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Chain Rule is one of the most important concepts in Differentiations That's why we will discuss it in upcoming concepts till then you can take some basic idea of this concept Thanking You !!
Case 1 :                  Case 2 :                =>    correct option (2)     Option 1) Option 2) Option 3) Option 4)
has exterme points at  So,  at    we can write   given that                                                Option 1) four irrational numbers. Option 2) four rational numbers. Option 3) two irrational and two rational numbers. Option 4) two irrational and one rational number.
Find shortest distance   the tangent at point P is parallel to So putiing the value of in curve                                              Shortest distance between two parallel lines or perpendicular distance from       Option 1) Option 2) Option 3)   Option 4)
Indeterminate forms - The form   are known as indeterminate form means they do not exist directly -     L - Hospital Rule - - wherein Given  and    This is in the form of L.D.E I.F. =        =  Now, So,  Option 1)exists and equals 0Option 2)does not existOption 3)exist and equals Option 4)exists and equals 4
@AMANDEEP KAUR SEKHON  a1, a2, a3, …. is an A.P. and b1, b2, b3, …… is a G.P. Then the sequence  is said to be an arithmetic-geometric progression. The sum of infinite term is
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use the concept    Arithmetic mean of two numbers (AM) - - wherein It is to be noted that the sequence a, A, b, is in AP where, a and b are the two numbers.       Geometric mean of two numbers (GM) - - wherein It is to be noted that a,G,b are in GP and a,b are two non - zero numbers.     Harmonic mean (HM) of two numbers a and b - -     multiply and divide by...
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It is not a function of only x- variable.    if we want to find differentiation of this function,we need partial differentiation w.r.t (x).      Hence, Final answer will be .
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As we have learned Lagrange's mean value theorem - If  a  function  f(x)  1.   is continuous in the closed interval [a,b] and  2.   is differentiable in the open interval [a, b] then  -   (A),(C), (D) are discontinous in [1,2] but (B) satisfies all condition of L.M.V.T           Option 1) Option 2) Option 3) Option 4)
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As we have learned Length of Tangent - - wherein Where      Length of tangent at (x,y)   Length =            Option 1) 1 Option 2) 2 Option 3) 3 Option 4) 4
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As we have learned Properties of differentiable functions - The composition of a differentiable function is a differentiable functions. -      is composition of sin x with sin x so it will  be diffrentiable      is compostion of  with  (both diffrentiable ) so it will be also diffrentiable    so sum of two diffrentiable function will also be diffrentiable for all            Option 1) 0 Option...
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As we have learned Irremovable discontinuity - A function f is said to possess irremovable discontinuity if at  x = a the left hand limit is not equal to the right hand limit so limit does not exist   -     Limit doesn't exist  , f(x) has irremovable discontinuty , so ffor no 'k' it will be continous     Option 1) If k= 1 , f(x ) becomes continous at x= 0 Option 2) If k= -1 , f(x ) becomes...
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As we have learned Removal discontinuity - A function  f is said to possess removable discontinuity if at x = a :     - wherein       f(0)=1 Limit exists but  not equal to f(0)  removable discontinuity at x= 0       Option 1) f(x ) is continous at x= 0 Option 2) f(x) has non - exsiting limit at x= 0   Option 3) f(x) has LHL=RHL = f(0) Option 4) f(x) has removable discontinuty at x=0
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As we have learned

Sum of n terms of a GP -

$S_{n}= \left\{\begin{matrix} a\frac{\left ( r^{n}-1 \right )}{r-1}, &if \: r\neq 1 \\ n\, a, & if \, r= 1 \end{matrix}\right.$

- wherein

$a\rightarrow$ first term

$r\rightarrow$ common ratio

$n\rightarrow$ number of terms

$S_{20}= 1 +\left ( \frac{4-1}{2} \right )+\left ( \frac{8-1}{4} \right )+\left ( \frac{16-1}{8} \right )+......$

$=1+2+2.......20times....\left ( 1/2+1/4+1/8+......20times \right )$

$39-1/2(\left ( \frac{1-(1/2)^{19}}{1-1/2} \right ))= 39-1+1/(2)^{19}$

$38+\frac{1}{2^{19}}$

Option 1)

$38+\frac{1}{2^{19}}$

This is correct

Option 2)

$38+\frac{1}{2^{20}}$

This is incorrect

Option 3)

$39+\frac{1}{2^{20}}$

This is incorrect

Option 4)

$39+\frac{1}{2^{19}}$

This is incorrect

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As we learnt in  Differentiation - Derivative  of a function  f(x) is defined as  f'(x) means  small increment    in x  corresponding increment in the value of y  be   - wherein               Now, Now, Put x=-1 From (i) and (ii) it is zero Option 1) This option is incorrect Option 2) 0 This option is correct Option 3) 1 This option is incorrect Option 4) This option is incorrect
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As we learnt in Condition for differentiable - A function  f(x) is said to be differentiable at    if      both exist and are equal otherwise non differentiable -      and      So f(x) is continuous at x = 0   So, g(x) is differentiable at        Option 1)  Both statements I and II are false.     This option is incorrect. Option 2)  Both statements I and II are true. This option is...
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As we learnt in  General term of a GP -   - wherein first term common ratio    Let first term is a andcommon ratio is r then Also   put in Option 1) 7290 Incorrect Option Option 2) 320 correct Option Option 3) 640 Incorrect Option Option 4) 2430 Incorrect Option
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As we learnt in Sum of n terms of an AP - and Sum of n terms of an AP - wherein first term common difference number of terms   Arithmetic mean of n numbers -   - wherein are the n numbers    Let the AP is given   Now                    Series is   upto terms mean Option 1)  26.5 Incorrect option Option 2)  28 Incorrect option Option 3) 29.5 Correct option Option...
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As we learnt in

Common ratio of a GP (r) -

The ratio of two consecutive terms of a GP

- wherein

eg: in 2, 4, 8, 16, - - - - - - -

r = 2

and in 100, 10, 1, 1/10 - - - - - - -

r = 1/10

${A}, {B},{C}$ are in  G.P where ${A}, {B},{C}$  are Terms of an A.P

Let first term is a and common difference is d, and common ratio be r then

A = a+d

B = a+4d

C= a+8d

$\therefore \: \frac{B}{A}=\frac{C}{B}=\frac{r}{1}$

$\frac{a+4d}{a+d} =\frac{a+8d}{a+4d}=\frac{r}{1}$

$\therefore \frac{a+4d+a+d}{a+4d-a-d}=\frac{r+1}{r-1}$

$\Rightarrow \frac{2a+5d}{3d}=\frac{r+1}{r-1}-----(i)$

$\frac{a+8d+a+4d}{a+8d-a-4d} =\frac{r+1}{r-1}$

$\Rightarrow \frac{2a+12d}{4d}=\frac{r+1}{r-1}------(ii)$

from (i) and (ii)

$\frac{2a+5d}{2a+12d}=\frac{3}{4}$

$8a+20d=6a+36d$

2a=16d

a=8d

$\Rightarrow \frac{r+1}{r-1}=\frac{2\times8d+5d}{3d}$$=\frac{16d+5d}{3d}$$=\frac{21}{3}=7$

$\therefore r+1=7r-7$

$8=6r$

$r=\frac{4}{3}$

Option 1)

Incorrect option

Option 2)

Correct option

Option 3)

Incorrect option

Option 4)

Incorrect option

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