Limits and Derivatives - Test Papers

 CBSE Test Paper 01

CH-13 Limits and Derivatives

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1. If G (x) = 25−x2−−−−−−√$\sqrt{25-x^2}$ then Ltx→1G(x)−G(1)x−1$\underset{x\to 1}{Lt}\frac{G(x)-G(1)}{x-1}$ has the value
   1. 124$\frac{1}{24}$
   2. −24−−√$-\sqrt{24}$
   3. −124√$\frac{-1}{\sqrt{24}}$
   4. 15$\frac{1}{5}$
2. Ltx→π3secx−2x−π3$\underset{x\to \frac{\pi }{3}}{Lt}\frac{\sec x-2}{x-\frac{\pi }{3}}$ is equal to
   1. 2
   2.  2+3–√$2+\sqrt{3}$
   3. 3–√$\sqrt{3}$
   4. 23–√$2\sqrt{3}$
3. The function, f(x)=(x−a)2cos1x−aforx≠a$f(x)=(x-a{)}^2\cos \frac{1}{x-a}forx\ne a$ and f (a) = 0, is
   1. continuous but not derivable at x = 0
   2. derivable at x = a
   3. not continuous at x = a
   4. neither continuous nor derivable at x=a 
4. Ltx→43−5+x√1−5−x√=$\underset{x\to 4}{Lt}\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}=$
   1. does not exist
   2. 0
   3. −13$-\frac{1}{3}$
   4. 13$\frac{1}{3}$
5. Ltx→0sinxn(sinx)m,n>m>0,$\underset{x\to 0}{Lt}\frac{\sin x^n}{{\left(\sin x\right)}^m},n>m>0,$ is equal to
   1. mn$\frac{m}{n}$
   2. 0
   3. 1
   4. nm$\frac{n}{m}$
6. Fill in the blanks:

   The value of given limit limx→0cosxπ−x$\underset{x\to 0}{lim}\frac{\cos x}{\pi -x}$ is ________.

7. Fill in the blanks:

   The value of limit limr→1πr2$\underset{r\to 1}{lim}\pi r^2$is ________.

8. font-family: Verdana font-size: 8px Evaluate limx→2x3−6x2+11x−6x2−6x+8$\underset{x\to 2}{lim}\frac{x^3-6x^2+11x-6}{x^2-6x+8}$

9. Find the derivative of x at x = 1

10. Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): ax4−bx2+cosx$\frac{a}{x^4}-\frac{b}{x^2}+\cos x$

11. Find the value of limx→0e3x−1x$\underset{x\to 0}{lim}\frac{e^{3x}-1}{x}$.

12. Find the derivative of 2x−34$2x-\frac{3}{4}$

13. Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): ax+bpx2+qx+r$\frac{ax+b}{p{x}^2+qx+r}$

14. Suppose f(x)=⎧⎩⎨⎪⎪a+bx,4,b−ax,x<1x=1x>1$f(x)=\{\begin{array}{cc}a+bx,& x<1\\ 4,& x=1\\ b-ax,& x>1\end{array}$ and if limx→1$\underset{x\to 1}{lim}$ f(x) = f(1), then what are the possible values of a and b?

15. The differentiation of sec x with respect to x is sec x tan x.

CBSE Test Paper 01
CH-13 Limits and Derivatives

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Solution

1. (c) −124√$\frac{-1}{\sqrt{24}}$

   Explanation: The equation is in the form of 0/0

   Using L'Hospital rule we have 1225−x2√⋅(−2x)1$\frac{\frac{1}{2\sqrt{25-x^2}}\cdot (-2x)}{1}$
   substituting x = 1 we get −124√$\frac{-1}{\sqrt{24}}$
2. (d) 23–√$2\sqrt{3}$

   Explanation: Using L'Hospital;

   Ltx→π3secxtanx1$\underset{x\to \frac{\pi }{3}}{Lt}\frac{\sec x\tan x}{1}$
   ⇒23–√$\Rightarrow 2\sqrt{3}$
3. (b) derivable at x = a
   Explanation: situation x - a = t; then the function will become
   ⇒Ltt→0t2cos1t$\Rightarrow \underset{t\to 0}{Lt}{t}^2\cos \frac{1}{t}$
   ⇒$\Rightarrow$ 0. Finite number = 0
   f(a) = 0
4. (c) −13$-\frac{1}{3}$

   Explanation: Equation is in the form of 0/0

   Using L'Hospital rule we get −125+x√125−x√$\frac{-\frac{1}{2\sqrt{5+x}}}{\frac{1}{2\sqrt{5-x}}}$
   Substituting x = 4 we get −13$\frac{-1}{3}$
5. (b) 0
   Explanation: litx→0sinxn(sinx)m⋅xm+nxm+n${\mathrm{lit}}_{x\to 0}\frac{\sin x^n}{(\sin x{)}^m}\cdot \frac{x^{m+n}}{x^{m+n}}$

   ⇒Ltx→0sinxnxn⋅xm(sinx)m⋅xnxm$\Rightarrow \underset{x\to 0}{Lt}\frac{\sin x^n}{x^n}\cdot \frac{x^m}{(\sin x{)}^m}\cdot \frac{x^n}{x^m}$
   ⇒1.1m.xn−m$\Rightarrow {1.1}^m.x^{n-m}$
   ⇒$\Rightarrow$ 1.0 = 0

6. 1π$\frac{1}{\pi }$

7. π$\pi$

8. When x = 2, the expression x3−6x2+11x−6x2−6x+8$\frac{x^3-6x^2+11x-6}{x^2-6x+8}$assume the form 00$\frac{0}{0}$. Therefore, (x-2) is factor common to numerator and denominator. Factorising the numerator and denominator, we have
   font - family : Verdana font - size : 8px ∴limx→2x3−6x2+11x−6x2−6x+8$\therefore \underset{x\to 2}{lim}\frac{x^3-6x^2+11x-6}{x^2-6x+8}$
   
   =limx→2(x−1)(x−2)(x−3)(x−2)(x−4)$=\underset{x\to 2}{lim}\frac{(x-1)(x-2)(x-3)}{(x-2)(x-4)}$
   
   =limx→2(x−1)(x−3)(x−4)=(2−1)(2−3)(2−4)=12$=\underset{x\to 2}{lim}\frac{(x-1)(x-3)}{(x-4)}=\frac{(2-1)(2-3)}{(2-4)}=\frac{1}{2}$

9. Here ddx$\frac{d}{dx}$(x) = 1
   ∴$\therefore$ Derivative of x at x = 1

10. Here f(x)=ax4−bx2+cosx=ax−4−bx−2+cosx$f(x)=\frac{a}{x^4}-\frac{b}{x^2}+\cos x=a{x}^{-}4-b{x}^{-2}+\cos x$
    ∴f'(x)=ddx[ax−4−bx2+cosx]$\therefore f\text{'}(x)=\frac{d}{dx}[a{x}^{-4}-b{x}^2+\cos x]$=addx(x−4)−bddx(x−2)+ddx(cosx)$=a\frac{d}{dx}(x^{-4})-b\frac{d}{dx}(x^{-2})+\frac{d}{dx}(\cos x)$
    −ax−5+2bx−3−sinx=−4ax5+2bx3−sinx$-a{x}^{-5}+2b{x}^{-3}-\sin x=\frac{-4a}{x^5}+\frac{2b}{x^3}-\sin x$
    −4ax−5+2bx−3−sinx=−4ax5+2bx3−sinx$-4a{x}^{-5}+2b{x}^{-3}-\sin x=\frac{-4a}{x^5}+\frac{2b}{x^3}-\sin x$

11. We have, limx→0e3x−1x=limx→0e3x−1x×33$\underset{x\to 0}{lim}\frac{e^{3x}-1}{x}=\underset{x\to 0}{lim}\frac{e^{3x}-1}{x}\times \frac{3}{3}$ [multiplying numerator and denominator by 3]
    =3limx→0e3x−13x$=3\underset{x\to 0}{lim}\frac{e^{3x}-1}{3x}$..... (i)
    Let h = 3x, as x →$\to$ 0, then h →$\to$ 0
    Then, from Eq. (i), we get
    limx→0e3x−1x=3limh→0eh−1h=3(1)$\underset{x\to 0}{lim}\frac{e^{3x}-1}{x}=3\underset{h\to 0}{lim}\frac{e^h-1}{h}=3(1)$[∵limx→0ex−1x=1]$\left[\because \underset{x\to 0}{lim}\frac{e^x-1}{x}=1\right]$
    = 3

12. Here f(x)=2x−34$f(x)=2x-\frac{3}{4}$
    ∴f'(x)=ddx(2x−34)$\therefore f\text{'}(x)=\frac{d}{dx}\left(2x-\frac{3}{4}\right)$
    =2ddx(x)−ddx(34)$=2\frac{d}{dx}(x)-\frac{d}{dx}\left(\frac{3}{4}\right)$
    = 2 × 1 - 0 = 2

13. f(x)=ax+bpx2+qx+r$f(x)=\frac{ax+b}{p{x}^2+qx+r}$
    ∴f'(x)=ddx[ax+bpx2+qx+r]$\therefore f\text{'}(x)=\frac{d}{dx}\left[\frac{ax+b}{p{x}^2+qx+r}\right]$
    =(px2+qx+r)ddx(ax+b)−(ax+b)ddx(px2+qx+r)(px2+qx+r)2$=\frac{(p{x}^2+qx+r)\frac{d}{dx}(ax+b)-(ax+b)\frac{d}{dx}(p{x}^2+qx+r)}{{(p{x}^2+qx+r)}^2}$
    =(px2+qx+r)(a)−(ax+b)(2px+q)(px2+qx+r)2$=\frac{(p{x}^2+qx+r)(a)-(ax+b)(2px+q)}{{(p{x}^2+qx+r)}^2}$
    =apx2+aqx+ar−2apx2−aqx−2bpx−bq(px2+qx+r)2$=\frac{ap{x}^2+aqx+ar-2ap{x}^2-aqx-2bpx-bq}{{(p{x}^2+qx+r)}^2}$
    =−apx2−2bpx+ar−bq(px2+qx+r)2$=\frac{-ap{x}^2-2bpx+ar-bq}{{(p{x}^2+qx+r)}^2}$

14. We  have,
    f(x)=⎧⎩⎨⎪⎪a+bx,4,b−ax,x>1x<1x=1$f(x)=\{\begin{array}{cc}a+bx,& x<1\\ 4,& x=1\\ b-ax,x>1& \end{array}$
    Now, LHL =limx→1−f(x)$=\underset{x\to 1^{-}}{lim}f(x)$
    =limx→1−(a+bx)=limh→0[a+b(1−h)]$=\underset{x\to 1^{-}}{lim}(a+bx)=\underset{h\to 0}{lim}[a+b(1-h)]$[putting x = 1 - h as x →$\to$ 1, then h →$\to$ 0]
    = a + b
    RHL =limx→1+f(x)$=\underset{x\to 1^{+}}{lim}f(x)$
    =limx→1+(b−ax)=limh→0[b−a(1+h)]$=\underset{x\to 1^{+}}{lim}(b-ax)=\underset{h\to 0}{lim}[b-a(1+h)]$[putting x = 1 + h as x →$\to$ 1, then h →$\to$ 0]
    = b - a
    Since, limx→1$\underset{x\to 1}{lim}$ f(x) = f(1)
    ∴$\therefore$ LHL = RHL = f(1)
    ⇒$\Rightarrow$ a + b = b - a = 4 [∵$\because$ f(1) = 4, given]
    ⇒$\Rightarrow$ a + b = 4 ..(i) and b - a = 4 ..(ii)
    On solving (i) and (ii), we get
    a = 0, b = 4

15. Let f(x) = sec x. Then, f(x + h) = sec (x + h)
    ∴$\therefore$ ddx$\frac{d}{dx}$ (f (x)) = limh→0f(x+h)−f(x)h$\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}$
    ⇒$\Rightarrow$ ddx$\frac{d}{dx}$ (f (x)) = limh→0sec(x+h)−secxh$\underset{h\to 0}{lim}\frac{\sec (x+h)-\sec x}{h}$
    ⇒$\Rightarrow$ ddx$\frac{d}{dx}$ (f (x)) = limh→01cos(x+h)−1cosxh$\underset{h\to 0}{lim}\frac{\frac{1}{\cos (x+h)}-\frac{1}{\cos x}}{h}$
    ⇒$\Rightarrow$ ddx$\frac{d}{dx}$ (f (x)) = limh→0cosx−cos(x+h)hcosxcos(x+h)$\underset{h\to 0}{lim}\frac{\cos x-\cos (x+h)}{h\cos x\cos (x+h)}$
    ⇒$\Rightarrow$ ddx$\frac{d}{dx}$ (f (x)) = limh→02sin(x+x+h2)sin(x+h−x2)hcosxcos(x+h)$\underset{h\to 0}{lim}\frac{2\sin \left(\frac{x+x+h}{2}\right)\sin \left(\frac{x+h-x}{2}\right)}{h\cos x\cos (x+h)}$ [∵cosC−cosD=2sin(C+D2)sin(D−C2)]$\left[\because \cos C-\cos D=2\sin \left(\frac{C+D}{2}\right)\sin \left(\frac{D-C}{2}\right)\right]$ 
    ⇒$\Rightarrow$ ddx$\frac{d}{dx}$ (f (x)) = limh→02sin(2x+h2)sin(h2)hcosxcos(x+h)$\underset{h\to 0}{lim}\frac{2\sin \left(\frac{2x+h}{2}\right)\sin \left(\frac{h}{2}\right)}{h\cos x\cos (x+h)}$
    ⇒$\Rightarrow$ ddx$\frac{d}{dx}$ (f (x)) = limh→0sin(2x+h2)cosxcos(x+h)×limh→0sin(h/2)(h/2)$\underset{h\to 0}{lim}\frac{\sin \left(\frac{2x+h}{2}\right)}{\cos x\cos (x+h)}\times \underset{h\to 0}{lim}\frac{\sin (h/2)}{(h/2)}$
    ⇒$\Rightarrow$ ddx$\frac{d}{dx}$ (f (x)) = sinxcosxcosx×$\frac{\sin x}{\cos x\cos x}\times$ 1 = tan x sec x. [∵limh→0sin(h/2)(h/2)=1]$\left[\because \underset{h\to 0}{lim}\frac{\sin (h/2)}{(h/2)}=1\right]$
    Hence, ddx$\frac{d}{dx}$ (sec x) = sec x tan x.