RD Sharma

### Find in each of the following:

Given ${{\mathbf{y}}^{\mathbf{3}}}-\text{ }\mathbf{3x}{{\mathbf{y}}^{\mathbf{2}}}~=\text{ }{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{y}$ Now we have to find...

Given sin-1 (2a/ 1+ a2) + sin-1 (2b/ 1+ b2) = 2 tan-1 x

### Prove that: (i) (ii)

$(i)$   $(ii)$

### Prove the following results: (ix) (x)

$(ix)$ $(x)$

### Prove the following results: (vii) (viii)

$(vii)$ $(viii)$

### Prove the following results: (v) (vi)

$(v)$ $(vi)$

### Prove the following results: (iii) (iv)

$(iii)$ $(iv)$

### Prove the following results: (i) (ii)

$(i)$ $(ii)$

### Evaluate the following: (iii) (iv)

$(iii)$ $(iv)$

### Evaluate the following: (i) (ii)

$(i)$ $(ii)$

Since, cos-1 (a/x) – cos-1 (b/x) = cos-1 (1/b) – cos-1 (1/a)

Given Cos (sin -1 3/5 + sin-1 5/13)

### Prove the following results: (iii)

$(iii)$

### Prove the following results: (i) (ii)

$(i)$ $(ii)$ LHS

### If , find

Since, $\cos ^{-1} x+\sin ^{-1} x=\pi / 2$ => $\cos ^{-1} x=\pi / 2-\sin ^{-1} x$ Substituting this in $\left(\sin ^{-1} x\right)^{2}+\left(\cos ^{-1} x\right)^{2}=17 \pi^{2} / 36$ $\left(\sin... read more$\cot \left(\cos ^{-1} 3 / 5+\sin ^{-1} x\right)=0$=>$\begin{array}{l} \left(\cos ^{-1} 3 / 5+\sin ^{-1} x\right)=\cot ^{-1}(0) \\ \left(\operatorname{Cos}^{-1} 3 / 5+\sin ^{-1} x\right)=\pi /...

Given sin-1 x + sin-1 y = π/3 ……. (i) And cos-1 x – cos-1 y = π/6 ……… (ii)

Since, cos-1 x + cos-1 y = π/4

### Evaluate: (v)

$(v)$ $=>0$

### Evaluate: (iii) for (iv) Cot

$(iii)$ $(iv)$

### Evaluate: (i) (ii) for

$(i)$ $(ii)$

### Evaluate: (iii) cot

$(iii)$

### Evaluate: (i) (ii) Sec

$(i)$ $(ii)$

### Evaluate each of the following: (ix)

$(ix)$ .

### Evaluate each of the following: (vii) Tan (viii)

$(vii)$ $(viii)$

### Evaluate each of the following: (v) (vi)

(v) \[\begin{array}{*{35}{l}} {}  \\ Let\text{ }co{{s}^{-1}}\left( 8/17 \right)\text{ }=\text{ }y  \\ cos\text{ }y\text{ }=\text{ }8/17\text{ }where\text{ }y\text{ }\in \text{ }\left[ 0,\text{ }\pi...

(iii) (iv)

### Evaluate each of the following: (i) (ii)

(i) \[\begin{array}{*{35}{l}} Given\text{ }sin\text{ }\left( si{{n}^{-1}}~7/25 \right)  \\ let\text{ }y\text{ }=\text{ }si{{n}^{-1}}~7/25  \\ sin\text{ }y\text{ }=\text{ }7/25\text{ }where\text{...

### Find the derivative of the function f defined by f (x) = mx + c at x = 0.

f(x) = mx + c, Checking the differentiability  at x = 0 This is the derivative of a function at x = 0, and also this is the derivative of this function at every value of x.

### If f (x) =, find f’ (4).

f(x) = x3 + 7x2 + 8x – 9, => Checking the differentiability  at x = 4

### If for the function Ø (x) =, Ø’ (5) = 97, find λ.

Finding  the value of λ given in the real function and we are given with the differentiability of the function f(x) = λx2 + 7x – 4 at x = 5 which is f ‘(5) = 97 =>

### Show that the derivative of the function f is given by f (x) =     , at x = 1 and x = 2 are equal.

We are given with a polynomial function f(x) = 2x3 – 9x2 + 12x + 9, and we have

### Discuss the continuity and differentiability of the function f (x) = |x| + |x -1| in the interval of (-1, 2).

Since, a polynomial and a constant function is continuous and differentiable everywhere => f(x) is continuous and differentiable for x ∈ (-1, 0) and x ∈ (0, 1) and (1, 2). Checking  continuity...

### Show that the function is defined as follows Is continuous at , but not differentiable thereat.

Since, LHL = RHL = f (2) Hence, F(x) is continuous at x = 2 Checking the differentiability at x = 2 $=> 5$ Since, (RHD at x = 2) ≠ (LHD at x = 2) Hence, f (2) is not differentiable at x =...

checking differentiability of given function at x = 3 =>  LHD (at x = 3) = RHD (at x = 3) = 12 Since, (LHD at x = 3) = (RHD at x = 3) Hence, f(x) is differentiable at x = 3.

### Show that f (x) = is not differentiable at x = 0.

Since, LHD and RHD does not exist at x = 0 Hence, f(x) is not differentiable at x = 0

(i) Let (ii) Let

(iii) Let

(i) Let (ii)

### Differentiate with respect to , if (i) (ii)

(i) Given sin-1 √ (1-x2) (ii) Given sin-1 √ (1-x2)

### Find dy/dx, when x = 2 t / 1+t^2 and y = 1-t^2 / 1+t^2.

Given, $x=2 t /\left(1+t^{2}\right)$ On differentiating $x$ with respect to t using quotient rule,  \begin{array}{l} \frac{\mathrm{dx}}{\mathrm{dt}}=\left[\frac{\left(1+\mathrm{t}^{2}\right)...

(v)   (vi)

(iv)

### Differentiate the following functions with respect to x:

Let y = (log x)cos x Taking log both the sides, we get

### Find dy/dx in each of the following:

differentiating the equation on both sides with respect to x, we get,

### Find dy/dx in each of the following:

differentiating the equation on both sides with respect to x, we get,

### Find dy/dx in each of the following:

differentiating the equation on both sides with respect to x, we get,

### Find dy/dx in each of the following:

differentiating the equation on both sides with respect to x, we get,

### Find dy/dx in each of the following:

differentiating the equation on both sides with respect to x, we get,

### Find dy/dx in each of the following:

differentiating the equation on both sides with respect to x, we get,

### Find dy/dx in each of the following:

differentiating the equation on both sides with respect to x, we get,

### Find dy/dx in each of the following:

differentiating the equation on both sides with respect to x, we get,

Let,

Let,

### Differentiate the following functions with respect to x: .

Let y = (log sin x)2

### Differentiate the following functions with respect to x: tan 5x

Let y = tan (5x°)

### Differentiate the following functions with respect to x:

Let y = sin2 (2x + 1) On differentiating y with respect to x, we get

Given tan2 x

### Differentiate the following functions from the first principles:

let f (x) = ecos x By using the first principle formula, we get,

### Differentiate the following functions from the first principles:

A real function f is said to be continuous at x = c, where c is any point in the domain of f if A function is continuous at x = c if Function is changing its nature (or expression) at x = 2, so we...

### If for , find the value which can be assigned to at so that the function becomes continuous everywhere in .

A real function f is said to be continuous at x = c, where c is any point in the domain of f if Where h is a very small positive number. i.e. left hand limit as x → c (LHL) = right hand limit as...

### The function is defined by If f is continuous on [0, 8], find the values of a and b.

A real function f is said to be continuous at x = c, where c is any point in the domain of f if h is a very small positive number. i.e. left hand limit as x → c (LHL) = right hand limit as x → c...

### The function Is continuous on . Find the most suitable values of and .

A real function f is said to be continuous at x = c, where c is any point in the domain of f if h is a very small positive number. i.e. left hand limit as x → c (LHL) = right hand limit as x → c...

### In the following, determine the value(s) of constant(s) involved in the definition so that the given function is continuous: (iii) (i v)

(iii) A real function f is said to be continuous at x = c, where c is any point in the domain of f if h is a very small positive number. i.e. left hand limit as x → c (LHL) = right hand limit as x →...

### In the following, determine the value(s) of constant(s) involved in the definition so that the given function is continuous: (i) (ii)

(i) A real function f is said to be continuous at x = c, where c is any point in the domain of f if h is a very small positive number. i.e. left hand limit as x → c (LHL) = right hand limit as x → c...

### Find the points of discontinuity, if any, of the following functions: (x i) (x i i)

(xi) A real function f is said to be continuous at x = c, where c is any point in the domain of f if h is a very small positive number. i.e. left hand limit as x → c (LHL) = right hand limit as x →...

### Find the points of discontinuity, if any, of the following functions: (i x) (x)

(ix) A real function f is said to be continuous at x = c, where c is any point in the domain of f if h is a very small positive number. i.e. left hand limit as x → c (LHL) = right hand limit as x →...

### Find the points of discontinuity, if any, of the following functions: (viii)

((vii) A real function f is said to be continuous at x = c, where c is any point in the domain of f if h is a very small positive number. i.e. left hand limit as x → c (LHL) = right hand limit as x...

### Find the points of discontinuity, if any, of the following functions: (i) (ii)

(i) A real function f is said to be continuous at x = c, where c is any point in the domain of f if h is a very small positive number. i.e. left hand limit as x → c (LHL) = right hand limit as x → c...

### Discuss the the continuity of the function

A real function f is said to be continuous at x = c, where c is any point in the domain of f if Since, h is a very small positive number. i.e. left hand limit as x → c (LHL) = right hand limit as x...

### Prove that the function is everywhere continuous.

A real function f is said to be continuous at x = c, where c is any point in the domain of f if A function is continuous at x = c if From definition of f(x), f(x) is defined for all real numbers....

### 40.

Solving L.H.S and divide the numerator and denominator with $(1-\cos A),$, we have $=\frac{(1-\cos A)(1-\cos A)}{(1+\cos A)(1-\cos A)}$ $=\frac{{{(1-\cos A)}^{2}}}{(1+{{\cos }^{2}}A)}$...

### 30.

Solving L.H.S, we get $LHS=\frac{\tan \theta }{1-\frac{1}{\tan \theta }}+\frac{\cot \theta }{1-\tan \theta }$ $=\frac{{{\tan }^{2}}\theta }{\tan \theta -1}+\frac{\cot \theta }{1-\tan \theta }$...

Solving L.H.S and using the trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1,$we get Multiplying by $(1-\cos \theta )$to Numerator and denominator $LHS=\frac{1+\sec \theta }{\sec... read more ### 28. Solve L.H.S$\frac{1+{{\tan }^{2}}\theta }{1+{{\cot }^{2}}\theta }$Using trigonometric identity${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1,and\cos e{{c}^{2}}\theta -{{\cot }^{2}}\theta =1$... read more ### 27. Firstly we will solve L.H.S=R.H.S Then use trigonometric identity$\sin \theta +\cos \theta =1,$, we get$LHS=\frac{{{(1+\sin \theta )}^{2}}+{{(1-\sin \theta )}^{2}}}{2{{\sec }^{2}}\theta }$... read more ### 26. Firstly we will solve L.H.S Using the trigonometric identity${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1,$, we get$LHS=\frac{1+\sin \theta }{\cos \theta }+\frac{\cos \theta }{1+\sin \theta }$... read more ### 25. Firstly we will solve L.H.S$LHS=\frac{1}{1+\sin A}+\frac{1}{1-\sin A}=\frac{(1-\sin A)+(1+\sin A)}{(1+\sin A)(1-\sin A)}=\frac{1-\sin A+1+\sin A}{1-{{\sin }^{2}}A}\because (1+\sin...
Solution: (i) Suppose $\sin ^{-1}\left(\cos \frac{3 \pi}{4}\right)=\mathrm{y}$ Therefore we can write the above equation as \$\sin \mathrm{y}=\cos \frac{3 \pi}{4}=-\sin \left(\pi-\frac{3...