M.C.Q (1 Marks)

MCQ 11 Mark

If $4a + 2b + c = 0,$ then the equation $3ax^2+ 2bx + c = 0$ has atleast one real root lying in the interval :

A
$(0, 1)$
B
$(1, 2)$
✓
$(0, 2)$
D
None of these.

Answer

Correct option: C.

$(0, 2)$

Let, $f(x)=a x^3+b x^2+c x+d$
$f(0)=d$
$f(2)=8 a+4 b+2 c+d$
$=2(4 a+2 b+c)+d$
$=2 \times 0+d$
$=0$
$f$ is continuous and differentiable on $(0,2)$
$f(0)=f(2)$
Using Rolle's theorem,
$f^{\prime}(x)=0 \text { for }(0,2)$
$3 a x^2+2 b x+c=0$
$f(x)$ has atleast one root in the interval $(0,2).$
Hence $f^{\prime}(x)$ must have root in the interval $(0,2)$.

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MCQ 21 Mark

For the function $\text{f}(\text{x})=\text{x}+1\text{x},\text{x}\in[1,3],$ the value of c for the Lagrange's mean value theorem is:

A
1
✓
$\sqrt3$
C
2
D
none of these

Answer

Correct option: B.

$\sqrt3$

We have$\text{f}(\text{x})=\text{x}+\frac{1}{\text{x}}=\frac{\text{x}^2+1}{\text{x}}$
Clearly, f(x) is continuous on [1, 3] and derivable on (1, 3).
Thus, both the conditions of Lagrange's theorem is satisfied.
Concequently there exists $\text{c}\in(1,3)$ such that
$\text{f}'(\text{c})=\frac{\text{f}(3)-\text{f}(1)}{3-1}=\frac{\text{f}(3)-\text{f}(1)}{2}$
Now, $\text{f}(\text{x})=\frac{\text{x}^2+1}{\text{x}}$
$\text{f}'(\text{x})=\frac{\text{x}^2-1}{\text{x}^2},\text{f}(1)=2,\text{f}(3)=\frac{10}{3}$
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(3)-\text{f}(1)}{2}$
$\Rightarrow\frac{\text{x}^2-1}{\text{x}^2}=\frac{4}{6}$
$\Rightarrow\frac{\text{x}^2-1}{\text{x}^2}=\frac{2}{3}$
$\Rightarrow3\text{x}^2-3=2\text{x}^2$
$\Rightarrow\text{x}=\pm\sqrt3$
Thus, $\text{c}=\sqrt3\in(1,3)$ such that $\text{f}'(\text{c})=\frac{\text{f}(3)-\text{f}(1)}{3-1}.$

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MCQ 31 Mark

The value of $c$ in Rolle's theorem for the function $\text{f}(\text{x})=\frac{\text{x}(\text{x}+1)}{\text{e}^{\text{x}}}$ defined on $[-1, 0]$ is :

A
$0.5$
B
$\frac{1+\sqrt5}{2}$
✓
$\frac{1-\sqrt5}{2}$
D
$-0.5$

Answer

Correct option: C.

$\frac{1-\sqrt5}{2}$

$f(x)=\frac{x(x+1)}{e^x} $ defined on $[-1,0]$
$\Rightarrow f(-1)=0 $ also $f(0)=0$
Now, $ f(x)=e^{-x}\left(x^2+x\right)$
$\Rightarrow f^{\prime}(x)=e^{-x}(2 x+1)-\left(x^2+x\right) e^{-x}$
$\Rightarrow f^{\prime}(x)=e^{-x}\left(2 x+1-x^2+x\right)$
$\Rightarrow f^{\prime}(x)=e^{-x}\left(-x^2+x-1\right)$
$\Rightarrow f^{\prime}(x)=0$
$\Rightarrow e^{-x}\left(-x^2+x-1\right)=0$
$\Rightarrow-x^2+x-1=0$
$\Rightarrow x^2-x+1=0$
$\Rightarrow x=\frac{1 \pm \sqrt{5}}{2}$
As $ x \in[-1,0]$
$x=\frac{1-\sqrt{5}}{2}$

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MCQ 41 Mark

When the tangent to the curve $\text{y}=\text{x}\log\text{x}$ is parallel to the chord joining the points (1, 0) and (e, e), the value of x is:

✓
$\text{e}^{\frac{1}{1}-\text{e}}$
B
$\text{e}^{(\text{e}-1)(2\text{e}-1)}$
C
$\text{e}^{\frac{2\text{e}-1}{\text{e}-1}}$
D
$\frac{\text{e}-1}{\text{e}}$

Answer

Correct option: A.

$\text{e}^{\frac{1}{1}-\text{e}}$

Given:$\text{y}=\text{f}(\text{x})=\text{x}\log\text{x}$
Differentiating the given function with respect to x, we get
$\text{f}'(\text{x})=1+\log\text{x}$
⇒ Slope of the tangent to the curve $=1+\log\text{x}$
Also,
Slope of the chord joining the points (1, 0) and (e, e), $(\text{m})=\frac{\text{e}}{\text{e}-1}$
The tangent to the curve is parallel to chord joining the points (1, 0) and (e, e).

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MCQ 51 Mark

The value of $c$ in Rolle's theorem when $f(x) = 2x^3 - 5x^2 - 4x + 3, \text{x}\in\Big[\frac{1}{3},3\Big]$ is:

✓
$2$
B
$-\frac{1}{3}$
C
$-2$
D
$\frac{2}{3}$

Answer

Correct option: A.

$2$

$f(x) = 2x^3 - 5x^2 - 4x + 3$
Differentiating the given function with respect to $x,$ we get
$f'(x) = 6x^2 - 10x - 4$
$\Rightarrow f'(c) = 6c^2 - 10c - 4$
$\therefore f'(c) =0$
$\Rightarrow 3c^2 - 5c - 2 = 0$
$\Rightarrow 3c^2 - 6c + c - 2 = 0$
$\Rightarrow 3c(c - 2) + c - 2 = 0$
$\Rightarrow (3c + 1)(c - 2) = 0$
$\Rightarrow\text{c}=2, \frac{-1}{3}$
$\therefore\ \text{c}=2\in\Big(\frac{1}{3},3\Big)$
Thus, $\text{c}=2\in\Big(\frac{1}{3},3\Big)$ for which Rolle's theorem holds.
Hence, the required value of $c$ is $2.$

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MCQ 61 Mark

Function $\text{f}(\text{x})=\cos\text{x}-2\lambda\text{x}$ is monotonic decreasing when:

A
$\lambda>\frac{1}{2}$
B
$\lambda<\frac{1}{2}$
✓
$\lambda<2$
D
$\lambda>2$

Answer

Correct option: C.

$\lambda<2$

$\text{f}(\text{x})=\cos\text{x}-2\lambda\text{x}$
$\text{f}'(\text{x})=-\sin\text{x}-2\lambda$
For f(x) to be decreasing, we must have
$\text{f}'(\text{x})<0$
$\Rightarrow-\sin\text{x}-2\lambda<0$
$\Rightarrow\sin\text{x}+2\lambda>0$
$\Rightarrow2\lambda>-\sin\text{x}$
We know that the maximum value of $-\sin\text{x}$ is 1.
$\Rightarrow2\lambda>1$
$\Rightarrow\lambda>\frac{1}{2}$

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MCQ 71 Mark

If from Lagrange's mean value theorem, we have $\text{f}\ '(\text{x}_1)=\frac{\text{f}(\text{b})-\text{f}(\text{a})}{\text{b}-\text{a}},$ then :

A
$\text{a} < \text{x}_1\leq\text{b}$
B
$\text{a}\leq\text{x}_1 < \text{b}$
✓
$\text{a} < \text{x}_1 < \text{b}$
D
$\text{a}\leq\text{x}_1\leq\text{b}$

Answer

Correct option: C.

$\text{a} < \text{x}_1 < \text{b}$

We have
$\text{f}(\text{x})=\text{x}+\frac{1}{\text{x}}=\frac{\text{x}^2+1}{\text{x}}$
In the Lagrange's mean value theorem, $\text{c}\in(\text{a},\text{b})$ such that $\text{f}\ '(\text{c})=\frac{\text{f}(\text{b})-\text{f}(\text{a})}{\text{b}-\text{a}}$
So, if there is $x_1$ such that $\text{f}\ '(\text{x}_1)=\frac{\text{f}(\text{b})-\text{f}(\text{a})}{\text{b}-\text{a}},$ then $\text{x}_1\in(\text{a},\text{b})$
$\Rightarrow\text{a} < \text{x}_1 < \text{b}$

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MCQ 81 Mark

If the polynomial equation $\text{a}_0\text{x}^{\text{n}}+\text{a}_{\text{n}-1}\text{x}^{\text{n}-1}+\text{a}_{\text{n}-2}\text{x}^{\text{n}-2}+...\text{a}_2\text{x}^2+\text{a}_1\text{x}+\text{a}_0=0$ n positive integer,has two different real roots $\alpha$ and $\beta,$ then between $\alpha$ and $\beta,$ the equation $\text{n}\text{a}_{\text{n}}\text{x}^{\text{n}-1}+(\text{n}-1)\text{a}_{\text{n}-1}\text{x}^{\text{n}-2}+...+\text{a}_1=0$ has:

A
Exactly one root.
B
Almost one root.
✓
At least one root.
D
No root.

Answer

Correct option: C.

At least one root.

We observe that, $\text{n}\text{a}_{\text{n}}\text{x}^{\text{n}-1}+(\text{n}-1)\text{a}_{\text{n}-1}\text{x}^{\text{n}-2}+...+\text{a}_1=0$ is the derivative of the polynomial $\text{a}_{\text{n}}\text{x}^{\text{n}}+\text{a}_{\text{n}-1}\text{x}^{\text{n}-1}+\text{a}_{\text{n}-2}\text{x}^{\text{n}-2}+...\text{a}_2\text{x}^2+\text{a}_1\text{x}+\text{a}_0=0$Polynomial function is continuous everywhere in R and concequently derivative in R.
Therefore, $\text{a}_{\text{n}}\text{x}^{\text{n}}+\text{a}_{\text{n}-1}\text{x}^{\text{n}-1}+\text{a}_{\text{n}-2}\text{x}^{\text{n}-2}+...\text{a}_2\text{x}^2+\text{a}_1\text{x}+\text{a}_0$ is continuous on $\alpha,\beta$ and derivative on $\alpha,\beta.$
Hence, it is satisfies the both the conditions of Rolle's theorem.
By algebric interpretation of Roll's theorem, we know that between any two roots of a function f(x), there exists atleast one root of its derivative.
Hence, the equation $\text{n}\text{a}_{\text{n}}\text{x}^{\text{n}-1}+(\text{n}-1)\text{a}_{\text{n}-1}\text{x}^{\text{n}-2}+...+\text{a}_1=0$ will have atleast one root between $\alpha$ and $\beta.$

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MCQ 91 Mark

The value of $c$ in Lagrange's mean value theorem for the function $f(x) = x(x - 2)$ when $\text{x}\in[1,2]$ is :

A
$1$
B
$\frac{1}{2}$
C
$\frac{2}{3}$
✓
$\frac{3}{2}$

Answer

Correct option: D.

$\frac{3}{2}$

We have
$f(x) = x(x - 2)$
It can be rewritten as $f(x) = x^2 - 2x$
We know that a polynomial function is everywhere continuous and differentiable.
Since, $f(x)$ is polynomial, it is continuous on $[1, 2]$ and differentiable on $[1, 2].$
So, there must exist at least one real number $\text{c}\in(1,2)$ such that
$\text{f}\ '(\text{c})=\frac{\text{f}(2)-\text{f}(1)}{2-1}$
$=\frac{\text{f}(2)-\text{f}(1)}{1}$
Now, $f(x) = x^2 - 2x$
$\Rightarrow f'(x) = 2x - 2$
and $f(1) = -1, f(2) = 0$
$\therefore\ \text{f}\ '(\text{x})=\frac{\text{f}(2)-\text{f}(1)}{2-1}$
$\Rightarrow\text{f}(\text{x})=\frac{0+1}{1}$
$\Rightarrow2\text{x}-2=1$
$\Rightarrow\text{x}=\frac{3}{2}$
$\therefore\ \text{c}=\frac{3}{2}\in(1,2)$

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MCQ 101 Mark

The value of $c$ in Rolle's theorem for the function $f(x) = x^3 - 3x$ in the interval $\big[0,\sqrt3\big]$ is:

✓
$1$
B
$-1$
C
$\frac{3}{2}$
D
$\frac{1}{3}$

Answer

Correct option: A.

$1$

$f(x)=x^3-3 x$ in the interval $[0, \sqrt{3}]$
$f(0)=0 $ and $ f(\sqrt{3})=0$
$f^{\prime}(x)=3 x^2-3$
$f^{\prime}(c)=3 c^2-3$
$f^{\prime}(c)=0$
$3 c^2-3=0$
$3 c^2=3$
$c^2=1$
$c= \pm 1$
$x \in[0, \sqrt{3}]$
Hence $x=1$

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MCQ 111 Mark

If $\text{f}(\text{x})=\text{e}^{\text{x}}\sin\text{x}$ in $[0,\pi],$ then c in Rolle's theorem is:

A
$\frac{\pi}{6}$
B
$\frac{\pi}{4}$
C
$\frac{\pi}{2}$
✓
$\frac{3\pi}{4}$

Answer

Correct option: D.

$\frac{3\pi}{4}$

$\text{f}(\text{x})=\text{e}^{\text{x}}\sin\text{x}$
$\text{f}'(\text{x})=\text{e}^{\text{x}}\cos\text{x}+\text{e}^{\text{x}}\sin\text{x}$
$\text{f}'(\text{c})=0$
$\text{e}^\text{c}(\cos\text{c}+\sin\text{c})=0$
$\cos\text{c}+\sin\text{c}=0$
$\cos\text{c}=-\sin\text{c}$
$\tan\text{c}=-1$
$\text{c}=\frac{3\pi}{4}\in(0,\pi)$

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MCQ 121 Mark

Rolle's theorem is applicable in case of $\phi(\text{x})=\text{a}^{\sin\text{x}},\text{a}>\text{a}$ in:

A
Any interval.
✓
Any interval $[0,\pi]$
C
Any interval $\Big[0,\frac{\pi}{2}\Big]$
D
None of these.

Answer

Correct option: B.

Any interval $[0,\pi]$

$\phi(\text{x})$ is continuous and differentiable function then using statement of Roll's theorem f(a) = f(b). Hence, here $\sin0=0$ also $\sin\pi=0.$

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Maths M.C.Q (1 Marks)

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