in each of the following: ![\[{{\mathbf{y}}^{\mathbf{3}}}-\text{ }\mathbf{3x}{{\mathbf{y}}^{\mathbf{2}}}~=\text{ }{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{y}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4f229b63c141fbbae52d5f6a5bd2a41d_l3.png "Rendered by QuickLaTeX.com")
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
(ii) 
$(i)$ $(ii)$
(x) 
$(ix)$ $(x)$
(viii) 
$(vii)$ $(viii)$
(vi) 
$(v)$ $(vi)$
(iv) 
$(iii)$ $(iv)$
(ii) 
$(i)$ $(ii)$
(iv) 
$(iii)$ $(iv)$
(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)
$(iii)$
(ii) 
$(i)$ $(ii)$ LHS
, 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
$(v)$ $=>0$
for 
(iv) Cot 
$(iii)$ $(iv)$
(ii) for 
$(i)$ $(ii)$
$(iii)$
(ii) Sec 
$(i)$ $(ii)$
$(ix)$ .
(viii) 
$(vii)$ $(viii)$
(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...
(iv) 
(iii) (iv)
(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{...
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.
, find f’ (4).f(x) = x3 + 7x2 + 8x – 9, => Checking the differentiability at x = 4
, 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 =>
![\[~\mathbf{2}{{\mathbf{x}}^{\mathbf{3}}}~\text{ }\mathbf{9}{{\mathbf{x}}^{\mathbf{2}~}}+\text{ }\mathbf{12}\text{ }\mathbf{x}\text{ }+\text{ }\mathbf{9}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c388eec2224700cda456bd12406ee5c9_l3.png "Rendered by QuickLaTeX.com")
We are given with a polynomial function f(x) = 2x3 – 9x2 + 12x + 9, and we have
– 4x + 7, show that f’ (5) = 2 f’ (7/2)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...
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.
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
with respect to 
, if, (i) 
(ii) 
(i) Let (ii) Let
with respect to 
if, (iii) 
(iii) Let
with respect to if, (i) 
(ii) 
(i) Let (ii)
with respect to 
, if (i) 
(ii) 
(i) Given sin-1 √ (1-x2) (ii) Given sin-1 √ (1-x2)
and 
, prove that 
.
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)...
(vii) (viii)
(v) (vi)
(iv)
Let y = (log x)cos x Taking log both the sides, we get
, prove that 
, prove that 
.
, prove that 
, prove that 
differentiating the equation on both sides with respect to x, we get,
differentiating the equation on both sides with respect to x, we get,
differentiating the equation on both sides with respect to x, we get,
differentiating the equation on both sides with respect to x, we get,
differentiating the equation on both sides with respect to x, we get,
differentiating the equation on both sides with respect to x, we get,
differentiating the equation on both sides with respect to x, we get,
differentiating the equation on both sides with respect to x, we get,
Let,
Let,
.Let y = (log sin x)2
![\[\mathbf{lo}{{\mathbf{g}}_{\mathbf{x}}}~\mathbf{3}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ff5c2c46f9e67b0239c75b89cbf3a0a4_l3.png "Rendered by QuickLaTeX.com")
Let y = tan (5x°)
![\[\mathbf{lo}{{\mathbf{g}}_{\mathbf{7}}}~\left( \mathbf{2x}\text{ }\text{ }\mathbf{3} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9548ffb8700a3c7168d42bef1698dab2_l3.png "Rendered by QuickLaTeX.com")
![\[~\mathbf{Si}{{\mathbf{n}}^{\mathbf{2}}}~\left( \mathbf{2x}\text{ }+\text{ }\mathbf{1} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b40c28b7926d22db3057d590b6cc73ca_l3.png "Rendered by QuickLaTeX.com")
Let y = sin2 (2x + 1) On differentiating y with respect to x, we get
![\[\mathbf{tan}\text{ }({{\mathbf{x}}^{\mathbf{o}}}~+\text{ }\mathbf{4}{{\mathbf{5}}^{\mathbf{o}}})\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-464d3400e1e5d6359b79f09c8ce8e182_l3.png "Rendered by QuickLaTeX.com")
Given tan2 x
let f (x) = ecos x By using the first principle formula, we get,
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...
for 
, find the value which can be assigned to 
at 
so that the function becomes continuous everywhere in ![[0, \pi / 2]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b8bf04cd2ca4b4771433b6b0371ba22f_l3.png "Rendered by QuickLaTeX.com")
.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...
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...
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...
(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 →...
(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...
(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 →...
(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 →...
(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...
(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...
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...
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....
![\left[\begin{array}{cc}7 & 1 \\ 4 & -3\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2c36dc19722d96911b3cd4f2fe953983_l3.png "Rendered by QuickLaTeX.com")
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...
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...
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...
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...
Solving L.H.S and divide the numerator and denominator with $(1-\cos A),$, we have $=\frac{(1-\cos A)(1-\cos A)}{(1+\cos A)(1-\cos A)}$ $=\frac{{{(1-\cos A)}^{2}}}{(1+{{\cos }^{2}}A)}$...
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...
(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...
(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...
(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...
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...
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...
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...
![\left[\begin{array}{cc}-3 & 5 \\ 2 & 4\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-375ad869b7946931aef27c709053c15e_l3.png "Rendered by QuickLaTeX.com")
![\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-79b4e1c9febd7f8cbe00652c013c9ec9_l3.png "Rendered by QuickLaTeX.com")
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...
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...
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...
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...
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...
Solve L.H.S $\frac{1+{{\tan }^{2}}\theta }{1+{{\cot }^{2}}\theta }$ Using trigonometric identity${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1,and\cos e{{c}^{2}}\theta -{{\cot }^{2}}\theta =1$...
Firstly we will solve L.H.S=R.H.S Then use trigonometric identity $\sin \theta +\cos \theta =1,$, we get $LHS=\frac{{{(1+\sin \theta )}^{2}}+{{(1-\sin \theta )}^{2}}}{2{{\sec }^{2}}\theta }$...
Firstly we will solve L.H.S Using the trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1,$, we get $LHS=\frac{1+\sin \theta }{\cos \theta }+\frac{\cos \theta }{1+\sin \theta }$...
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...