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...
$\cot \left(\cos ^{-1} 3 / 5+\sin ^{-1} x\right)=0$ => $\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...
Evaluate each of the following: (iii) (iv)
(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
If f is defined by f (x) = – 4x + 7, show that f’ (5) = 2 f’ (7/2)
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
Show that f (x) = |x – 3| is continuous but not differentiable at x = 3.
Differentiate with respect to , if, (i) (ii)
(i) Let (ii) Let
Differentiate with respect to if, (iii)
(iii) Let
Differentiate with respect to if, (i) (ii)
(i) Let (ii)
Differentiate with respect to , if (i) (ii)
(i) Given sin-1 √ (1-x2) (ii) Given sin-1 √ (1-x2)
If and , prove that .
Find dy/dx, when
Find dy/dx, when
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)...
Find dy/dx, when
Find dy/dx, when
Find dy/dx, when x = 3 a t / 1+t^2 and y = 3 a t^2/1+t^2
Find dy/dx, when
Find dy/dx, when
Find dy/dx, when
Find dy/dx, when
Find dy/dx, when
Find dy/dx, when
Find dy/dx, when
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
(vii) (viii)
Differentiate the following functions with respect to x:
(v) (vi)
Differentiate the following functions with respect to x:
(iv)
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Let y = (log x)cos x Taking log both the sides, we get
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
If , prove that
, prove that .
If , prove that
If , prove that
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:
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:
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:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Let,
Differentiate the following functions with respect to x:
Let,
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x: log (cosec x – cot x)
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x: x
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x: .
Let y = (log sin x)2
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x: tan 5x
Let y = tan (5x°)
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Let y = sin2 (2x + 1) On differentiating y with respect to x, we get
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x: Sin (log x)
Differentiate the following functions with respect to x:
Differentiate the following functions with respect to x:
Given tan2 x
Differentiate the following functions from the first principles:
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:
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...
Find the values of a and b so that the function f (x) defined by
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....
Find the inverse of the following matrices by using elementary row transformations:
Solution: For row transformation $\begin{array}{l} A=I A \\ \Rightarrow\left[\begin{array}{cc} 7 & 1 \\ 4 & -3 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1...
43.
Assuming the L.H.S and taking L.C.M and on simplifying we will get, $=\frac{(\cos ecA)(\cos ecA+1+\cos ecA-1)}{(\cos e{{c}^{2}}A-1)}$ $=\frac{(2\cos e{{c}^{2}}A)}{{{\cot }^{2}}A}$ $=\frac{2{{\sin...
42.
Solving the L.H.S, we will get $=\frac{\cos A}{1-\tan A}+\frac{\sin A}{1-\cot A}$ $=\frac{\cos A}{1-\frac{\sin A}{\cos A}}+\frac{\sin A}{1-\frac{\cos A}{\sin A}}$ $=\frac{{{\cos }^{2}}A}{\cos A-\sin...
41.
Assuming L.H.S and taking L.C.M and on simplifying we will get, $=\frac{\sec A+1+\sec A-1}{(\sec A+1)(\sec A-1)}$ $=\frac{2\sec A}{({{\sec }^{2}}A-1)}$ $=\frac{2{{\cos }^{2}}A}{(\cos A{{\sin...
40.
Solving L.H.S and divide the numerator and denominator with $(1-\cos A),$, we have $=\frac{(1-\cos A)(1-\cos A)}{(1+\cos A)(1-\cos A)}$ $=\frac{{{(1-\cos A)}^{2}}}{(1+{{\cos }^{2}}A)}$...
39.
Solving $LHS={{(\sec A-\tan A)}^{2}}$, we get $={{\left[ \frac{1}{\cos A}-\frac{\sin A}{\cos A} \right]}^{2}}$ $=\frac{{{(1-\sin A)}^{2}}}{{{\cos }^{2}}A}$ $=\frac{{{(1-\sin A)}^{2}}}{1-{{\sin...
38. Prove that:(iii)(iv)
Solving L.H.S and dividing the numerator and denominator with its respective conjugates, we have $=\sqrt{\frac{(1-\cos \theta )(1-\cos \theta )}{(1+\cos \theta )(1-\cos \theta...
38. Prove that: (i)(ii)
Solving L.H.S and divide the numerator and denominator with its respective conjugates, we have $=\sqrt{\frac{(\sec \theta -1)(\sec \theta -1)}{(\sec \theta +1)(\sec \theta -1)}}+\sqrt{\frac{(\sec...
37. (i) (ii)
Solving L.H.S and dividing the numerator and denominator with $\sqrt{(1+\sin A)},$we have $=\sqrt{\frac{(1+\sin A)(1+\sin A)}{(1-\sin A)(1+\sin A)}}=\sqrt{\frac{{{(1+\sin A)}^{2}}}{1-{{\sin...
36.
Solving L.H.S $LHS=\frac{1+\cos A}{\sin A}$ Multiply the numerator and denominator by $(1-\cos A)$ we will have $=\frac{(1+\cos A)(1-\cos A)}{\sin A(1-\cos A)}$ $=\frac{1-{{\cos }^{2}}A}{\sin...
35.
Solving L.H.S $LHS=\frac{\sec A-\tan A}{\sec A+\tan A}$ Dividing the denominator and numerator with $(\sec A+\tan A)$ and using ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1,$we have, $=\frac{{{\sec...
34.
Solving L.H.S and using the trigonometric identity ${{\sin }^{2}}A+{{\cos }^{2}}A=1,$, we have ${{\sin }^{2}}A=1-{{\cos }^{2}}A$ $\Rightarrow {{\sin }^{2}}A=(1-\cos A)(1+\cos A)$ $LHS=\frac{1+\cos...
Find the adjoint of each of the following matrices:
(i)
(ii) Verify that for the above matrices.
Solution: (i) Suppose $A=\left[\begin{array}{cc}-3 & 5 \\ 2 & 4\end{array}\right]$ Cofactors of $A$ are $C_{11}=4$ $C_{12}=-2$ $C_{21}=-5$ $C_{22}=-3$ Since, adj...
33.
First solve L.H.S and using the trigonometric identity we all know that ${{\sec }^{2}}\theta {{\tan }^{2}}\theta =1\Rightarrow 1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta $ $LHS=\frac{{{\sec...
32.
Using the trigonometric identity we get $\cos e{{c}^{2}}\theta +{{\cot }^{2}}\theta =1$ cubing it on both side ${{(\cos e{{c}^{6}}\theta +{{\cot }^{2}}\theta )}^{3}}=1$ $\cos e{{c}^{6}}-{{\cot...
31.
Using trigonometric identity, ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$ Cubing it on both side ${{({{\sec }^{2}}\theta -{{\tan }^{2}}\theta )}^{2}}=1$ ${{\sec }^{6}}\theta -{{\tan }^{6}}\theta...
30.
Solving L.H.S, we get $LHS=\frac{\tan \theta }{1-\frac{1}{\tan \theta }}+\frac{\cot \theta }{1-\tan \theta }$ $=\frac{{{\tan }^{2}}\theta }{\tan \theta -1}+\frac{\cot \theta }{1-\tan \theta }$...
29.
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...
28.
Solve L.H.S $\frac{1+{{\tan }^{2}}\theta }{1+{{\cot }^{2}}\theta }$ Using trigonometric identity${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1,and\cos e{{c}^{2}}\theta -{{\cot }^{2}}\theta =1$...
27.
Firstly we will solve L.H.S=R.H.S Then use trigonometric identity $\sin \theta +\cos \theta =1,$, we get $LHS=\frac{{{(1+\sin \theta )}^{2}}+{{(1-\sin \theta )}^{2}}}{2{{\sec }^{2}}\theta }$...
26.
Firstly we will solve L.H.S Using the trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1,$, we get $LHS=\frac{1+\sin \theta }{\cos \theta }+\frac{\cos \theta }{1+\sin \theta }$...
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...
Find the principal value of the following:
(i)
(ii)
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...
Find the principal value of the following:
(i)
(ii)
Solution: (i) It is given that functions can be written as $\sin ^{-1}\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}\right)=\sin ^{-1}\left(\frac{\sqrt{3}}{2 \sqrt{2}}-\frac{1}{2 \sqrt{2}}\right) $Taking $1 /...