\displaystyle x\). \[m\ln x+n\ln y=(m+n)\ln(x+y) \] \[\frac mx+\frac ny\frac{dy}{dx}=\frac{m+n}{x+y}\left(1+\frac{dy}{dx}\right) \] \[\frac{dy}{dx}\left[\frac ny-\frac{m+n}{x+y}\right]=\frac{m+n}{x+y}-\frac mx \] \[\frac{dy}{dx}\cdot\frac{nx-my}{y(x+y)}=\frac{nx-my}{x(x+y)} \] Cancel \(\displaystyle nx-my\); it is not zero, since \(\displaystyle nx=my\) would force \(\displaystyle m^mn^n=(m+n)^{m+n}\), impossible for \(\displaystyle m,n>0\). \[\text{(i)}\quad\frac{dy}{dx}=\frac yx \] \[\text{(ii)}\quad\frac{d^2y}{dx^2}=\frac{x\dfrac{dy}{dx}-y}{x^2}=\frac{y-y}{x^2}=0 \] Answer: \(\displaystyle \frac{dy}{dx}=\frac yx\) and \(\displaystyle \frac{d^2y}{dx^2}=0\).